CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Odisha State Board CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning Questions and Answers.

CHSE Odisha 12th Class Education Chapter 6 Question Answer Meaning, Nature, and Factors of Learning

Group – A

Short type Questions with Answers
I. Answer with in Two/Three sentence:

Question 1.
What is learning?
Answer:
Learning is the process through which individuals acquire knowledge, attitudes, and skills necessary to meet the demands of life; It involves a change in behavior or behavior potential brought about by experiences.

Question 2.
Give an example of vicarious learning.
Answer:
Observing someone else and learning from that observation without being directly involved in the experience is vicarious learning. For instance, a child learns how to clap hands by watching someone else do it.

Question 3.
According to John Dewey, how does learning occur?
Answer:
John Dewey described learning as an active and social process where individuals construct meaning through experiences. He emphasized hands-on, experiential learning for genuine understanding.

Question 4.
What did Lev Vygotsky contribute to the understanding of learning?
Answer:
Lev Vygotsky’s sociocultural theory highlights the significance of social interaction in learning. He introduced the Zone of Proximal Development (ZPD), where learning is most effective with guidance from a knowledgeable other.

Question 5.
How did B.F. Skinner define learning?
Answer:
B.F. Skinner, associated with behaviorism, defined learning as a change in behavior due to reinforcement or punishment. He focused on observable behaviors and the role of the environment in shaping behavior.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 6.
According to Carl Rogers, what is essential for meaningful education?
Answer:
Carl Rogers emphasized learner-centered education, stating that creating a supportive and non-judgmental. environment is essential. He believed that self-directed learning, driven by personal experiences and interests, is crucial for meaningful education.

Question 7.
How does the family influence a child’s learning process?
Answer:
The family shapes a child’s behavior patterns, social skills, and value orientation, serving as the primary site for learning cultural norms, interpersonal skills, and the do’s and don’ts of their community.

Question 8.
What role do peer relationships play in classroom learning?
Answer:
Healthy peer relationships create a tension-free environment, facilitating student learning and competition, while poor peer relationships can adversely affect the learning climate, emphasizing the importance of promoting positive interactions.

Question 9.
How does the socio-economic status of a group impact learning opportunities?
Answer:
The socio-economic status directly links to the degree of stimulation or enrichment available in the learning environment, with adequately enriching environments offering more learning opportunities compared to impoverished or needy environments.

Question 10.
What influence do caste, class, and religion have on identity and attitudes?
Answer:
Caste, class, and religion play a predominant role in shaping identity, self-concept, attitudes, and value orientation, contributing to variations in goals and achievement patterns across different groups.

Question 11.
Why is it recommended to organize free discussions during the teaching-learning process?
Answer:
Organizing free discussions helps improve the classroom learning climate by fostering a sound peer relationship, allowing students to meet and interact freely, and addressing any misunderstandings promptly to maintain a healthy and cordial environment.

II. Answer with in Five/Six sentence :

Question 1.
What is the essence of learning, and how does it manifest through experiences?
Answer:
Learning is the process through which individuals acquire knowledge, attitudes, and skills necessary for navigating life. It involves, a change in behavior or behavior potential resulting from experiences. For example, a child learns to avoid touching a burning candle after getting burnt. Learning can occur through both vicarious experiences, where individuals observe and learn from others, and direct experiences, involving personal engagement. This transformative process is crucial for adapting to the demands of life.

Question 2.
How do various educational theorists define learning, and what are their key perspectives?
Answer:
Educational theorists like John Dewey, Jean Piaget, Lev Vygotsky, Maria MontessOri, Jerome Bruner, B.F. Skinner, Howard Gardner, Albert Bandura, Ivan Pavlov, and Carl Rogers offer diverse definitions of learning. Dewey emphasizes experiential learning, while Piaget focuses on cognitive development. Vygotsky highlights social interaction, and Montessori emphasizes self- directed, hands-on learning. Skinner, with behaviorism, emphasizes reinforcement, and Gardner introduces multiple intelligences. Bandura stresses observation, and Pavlov associates learning with conditioned responses. Rogers advocates learner-centered education.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 3.
What are the key characteristics and the nature of learning?
Answer:
Learning is a lifelong, continuous process that involves the acquisition of knowledge, skills, and attitudes through experiences. It is purposive, goal-directed, and typically leads to some degree of permanence in behavior change. Learning is universal, occurring across ages, sexes, races, and cultures. It prepares individuals for adjustment to new situations and is comprehensive, covering cognitive, affective, and psychomotor domains. Learning involves organizing experiences, resulting in changes in responses or behavior, which may be favorable or unfavorable.

Question 4.
How does the concept of learning differ when viewed as a process versus a product?
Answer:
Viewing learning as a process emphasizes its internal and personal nature, contrasting with a product-oriented perspective where knowledge is seen as external and acquired. The process- oriented approach aligns with understanding the real world and using knowledge as a tool for survival. In contrast, the product-oriented view, criticized by Paulo Freire, likens education to a banking concept,, treating students as depositories and the teacher as the depositor.

Question 5.
Why is learning considered purposive, and what role does it play in the overall development of an individual?
Answer:
Learning is considered purposive because it is not an aimless activity; it is goal-directed. True learning is purposeful and contributes to the development of an individual. It involves adapting to new situations, achieving goals, and bringing about changes in behavior. Learning prepares individuals for effective adjustment to the challenges they encounter, ensuring a continuous, never- ending process that spans the entire lifespan.

Question 6.
How does the family contribute to a child’s learning, process?
Answer:
The family serves as the primary socialization agent, shaping a child’s behavior patterns, social skills, attitudes, and cultural norms. The child learns survival skills, interpersonal dynamics, and value orientations within the family context. Positive parent-child relationships, based on mutual respect, create a conducive learning atmosphere, while unhealthy environments can hinder learning, leading to resistance and difficulties in coping.

Question 7.
What role does the neighborhood and community play in a child’s learning?
Answer:
The neighborhood and community influence a child’s attitudes, habits, beliefs, and social roles through interactions with peers, age-mates, and elders. Conditioning, social learning, and modeling contribute to the child’s understanding of various aspects of life. Positive peer relationships enhance the learning environment, creating a tension-free atmosphere, while negative relationships can adversely affect learning outcomes.

Question 8.
What is the distinction between school culture and school climate, and how do they influence learning?
Answer:
School culture represents shared beliefs and attitudes at the district-wide level, guiding the organization’s identity, while school climate characterizes the feel of a school at the building and classroom levels. Culture is deeply ingrained and unconscious, reflected in artifacts, values, and assumptions. In contrast, climate focuses on feelings and attitudes, susceptible to change, and crucial for creating preconditions for effective teaching and learning. Changes in culture can impact climate positively or adversely. Both culture and climate are vital for shaping expected behaviors and improving academic performance.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 9.
What are the key components of school climate, and why is caring often considered a core element?
Answer:
School climate encompasses areas like appearance, faculty relations, student interactions, leadership, learning environment, attitude, and school-community relations. While there’s no consistent agreement on components, caring is often emphasized as a core element. A positive climate contributes to academic success, addresses issues like bullying and conflicts, and ensures a safe, orderly environment conducive to learning.

Question 10.
How do physical, social, affective, and academic environments collectively contribute to school climate?
Answer:
The physical environment, promoting social interaction and a warm, affective atmosphere, collectively contributes to the academic environment. These aspects of school climate are interrelated; for instance, a welcoming physical environment encourages social interaction, fostering a positive affective environment. Collectively, these elements impact and are impacted by the overall academic environment, influencing the learning experience for Students.

Question 11.
What role do school resources and policies play in influencing student learning?
Answer:
School resources, including learning materials, facilities, and infrastructure, are crucial in determining the effectiveness of learning. Proper management of physical resources, such as classrooms, libraries, laboratories, and sports facilities, contributes to a conducive learning environment. Additionally, school policies, administration nature, distribution of responsibilities, leadership, and disciplinary measures among staff and students are decisive factors influencing the learning outcomes for students.

Group – B

Long Type Questions With Answers

Question 1.
What is learning, and how does it take place? Provide examples to illustrate the learning process.
Answer:
• Learning is a multifaceted process through which individuals acquire new knowledge, skills, attitudes, or behaviors. It is a dynamic and continuous journey that occurs throughout life, influenced by various factors and experiences. The mechanisms of learning are diverse, encompassing formal education, observation, trial and error, and interactions with the environment. Understanding the intricacies of learning involves exploring its psychological, cognitive, and behavioral dimensions.

• At its core, learning involves the acquisition of information and the ability to apply that knowledge in different contexts. One prominent model of learning is the behaviorist perspective, which emphasizes observable behaviors and the role of external stimuli in shaping these behaviors. For instance, a child learning to say “please” and “thank you” after receiving positive reinforcement from parents or caregivers exemplifies behaviorist learning.

• Cognitive theories, on the other hand, delve into the mental processes involved in learning. Jean Piaget’s theory of cognitive development highlights how individuals construct knowledge through interaction with their environment. In a classroom setting, a student might learn mathematical concepts by assimilating new information into existing mental schemas.

• Social constructivism, championed by theorists like Lev Vygotsky, emphasizes the social and cultural influences on learning. It posits that learning is not only an individual endeavor but also a social process. For instance, a group of students working collaboratively on a project learns not only from the content but also from each other’s perspectives and contributions.

• Learning takes place through formal education systems, where structured curricula and instructional methods guide students in acquiring specific knowledge and skills. In a science class, students might learn about the principles of photosynthesis through lectures, laboratory experiments, and discussions. This formalized approach provides a systematic framework for learning and assessment.

• Observational learning, as proposed by Albert Bandura’s social learning theory, occurs through watching and imitating others. Children often learn behaviors and social norms by observing role models, such as parents, teachers, or peers. A child imitating a sibling’s mannerisms or a student emulating a successful classmate’s study habits exemplifies observational learning.

• Trial and error is a fundamental aspect of learning, especially in acquiring practical skills. When learning to ride a bicycle, individuals make adjustments based on their experiences – discovering the right balance, pedaling techniques, and steering control through a series of attempts. Trial and error contribute to skill refinement and the development of problem-solving capabilities.

• Interactions with the environment foster experiential learning, where individuals learn through direct engagement with real-world situations. An aspiring chef, for example, learns to cook by experimenting with ingredients, following recipes, and adjusting techniques based on taste end feedback. Experiential learning emphasizes hands-on experiences, encouraging a deeper understanding of concepts.

• The process of learning is not limited to a specific age or setting; it extends to lifelong learning. Adults, for instance, might engage in continuous learning through professional development courses, workshops, or self-directed exploration of new subjects. Lifelong learning reflects the adaptability of individuals to evolving knowledge and skills throughout their lives.

• In conclusion, learning is a dynamic and multifaceted process involving the acquisition of knowledge, skills, attitudes, and behaviors. It occurs through diverse mechanisms, including formal education, observation, trial and error, and interactions with the environment. Understanding the various dimensions of learning provides insights into how individuals navigate the complexities of acquiring and applying information throughout their lives.

Question 2.
What are the general characteristics of learning, and how do these characteristics contribute to the overall process of knowledge acquisition and skill development ? Answer: Learning, as a fundamental aspect of human development, exhibits several key characteristics that shape the overall process of knowledge acquisition and skill development. Understanding these characteristics provides insights into the dynamics of learning and its impact on individual growth and adaptation.

Adaptability: Learning is inherently adaptable, allowing individuals to adjust their behaviors, thoughts, and actions based on new information and experiences. This characteristic enables learners to navigate diverse situations, fostering flexibility and resilience in response to changing environments.

Individuality : Learning is a unique and individualized process. Each person brings their own background, prior knowledge, and cognitive abilities to the learning journey. Recognizing and respecting individual differences is essential for effective teaching and personalized learning experiences.

Active Engagement : Learning is most effective when individuals actively engage with the content or task. Active participation enhances understanding and retention, as learners process information through interactions, discussions, and hands-on experiences. This characteristic emphasizes the importance of learner involvement in the educational process.

Continuous and Lifelong : Learning is a continuous and lifelong endeavor that extends beyond formal education. Individuals engage in learning throughout their lives, adapting to new challenges, acquiring new skills, and staying abreast of evolving knowledge. Lifelong learning promotes personal and professional development.

Meaning-Making : Learning involves the construction of meaning from experiences and information. Learners interpret and relate new knowledge to their existing understanding, creating a personalized understanding of the subject matter. Meaning-making is essential for retention and application.

Social Influence: Learning is not only an individual process but is also influenced by social interactions. Collaborative learning, peer discussions, and mentorship contribute to a richer learning experience. Social interactions provide diverse perspectives, fostering a deeper understanding of concepts.

Multifaceted Modalities : Learning occurs through various modalities, encompassing visual, auditory, kinesthetic, and other sensory channels. Recognizing and accommodating diverse learning styles ensures that educational strategies cater to the varied needs of learners.

Feedback and Reflection : Learning benefits from timely feedback and reflection. Constructive feedback informs learners about their progress, highlights areas for improvement, and reinforces successful strategies. Reflection allows individuals to internalize learning experiences, enhancing metacognitive skills.

These characteristics collectively contribute to the overall process of knowledge acquisition and skill development. Adaptability enables individuals to apply learning in different contexts, while active engagement ensures a deeper understanding of the subject matter. Recognizing the individuality of learners and incorporating various modalities enhances accessibility and inclusivity in education.

Continuous and lifelong learning fosters a culture of curiosity and growth, promoting personal and professional development. Meaning-making and social influence emphasize the importance of meaningful connections and collaborative learning experiences.

In conclusion, the general characteristics of learning create a dynamic’ and versatile framework for acquiring knowledge and developing skills. Understanding and leveraging these characteristics in educational settings enhance the effectiveness of teaching methods and contribute to the holistic growth of individuals.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 3.
Examine various approaches to learning and assess the effectiveness of each. Which method stands out as the most effective, and what factors contribute to its superiority?
Answer:
Learning is a nuanced process, and various approaches cater to diverse preferences and objectives. Evaluating the effectiveness of each method involves considering factors such as engagement, retention, and adaptability. While individual preferences may influence the perceived effectiveness of a particular method, a comprehensive analysis sheds light on the multifaceted nature of learning.

Traditional Classroom Learning : Traditional classroom learning involves face-to-face interactions with instructors and peers. This method proyides structure, guidance, and immediate feedback. However, its effectiveness can be hindered by passive engagement, limited individualization, and challenges in adapting to diverse learning styles.

Online and E-Learning : The digital age has ushered in online and e-leaming platforms, offering flexibility and accessibility. These methods can cater to various learning preferences, allowing self-paced study and multimedia integration. However, concerns about limited interpersonal interactions and potential distractions may impact engagement and social aspects of learning.

Experiential Learning : Experiential learning emphasizes hands-on experiences, fostering active engagement and practical application. Field trips, simulations, and real-world projects enhance understanding and retention. However, logistical challenges, resource requirements, and potential time constraints can limit the widespread implementation of experiential learning.

Collaborative Learning: Collaborative learning encourages group interactions, discussions, and joint problem-solving. This approach enhances social skills, provides diverse perspectives, and promotes a sense of shared responsibility. However, challenges in group dynamics, potential inequalities in contribution, and individual preferences for autonomy may affect its effectiveness.

Self-Directed Learning: Self-directed learning empowers individuals to take control of their learning journey. This approach promotes autonomy, critical thinking, and personalized exploration. However, it requires strong self-discipline, motivation, and the ability to set and achieve goals, making it more suitable for certain learners.

Blended Learning: Blended learning combines traditional and online methods, offering a hybrid approach. It provides flexibility while maintaining some face-to-face interactions. Blended learning, addresses the limitations of both traditional and online methods, catering to a broader range of learning preferences.

Effectiveness Assessment: Determining the most effective learning method is subjective and contingent on various factors. Considerations include the nature of the subject matter, individual learning styles, technological infrastructure, and the availability of resources. Additionally, a combination of methods tailored to specific learning objectives may yield optimal results.

Conclusion : No single learning method universally stands out as the most effective. The suitability of an approach depends on the context and learner preferences. Blending methods to create a diverse and adaptive learning environment might be the key to addressing the varied needs of today’s learners. Ultimately, the effectiveness of a learning method is a dynamic and evolving concept that requires continual evaluation and adjustment.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 4.
Examine different elements impacting the learning process. Is it possible to enhance the learning’ experience?
Answer:
The learning process is intricately woven into a web of various elements that significantly influence its efficacy. Understanding these factors provides insights into potential avenues for improving the overall learning experience.

Socio-Cultural Environment : The socio-cultural context, including family dynamics, community influences, and cultural backgrounds, shapes an individual’s learning journey. Creating a supportive and inclusive environment that respects diverse perspectives and values can positively impact the learning experience.

Educational Environment: The school or educational setting plays a pivotal role in learning. Factors such as classroom climate, teacher-student relationships, and the availability of resources contribute to the overall educational experience. An engaging and well-equipped educational environment fosters a positive atmosphere for learning.

Individual Characteristics : Learner-specific traits, including prior knowledge, cognitive abilities, and motivation levels, significantly impact how information is processed and retained. Recognizing and addressing individual differences through personalized learning approaches can enhance overall comprehension and engagement.

Teaching Methods and Pedagogy : The choice of teaching methods and pedagogical approaches directly influences the effectiveness of knowledge transfer. Incorporating diverse instructional strategies, including experiential learning, collaborative activities, and technology integration, caters to varied learning styles and enhances the overall learning experience.

Technology Integration : In the digital age, technology has become a potent tool for learning. Utilizing educational technology, such as interactive simulations, online resources, and virtual classrooms, can enhance accessibility, engagement, and the overall effectiveness of the learning process.

Feedback Mechanisms : Regular and constructive feedback is crucial for learner growth. Implementing effective feedback mechanisms, including assessments, evaluations, and peer reviews, provides learners with valuable insights into their progress, fostering a culture of continuous improvement.

Lifelong Learning Opportunities : Creating a culture of lifelong learning beyond formal education expands opportunities for skill development and knowledge acquisition. Encouraging individuals to engage in continuous learning through workshops, online courses, and professional development programs contributes to ongoing personal and professional growth.

Inclusive and Diverse Curriculum : A curriculum that reflects diverse perspectives, cultures, and disciplines enriches the learning experience. Inclusivity in educational content ensures that learners encounter a broad range of ideas, fostering critical thinking and a more comprehensive understanding of the subject matter.

Enhancing the Learning Experience : Yes, learning can be improved by strategically addressing these factors. Implementing evidence-based teaching practices, fostering a positive socio-cultural and educational environment, leveraging technology, and adopting inclusive pedagogies contribute to an enriched learning experience.

In conclusion, the learning process is a complex interplay of multiple elements, each contributing to the overall efficacy of education. Recognizing these factors and actively seeking opportunities to enhance the learning experience empowers educators, institutions, and learners to create a more dynamic and impactful educational journey.

Group – C

Objective type Questions with Answers
II. Multiple Choice Questions with Answers :

Question 1.
What is the role of senses in the learning process?
(i) Senses have no significant impact on learning.
(ii) Learning is enhanced when the maximum number of senses is utilized.
(iii) Senses only play a role in theoretical subjects.
(iv) Learning is not influenced by sensory engagement.
Answer:
(ii) Learning is enhanced when the maximum number of senses is utilized.

Question 2.
How does revision and practice contribute to learning outcomes?
(i) They have no impact on learning.
(ii) They hinder the learning process.
(iii) They lead to better retention, reproduction, and utilization of learning.
(iv) Revision and practice only benefif certain subjecfs.
Answer:
(iii) They lead to better retention, reproduction, and utilization of learning.

Question 3.
What is emphasized regarding teaching methods in enhancing learning?
(i) Teacher-dominated methods are always more effective.
(ii) Lecture-based approaches are the most successful.
(iii) Learner-centered methods with teacher-pupil interaction are more helpful.
(iv) Teaching methods have no influence on learning outcomes.
Answer:
(iii) Learner-centered methods with teacher-pupil interaction are more helpful.

Question 4.
What are the two broad categories of media in the context of learning?
(i) Visual and auditory media.
(ii) Print and non-print media.
(iii) Digital and analog media.
(iv) Traditional and modem media.
Answer:
(ii) Print and non-print media.

Question 5.
How can non-print media contribute to learning more efficiently than print media?
(i) They are less engaging for students.
(ii) Non-print media lacks diversity.
(iii) They facilitate faster learning and meet diverse objectives.
(iv) Print media is always superior.
Answer:
(iii) They facilitate faster learning and meet diverse objectives.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 6.
What is school culture based on?
(i) Recent developments
(ii) Future aspirations
(iii) Shared beliefs and attitudes
(iv) Individual preferences
Answer:
(iii) Shared beliefs and attitudes

Question 7.
How does school climate differ from school culture?
(i) School climate is based on past experiences, while school culture is about future actions.
(ii) School climate is more about physical aspects, while school culture focuses on psychological aspects.
(iii) School climate characterizes the district-wide organization, while school culture is at the school building and classroom level.
(iv) There is no difference; the terms are used interchangeably.
Answer:
(ii) School climate is more about physical aspects, while school culture focuses on psychological aspects.

Question 8.
What does school climate mainly emphasize?
(i) Shared beliefs
(ii) Safety
(iii) School policies
(iv) Future aspirations
Answer:
(ii) Safety

Question 9.
What are the four aspects defining school climate in the comprehensive view?
(i) Physical, social, academic, and administrative
(ii) Emotional, cultural, learning, and environmental
(iii) Financial, technological, disciplinary, and leadership
(iv) Welcoming, communicative, affective, and academic
Answer:
(i) Physical, social, academic, and administrative

Question 10.
Which of the following is considered a physical resource influencing learning in school?
(i) School policies
(ii) Leadership
(iii) Library facilities
(iv) Accountability
Answer:
(iii) Library facilities

Question 11.
According to the social constructivist view, how is learning oriented and guided?
(i) Individually focused
(ii) Culturally oriented
(iii) Technologically driven
(iv) Biologically influenced
Answer:
(ii) Culturally oriented

Question 12.
What is the primary influence in the socialization process of an individual, shaping behavior patterns, attitudes, and values?
(i) School
(ii) Neighbourhood
(iii) Family
(iv) Peers
Answer:
(iii) Family

Question 13.
How does a healthy peer relationship impact learning?
(i) It hinders the learning process.
(ii) It creates a competitive and stressful environment.
(iii) It provides a tension-free environment for learning.
(iv) It has no significant influence on learning.
Answer:
(iii) It provides a tension-free environment for learning.

Question 14.
What role do caste, class, and religion play in shaping identity, self-concept, and attitudes?
(i) No significant role
(ii) Minor influence
(iii) Predominant role
(iv) Limited impact
Answer:
(iii) Predominant role

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 15.
How does the socio-economic status of a group influence learning opportunities?
(i) No influence
(ii) Negligible impact
(iii) Directly linked with learning opportunities
(iv) Indirectly related to learning experiences
Answer:
(iii) Directly linked with learning opportunities

Question 16.
What is intelligence, according to Alfred Binet?
(i) The ability to deal with cognitive complexity
(ii) Judgment, practical sense, initiative
(iii) Goal-directed adaptive behavior
(iv) A set of problem-solving skills
Answer:
(ii) Judgment, practical sense, initiative. ,

Question 17.
According to Howard Gardner, how many distinct abilities are there in his theory of multiple intelligences?
(i) 5
(ii) 7
(iii) 8
(iv) 10
Answer:
(iii) 8

Question 18.
Which theory of intelligence describes three fundamental aspects: Analytic intelligence, Creative intelligence, and Practical intelligence?
(i) Cattell-Hom-Carroll theory
(ii) Howard Gardner’s multiple intelligences
(iii) Sternberg’s triarchic theory
(iv) Alfred Binet’s theory
Answer:
(iii) Sternberg’s triarchic theory

Question 19.
What is the aggregate or global capacity of an individual, according to David Wechsler?
(i) Logical-mathematical intelligence
(ii) Emotional intelligence
(iii) Practical intelligence
(iv) The capacity to act purposefully, think rationally, and deal effectively with the environment.
Answer:
(iv) The capacity to act’purposefully, think rationally, and deal effectively with the environment.

Question 20.
Which factor is at the top of the hierarchy in the Cattell-Horn-Carroll theory of intelligence?
(i) Fluid intelligence (Gf)
(ii) Crystallized intelligence (Gc)
(iii) General intelligence factor ‘g’
(iv) Quantitative reasoning (Gq)
Answer:
(iii) General intelligence factor ‘g’

Question 21.
What is the main role of motivation in the learning process?
(i) Hindering learning
(ii) Initiating learning
(iii) Stagnating learning
(iv) Irrelevant to learning
Answer:
(ii) Initiating learning

Question 22.
What is the term for an internal state that arouses, directs, and maintains behavior, according to the definition given in the passage?
(i) Emotion
(ii) Interest
(iii) Maturation
(iv) Motivation
Answer:
(iv) Motivation

Question 23.
According to the passage, what is maturation in the context of readiness to learn?
(i) Biological aging
(ii) Acquiring knowledge and skills
(iii) A natural process of unfolding development stages
(iv) Cognitive ability
Answer:
(iii) A natural process of unfolding development stages

Question 24.
How are intrinsic motivation and extrinsic motivation different?
(i) Intrinsic motivation is driven by external factors, while extrinsic is driven by internal factors.
(ii) Intrinsic motivation is for long-term goals, while extrinsic is for short-term goals.
(iii) Intrinsic motivation is internal to the individual, driven by the task itself, while extrinsic is driven by external rewards or punishment.
(iv) Intrinsic motivation is for learning, while extrinsic is for performance.
Answer:
(iii) Intrinsic motivation is internal to the individual, driven by the task itself, while extrinsic is driven by external rewards or punishment.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 25.
What does the triarchic theory of intelligence propose?
(i) Intelligence is a single, general ability.
(ii) Intelligence is comprised of three aspects: Analytic intelligence, Creative intelligence, and Practical intelligence.
(iii) Intelligence is solely based on cognitive abilities.
(iv) Intelligence is fixed and cannot be developed.
Answer:
(ii) Intelligence is comprised of three aspects: Analytic intelligence, Creative intelligence, and Practical intelligence.

Question 26.
What is intelligence, according to Alfred Binet?
(i) The ability to deal with cognitive complexity
(ii) Judgment, practical sense, initiative
(iii) Goal-directed adaptive behavior
(iv) A set of problem-solving skills
Answer:
(ii) Judgment, practical sense, initiative

Question 27.
According to the Cattell-Hom-Carroll theory, what is at the top of the hierarchy of factors related to intelligence ?
(i) Fluid intelligence (Gf)
(ii) Crystallized intelligence (Gc)
(hi) General intelligence factor ‘g’
(iv) Quantitative reasoning (Gq)
Answer:
(iii) General intelligence factor ‘g’

Question 28.
What are the three fundamental aspects of intelligence in Sternberg’s triarchic theory?
(i) Fluid, crystallized, and quantitative intelligence
(ii) Analytic, creative, and practical intelligence
(iii) Visual, auditory, and processing speed intelligence
(iv) Short-term memory, long-term storage, and retrieval intelligence
Answer:
(ii) Analytic, creative, and practical intelligence

Question 29.
What does the Theory of Successful Intelligence by Sternberg emphasize?
(i) The importance of fluid intelligence
(ii) The need for high IQ scores
(iii) The integration of analytic, creative, and practical intelligence
(iv) The dominance of crystallized intelligence
Answer:
(iii) The integration of analytic, creative, and practical intelligence

Question 30.
According to Howard Gardner, how many distinct abilities are there in his theory of multiple intelligences?
(i) Six
(ii) Seven
(iii) Eight
(iv) Ten
Answer:
(iii) Eight

Question 31.
What is motivation defined as formally?
(i) External stimuli that lead to behavior
(ii) An internal state that arouses, directs, and maintains behavior
(iii) The ability to adapt to a new environment
(iv) The aggregate capacity of the individual to act purposefully
Answer:
(ii) An internal state that arouses, directs, and maintains behavior

Question 32.
Which type of motivation is exhibited when a learner engages in an activity as a means to an end?
(i) Extrinsic motivation
(ii) Intrinsic motivation
(iii) Cognitive motivation
(iv) Emotional motivation
Answer:
(i) Extrinsic motivation

Question 33.
What is situated motivation?
(i) Motivation based on external rewards
(ii) Motivation influenced by the learner’s internal state
(iii) Motivation arising from environmental conditions
(iv) Motivation focused on personal goals
Answer:
(iii) Motivation arising from environmental conditions

Question 34.
How does maturation affect readiness to learn?
(i) Maturation accelerates the learning process
(ii) Maturation is unrelated to readiness
(iii) Maturation determines the appropriate time for learning
(iv) Maturation is a hindrance to learning
Answer:
(iii) Maturation determines the appropriate time for learning

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 35.
What role do emotions play in learning?
(i) Emotions have no impact on learning
(ii) Emotions only affect learning negatively
(iii) Emotions have a bi-directional and complex relationship with learning
(iv) Emotions are unrelated to cognitive processes
Answer:
(iii) Emotions have a bi-directional and complex relationship with learning

Question 36.
How are interests related to learning?
(i) Interests have no impact on learning
(ii) Interests are determined by age only
(iii) Interests influence what one learns and how well
(iv) Interests are external factors in the learning process
Answer:
(iii) Interests influence what one learns and how well

II. Fill in the blanks :

Question 1.
Intelligence is positively related to the learning ability of children. Vandana, Iqbal, Sanjay, Elizabeth, and Vishal showcase different _____ .
Answer:
Talents

Question 2.
Psychometric testing is widely used to measure intelligence, with IQ tests reflecting the Cattell-Hom-Carroll theory’s hierarchy of factors, where ‘g’ stands for general _____.
Answer:
Intelligence

Question 3.
According to Howard Gardner, there are eight distinct areas of multiple _____ .
Answer:
Intelligence

Question 4.
Motivation is an internal state that arouses, directs, and maintains _____ .
Answer:
Behavior

Question 5.
Mr. Dey, the History teacher, struggles with his disinterested class, emphasizing the importance
of fostering _____ for learning.
Answer:
Motivation

Question 6.
Maturation is a relatively permanent change in an individual that occurs as a result of biological _____
Answer:
Aging

Question 7.
Emotions can affect learning both positively and _____.
Answer:
Negatively

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 8.
Interests are determined by the need structure of an individual and influence what one learns best, serving as a key factor in _____.
Answer:
Motivation

Question 9.
Attitudes, whether positive or negative, influence the direction, intensity, and commitment with which one engages in an _____.
Answer:
Activity

Question 10.
Prejudice is a negative attitude towards an object or event, while stereotypes represent a mind set or readiness to react in a certain way to a _____.
Answer:
Stimulus situation

Question 11.
The limbic system, located in the middle of the brain, interprets and directs _____ and behavior.
Answer: Emotions

Question 12.
The triarcbic theory of intelligence proposed by Robert Sternberg includes Analytic, Creative, and Practical _____.
Answer:
Intelligence

Question 13.
In the Cattell-Horn-Carroll theory, ‘g’ represents the general _____ factor at tjie top of the hierarchy.
Answer:
Intelligence

Question 14.
Motivation plays a pivotal role in activating, guiding, and maintaining _____.
Answer:
Learning

Question 15.
Adequate motivation not only sets activities in motion but also sustains and directs these activities, serving as an indispensable factor in promoting _____.
Answer:
Learning

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 16.
Family has a significant impact on a child’s, _____ influencing behavior patterns, social skills, and cultural norms.
Answer:
Learning

Question 17.
A congenial atmosphere based on mutual respect and faith in the child-parent relationship can positively impact the student’s _____.
Answer:
Learning

Question 18.
The _____ and community contribute to shaping a child’s attitudes, habits, beliefs, and social roles through direct and indirect experiences.
Answer:
Neighbourhood

Question 19.
A sound peer relationship in the classroom creates a tension-free environment, facilitating better _____ and competition among students.
Answer:
Learning

Question 20.
Socio-economic status is directly linked to the degree of _____ available in the learning environment, influencing learning opportunities.
Answer:
Stimulation

Question 21.
The term ‘school environment’ includes both ‘school culture’ and ‘school climate,’ which affect the behavior of _____ and _____
Answer:
teachers, students.

Question 22.
_____ reflects the shared beliefs and attitudes that characterize the districtwide organization and establish boundaries for its constituent units.
Answer:
School culture.

Question 23.
School culture is based on past experience and provides a template for future action based on “how we do things in this _____.”
Answer:
organization.

Question 24.
School climate focuses on the feelings and attitudes about a school expressed by ______, _____, _____, and _____.
Answer:
students, teachers, staff, parents.

Question 25.
The physical and psychological aspects of the school that are more susceptible to change and provide the preconditions ‘necessary for teaching and learning to take place are part of the _____.
Answer:
school climate.

Question 26.
Learning, according to John Dewey, is an _____ process where individuals construct meaning through experiences.
Answer:
active and social

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 27.
According to Lev Vygotsky, learning is most effective in the Zone of Proximal Development (ZPD) with guidance from a knowledgeable _____.
Answer:
other

Question 28.
B.F. Skinner, associated with behaviorism, defined learning as a change in behavior due to _____ or punishment.
Answer:
reinforcement

Question 29.
Learning is a process that involves some degree of _____ as temporary changes in behavior are not considered true learning.
Answer:
permanence

Question 30.
Learning is _____ and continuous, applicable to every creature throughout their life.
Answer:
universal

Question 31.
Learning is a comprehensive process that covers all domains of human behavior, including Cognitive, Affective, and _____ domains.
Answer:
Psychomotor

Question 32.
Changes in behavior due to native response tendencies like instincts and reflexes are not considered part of the process of _____.
Answer:
Learning

III. Correct the Sentences:

Question 1.
Learning is the process by which an individual acquires knowledge, attitudes, and skills that are necessary to meet the demands of life.
Answer:
Learning is the process through which individuals acquire knowledge, attitudes, and skills essential to meet the demands of life.

Question 2.
He emphasized hands-on, experientialieaming as essential for genuine understanding. Answer: Dewey emphasized hands-on, experiential learning as essential for genuine understanding.

Question 3.
For example, a child learns how to clap hands by moving someone else do it.
Answer:
For example, a child learns how to clap hands by seeing someone else do it.

Question 4.
Learning occupies a very important place in our life.
Answer:
Learning occupies a crucial place in our life.

Question 5.
Attitudes, fears, gestures, motor skills, language skills, etc. are the products of learning Answer: Attitudes, fears, gestures, motor skills, language skills, etc., are the outcomes of learning.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 6.
Learning is an aimless activity.
Answer:
Learning is not an aimless activity.

Question 7.
Learning helps the individual to adjust herself/himself adequately and adapt to the changes that may be necessary to the new situations.
Answer:
Learning helps the individual to adjust themselves adequately and adapt to the changes that may be necessary in new situations.

Question 8.
Carryinging the content matter by a learner for examination and forgetting it after sometime does not bring any change (to some extent to permanence) in the total behaviour pattern of the learner.
Answer:
Cramming the content matter by a learner for examination and forgetting it after soihe time does not bring any lasting change in the
total behavior pattern of the learner.

Question 9.
Learning is comprehensive.The scope of leaning is spread over each and every dimension of life.
Answer:
Learning is comprehensive: The scope of learning is spread over each and every dimension of life.

Question 10.
Learning leads to changes in behavior but this does not necessarily mean that these changes always bring about improvement or negative development.
Answer:
Learning leads to changes in behavior, but this does not necessarily mean that these changes always bring about improvement or positive development.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 11.
Changes in behaviour on the basis of native response tendencies like instructs and reflexes (e.g. infant’s sucking behaviour, blinking at bright lights) cannot be attributed to learning. Answer: Changes in behavior based on native response tendencies like instincts and reflexes (e.g., infant’s sucking behavior, blinking at bright lights) cannot be attributed to learning.

Question 12.
Varied teaching methods, catering to different leaning styles, enhance understanding and interest.
Answer:
Varied teaching methods, catering to different learning styles, enhance understanding and interest.

Question 13.
Employing a mix of instructional techniques ensures that students with same learning preferences can access and benefit from the material.
Answer:
Employing a mix of instructional techniques ensures that students with diverse learning preferences can access and benefit from the material.

II. Answer the following questions in one word :

Question 1.
What is learning?
Answer:
Learning is the process of acquiring knowledge, attitudes, and skills through experiences, leading to a change in behavior.

Question 2.
How can learning occur?
Answer:
Learning can result from both direct experiences and observing others (vicarious experiences).

Question 3.
According to John Dewey, what is the nature of learning?
Answer:
John Dewey viewed learning as an active and social process where individuals construct meaning through experiences.

Question 4.
What does Piaget’s theory emphasize about learning?
Answer:
Piaget’s theory highlights learning as a process of adaptation and assimilation, with stages in intellectual growth.

Question 5.
According to Vygotsky, what is crucial for effective learning?
Answer:
Vygotsky emphasized the significance of social interaction in learning, introducing the Zone of Proximal Development (ZPD).

Question 6.
How did B.F. Skinner define learning?
Answer:
Skinner defined learning as a change in behavior due to reinforcement or punishment, focusing on observable behaviors.

Question 7.
What is the Zone of Proximal Development (ZPD) in learning?
Answer:
ZPD, proposed by Vygotsky, is the range where learning is most effective with guidance from a knowledgeable other.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 8.
What did Maria Montessori emphasize about learning?
Answer:
Montessori viewed learning as a natural process unfolding in a prepared environment, promoting self-directed, hands-on learning.

Question 9.
How did Ivan Pavlov contribute to the understanding of learning?
Answer:
Pavlov, known for classical conditioning, viewed learning as a process of associating stimuli with responses.

Question 10.
According to Carl Rogers, what is essential for meaningful education?
Answer:
Rogers emphasized learner-centered education, stating that a supportive and.non- judgmental environment is crucial for meaningful learning.

Question 11.
What is intelligence, according to Howard Gardner’s theory?
Answer:
Intelligence, according to Gardner, involves eight distinct abilities, including linguistic, logical-mathematical, spatial, musical, bodily-kinesthetic, interpersonal, intrapersonal, and naturalist intelligence. .

Question 12.
How does motivation impact learning, according to the text?
Answer:
Motivation serves to activate, guide, and maintain behavior; it directs behavior toward goals, increases effort and energy, and affects cognitive processes.

Question 13.
What is extrinsic motivation?
Answer:
Extrinsic motivation is when individuals are driven by external factors, such as grades, money, or recognition, rather than finding inherent pleasure or interest in the task itself.

Question 14.
What does maturation refer to in the context of learning readiness?
Answer:
Maturation is the relatively permanent change in an individual’s cognitive, emotional, or physical aspects due to biological aging, influencing their readiness to learn specific skills or concepts.

Question 15.
Why is it important for teachers to consider maturation in their teaching approach?
Answer:
Maturation determines the readiness for learning, guiding teachers in deciding what, how, and when to teach based on the developmental stages of their students.

CHSE Odisha Class 12 Education Solutions Chapter 6 Meaning, Nature, and Factors of Learning

Question 16.
What is the primary influence of the family in a child’s learning process?
Answer:
The family shapes behavior patterns, social skills, and cultural norms for the child’s learning.

Question 17.
How does a healthy peer relationship impact learning in a classroom?
Answer:
It creates a tension-free environment, promoting better learning and competition among students.

Question 18.
What role do caste, class, and religion play in shaping identity and attitudes in our country?
Answer:
They significantly influence self-concept, attitudes, and value orientation in individuals.

Question 19.
Why is a distorted and unhealthy family environment detrimental to a student’s learning?
Answer:
It adversely affects the student’s ability to cope and resist learning.

Question 20.
According to the social constructivist view, what guides all learning?
Answer:
Learning is culturally oriented and guided according to the social constructivist view.

Introduction

Learning is the process by which an individual acquires knowledge, attitudes and skills that are necessary to meet the demands of life. While touching a burning candle, a child gets burnt and he withdraws the fingers. When he faces a similar situation again he withdraws his fingers faster. Gradually he learns to avoid not only the burning candle but also other burning things. The behaviour of an individual is thus changed through experiences. This change in behaviour brought about by experiences is commonly known as learning.

Thus, Learning means change in behaviour or behaviour potential that occurs as a result of experience. Learning can result from both vicarious and direct experiences. Vicarious means observing someone and learning from that observation and not being directly involved in the experience. For example, a child learns how to clap hands by seeing someone else do it. Learning also takes place through direct experiences. For example, a child learns to write by practicing writing. A child normally learns from his parents, teachers and the environment.

Definitions of Learning :
• John Dewey (1859-1952): Learning, according to Dewey, is an active and social process where individuals construct meaning through experiences. He emphasized hands-on, experiential learning as essential for genuine understanding.

• Jean Piaget (1896-1980) : Piaget’s theory of cognitive development posits that learning is a process of adaptation and assimilation. He highlighted the role of stages in intellectual growth, where individuals actively construct knowledge through interactions with their environment.

• Lev Vygotsky (1896-1934) : Vygotsky’s sociocultural theor/ emphasizes the significance of social interaction in learning. He proposed the concept of the Zone of Proximal Development (ZPD), where learning is most effective with guidance from a knowledgeable other.

• Maria Montessori (1870-1952) : Montessori viewed learning as a natural process that unfolds in a prepared environment. She emphasized the importance of self-directed, hands- on learning and believed that children learn best through exploration and discovery.

• Jerome Bruner (1915-2016) : Bruner stressed the importance of the active organization of knowledge. His constructivist theory posited that learning involves a spiral curriculum, where complex ideas are presented at a basic level and revisited as learners develop a deeper understanding.

• B.F. Skinner (1904-1990) : Skinner, associated with behaviorism, defined learning as a change in behavior due to reinforcement or punishment. He focused on observable behaviors and the role of the environment in shaping and controlling behavior.

• Howard Gardner (b. 1943) : Gardner proposed the theory of multiple intelligences, suggesting that learning is diverse and individuals have varying strengths in different forms of intelligence. He identified several intelligences, including linguistic, logical-mathematical, and interpersonal.

• Albert Bandura (b. 1925) : Bandura’s social cognitive theory posits that learning occurs through observation, imitation, and modeling. He emphasized the role of self-efficacy, where individuals ’ beliefs about their abilities influence their learning and behavior.

• Ivan Pavlov (1849-1936): Pavlov, known for classical conditioning, viewed learning as a process of associating stimuli with responses. He conducted experiments with dogs, demonstrating how behaviors could be elicited through conditioned responses to stimuli.

• Carl Rogers (1902-1987) : Rogers emphasized learner-centered education, where learning is facilitated by creating a supportive and non-judgmental environment. He believed that self-directed learning, driven by personal experiences and interests, is essential for meaningful education.

Nature of learning:
Learning occupies a very important place in our life. It provides a key to the structure of our personality and behaviour. Experience, direct or indirect, plays a very important and dominating role in moulding and shaping the behaviour of the individual from the very beginning. When a child touches a hot pan and gets burnt, s/he immediately withdraws her/his hand and learns to touch such vessels carefully. S/he concludes that if one touches a hot vessel, one gets burnt.

In the same way from other experiences, in her/his day to day life, s/he derives different conclusions and modifies her/his behaviour. These changes in behaviour brought about by experience are commonly known as learning and this process of gaining experiences, drawing conclusions, and changing behaviour goes on from womb to tomb.

• Learning is a process and not a product : Learning is a fundamental and life-long process. Attitudes, fears, gestures, motor skills, language skills, etc. are the products of learning. They are not learning themselves. In a classroom, when learning is viewed as a product then it is viewed as something external. Something like shopping- people go out and buy knowledge and then it becomes their possession.

Paulo Freire in his book ‘Pedagogy of the Oppressed’ criticizes this and says that education thus becomes an act of depositing, in which the students are the depositories and the teacher is the depositor. In this ‘banking’concept of education, the teacher is the subject of the learning process, while the pupils are mere objects. Whereas, when learning is viewed as a process, it is viewed as something internal or personal. It is something that a child does in order to understand the real world and uses it as a tool for survival.

• Learning is purposive or goal directed : Learning is not an aimless activity. All true learning is based on purpose. We do not learn anything and everything that comes in our way in a haphazard manner. However, some experts argue that sometimes learning is unintended.

• Learning generally involves some degree of permanence : Activities bringing temporary change in behaviour and not lasting do not come under learning. For example, cramming the content matter by a learner for examination and forgetting it after sometime does not bring any change (to some extent to permanence) in the total behaviour pattern of the learner and thus this type of learning cannot be said as true learning.

• Learning is universal and continuous : Every creature till it lives, learns. In human beings it is not restricted to any particular age, sex, race or culture. It is a continuous never- ending process which starts from birth and continues till death.

• Learning prepares for adjustment: Learning helps the individual to adjust herself/himself adequately and adapt to the changes that may be necessary to the new situations. We meet with new situations which demand solutions. Repeated efforts are required react to them effectively. These experiences leave behind some effects in the mental structure and modify our behaviour.

• Learning is comprehensive : The scope of learning is spread over each and every dimension of life. It is a very comprehensive process which covers all domains – Cognitive, Affective and Psychomotor- of human behaviour.

• Learning is change in response or behaviour may be favourable or unfavourable : Learning leads to changes in behavior but this does not necessarily mean that these changes always bring about improvement or positive development. There are chances to drift to the negative side too.

• Learning is organizing experience : Learning involves all those experience and training of an individual (right from birth) which help her/him to produce changes in behaviour. It is not mere addition to knowledge or mere acquisition of facts. It is the reorganization of experience which may also include unlearning.

• Instincts and reflexes are not learning : Changes in behaviour on the basis of native response tendencies like instincts and reflexes (e.g. infant’s sucking behaviour, blinking at bright lights)cannot be attributed to learning.

• Learning does not include changes in behaviour on account of maturation, fatigue, illness, or drug etc.

Types of Learning :
Understanding that learning encompasses a broad spectrum of skills, attitudes, and competencies is crucial for enhancing efficiency and effectiveness in our environment. These learnings can be direct or indirect, acquired through experience, conscious effort, critical thinking, reasoning, and even emotions. Let’s explore some of the various types of learning that individuals engage in:

1. Cognitive Learning: Cognitive learning, also known as intellectual learning, involves gaining information, memorization, and understanding relationships between ideas. School and college education primarily focuses on cognitive learning, but it is not limited to formal settings. Everyday activities contribute to cognitive learning as individuals gradually comprehend and improve their skills through insight and understanding.

2. Psychomotor Learning : Psychomotor learning pertains to the development of physical skills and muscular coordination, such as riding a bicycle, stitching, or playing sports. Skills involving eye-hand coordination or gross and fine motor skills fall under psychomotor learning. Mastery of these physical abilities contributes to overall skill development.

3. Affective Learning : Affective learning involves emotions and feelings, including fear, love, passion, and dislikes. Influenced by socio-cultural and emotional surroundings, affective learning encompasses both positive and negative emotions. Children, for instance, absorb values and attitudes from their family, and individuals often unconsciously adopt behaviors and sentiments from their environment.

4. Collateral or Concomitant Learning : Unconscious learning from the environment, acquiring attitudes, values, and preferences. Children observing their parents’ behavior learn values and attitudes. Teachers, while primarily focused on teaching subjects, also influence students’ behavior, leading to collateral or concomitant learning.

5. Discrimination Learning : Discrimination learning involves distinguishing between similar or related concepts, emphasizing differences. Gagne places discrimination learning before concept learning, highlighting the importance of understanding subtle differences between concepts. This process contributes to effective concept learning.

6. Principles Learning: Similar to concept learning, principles learning involves understanding relationships within situations, noting common features and relationships. Principles learning focuses on identifying and generalizing common relationships. Principles have applicability in problem-solving and can be applied deductively to parallel situations.

7. Factors Affecting Learning : Learning is a complex process influenced by a myriad of factors that shape individuals’ abilities, preferences, and outcomes. Understanding these factors is crucial for educators, policymakers, and learners themselves to create effective and inclusive learning environments. In this extensive exploration, we delve into the multifaceted dimensions of the factors affecting learning, categorizing them into learner-related, teacher-related, task-related, and environment-related factors.

Learner-Related Factors :
• Maturation : Maturation plays a pivotal role in determining a learner’s readiness to engage in and benefit from the learning process. Physical, intellectual, and socio-emotional maturation are essential components. For instance, a one-year-old might not be developmentally ready to write, emphasizing the importance of aligning educational expectations with the individuai’s stage of maturation.

• Motivation : Motivation is a driving force behind learning. Intrinsic motivation, derived from internal satisfaction and interest, contrasts with extrinsic motivation, driven by external rewards or consequences. Understanding what motivates an individual is crucial for educators. For example, a student’s intrinsic motivation for a particular subject, like Ashok’s passion for cricket, can significantly impact the depth and sustainability of learning.

• Psychological Needs : Psychological needs, encompassing desires unique to individuals, greatly influence learning preferences and engagement. These needs include love, belongingness, social prestige, esteem, recognition, affiliation, self-expression, creativity, aesthetics, and higher cognitive quests. Recognizing and addressing these needs can enhance the relevance and meaningfulness of the learning experience.

• Self-Concept: Self-concept, the beliefs individuals hold about their talents, potentialities, weaknesses, and strengths, shapes their behavior and learning orientations. Positive self-concept fosters effective learning, while a negative self-image may hinder academic performance. Understanding and nurturing a positive self-concept is crucial for educators.

Levels of Aspiration : The extent to which an individual strives to achieve, known as their levels of aspiration, influences the dedication and effort put into learning. Setting high aspirations can propel learners to engage more intensively with the learning process, fostering a sense of purpose and direction.

• Interests : Interests are activities or pastimes voluntarily engaged in for enjoyment. Individuals are more likely to invest time and energy in areas of personal interest. Recognizing and incorporating students’ interests into the learning curriculum can enhance motivation and facilitate a more enjoyable learning experience.

• Study and Work Habits : Effective study and work habits are instrumental in determining the efficiency of the learning process. Organizing material, managing time judiciously, and employing strategic learning methods contribute to better performance. Educators play a crucial role in guiding students to develop effective study habits. Understanding and addressing these learner-related factors require personalized approaches that acknowledge the diversity of individuals and their unique learning profiles.

Teacher-Related Factors :
Personality: The teacher’s personality traits significantly influence the classroom atmosphere and student engagement. Modelling, enthusiasm, caring, and positive expectations create a conducive learning environment. Teachers who exhibit these traits can inspire and motivate students to actively participate in the learning process.

Communication : Effective communication is paramount in facilitating understanding and engagement. Clear and engaging communication enhances students’ comprehension and involvement. Teachers must convey information in a manner that resonates with diverse learning styles, ensuring that all students have access to the material.

Leadership Style : The teacher’s leadership style shapes the classroom dynamics and student response. Whether adopting an authoritarian, democratic, benevolent, or indifferent approach, the teacher’s leadership style influences the freedom of communication and student motivation. Creating an environment that encourages open dialogue and collaboration fosters a positive learning atmosphere.

Interpersonal Relationships : The relationships teachers cultivate with their students significantly impact the emotional tone and motivation within the classroom. Positive relationships, characterized by respect, empathy, and open communication, create a comfortable and motivated learning environment. Conversely, negative relationships may induce fear and hinder the learning experience.

Teaching Styles : The manner in which teachers present information, known as their teaching styles, can vary widely. Some teachers may adopt a lecture-based approach, while others favor interactive or hands-on methods. Recognizing and incorporating diverse teaching Styles can accommodate different learning preferences, ensuring a more inclusive educational experience.

Expectations : Teacher expectations play a crucial role in shaping student behavior and outcomes. Positive teacher expectations communicate that all students, can learn to their fullest potential. Teachers who set reasonable expectations and create a trusting and supportive learning environment can inspire students to meet and exceed those expectations.

Task Interpretation : How teachers interpret and present tasks to students influences the perceived meaning, purpose, and relevance of the learning material. Aligning tasks with real-world implications and demonstrating the applicability of knowledge enhances students’ understanding and motivation. Task interpretation is central to bridging the gap between academic content and its practical significance. Understanding and leveraging these teacher-related factors require ongoing professional development and a commitment to creating a positive and inclusive learning environment.

Task-Related Factors :
1. Methods Used : Teaching methods and strategies impact the comprehension and engagement of students. Varied teaching methods, catering to different learning styles, enhance understanding and interest. Employing a mix of instructional techniques ensures that students with diverse learning preferences can access and benefit from the material.

2. Nature of Task : The complexity and relevance of tasks significantly influence student motivation and learning outcomes. Tasks aligned with real-world implications and applicability enhance meaning and purpose in learning. Educators must carefully design tasks that challenge students while fostering a sense of value and accomplishment.

3. Teaching Aids Used : Visual aids and resources contribute to the effectiveness of teaching and student understanding. Incorporating multimedia presentations, interactive tools, and other visual aids enhances the visual and experiential aspects of learning. Utilizing a variety of teaching aids caters to diverse learning styles and reinforces key concepts.

4. Structure of the Discipline : The organization and presentation of subject matter impact students’ grasp of concepts. A well-structured curriculum facilitates systematic learning and comprehension. Educators must carefully consider how content is organized and presented to ensure that students can build a coherent understanding of the discipline. Task-related factors require educators to be flexible and adaptive, tailoring instructional approaches to the specific needs and preferences of their students.

Environment-Related Factors :
1. Physical Environment: The physical setting of the classroom, including factors such as seating arrangement, lighting, and overall comfort, can impact the learning experience. A well- organized, comfortable classroom promotes focus and engagement. Educators must consider the physical environment’s influence on student well-being and concentration.

2. Socio-Emotional Climate : The interpersonal feeling tones within the classroom, known as the socio-emotional climate, significantly influence student receptivity and productivity. Positive
classroom climate, characterized by respect, empathy, and open communication, enhances motivation and learning outcomes. Fostering a supportive and inclusive socio-emotional climate is essential for creating a conducive learning environment.

3. Attention to Environmental Factors : Recognition of external factors, such as climate and comfort, is crucial for understanding their impact on learning. Teachers and administrators must consider the influence of physical comfort and socio-emotional dynamics on student engagement and performance. Addressing environmental factors ensures a more holistic approach to supporting learners.

Conclusion : In conclusion, learning is a multifaceted process intricately woven with various factors that influence its dynamics. Learner-related factors emphasize the importance of recognizing individual differences and tailoring educational approaches to diverse needs. Teacher-related factors underscore the critical role educators play in creating a positive and inclusive learning environment through their personalities, communication, and instructional strategies.

Task-related factors highlight the need for well-designed and relevant learning tasks that challenge students while fostering a sense of value and accomplishment. Finally, environment- related factors encompass the physical and socio-emotional aspects of the learning environment, emphasizing the need for comfortable and supportive surroundings.

Understanding and addressing these factors require collaboration among educators,policymakers, and learners themselves. Ongoing research and professional development are essential to stay abreast of evolving educational practices that promote effective and inclusive learning experiences. By acknowledging the intricate interplay of these factors, we pave the way for a more enriching and equitable educational landscape.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a)

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(a)

Question 1.
Multiply (2√-3 + 3√-2) by (4√-3 – 5√-2)
Solution:
(2√-3 + 3√-2) by (4√-3 – 5√-2)
= (2√3i + 3√2i) (4√3i – 5√2i)
= i2 (2√3 + 3√2) (4√3 – 5√2)
= – 1(24 – 10√6 + 12√6 – 30)
= – 1(- 6 + 2√6) = 6 – 2√6

Question 2.
Multiply (3√-7 – 5√-2) (3√-2 + 5√-2)
Solution:
(3√-7 – 5√-2) (3√-2 + 5√-2)
= (3√7i – 5√-2i) (3√2i + 5√2i)
= i2 (3√7- 5√2) (3√2 + 5√2)
= (- 1)8√2(3√7 – 5√2 )
= – 24√14 + 80

Question 3.
Multiply (√-1 +√-1) (a – b√-1)
Solution:
(√-1 +√-1) (a – b√-1)
= (i + i) (a – bi) = 2i(a – bi)
= 2ai – 2bi2 = 2ai + 2b

Question 4.
Multiply (x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
Solution:
(x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
= (x + \(\frac{-1-i \sqrt{3}}{2}\)) (x + \(\frac{-1+i \sqrt{3}}{2}\))
= (x + ω) (x + ω2) = x2 + ω2x + ωx + ω3
= x2 + x (ω 2 + ω) + 1 = x2 –  x + 1

Question 5.
Express with rational denominator. \(\frac{1}{3-\sqrt{-2}}\)
Solution:
\(\begin{aligned}
& \frac{1}{3-\sqrt{-2}}=\frac{1}{3-\sqrt{2} \mathrm{i}}=\frac{3+\sqrt{2} \mathrm{i}}{(3-\sqrt{2} \mathrm{i})(3+\sqrt{2} \mathrm{i})} \\
& =\frac{3+\sqrt{2} \mathrm{i}}{9-2 \mathrm{i}^2}=\frac{3+\sqrt{2} \mathrm{i}}{9+2}=\frac{3+\mathrm{i} \sqrt{2}}{11}
\end{aligned}\)

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a)

Question 6.
\(\frac{3 \sqrt{-2}+2(-5)}{3 \sqrt{-2}-2 \sqrt{-2}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a)

Question 7.
\(\frac{3+2 \sqrt{-1}}{2-5 \sqrt{-1}}+\frac{3-2 \sqrt{-1}}{2+5 \sqrt{-1}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 1

Question 8.
\(\frac{a+x \sqrt{-1}}{a-x \sqrt{-1}}-\frac{a-x \sqrt{-1}}{a+x \sqrt{-1}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 2

Question 9.
\(\frac{(x+\sqrt{-1})^2}{x-\sqrt{-1}}-\frac{\left(x-\sqrt{-1^2}\right)}{x+\sqrt{-1}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 3

Question 10.
\(\frac{(a+\sqrt{-1})^3-(a-\sqrt{-1})^3}{(a+\sqrt{-1})^2-(a-\sqrt{-1})^2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 4

Question 11.
Find the value of (- i)4n+3; when n is positive.
Solution:
(- i)4n+3
= (-i4n) (-i)3 = 1(- i3) = – (-i) = i

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a)

Question 12.
Find the square root of (a + 40i) + \(\sqrt{9-40 \sqrt{-i}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 5

Question 13.
Express in the form of a + ib:
(i) \(\frac{3+5 i}{2-3 i}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 6

(ii) \(\frac{\sqrt{3}-i \sqrt{2}}{2 \sqrt{3}-i \sqrt{3}}\)
Solution
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 7

(iii) \(\frac{(\mathrm{I}+\mathrm{i})^2}{3-\mathrm{i}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 8

(iv) \(\frac{(a+i b)^3}{a-i b}-\frac{(a-i b)^3}{a+i b}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 9

(v) \(\frac{1+i}{1-i}\)
Solution:
\(\frac{1+i}{1-i}\) = \(\frac{(1+i)^2}{2}=\frac{1-1+2 i}{2}\)
= i = 0 + i

Question 14.
Express the following points geometrically in the Argand plane.
(i) 1
Solution:
1 = 1 + i0 = (1, 0)

(ii) 3i
Solution:
3i = 0 + 3i = (0, 3)

(iii) – 2
Solution:
– 2 = – 2 + i0 = (- 2, 0)

(iv) 3 + 2i
Solution:
3 + 2i = (3, 2)

(v) – 3 + i
Solution:
– 3 + i = (- 3, 1)

(vi) 1-i
Solution:
1 – i = (1, – 1)

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a)

Question 15.
Show that the following numbers are equidistant from the origin:
√2 +i, 1 + i√2, i√3
Solution:
|√2 + i| = \(\sqrt{(\sqrt{2})^2+1^2}\) = √3
|1 + i√2| = \(\sqrt{1^2+(\sqrt{2})^2}\) = √3
and |i√3| = √3
∴ The points are equidistant from the origin.

Question 16.
Express each of the above complex numbers in the polar form.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) 10

Question 17.
If 1, ω, ω2 are the three cube roots of unity, prove that (1 + ω2)4 = ω
Solution:
L.H.S. = (1 + ω2)4 = (- ω)4 = ω4
= ω3 .ω = 1. ω = ω
= R.H.S. (Proved)

Question 18.
(1 – ω+ ω2 ) (1 + ω – ω2 ) = 4
Solution:
L.H.S. = (1 – ω+ ω2 ) (1 + ω – ω2 )
= (- ω – ω )(- ω2 – ω2 ) (∴ 1 + ω + ω2 = 0)
= (- 2ω2 – 2ω2) = 4ω3 = 4 = R.H.S.

Question 19.
(1 – ω) (1 – ω)2 (1 – ω4) (1 – ω5) = 9
Solution:
L.H.S. =
(1 – ω) (1 – ω)2 (1 – ω4) (1 – ω5)
= (1 – ω) (1 – ω2) (1 – ω) (1 – ω2)
= (1 – ω)2 (1 – ω2)2
= {(1 – ω) (1 – ω2)}2
= (1 – ω2 – ω + ω3)2
= {3 – (ω2 + ω + 1)}2
= (3)2 = 9 = R.H.S.

Question 20.
(2 + 5ω + 2ω2 )6 = (2 + 2ω + 5ω2 )6 =729
Solution:
L.H.S. = (2 + 5ω + 2ω2 )6
(2 + 2ω2 + 5ω)6 = {2(1 + ω2 ) + 5ω}6
(- 2ω + 5ω)6 = (3ω)6 = 729ω6 = 729
Again, (2 + 2ω + 5ω2 )6
= {2(1 + ω) + 5ω2 )6
= (- 2ω2 + 5ω2 )6 = (3ω2)6
= 729ω12 =729
∴ (2 + 5ω + 2ω2 )6 = (2 + 2ω + 5ω2 )6 =729

Question 21.
(1 – ω + ω2 ) (1 – ω2 + ω4) (1 – ω42)  ….to 2n factors = 22n
Solution:
L.H.S. = (1 – ω + ω2 ) (1 – ω2 + ω4) (1 – ω42)  ….to 2n factors = 22n
= (- ω – ω) (1 – ω2 + ω) (1 – ω + ω2 )  ….to 2n factors = 22n
= (- 2ω) (- ω2 – ω2 ) (- ω – ω)  ….to 2n factors = 22n
= [(- 2ω)(- 2ω) … to n factors] × [(- 2ω2)(- 2ω2) …. to n factors]
= (- 2ω)n × (- 2ω2)n = (4ω3)n = 4n = 22n
R.H.S. (Proved)

Question 22.
Prove that x3 + y3 + z3– 3xyz
= (x + y + z) (x + ωy + ω2z) (x + yω2 + zω)
Solution:
R.H.S. = (x + y + z) (x + ωy + ω2z) (x + yω2 + zω)
= (x + y + z) (x + xyω2 + zxω + xyω + y2ω3+ yz ω2 + zxω2 + yzω4 + z2ω3)
= (x + y + z) [x2 + y2 + z2 + xy (ω2 + ω) +yz (ω2 + ω) + zx(ω2 + ω)]
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = L.H.S. (Proved)

Question 23.
If x = a + b, y = aω + b ω2, z = aω2 + bω show that
(1) xyz = a3 + b3
Solution:
L.H.S. = xyz
= (a + b) (a ω + b ω2) (a ω2 + b ω)
= (a + b) (a2 ω3 + ab ω2 + ab ω4 + b2 ω3)
= (a + b) {a2 + b2 + ab(ω2 + ω)}
= (a + b) (a2 – ab + b2) = a3 + b3 = R.H.S. (Proved)

(2) x2 + y2 + z2 = 6ab
Solution:
L.H.S. = x2 + y2 + z2
= (a + b)2 + (a ω + b ω2)2 (a ω2 + b ω)2
= a2 + b2 + 2ab + a2 ω2 + b2 ω4 + 2ab ω3 + a2 ω4 + b2 ω2 + 2ab ω3
= a2 + a2 ω2 + a2 ω + b2 + b2 ω + b2 ω2 + 2ab + 2ab + 2ab
= a2(1 + ω2 + ω) + b2 (1 + ω + ω2) + 6ab
= 0 + 0 + 6ab = 6ab = R.H.S. (Proved)

(3) x3 + y3 + z3 = 3(a3 + b3)
Solution:
L.H.S. = x3 + y3 + z3
= (a + b)3 + (a ω + b ω2)3 + (aω2 + b2)3
= a3 + 3a2b + 3ab2 + b3 + a3 ω3 + 3a2b22 + 3a ωb2 ω4 + b3 ω6 + a3 ω6 + 3a2 ω4bω + 3a ω2b2 ω2 + b3 ω3
= a3 + a3 + a3 + b3 + b3 + b3 + 3a2b(1 + ω4 + ω5) + 3ab2 (1 + ω5 + ω4)
= 3a3 + 3b3 + 0 + 0 = 3 (a3 + b3) = R.H.S. (Proved)

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a)

Question 24.
If ax + by + cz = X, cx + by + az = Y, bx + ay + cz = Z
show that (a2 + b2 + c2 – ab – bc- ca) (x2 +y2 + z2 – xy – yz – zx) = X2 + Y2 + Z2 – YZ – ZX – XY
Solution:
L.H.S.
= (a2 + b2 + c2 – ab – bc – ca) (x2 + y2 + z2 – xy – yz – zx)
= (a + b ω + cω2) (a + bω2 + cω) (x + yω + zω)2 (x + yω2 + z ω) (Refer Q.No.22)
= {(a + bω + cω2) (x + yω + zω2)} {(a + bω2 + cω) (x + yω2 + zω)}
= (ax + ayω + azω2 + bx ω + byω2 + bzω3 + cxω2 + cyω3 + czω4) × (ax + ay ω2 + azω + bxω2 + byω4 + bzω3 + cxω + cyω3 + czω2)
= {(ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω2} x (ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω2}
= (X + Yω2 + Zω) (X + Yω + Zω2)
= X2 + Y2 + Z2 – XY – YZ – ZX (Refer Q. No. 22)
(Where X = ax + cy + bz.
Y = cx + by + az.
Z = bx + ay + cz).

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(a)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(a)

Question 1.
State which of the following is positive.
(i) cos 271°
Solution:
cos 271° is + ve as 271° lies in 4th quadrant.

(ii) sec 73°
Solution:
sec 73° is + ve as sec +ve in the 1st quadrant.

(iii) sin 302°
Solution:
sin 302° is- ve as sin is -ve in the 4th quadrant

(iv) cosec 159°
Solution:
cosec 159° is + ve as 159° lies in 2nd quadrant and cosec is +ve there.

(v) sec 199°
Solution:
sec 199° is – ve as 199° lies in the 3rd quadrant and sec is -ve there.

(vi) cosec 126°
Solution:
cosec 126° is + ve as cosec is +ve in 2nd quadrant.

(vii) cos 315°
Solution:
cos 315° is +ve as 315° lies in 4th quadrant and cos is +ve there.

(viii) cot 375°
Solution:
cot 375° is +ve as 375° lies in 1st quadrant.

Question 2.
Express the following as trigonometric ratios of some acute angles.
(i) sin 1185°
Solution:
sin 1185° = sin\(\left(13 \frac{\pi}{2}+15^{\circ}\right)\)
=(- 1) \(\frac{13-1}{2}\) cos 15° = cos 15°

(ii) tan 235°
Solution:
tan 235° = tan (180° + 45°) = tan 45°

(iii) sin (- 3333°)
Solution:
sin (-3333°) – -sin 3333°
= – sin\(\left(37 \frac{\pi}{2}+3^{\circ}\right)\)
= – (- 1) \(\frac{27-1}{2}\) cos 3° =- cos 3°

(iv) cot (- 3888°)
Solution:
cot (-3888°) = – cot 3888°
= – cot\(\left(43 \frac{\pi}{2}+18^{\circ}\right)\)
= – (- tan 18°) = tan 18°

(v) tan 458°
Solution:
tan 458° = tan\(\left(5 \frac{\pi}{2}+8^{\circ}\right)\) = – cot 8°

(vi) cosec (- 60°)
Solution:
cosec (- 60°) = – cosec 60°

(vii) cos 500°
Solution:
cos 500° = cos\(\left(5 \frac{\pi}{2}+50^{\circ}\right)\)
= – (-1) \(\frac{5+1}{2}\) sin 55° – sin 50°

(viii)sec 380°
Solution:
sec 380° = sec (360° + 20°)
= sec 20°

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(a)

Question 3.
Find the domain of tangent and cotangent functions.
Solution:
Domain of tan x is R – \(\left\{\frac{(2 n+1) \pi}{2}, n \in Z\right\}\) as tangent is not defined for
x = \(\frac{(2 n+1) \pi}{2}\)
The domain of cot x is R – {nπ, n ∈ Z} as cotangent is not defined for x = nπ.

Question 4.
Determine the ranges of sine and cosine functions.
Solution:
The maximum and minimum values of sine and cosine are 1 and -1, respectively.
∴ Ranges of sine and cosine are [-1, 1].

Question 5.
Find a value of A when cos 2A = sin 3A
Solution:
cos 2A = sin 3A = cos (90° – 3A)
or, 2A = 90° – 3A
or, 5A = 90° or, A = 18°

Question 6.
Find the value of
cos 1°. cos 2° …..cos 100°
Solution:
cos 1° cos 2° …..cos 100°
= 0 as cos 90° is there which is zero.

Question 7.
Find the value of
cos 24° + cos 5° + cos 175° + cos 204° + cos 300°
Solution:
cos 24° + cos 5° + cos 175° + cos 204° + cos 300°
= cos 24° + cos 5° + cos (180° – 5°) + cos (180° + 24°) + cos (360°- 60°)
= cos 24° + cos 5° – cos 5° – cos 24° + cos 60° = cos 60° = 1/2

Question 8.
Evaluate
tan\(\frac{\pi}{20}\).tan\(\frac{3 \pi}{20}\).tan\(\frac{5 \pi}{20}\).tan\(\frac{7 \pi}{20}\).tan\(\frac{9 \pi}{20}\)
Solution:
tan\(\frac{\pi}{20}\).tan\(\frac{3 \pi}{20}\).tan\(\frac{5 \pi}{20}\).tan\(\frac{7 \pi}{20}\).tan\(\frac{9 \pi}{20}\)
= tan 9° tan 27° tan 45° tan 63° tan 81°
= tan 9°. tan 27°. 1 tan (90° – 27°). tan (90° – 9°)
= tan 9° tan 27° cot 27° cot 9°
= (tan 9°. cot 9°) x (tan 27°. cot 27°)
=1 × 1=1

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(a)

Question 9.
Show that
\(\frac{\sin ^3\left(180^{\circ}+\mathbf{A}\right) \cdot \tan \left(360^{\circ}-\mathbf{A}\right) \sec ^2\left(180^{\circ}-\mathbf{A}\right)}{\cos ^2\left(90^{\circ}+\mathbf{A}\right){cosec}^2 A \cdot \sin \left(180^{\circ}-A\right)}\) = tan3 A
Solution:
L.H.S
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(a) 1
= tan3A      (R.H.S)

Question 10.
If A = cos2 θ + sin4 θ then prove that for all values of θ, 3/4 ≤ A ≤ 1.
Solution:
A = cos2 θ + sin4 θ =1 – sin2 θ sin4 θ
or, sin4 θ – sin2 θ + (1 – A) = 0 …(1)
Eqn. (I) is quadratic in sin2 θ.
∴ sin2 θ = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
\(=\frac{1 \pm \sqrt{1-4(1-\mathrm{A})}}{2 \times 1}\)
Where a=1, b = – 1, c = 1 – A
∴ sin2 θ = \(\frac{1 \pm \sqrt{4 A-3}}{2}\)
We know that sin20 is not negative and lies in [0, 1]
So, \(\sqrt{4 \mathrm{~A}-3}\) ≤ 1
⇒ 4A – 3 ≤ 1 ⇒ 4A ≤ 4 ⇒ A ≤ 1  …(2)
Again, since sin2 θ is real,
b2 – 4ac must be +ve
i.e., 4A – 3 ≥ 0 ⇒ A ≥ 3/4
∴ From (2) and (3),
We have 3/4 ≤ A ≤ 1          (Proved)

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(a)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(a)

Question 1.
What is the total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3}?
Solution:
The total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3} is 32 = 9.

Question 2.
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face is thrown twice in succession, what is the total number of outcomes thus obtained?
Solution:
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face, is thrown twice in succession.
∴ The total number of outcomes is 62 = 36.

Question 3.
Five cities A, B, C, D, and E are connected to each other by straight roads. What is the total number of such roads?
Solution:
Five cities A, B, C, D, and E are connected to each other by straight roads.
∴ The number of such roads is \(\frac{{ }^5 \mathrm{P}_2}{2 !}\) = 10

Question 4.
What is the total number of different diagonals of a given pentagon?
Solution:
The total number of different diagonals of a given pentagon is
\(\frac{{ }^5 \mathrm{P}_2}{2 !}\) – 5 = 10 – 5 = 5

Question 5.
There are two routes joining city A to city B and three routes joining B to city C. In how many ways can a person perform a journey from A to C?
Solution:
There are two routes joining city A to city B and three routes joining B to city C.
∴ By the fundamental principle of counting, the total number of journeys in which a person can perform from A to C is 2 × 3 = 6.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(a)

Question 6.
How many different four-lettered words can be formed by using the four letters a, b, c, and d, while the letters can be repeated?
Solution:
The number of different words that can be formed by using the four letters a, b, c, and d, while the letters can be repeated is 44 = 256.

Question 7.
What is the sum of all three-digit numbers formed by using the digits 1, 2, and 3?
Solution:
The 3-digit numbers that can be formed by using the digits 1, 2, and 3 are 123, 132, 213, 231, 312, and 321.
∴ Their sum = 1332.

Question 8.
How many different words with two letters can be formed by using the letters of the word JUNGLE, each containing one vowel and one consonant?
Solution:
The word ‘Jungle’ contains 2 vowels and 4 consonants. Each word contains one vowel and one consonant.
The number of different words formed.
2 × 4P1 × 2P1 = 2 × 4 × 2 = 16

Question 9.
There are four doors leading to the inside of a cinema hall. In how many ways can a person enter into it and come out?
Solution:
There are four doors leading to the inside of a cinema hall. A person can enter into it and come out in 42 = 16 different ways. (By the principle of counting.)

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 5 Principles Of Mathematical Induction Ex 5 Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Exercise 5

Prove the following by induction.

Question 1.
1 + 2 + 3 + …… + n = \(\frac{n(n+1)}{2}\)
Solution:
Let pn be the given statement
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5

Question 2.
12 + 22 + …… + n2 = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let pn be the given statement
when n = 1
12 =1 = \(\frac{1(1+1)(2 \times 1+1)}{6}\)
P1 is true
Let Pk be true.
i.e. 12 + 22 + … + k2 = \(\frac{k(k+1)(2 k+1)}{6}\)
we shall prove Pk + 1 is true i.e., 12 + 22 + … + k2 + (k + 1)2
\(=\frac{(k+1)(k+2)(2 k+3)}{6}\)
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 1
∴ Pk+1 is true
∴ Pn is true of all values of n∈N.

Question 3.
1 + r + r2+ …. + rn = \(\frac{r^{n+1}-1}{r-1}\)
Solution:
Let Pn be the given statement
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 2
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 3

Question 4.
5n – 1 is divisible by 4.
Solution:
Let Pn = 5n – 1
When n = 1,
51 – 1 = 4 is divisible by 4.
∴ P1, is true.
Let Pk be true i.e.,
5k – 1 is divisible by 4.
Let 5k+1 – 1 = 4m, me Z
Now 5k + 1 – 1 = 5k. 5 – 5 + 4
= 5 (5k – 1) + 4
= 5 × 4m + 4 = 4 (5m + 1)
which is divisible by 4.
∴ Pk+1 is true.
∴ Pn is true for all values of n ∈ N

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5

Question 5.
72n + 23n-3  3n-1 is divisible by 25 for any natural number n > 1.
Solution:
Let 72n + 23n-3 . 3n-1
when n = 1, 71 + 20 . 30 ⇒ 49 + 1 = 50
which is divisible by 25.
∴ P1 is true. Let Pk be true.
i.e., 72k + 23k-3 3k-1 is divisible by 35
Now \(7^{2 \overline{k+1}}+2^{3 \overline{k+1}-3} 3^{\overline{k+1}-1}\)
=72k+2 + 23k. 3k
= 72k. 72 + 23k-3 23. 3k-1. 31
= 72k. 49 + 23k-3. 3k-1. 24
= 72k (25 + 24) + 24. 23k-3. 3k-1
= 72k. 25 + 24 (72k + 23k-3. 3k-1)
= 72k. 25 + 24 × 25m
Which is divisible by 25 (∵ Pk is true)
∴ Pk+1 is true
∴ Pn is true for all values of n > 1.

Question 6.
7. 52n-1 + 23n+1 is divisible by 17 for every natural number n ≥ 1.
Solution:
Let Pn = 7. 52n – 1 + 23n+1.
When n = 1, 7.5 + 24 = 35 + 16 = 51
Which is divisible by 17.
P1, is true.
Let Pk be true i.e., 7.52k-1 + 23k+1  is divisible by 17.
Let 7.52k-1 + 23k+1 = 17 m, m ∈ Z
Now, \(7.5^{2 \overline{k+1}-1}+2^{3 \overline{k+1}+1}\)
= 7.52k-1 + 23k+4
= 7.52k-1 . 52 + 23k+1 . 23
= 25. 7. 52k-1 + 8. 23k+1
= (17 + 8) 7.52k-1 + 8. 23k+1
= 17. 7. 52k-1 + 8 (7. 52k-1 + 23k+1)
= 17 × 7 × 52k-1 + 8 × 17m
Which is divisible by 17.
Hence Pk+1. is true.
∴ Pn is true for all values of n ≥ 1.

Question 7.
4n+1 + 15n + 14 is divisible by 9 for every natural number n ≥ 0.
Solution:
Let Pn = 4n+1+ 15 n + 14
when n = 1, 42 + 15 + 14 = 45 is divisible by 9.
∴ P1 is true. Let Pk be true.
i.e., 4k+1 + 15k + 14 is divisible by 9.
Now, 4k+1+1 + 15 (k + 1) + 14
= 4k+2 + 15k + 29
= 4k+1. 4 + 60k + 56 – 45k – 27
= 4 (4k+1 + 15k + 14) – 9 (5k + 3)
Which is divisible by 9.
∴ Pk+1, is true.
∴ Pn is true for all values of n ≥ 0.

Question 8.
3(2n-1) + 7 is divisible by 9 for every natural number n ≥ 2.
Solution:
Let Pn = 32(n-1) + 7
When n = 2. 32 + 7 = 16 is divisible by 8.
∴ P2 is true.
Let Pk be true.
i.e., 52(k-1) + 7 is divisible by 8.
Let 32k-2 + 7 = 8m. m ∈ Z.
Now \(3^{2(\overline{k+1}-1)}\) + 7 = 32k + 7
= 32k-2. 32 + 63 – 56
= 9(32k-2 + 7) – 56
= 9 × 8m – 56 = 8 (9m – 7)
Which is divisible by 8.
Pk+1 is true.
Pn is true for all values on n ≥ 2.

Question 9.
5(2n-4)  – 6n + 32 is divisible by 9 for every natural number n ≥ 5.
Solution:
Let P = 52(n-4) – 6n + 32
For n = 5, P5 = 52 – 6. 5 + 32
= 25 – 30 + 32 = 27
Which is divisible by 9.
Hence P5 is true.
Let Pk is true.
Let Pk is divisible by 9.
Let Pk = 52(k-4) – 6k + 32 = 9., m ≥ Z
52k+2 = 576 m + 24k  + 25 … (1)
we shall prove that Pk+1 is true.
Now 52(K+1)+2 – 24(k+1) – 25
= 52 (52k+2) – 24k – 24 – 25
= 52[576m + 24k + 25] – 24k – 24 – 25
= 25 × 576 m + 25 × 24k + 25 × 25 – 24k – 24 – 25
= 25 × 576 m + 576 k + 576
= 576 [25 m + k + 1]
which is divisible by 576
∴ Pk+1 is true.
So by the method of induction Pn is true for all n.
i.e., 52n+2  – 24n – 25 is divisible by 576 for all n ∈ N.
Hence Pk+1 is true.
So by methods of induction Pn is true.
i.e., 52n+2 – 24n – 25 is divisible by 576 for all n.

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5

Question 10.
\(\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}\)
Solution:
when n = 1,
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 4
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 5

Question 11.
1.3 + 2.4 + 3.5 + …….. + n(n + 2) = \(\frac{n(n+1)(2 n+7)}{6}\)
Solution:
when n = 1,
we have 1.3 = 3 = \(\frac{3 \times 6}{6}\)
\(=\frac{1 \times 2 \times 9}{6}=\frac{1(1+1)(2 \times 1+7)}{6}\)
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 6
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 7

Question 12.
xn – yn = (x – y)(xn-1 + xn-2 y + … + xyn-2 + yn-1); x, y ∈ R [Hint : Write xn+1 – yn+1 = x(xn – yn) + yn(x – y)]
Solution:
Let p(n) is
xn – yn = (x – y)(xn-1 + xn-2 y + … + xyn-2 + yn-1); x, y ∈ R
Step – 1:
For n = 2
x2 – y2 = (x – y) (x + y) (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., xk – yk = (x – y)(xk-1 + xk-2y + … +xyk-2 + yk-1)
Step – 3:
Let us prove Pk+1 is true.
i.e., xk+1 – yk+1 = (x – y) (xk + xk-1y + … (xyk-1 + yk)
L.H.S. = xk+1 – yk+1
= xk+1 – xyk + xyk – yk+1
= x(xk – yk) + yk (x – y)
= x(x – y)(xk-1 + xk-2 y + … + xyk-2 + yk-1) + yk(x – y) [by (1)]
= (x – y) [xk + xk-1 y + … + xyk-2 + xyk-1 + yk]
= R.H.S.
∴ P(k+1) is true.
Step – 4:
By Principle Of Mathematical Induction P(n) is true for all n ∈ N.

Question 13.
1 + 3 + 5 + ……. +(2n – 1) = n2
Solution:
Let P(n) is : 1 + 3 + 5 + ……. +(2n – 1) = n2.
Step – 1:
For n = 2
L.H.S. = 1 + 3 = 4 = 22 (R.H.S)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., 1 + 3 + 5 … + (2k – 1) = k2 …(1)
Step – 3:
We will prove that P(k + 1) is true
i.e., we want to prove.
1+ 3 + 5 + … + (2k – 1) + (2k + 1) = (k + 1)2
L.H.S. = 1 + 3 + 5 + … + (2k – 1) + (2k + 1)
= k2 + 2k + 1          [By – (1)]
= (k + 1)2 = R.H.S.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 1 + 3 + 5 ….+ (2n – 1) = n2

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5

Question 14.
n > n; n is a natural number.
Solution:
Let P(n) is 2n > n
Step – 1:
21 > 1 (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
⇒ 2k > k
Step – 3:
We shall prove that P(k + 1) is true
i.e., 2k+1 > k + 1
Now 2k+1 = 2.2k > 2k ≥ k + 1 for k ∈ N.
∴ 2k+1 > k + 1
⇒ P(k + 1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 2n > n for n ∈ N

Question 15.
(1, 2, 3 … n)3  > 8 (13 + 23 + 33 + … + n3), for n > 3.
Solution:
Let P(n) is
(1, 2, 3 … n)3  > 8 (13 + 23 + 33 + … + n3), for n > 3.
Step – 1:
For n = 4
(1. 2. 3. 4)3 = 243 = 13824
8(13+ 23 + 33 + 43) = 808
∴ (1. 2 . 3 . 4)3 > 8(13 + 23 + 33 + 43)
∴ P(4) is true.
Step- 2:
Let P(k) is true.
(1. 2. 3…….k)3 > 8(13 + 23 + 33 + …… + k3)
Step – 3:
We shall prove that P(k+1) is true.
i.e., (1. 2. 3. …….. k(k+1))3 > 8(13 + 23 + … + k3 + (k + 1)3)
Now (1. 2. 3. …….. k(k + 1)3)
= (1. 2. 3 … k)3 (k + 1)3
> 8 (13 + 23 + … k3) (k + 1)3
> 8 (13 + 23 + … k3) + 8(k + 1)3
= 8 (13 + 23 + … + k3 + (k + 1)3)
P(k+1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N and n > 3.

Question 16.
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\) for every positive integer n.
Solution:
Let P(n) is
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\)
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 8
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 9
CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Ex 5 10
Step-4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(b)

Question 1.
Fill in the blanks in each of the following, using the answers given against each of them :
(a) The slope and x-intercept of the line 3x – y + k = 0 are equal if k = _________ . (0, -1, 3, -9)
Solution:
-9

(b) The lines 2x – 3y + 1 = 0 and 3x + ky – 1=0 are perpendicular to each other if k = ___________ . (2, 3, -2, -3)
Solution:
2

(c) The lines 3x + ky – 4 = 0 and k – Ay – 3x = 0 are coincident if k = _____________. (1, -4, 4, -1)
Solution:
4

(d) The distance between the lines 3x – 1 = 0 and x + 3 = 0 is _________ units. (4, 2, \(\frac{8}{3}\), \(\frac{10}{3}\))
Solution:
\(\frac{10}{3}\)

(e) The angle between the lines x = 2 and x – √3y + 1 = 0 is _________. (30°, 60°, 120°, 150°)
Solution:
60°

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 2.
State with reasons which of the following are true or false :
(a) The equation x = k represents a line parallel to x – axis for all real values of k.
Solution:
False. As the line x = k is parallel to y- axis for all values of k.

(b) The line, y + x + 1 = 0 makes an angle 45° with y – axis.
Solution:
y + x + 1 = 0
∴ Its slope = -1 = tan 135°
∴ It makes 45° with y – axis, as it makes 135° with x – axis. (True)

(c) The lines represented by 2x – 3y + 1 = 0 and 3x + 2y – k = 0 are perpendicular to each other for positive values of k only.
Solution:
2x – 3y + 1 = 0, 3x + 2y – k = 0
∴ \(m_1 m_2=\frac{2}{3} \times \frac{(-3)}{2}=-1\)
∴ The lines are perpendicular to each other for + ve values of k only. (False)

(d) The lines represented by px + 2y – 1 = 0 and 3x + py + 1 = 0 are not coincident for any value of ‘p’.
Solution:
px + 2y – 1 = 0, 3x + py + 1=0
∴ \(\frac{p}{3}=\frac{2}{p}=\frac{-1}{1} \Rightarrow p^2=6\)
and p = -3 or -2
There is no particular value of p for which \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) (True)

(e) The equation of the line whose x and y – intercepts are 1 and -1 respectively is x – y + 1 = 0.
Solution:
Equation of the line whose intercepts 1 and -1 is \(\frac{x}{1}+\frac{y}{-1}\) = 1
or, x – y = 1 (False)

(f) The point (-1, 2) lies on the line 2x + 3y – 4 = 0.
Solution:
Putting x = – 1, y = 2
we have 2 (- 1) + 3 × 2 – 4
= -2 + 6 – 4 = 0
∴ The point (-1, 2) lies on the line 2x + 3 – 4 = 0 (True)

(g) The equation of a line through (1, 1) and (-2, -2) is y = – 2x.
Solution:
The equation of the line through (1, 1) and (-2, -2) is y – y1 = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x1)
or, y – 1 = \(\frac{-2-1}{-2-1}\) (x – 1)
or, y – 1 = x – 1
or, x – y = 0 (False)

(h) The line through (1, 2) perpendicular to y = x is y + x – 2 =0.
Solution:
The slope of the line y = x is 1.
∴ The slope of the line perpendicular to the above line is -1.
∴ The equation of the line through (1, 2) having slope – 1 is y – y1 = m(x – x1)
or, y – 2 = -1 (x – 1)
or, y – 2= -x + 1
or, x + y = 3 (False)

(i) The lines \(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1 are intersecting but not perpendicular to each other.
Solution:
\(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1
∴ \(m_1 m_2=\frac{\left(-\frac{1}{a}\right)}{\frac{1}{b}} \times \frac{\left(-\frac{1}{b}\right)}{\left(-\frac{1}{a}\right)}=-1\)
∴ The lines intersect and are perpendicular to each other. (False)

(j) The points (1, 2) and (3, – 2) are on the opposite sides of the line 2x + y = 1.
Solution:
2x + y = 1
Putting x = 1, y = 2,
we have 2 × 1 + 2 = 4 > 1
Putting x – 3, y = -2,
we have 2 × 3 – 2 – 4 > 1
∴ Points (1, 2) and (3, – 2) lie on the same side of the line 2x + y = 1 (False)

Question 3.
A point P (x, y) is such that its distance from the fixed point (α, 0) is equal to its distance from the y – axis. Prove that the equation of the locus is given by, y2 = α (2x – α).
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 4.
Find the locus of the point P (x, y) such that the area of the triangle PAB is 5, where A is the point (1, -1) and B is the tie point (5, 2).
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 1
= \(\frac{1}{2}\) (-3x + 4y + 7) = 5
or, – 3x + 4y + 7 = 10
or, 3x – 4y + 3 =0 which is the locus of the point P (x, y).

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 5.
A point is such that its distance from the point (3, 0) is twice its distance from the point (-3, 0). Find the equation of the locus.
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 2

Question 6.
Obtain the equation of straight lines:
(a) Passing through (1, – 1) and making an angle 150°.
Solution:
The slope of the line
= tan 150° = –\(\frac{1}{\sqrt{3}}\)
∴ The equation of the line is y – y1 = m(x – x1)
or, y + 1 = –\(\frac{1}{\sqrt{3}}\) (x – 1)
or, y√3 + √3 = -x + 1
or, x + y√3 + √3 – 1 = 0

(b) Passing through (-1, 2) and making intercept 2 on the y-axis.
Solution:
Let the equation of the line be
y – mx + c or, y = mx + 2
∴ As the line passes through (-1, 2)
we have 2 = – m + 2, or, m = 0
∴ Equation of the line is y = 2.

(c) Passing through the points (2, 3) and (-4, 1).
Solution:
The equation of the line is
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 3

(d) Passing through (- 2, 3) and a sum of whose intercepts in 2.
Solution:
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}\) = 1 where a + b = 2     …….(1)
Again, as the line passes through the point (-2, 3), we have \(\frac{-2}{a}+\frac{3}{b}\) = 1      ………(2)
From (1), we have a= 2 – b
∴ From (2) \(\rightarrow \frac{-2}{2-b}+\frac{3}{b}=1\)
or, – 2b + 6 – 3b = (2 – b)b
or, 6 – 5b = 2b – b2
or, b2 – 7b + 6 = 0
or, (b – 6)(b – 1) = 0
∴ b = 6, 1
∴ a =2 – b = 2 – 6 = -4
or, 2 – 1 = 1
∴ Equation of the lines are \(\frac{x}{-4}+\frac{y}{6}\) = 1 or \(\frac{x}{1}+\frac{y}{1}\) = 1
i, e. -3x + 2y = 12 or, x + y = 1

(e) Whose perpendicular distance from the origin is 2 such that the perpendicular from the origin has indication 150°.
Solution:
Here p = 2, α = 150°
The equation of the line in normal form is x cos α + y sin α = p
or, x cos 150° + y sin 150° = 2
or, \(\frac{-x \sqrt{3}}{2}+y \cdot \frac{1}{2}\) = 2
or, -x √3 + y = 4
or, x√3 – y + 4 = 0

(f) Bisecting the line segment joining (3, – 4) and (1, 2) at right angles.
Solution:
The slope of the line \(\overline{\mathrm{AB}}\) is
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 4

(g) Bisecting the line segment joining, (a, 0) and (0, b) at right angles.
Solution:
Refer to (f)

(h) Bisecting the line segments joining (a, b), (a’, b’) and (-a, b), (a’, -b’).
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 5

(i) Passing through the origin and the points of trisection of the portion of the line 3x + y – 12 = 0 intercepted between the coordinate axes.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 6
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 7

(j) Passing through (-4, 2) and parallel to the line 4x – 3y = 10.
Solution:
Slope of the line 4x – 3y = 10 is \(\frac{-4}{-3}=\frac{4}{3}\)
∴ The slope of the line parallel to the above line is \(\frac{4}{3}\).
∴ Equation of the line through (- 4, 2) and having slope \(\frac{4}{3}\) is y – y1 = m(x – x1)
or, y – 2 = \(\frac{4}{3}\) (x + 4)
or, 3y – 6 = 4x + 16
or, 4x – 3y + 22 = 0

(k) Passing through the point (a cos3 θ, a sin3 θ) and perpendicular to the straight line x sec θ + y cosec θ = α.
Solution:
The slope of the line x sec θ + y cosec θ = a is \(\frac{-\sec \theta}{{cosec} \theta}\) = -tanθ
∴ Slope of the required line  = cot θ
∴ Equation of the line through (a cos3 θ, a sin3 θ) is y – y1 = m(x – x1)
or, y – a sin3 θ = cot θ(x – a cos3 θ)
or, y – a sin3 θ = \(\frac{\cos \theta}{\sin \theta}\) (x – a cos3 θ)
or y sin θ – a sin4 θ = x cos  θ – a cos4 θ
or (x cos θ – y sin θ) + a(sin4 θ – cos4 θ) = 0

(l) Which passes through the point (3, -4) and is such that its portion between the axes is divided at this point internally in the ratio 2: 3.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 8
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 9

(m) which passes through the point (α, β) and is such that the given point bisects its portion between the coordinate axis.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 10
x = 2α , y = 2β
∴ Equation of the line \(\overleftrightarrow{\mathrm{AB}}\) is \(\frac{x}{2 \alpha}+\frac{y}{2 \beta}\) = 1(Intercept form)

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 7.
(a) Find the equation of the lines that is parallel to the line 3x + 4y + 7 = 0 and is at a distance 2 from it.
Solution:
3x + 4y + 7 = 0
or, \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) = 0(Normal form)
∴ Equation of the lines parallel to the above line and 2 units away from it are \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) ± 2 = 0
or, 3x + 4y + 7 ± 10 = 0
∴ 3x + 4y + 17 = 0 and 3x + 4y – 3 = 0

(b) Find the equations of diagonals of the parallelogram formed by the lines ax + by = 0, ax + by + c = 0, lx + my = 0, and lx + my + n = 0. What is the condition that this will be a rhombus?
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 11
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 12
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 13
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 14
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 15
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 16

(c) Find the equation of the line passing through the intersection of 2x – y – 1 = 0 and 3x – 4y + 6 = 0 and parallel to the line x + y – 2 = 0.
Solution:
Let the equation of the required line be (2x – y – 1) + λ(3x – 4y + 6) = 0
or, x(2 + 3λ) + λ(-1 – Aλ) + 6λ – 1 = 0
As this line is parallel to the line x + y – 2 = 0
we have their slopes are equal.
∴ \(-\left(\frac{2+3 \lambda}{-1-4 \lambda}\right)=\frac{-1}{1}\)
or, 2 + 3λ = -1 – 4λ
or, 7λ = -3 or, λ = \(\frac{-3}{7}\)
∴ Equation of the line is (2x – y – 1) – \(\frac{3}{7}\) (3x – 4y + 6) = 0
or, 14x – 7y – 1 – 9x + 12y – 18 = 0
or, 5x + 5y – 25= 0
or, x + y = 5

(d) Find the equation of the line passing through the point of intersection of lines x + 3y + 2 = 0 and x – 2y – 4 = 0 and perpendicular to the line 2y + 5x – 9 = 0.
Solution:
Let the equation of the line be (x + 3y + 2) + λ(x – 2y – 4) = 0
or, x(1 + λ) + y(3 – 2λ) + 2 – 4λ = 0
As this line is perpendicular to the line 2y + 5x – 9 = 0.
We have the product of their slopes is -1.
∴ \(\frac{1+\lambda}{3-2 \lambda} \times \frac{5}{2}\) = -1
or, 5 + 5λ = – 6 + 4λ
or, λ = -6 – 5 = -11.
∴ Equation of the required line is (x + 3y + 2) – 11(x – 2y – 4) = 0
or, x + 3y + 2- 11x + 22y + 44 = 0
or, – 10x + 25y + 46 = 0
or, 10x – 25y – 46 = 0

(e) Find the equation of the line passing through the intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0 and the centroid of the triangle whose vertices are the points (3, -1) (1, 3) and (2, 4).
Solution:
Let the equation of the required line (x + 3y – 1) + λ(3x – y + 1) = 0   … (1)
Again, the centroid of the triangle with vertices (3, – 1), (1, 3), and (2, 4) is \(\left(\frac{3+1+2}{3}, \frac{-1+3+4}{3}\right)\) = (2, 2)
As line (1) passes through (2, 2), we have (2 + 6 – 1) +1(6 – 2 + 1) = 0
or, 7 + 5λ = 0 or, λ = \(\frac{-7}{5}\)
∴ Equation of the line (x + 3y – 1) – \(\frac{7}{5}\) (3x – y + 1) = 0
or, 5x + 15y – 5 – 21x + 7y – 7 = 0
or, 22y – 16x – 12 = 0
or, 11y – 8x – 6 = 0
or, 8x – 11y + 6 = 0

Question 8.
If lx + my + 3 = 0 and 3x – 2y – 1 = 0 represent the same line, find the values of l and m.
Solution:
lx + my + 3 = 0 and 3x – 2y – 1 = 0 represents the same line
∴ \(\frac{l}{3}=\frac{m}{-2}=\frac{3}{-1}\)
∴ l = -9, m = 6

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 9.
Find the equation of sides of a triangle whose vertices are at (1, 2), (2, 3), and (-3, -5).
Solution:
Equation of \(\overline{\mathrm{AB}}\) is \(y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\)
\(y-2=\frac{3-2}{2-1}(x-1)\)
or, y – 2 = x – 1
or, x – y + 1 = 0
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 17
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 18

Question 10.
Show that origin is within the triangle whose sides are given by equations, 3x – 2y = 1, 5x + 3y + 11 = 0, and x – 7y + 25 = 0.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 19
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 20
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 21
∴ The origin lies within the triangle ABC.

Question 11.
(a) Find the equations of straight lines passing through the point (3, -2) and making an angle 45° with the line 6x + 5y = 1.
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 22
or, 11x – y = 35, x + 11y + 19 = 0

(b) Two straight lines are drawn through the point (3, 4) inclined at an angle 45° to the line x – y – 2 = 0. Find their equations and obtain area included by the above three lines.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 23
Slope of L2 = 0
Then as L2 ⊥ L3
Slope of L3 = ∞
∴ Equation of L2 is
y – y1 =m(x – x1)
or, y – 4 = 0(x – 3) = 0
or, y = 4
∴ Equation of L3 is y – 4 = ∞ (x – 3)
or, x – 3 = 0 or, x = 3
∴ Sloving L1 and L2, we have
x – y – 2 = 0, y = 4
or, x = 6
The coordinates of A are (6, 4).
Again solving L1 and L3, we have
x – y – 2 = 0, x = 3
or, y = x – 2 = 3 – 2 = 1
∴ The coordinates of B are (3, 1).
Area of the triangle PAB is
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 24

(c) Show that the area of the triangle formed by the lines given by the equations y = m1x + c1,y = m2x + c2, and x = 0 is \(\frac{1}{2} \frac{\left(c_1-c_2\right)^2}{\left[m_2-m_1\right]}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 25
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 26

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 12.
Find the equation of lines passing through the origin and perpendicular to the lines 3x + 2y – 5 = 0 and 4x + 3y = 7. Obtain the coordinates of the points where these perpendiculars meet the given lines. Prove that the equation of a line passing through these two points is 23x + 11y – 35 = 0.
Solution:
The slopes of the line 3x + 2y – 5 = 0 and 4x + 3y = 7 are \(\frac{-3}{2}\) and \(\frac{-4}{3}\)
∴ Slopes of the lines perpendicular to the above lines are \(\frac{2}{3}\) and \(\frac{3}{4}\)
∴ Equation of the lines through the origin and having slopes \(\frac{2}{3}\) and \(\frac{3}{4}\)
y = \(\frac{2x}{3}\) and y = \(\frac{3x}{4}\)
Now solving 3x + 2y – 5 = 0 and y = \(\frac{2x}{3}\)
we have 3x + \(\frac{4x}{3}\) – 5 = 0
or, 9x + 4x – 15 = 0
or, x = \(\frac{15}{13}\)
∴ y = \(\frac{2 x}{3}=\frac{2}{3} \times \frac{15}{13}=\frac{10}{13}\)
∴ The perpendicular y = \(\frac{2 x}{3}\) meets the line 3x + 2y – 5 = 0 at \(\left(\frac{15}{13}, \frac{10}{13}\right)\)
Again, solving 4x + 3y = 7 and y = \(\frac{3 x}{4}\)
we have 4x + 3 × \(\frac{3 x}{4}\) = 7
or, 16x + 9x = 28 or, x = \(\frac{28}{25}\)
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 27
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 28

Question 13.
(a) Find the length of a perpendicular drawn from the point (-3, -4) to the straight line whose equation is 12x – 5y + 65 = 0.
Solution:
The length of the perpendicular drawn from the point (- 3, -4) to the straight line 12x – 5y + 65 = 0 is
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 29

(b) Find the perpendicular distances of the point (2, 1) from the parallel lines 3x – 4y + 4 = 0 and 4y – 3x + 5 = 0. Hence find the distance between them.
Solution:
The distance of the point (2, 1) from the line 3x – 4y + 4 = 0 is \(\left|\frac{3 \times 2-4 \times 1+4}{\sqrt{9+16}}\right|=\frac{6}{5}\)
Again distance of the point (2, 1) from the line 4y – 3x + 5 = 0 is
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 30

(c) Find the distance of the point (3, 2) from, the line x + 3y – 1 = 0, measured parallel to the line 3x – 4y + 1 = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 31
Let the coordinates of M be (h, k).
As \(\overline{\mathrm{PM}} \| \mathrm{L}_1\), We have their slopes are equal.
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 32

(d) Find the distance of the point (-1, -2) from the line x + 3y – 7 = 0, measured parallel to the line 3x + 2y – 5 = 0
Solution:
Slope of the line 3x + 2y – 5 = 0 is \(\left(-\frac{3}{2}\right)\)
Equation of the line through (-1, -2) and parallel to this line is y + 2 = – \(\frac{3}{2}\) (x + 1)
⇒ 2y + 4 = -3x – 3
⇒ 3x + 2y + 7 = 0 …(1)
Given line is : x + 3y – 7 = 0    ….(2)
from (1) and (2) we get 7y – 28 = 0 Py = 4 and x = \(-\frac{35}{7}\) = -5
Thus the required distance is \(\sqrt{(-1+5)^2+(-2-4)^2} \quad=\sqrt{16+36}\)
= √52 = 2√3 units.

(e) Fine the distance of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) from the origin.
Solution:
The equation of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) is y – y1 = \(\frac{y_2-y_1}{x_2-x_1}\) x – x1
or, y – a sin α
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 33
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 34

Question 14.
Find the length of perpendiculars drawn from the origin on the sides of the triangle whose vertices are A( 2, 1), B (3, 2), and C (- 1, -1).
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 35
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 36

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 15.
Show that the product of perpendicular from the points \(\left(\pm \sqrt{a^2-b^2}, 0\right)\) upon the straight line \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 is b2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 37
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 38

Question 16.
Show that the lengths of perpendiculars drawn from any point of the straight line 2x + 11y – 5 = 0 on the lines 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0 are equal to each other.
Solution:
Let P(h, k) is any point on the line
2k + 11y – 5 = 0
∴ 2h + 11k – 5 = 0
Now the length of the perpendicular from P on the line 24x + 7y – 20 = 0 is
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 39
Clearly d1 = d2

Question 17.
If p and p’ are the length of perpendiculars drawn from the origin upon the lines x sec α + y cosec α = 0 and x cos α – y sin α – a cos 2α = 0
Prove that 4p2 + p’2 = a2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 40

Question 18.
Obtain the equation of the lines passing through the foot of the perpendicular from (h, k) on the line Ax + By + C = 0 and bisect the angle between the perpendicular and the given line.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 41
Slope of the line L is \(\frac{-\mathrm{A}}{\mathrm{B}}\)
∴ Slope of the line \(\overline{\mathrm{PM}} \text { is } \frac{\mathrm{B}}{\mathrm{A}}\)
∴ Equation of the line \(\overline{\mathrm{PM}}\) is y – y1 = m(x – x1)
or, y – k = \(\frac{\mathrm{B}}{\mathrm{A}}\) (x – h)
or, Ay – Ak =Bx – Bh
or, Bx – Ay + Ak – Bh = 0
∴ Equation of the bisectors of the angles between the lines L and \(\overline{\mathrm{PM}}\) is
\(\frac{\mathrm{A} x+\mathrm{B} y+\mathrm{C}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\pm \frac{\mathrm{B} x-\mathrm{A} y+\mathrm{A} k-\mathrm{B} h}{\sqrt{\mathrm{B}^2+\mathrm{A}^2}}\)
or, Ax + By + C = ± (Bx – Ay +Ak – Bh)

Question 19.
Find the direction in which a straight line must be drawn through the point(1, 2) such that its point of intersection with the line x + y – 4 = 0 is at a distant \(\frac{1}{3} \sqrt{6}\) from this point.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 42
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 43
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 44

Question 20.
A triangle has its three sides formed by the lines x + y = 3, x + 3y = 3, and 3x + 2y = 6. Without solving for the vertices, find the equation of its altitudes and also calculate the angles of the triangle.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 45
or 1 + 3λ = – 3 – 6λ
or 9λ = – 4 or , λ = \(\frac{-4}{9}\)
∴ Equation of \(\overline{\mathrm{AD}}\) is (x + y – 3) – \(\frac{4}{9}\) (3x + 2y – 6) = 0
or, 9x + 9y – 27 – 12x – 8y + 24 = 0
or, -3x + y – 3 = 0
or, 3x – y + 3 = 0
Let the equation of \(\overline{\mathrm{BE}}\) be (x + y – 3) + 1(x + 3y – 3) = 0
or, x(1 + λ) + y( 1 + 3λ) – 3 – 3λ = 0
As \(\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}\)
we have \(\frac{1+\lambda}{1+3 \lambda} \times \frac{3}{2}\) = -1
or, 3 + 3λ = -2 – 6λ
or, 9λ = – 5 or λ = \(\frac{-5}{9}\)
∴ Equation of \(\overline{\mathrm{BE}}\) is (x + y – 3) – \(\frac{5}{9}\) (x + 3y – 3) = 0
or 9x + 9y – 27 – 5x – 15y + 15 = 0
or, 4x – 6y – 12 = 0
or, 2x – 3y – 6 = 0
Let the equation of \(\overline{\mathrm{CE}}\) be (3x + 2y – 6) + λ (x + 3y – 3) = 0
x(3 + λ) + y (x + 3λ) – 6 – 3λ = 0
As \(\overline{\mathrm{CF}} \perp \overline{\mathrm{AB}} .\)
we have \(\frac{3+\lambda}{2+3 \lambda}\) × 1 = -1
or, 3 + λ = -2 – 3λ
or, 4λ = -2 – 3 = -5
or, λ = \(\frac{-5}{4}\)
∴ Equation of is \(\overline{\mathrm{CF}}\) (3x + 2y – 6) \(\frac{-5}{4}\) (x + 3y – 3) = 0
or, 12x + 8y – 24 – 5x – 15y + 15 =0
or, 7x – 7y – 9 = 0
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 46
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 47

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 21.
A triangle has its vertices at P(1, -1), Q(3, 4) and R(2, 5). Find the equation of altitudes through P and Q and obtain the coordinates of their point of intersection. (This point is called the ortho-center of the triangle.)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 48
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 49

Question 22.
(a) Show that the line passing through (6, 0) and (-2, -4) is concurrent with the lines
2x – 3y – 11 = 0 and 3x – 4y = 16
Solution:
The equation of the line through (6,0) and (-2, -4) is
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 50

(b) Show that the lines lx + my + n = 0 mx + ny + 1 = 0 and nx + ly + m = 0 are concurrent, l + m + n = 0
Solution:
As the lines
lx + my + n = 0
mx + ny + 1 = 0
and nx + ly + m = 0 are concurrent.
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 51

Question 23.
Obtain the equation of the bisector of the acute angle between the pair of lines.
(a) x + 2y = 1, 2x + y + 3 = 0
Solution:
Equation ofthe bisectors ofthe angles between the lines x + 2y – 1 = 0 and 2x + y + 3 = 0 are \(\frac{x+2 y-1}{\sqrt{1^2+2^2}}=\pm \frac{2 x+y+3}{\sqrt{2^2+1^2}}\)
or, x + 2y – 1 = ± (2x + y + 3)
∴ x + 2y – 1 = 2x + y + 3 and
x + 2y – 1= -2x – y – 3
∴ x – y + 4 = 0 and 3x + 3y + 2 = 0
Let θ be the angle between x + 2y – 1 = 0 and x – y + 4 = 0
∴ tan θ = \(\frac{a_1 b_2-a_2 b_1}{a_1 a_2+b_1 b_2}\)
\(=\frac{1 \cdot(-1)-(+1) \cdot 2}{1 \cdot 1+2(-1)}\)
\(=\frac{-1-2}{1-2}=\frac{-3}{-1}=3\)
sec2 θ = 1 + tan2 θ = 1 + 9 = 10
cos2 θ = 1/10
cos θ = \(\frac{1}{\sqrt{10}}<\frac{1}{\sqrt{2}}\) ⇒ θ > 45°
∴ x – y + 4 = 0 is the obtuse angle bisector.
⇒ 3x + 3y + 2 = 0 is acute angle bisector.

(b) 3x – 4y = 5, 12y – 5x = 2
Solution:
Given equation of lines are
3x – 4y – 5 = 0    …..(1) and  5x – 12y + 2 = 0     ……(2)
Equation of bisectors of angles between these Unes are:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 52

Question 24.
(a) Find the coordinates of the center of the inscribed circle of the triangle formed by the line x cos α + y sin α = p with the coordinate axes.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 53
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 54
\(\left(\frac{p}{\sin \alpha+\cos \alpha+1}, \frac{p}{\sin \alpha+\cos \alpha+1}\right)\)

(b) Find the coordinates of the circumcentre and incentre of the triangle formed by the lines 3x – y = 5, x + 2y = 4, and 5x + 3y + 1 = 0.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 55
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 56
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 57
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 58

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 25.
The vertices B, and C of a triangle ABC lie on the lines 3y = 4x and y = 0 respectively, and the side \(\overline{\mathbf{B C}}\) passes through the point (2/3, 2/3). If ABOC is a rhombus, where O is the origin, find the equation of \(\overline{\mathbf{B C}}\) and also the coordinates of A.
Answer:
Let the coordinates of C be (a, 0) so that the length of the side of the rhombus is ‘a’
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 59
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 60
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 61

Question 26.
Find the equation of the lines represented by the following equations.
(a) 4x2 – y2 = 0
Solution:
4x2 – y2 = 0
or, (2x + y)(2x – y) = 0
∴ 2x + y = 0 and 2x – y = 0 are the two separate lines.

(b) 2x2 – 5xy – 3y2
Solution:
2x2 – 5xy – 3y2
or, 2x2 – 6xy + xy – 3y2 = 0
or, 2x(x – 3y) + y(x – 3y) = 0
or, (x – 3y)(2x + y) = 0
∴ x – 3y = 0 and 2x + y = 0 are the two separate lines.

(c) x2 + 2xy sec θ + y2 = 0
Solution:
x2 + 2xy sec θ + y2 = 0
∴ a = 1, b = 2y sec θ, c = y2
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{2 y \sec \theta \pm \sqrt{4 y^2 \sec ^2 \theta-4 y^2}}{2}\)
= \(\frac{-2 y \sec \theta \pm 2 y \tan \theta}{2}\)
= y (- sec θ ± tan θ)
∴ x = y (- sec θ + tan θ) and x – y (- sec θ – tan θ) are the two separate lines.

(d) 3x2 + 4xy = 0
Solution:
3x2 + 4xy = 0
or x (3x + 4y) = 0
∴ x = 0 and 3x + 4y = 0 are the two separate lines.

Question 27.
From the equations which represent the following pair of lines.
(a) y = mx; y = nx
Solution:
y – mx = 0, y – nx = 0
or (y – mx) (y – nx) = 0
or, y2 – nxy – mxy + mnx2 = 0
or, y2 – xy (m + n) + mnc2 = 0 which is the equation of a pair of lines.

(b) y – 3x = 0 ; y + 3x = 0
Solution:
y – 3x = 0, y + 3x = 0
∴ (y – 3x) (y + 3x) = 0
or, y2 – 9x2 = 0 which is the equation of a pair of lines.

(c) 2x – 3y + 1 = 0 ; 2x + 3y + 1 = 0
Solution:
2x – 3y + 1 = 0; 2x + 3y + 1 =0
or, (2x – 3y + 1)(2x + 3y + 1) = 0
or, (2x + 1)2 – 9y2 = 0
or, 4x2 + 1 + 4x – 9y2 = 0
or, 4x2 – 9y2 + 4x + 1= 0 which represents a pair of lines.

(d) x = y. x + 2y + 5 = 0
Solution:
x = y, x + 2y + 5 = 0
∴ (x – y) (x + 2y + 5) = 0
or, x2 + 2xy + 5x – xy – 2y2 – 5y =0
or, x2 – 2y2 + xy + 5x – 5y = 0 which represents a pair of lines.

Question 28.
Which of the following equations represents a pair of lines?
(a) 2x2 – 6y2 + 3x +  y + 1 = 0
Solution:
a = 2, b = -6, 2g = 3
2f = 1, c = 1
∴ g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), h = 0
∴ abc + 2fgh – ah2 – bg2 – ch2
= 2(-6). 1 + 2 × \(\frac{1}{2}\) × \(\frac{3}{2}\) – 0 – (-6) × \(\frac{9}{4}\) – 1 × 0
= -12 + \(\frac{3}{2}\) + \(\frac{27}{2}\) = \(\frac{6}{2}\) = 3 ≠ 0
∴ The given equation does not represent a pair or lines.

(b) 10x2 – xy – 6y2 – x + 5y – 1 = 0
Solution:
a = 10. 2h = 1
B = -6, 2g = -1
2f = 5. C= -1
∴ h = –\(\frac{1}{2}\), g = –\(\frac{1}{2}\) , f = \(\frac{5}{2}\)
∴ abc + 2fgh – ah2 – bg2 – ch2
= 10(-6)(-1) + 2 × \(\frac{5}{2}\) × (-\(\frac{1}{2}\)) × (-\(\frac{5}{2}\)) – 10 × \(\frac{2.5}{4}\) – (-6)\(\frac{1}{4}\) – (-1)\(\frac{1}{4}\)
= 60 + \(\frac{5}{4}\) – \(\frac{250}{4}\) + \(\frac{6}{4}\) + \(\frac{1}{4}\)
= \(\frac{240+5+6-250+1}{4}=\frac{2}{4}\)
∴ The given equation does not represent a pair of lines.

(c) xy + x + y + 1 = 0
Solution:
xy + x + y + 1= 0
or, x(y + 1) + 1(y + 1 ) = 0
or (y + 1 )(x + 1) =0
∴ x + 1 = 0
and y + 1 = 0 are the two separate lines,
∴ The given equation represents a pair of lines.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 29.
For what value of λ do the following equations represent pair of straight lines?
(a) λx2 + 5xy – 2y2 – 8x + 5y – λ = 0
Solution:
λx2 + 5xy – 2y2 – 8x + 5y – λ = 0
∴ a = λ, 2h = 5, b = -2, 2g = -8
2f = 5, c = -1
∴ h = \(\frac{5}{2}\), g = -4, f = \(\frac{5}{2}\)
As the given equation represent a pair of lines, we have abc + 2fgh – ah2 – bg2 – ch2 = 0
or, λ(-2)(-λ) + 2. \(\frac{5}{2}\) (-4). \(\frac{5}{2}\) -λ × \(\frac{25}{4}\) – (-2) (-4)2 – (-λ) × \(\frac{25}{4}\) = 0
or, 2λ2 – 50 – \(\frac{25 λ }{4}\) + 32 + \(\frac{25 λ }{4}\) = 0
or, 2λ2 = 18 or, λ2 = 9
λ = ±3

(b) x2 – 4xy – y2 +6x + 8y + λ = 0
Solution:
Here a = 1, 2h = -1, b = -1, 2g = 6, 2f = 8, c = τ
As the given equation represent a pair of lines, we have
abc + 2fgh – af2 – bg2 – ch2 = 0
⇒ (-1) τ + 2.4.3 (-2) – 1. 42 – (-1). 32 – τ(-2)2 = 0
⇒ -τ – 48 – 16 + 9 – 4τ = 0
⇒ -5τ – 55 = 0 ⇒ τ = -11

Question 30.
(a) Obtain the value of λ for which the pair of straight lines represented by 3x2 – 8xy + λy2 = 0 are perpendicular to each other.
Solution:
3x2 – 8xy + λy2 = 0
∴ a = 3. 2h = -8, b = λ
As the pair of lines are perpendicular to each other, we have a + b = 0.
or, 3 + λ = 0 – or, λ = -3

(b) Prove that a pair of lines through the origin perpendicular to the pair of lines represented by px2 – 2qxy + ry2 = 0 is given by rx2 – 2qxy + py2 = 0
Solution:
px2 – 2qxy + ry2 = 0
∴ a = p, b = 2qy, c = ry2
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 62

(c) Obtain the condition that a line of the pair of lines ax2 + 2hxy + by2 = 0,
(i) Coincides with
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 63

(ii) is perpendicular to, a line of the pair of lines px2 + 2qxy + ry2 = 0
Solution:

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 64

Question 31.
Find the acute angle between the pair of lines given by :
(a) x2 + 2xy – 4y2 = 0
Solution:
x2 + 2xy – 4y2 = 0
∴ a=1, 2h = 2, b = -4
∴ tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}=\frac{\pm 2 \sqrt{1+4}}{1-4}\)
\(=\pm \frac{2 \sqrt{5}}{-3}=\mp \frac{2 \sqrt{5}}{3}\)
∴ The acute angle between the pair of lines is tan-1 \(\frac{2 \sqrt{5}}{3}\)

(b) 2x2 + xy – 3y2 + 3x + 2y + 1 = 0
Solution:
2x2 + xy – 3y2 + 3x + 2y + 1 = 0
∴ a = 2, 2h = 1, b = -3, 2g = 3 2f= 2, c = 1.
tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}\)
\(=\pm \frac{2 \sqrt{\frac{1}{4}+6}}{2-3}=\pm \frac{2 \times 5}{2(-1)}\) = ± 5
∴ The acute angle is tan-1 5

(c) x2 + xy – 6y2 – x – 8y – 2 = 0
Solution:
Given Equation is x2 + xy – 6y2 – x – 8y – 2 = 0
here a = 1, 2h = 1, b = -6 thus if 0 is the acute angle between two lines then
tan θ = \(=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
= \(\left|\frac{2 \times 5}{-10}\right|\) = 1
∴ θ = 45°

Question: 32.
Write down the equation of the pair of bisectors of the following pair of lines :
(a) x2 – y2 = 0 ;
Solution:
x2 – y2 = 0
∴ a = 1, b = -1, h = 0
∴ The equation of the bisectors of the angles between the pair of lines are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{1+1}=\frac{x y}{0}\)
or, xy = 0

(b) 4x2 – xy – 3y2 = 0
Solution:
4x2 – xy – 3y2 = 0
∴ a = 4, 2h = -1, b = -3
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{(a-b)}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{7}=\frac{x y}{\left(-\frac{1}{2}\right)}\)
or, x2 – y2 = -14xy
or, x2 + 14xy – y2 = 0

(c) x2 cos θ + 2xy – y2 sin θ = 0
Solution:
x2 cos θ + 2xy – y2 sin θ = 0
∴ a = cos θ, 2h = 2, b = – sin θ
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{\cos \theta+\sin \theta}=\frac{x y}{1}\)
or, x2 – y2 = xy(cos θ + sin θ)

(d) x2 – 2xy tan θ – y2 = 0
Solution:
x2 – 2xy tan θ – y2 = 0
∴ a = 1, 2h = -2 tan θ, b = -1
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{2}=\frac{x y}{-\tan \theta}\)
or, x2 – y2 = 2xy cot θ
or, x2 + 2xy cot θ – y2 = 0

Question 33.
If the pair of lines represented by x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 be such that each pair bisects the angle between the other pair, then prove that pq = -1.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b) 65

Question 34.
Transform the equation: x2 + y2 – 2x – 4y + 1 = 0 by shifting the origin to (1, 2) and keeping the axes parallel.
Solution:
x2 + y2 – 2x – 4y + 1 = 0     ……(1)
Let h = 1, k = 2
Taking x’ + h and y = y’ + k we have
(x’ + h)2 + (y’ + k)2 – 2(x’ + j) -4(y’ + k) + 1=0
or, (x + 1)2 + (y’ + 2)2 – 2(x’ + 1) – 4(y’+ 2) + 1=0
or, x‘2 + 1 + 2x’ + y‘2 + 4 – 4y’ – 2x’ – 2 – 4y’- 8 + 1 = 0
or, x‘2 + y’2 – 4 = 0
∴ The transformed equation is x2 + y2 = 4

Question 35.
Transform the equation: 2x2 + 3y2 + 4xy – 12x – 14y + 20 = 0. When referred to parallel axes through(2, 1).
Solution:
2x2 + 3y2 + 4xy – 12x – 14y + 20 = 0
Let h = 2, k = 1
Taking x = x’ + 1 and y = y’ + 1
we have
2(x’ + 2)2 + 3(y’ + 1)2 + 4(x’ + k)(y’ + 1) – 12 (x’ + 2)- 14 (y’ + 1) + 20 = 0
or, 2x‘2 + 8 + 8x’ + 3 + 6y’ + 3y’2 + 4x’y’ + 4x’ + 8y’ + 8 – 12x’ – 14y’ – 18 = 0
or, 2x‘2 + 3y’2 + 4x’y’ + 1=0
The transformed equation is
2x2 + 3y2 + 4xy + 1 = 0

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Ex 11(b)

Question 36.
Find the measure of rotation so that the equation x2 – xy + y2 = 5 when transformed does not contain xy- term.
Solution:
x2 – xy + y2 = 5
Taking x = x’ cos α – y’ sin α
y = x’ sin α – y’ cos α
We get (x’ cos α – y’ sin α)2 – (x’ cos α – y’ cos α) (x’ sin α + y’ cos α) + (x’ sin α + y’ cos α)2 = 5
⇒ x‘2 cos2 α + y‘2 sin2 α – 2x’y sin α.
cos α – x‘2 sin α. cos α – x’y cos2 α + x’y’ sin2 α + y‘2 sin α. cos α + x‘2 sin2 α + y‘2 cos2 α + 2x’y’ sin α cos α = 5
Given that the transformed equation does not xy term.
Hence the co-efficient of x’y’ is zero.
That is sin2 α – cos2 α = 0
⇒ sin2 α = cos2 α
⇒ tan2 α = 1 ⇒ tan α = 1 ⇒ α= 45°

Question 37.
What does the equation x + 2y – 10 =0 become when the origin is changed to (4, 3)?
Solution:
x + 2y – 10 = 0
Let h = 4, k = 3
Taking x = x’ + 4, y = y’ + 3
we have x’ + 4 + 2 (y’ + 3) – 10 = 0
or, x + 2y’ = 0
∴ The transformed equation is x + 2y = 0.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(c)

Question 1.
Compute the following :
(i) 12C3
Solution:
12C3 = \(\frac{(12) !}{3 ! 9 !}=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2}\) = 220

(ii) 15C12
Solution:
15C12 = \(\frac{(15) !}{(12) ! 3 !}=\frac{15 \cdot 14 \cdot 13}{3 \cdot 2}\)
= 5.7.13 = 455

(iii) 9C4 + 9C5
Solution:
9C4 + 9C5 = \(\frac{9 !}{4 ! 5 !}+\frac{9 !}{5 ! 4 !}\)
\(=\frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}\) × 2 = 252

(iv) 7C3 + 6C4 + 6C3
Solution:
7C3 + 6C4 + 6C3 = 7C3 + 6C4 + 6C4-1
= 7C3 + 6+1C4 = 7C3 + 7C4
(∴ ncr + nCr-1– = n+1cr)
= 7C4 + 7C4-1 = 7+1C4
= 8C4 = \(\frac{8 !}{4 !(8-4) !}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}\) = 70

(v) 8C0 + 8C1 + …….. + 8c8
Solution:
8C0 + 8C1 + …….. + 8c8 = 28 = 256

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c)

Question 2.
Solve :
(i) nC4 = nC11 ;
Solution:
nC4 = nC11 ;  (∴ n = 4 + 11 = 15)

(ii) 2nC3 : nC3 = 44: 5
Solution:
2nC3 : nC3 = \(\frac{44}{5}\)
⇒ \(\frac{2 n !}{2 n-3 !} / \frac{n !}{n-3 !}=\frac{44}{5}\)
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c)

Question 3.
Find n and r if nPr = 1680, nCr = 70.
Solution:
nPr = 1680, nCr = 70
∴ \(\frac{{ }^n \mathrm{P}_r}{{ }^n \mathrm{C}_r}=\frac{1680}{70}\)
or, r ! = 24 = 4!
∴ r = 4
Again, nCr = 70 or nC4 = 70
or, \(\frac{n !}{4 !(n-4) !}=70\)
or, n(n – 1) (n – 2) (n – 3)
= 70 × 4! = 7 × 10 × 4 × 3 × 2
= 8 × 7 × 6 × 5
or, n(n – 1) (n – 2) (n – 3)
or, 8(8 – 1) (8 – 2) (8 – 3)
∴ n = 8

Question 4.
How many diagonals can an n-gon(a polygon with n sides) have?
Solution:
A polygon of n – sides has n vertices.
∴ The number of st. lines joining the n-vertices is nC2.
∴ The number of diagonals is nC2 – n
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c) 1

Question 5.
If a set A has n elements and another set B has m elements, what is the number of relations from A to B?
Solution:
If |A| = n, |B| = m
then |A × B| = mn
∴ The number of possible subsets of
A × B = 2mn
∴ The number of relations from A to B is 2mn.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c)

Question 6.
From five consonants and four vowels, how many words consist of three consonants and two vowels?
Solution:
Words of consisting of 3 consonants and 2 vowels are to be formed from five consonants and 4 vowels.
∴ The number of ways = 5C3 × 4C2
Again, 5 letters can be arranged among themselves in 5! ways.
∴ The total number of ways
= 5C3 × 4C2 × 5! = 10 × 6 × 120 = 7200.

Question 7.
In how many ways can a committee of four gentlemen and three ladies be formed out of seven gentlemen and six ladies?
Solution:
A committee of 4 gentlemen and 3 ladies is to be formed out of 7 gentlemen and 6 ladies.
∴ The number of ways in which the committee can be formed.
7C4 × 6C3 = \(\frac{7 \cdot 6 \cdot 5}{3.2} \times \frac{6 \cdot 5 \cdot 4}{3 \cdot 2}\) = 700

Question 8.
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily. In how many ways can this be done? Find also the number of ways such that at least 3 black balls can be drawn.
Solution:
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily.
∴ The number of ways in which balls are drawn \({ }^9 \mathrm{C}_6=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}\) = 84 as the total number of balls is 9. If at least 3 black balls are drawn, then the drawing can be made as follows.

Black(4) White(5)
3 3
4 2

The number of ways in which at least 3 black balls are drawn
= (4C3 × 5C3) + (4C4 × 5C2)
= (4 × 10) + (1 × 10) = 50

Question 9.
How many triangles can be drawn by joining the vertices of a decagon?
Solution:
A decagon has 10 vertices and 3 noncollinear points are required to be a triangle.
∴ The number of triangles formed by the joining of the vertices of a decagon is
10C3 = \(\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}\) = 120

Question 9.
How many triangles can be drawn by joining the vertices and the center of a regular hexagon?
Solution:
A regular hexagon has six vertices. Triangles are to be formed by joining the vertices and center of the hexagon. So there is a total of 7 points. So the number of triangles formed.
7C3 = \(\frac{7 !}{3 ! 4 !}=\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\) = 35
As the hexagon has 3 main diagonals, which pass through the center hence can not form 3 triangles.
∴ The required number of triangles 35 – 3 = 32

Question 11.
Sixty points lie on a plane, out of which no three points are collinear. How many straight lines can be formed by joining pairs of points?
Solution:
Sixty points lie on a plane, out of which no. 3 points are collinear. A straight line required two points. The number of straight lines formed by joining 60 points is
60C2 = \(\frac{60 !}{2 \times 58 !}=\frac{60 \times 59}{2}\) = 1770

Question 12.
In how many ways can 10 boys and 10 girls sit in a row so that no two boys sit together?
Solution:
10 boys and 10 girls sit in a row so that no two boys sit together. So a boy is to be seated between two girls or at the two ends of the row. So the boys are to be sitted in 11 positions in 11C10 ways. Again 10 boys and 10 girls can be arranged among themselves in 10! and 10! ways respectively.
∴ The total number of ways = 11C10 × 10! × 10! = (11)! × (10)!

Question 13.
In how many ways can six men and seven girls sit in a row so that the girls always sit together?
Solution:
Six men and seven girls sit in a row so that the girls always sit together. Considering the 7 girls as one person, there are a total of 7 persons who can sit in 7! ways. Again the 7 girls can be arranged among themselves in 7! ways.
∴ The total number of arrangements
= 7! × 7!
= (7!)2

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c)

Question 14.
How many factors does 1155 have that are divisible by 3?
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c) 2
∴ In order to be a factor of 1155 divisible by 3, we have to choose one or two of 5, 7, and 11 along with 3 or 3 alone.
∴ The number of ways = 3C1 + 3C2 + 3C0 = 23 – 1 = 7
∴ The number of factors is 7 excluding 1155 itself.

Question 15.
How many factors does 210 have?
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c) 3
∴ We can choose at least one 2, 3, 5, or 7  to be a factor of 210.
∴ The number of factors.
= 4C1 + 4C2 + 4C3 + 4C4 = 24 – 1 = 15
∴ The number of factors is 210 is 15. (Including 215 itself and excluding 1).

Question 16.
If n is a product of k distinct primes what is the total number of factors of n?
Solution:
n is a product of k distinct primes.
∴ In order to be a factor, of n, we have chosen at least one of k distinct primes.
∴ The number of ways = kC1 + kC2 + ……… kCk-1 = 2k  – 1 – 1
∴ The number of factors of n is 2k – 2.
(Excluding 1 as 1 is not prime. It is also not include n.)

Question 17.
If m has the prime factor decomposition P1r1, P2r2 ….. Pnrn, what is the total number of factors of m (excluding 1)?
Solution:
m has the prime factor decomposition P1r1, P2r2 ….. Pnrn,
∴ m = P1r1, P2r2 ….. Pnrn,
P1 is a factor of m which occurs r1 times. Each of the factors P1r1 will give rise to (r1 + 1) factors.
Similarly
P2r2 gives (r2 + 1) factors and so on.
∴ The total number of factors (r1 + 1) (r2 + 1) ….. (rn + 1) – 1 (including m).

Question 18.
If 20! were multiplied out, how many consecutive zeros would it have on the right?
Solution:
If 20! were multiplied out, then the number of consecutive zeros on the right is 4. due to the presence of 4 x 5, 10, 14 x 15,20.

Question 19.
How many factors of 10,000 end with a 5 on the right?
Answer:
We have 1000 = 24 × 54 The factors of 10000 ending with 5 are 5, 5 × 5 = 25, 5 × 5 × 5 = 125
5 × 5 × 5 × 5 = 625
∴ There are 4 factors ending With 5.

Question 20.
A man has 6 friends. In how many ways can he invite two or more to a dinner party?
Solution:
A man has 6 friends. He can invite 2 or more of his friends to a dinner party.
∴ He can invite 2, 3, 4, 5, or 6 of his friends in
6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 266C06C1
= 64 – 7 = 57 ways.

Question 21.
In how many ways can a student choose 5 courses out of 9 if 2 courses are compulsory?
Solution:
A student is to choose 5 courses out of 9 in which 2 courses are compulsory. as 2 courses are compulsory, he is to choose 3 courses out of 7 courses in 7C3 = 35 ways.

Question 22.
In how many ways can a student choose five courses out of the courses? C1, C2, …………. C9 if C1, C2 are compulsory and C6, C8 cannot be taken together?
Solution:
A student chooses five courses out of the courses? C1, C2, …………. C9 if C1, C2 are compulsory and C6, C8 cannot be taken together.
∴ He is to choose 3 courses out of C3, C4, ………… ,C8, C9.
Without taking any restrictions 3 courses out of C3, C4, …… C9, i.e. from 7 courses in 7C3 ways. If C6, C8 are taken together then one course only to be choosen from C3, C4, C5, C7, C9 by 5C1 ways. Hence required number of ways.
= 7C35C1
= \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\) – 5
= 35 – 5 = 30 ways.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c)

Question 23.
A cricket team consisting of 11 players is to be chosen from 8 batsmen and 5 bowlers. In how many ways can the team be chosen so as to include at least 3 bowlers?
Solution:
A cricket team consisting of 1 1 player is to be chosen from 8 batsmen and 5 bowlers, as to include at least 3 bowlers.
The selection can be made as follows :

Batsmen(8) Bowlers(5)
8 3
7 4
6 5

The number of selections is
(8C8 × 5C3) + (8C7 × 5C4) + (8C6 × 5C5)
= (1 × 10) + (8 × 5) + (28 × 1)
= 10 + 40 + 28 = 78

Question 24.
There are n + r points on a plane out of which n points lie on a straight line L and out of the remaining r points that lie outside L, no three points are collinear. What is the number of straight lines that can be formed by joining pairs of their points?
Solution:
There are n + r points on a plane out of which n points are collinear and out of which r points are not collinear.
∴ We can form a straight line by joining any two points.
n-collinear points form one line and r-non-collinear points form rC2 lines.
Again, each of the r non-collinear points when joined to each of the noncollinear points, forms n lines.
∴ The number of such limes is r x n.
∴ The total number of lines
CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Ex 8(c) 4

Question 25.
There are 10 books in a shelf with different titles; five of these have red covers and others have green covers. In how many ways can these be arranged so that the red books are placed together?
Solution:
There are 10 books in a shelf with different titles, 5 of these are red covers and others are green covers considering 5 red-covered books as one book, we have a total of 6 books which can be arranged in 6! ways. The five red cover books are arranged among themselves in 5! ways.
∴ The total number of arrangements
= 5! x 6!

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

Odisha State Board CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम् Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Sanskrit Poem 2 भाति मे भारतम् Question Answer

GROUP – A

(क) बन्धनीमध्यात् शुद्धम् उत्तरं चित्वा लिखत-

1. _________ भाति मेऽनारतं भारतम् ।
(मानसे, भूतले, मण्ड़ले)
Answer:
भूतले

2. भूतले भाति मेऽनारतं _________ ।
(भारतम्, कोशल:, गुर्जर:)
Answer:
भारतम्

3. पावनं _________ उद्घोषयत् भारतं भाति ।
( चैतन्यम्, भावनम्, विश्वबन्धुत्वम् )
Answer:
विश्वबन्धुत्वम्

4. विश्ववन्द्यैः चरित्रैः _________ पावयत् भारतं भाति ।
(भुबं, दिवं, जगत्)
Answer:
जगत्

5. विश्ववन्द्यैः _________ जगत् पावयत् भारतं भाति ।
(चरित्रैः, पवित्रैः, महद्भिः )
Answer:
चरित्रैः

6. विश्वमेकं _________ समालोकयद् ।
(संसारं, परिवार, कुटुम्ब )
Answer:
कुटुम्ब

7. यद्धि. _________ सदप्यस्त्यभिन्नं सदा ।
(अन्नं, भिन्नं, अन्यं)
Answer:
भिन्नं

8. _________ मार्गं भणन्ति आगमाः ।
( धर्मस्य, अर्थस्य, मोक्षस्य )
Answer:
मोक्षस्य

9. मोक्षस्य मार्गं भणन्ति _________ ।
(निगमा:, आगमाः, सुगमा : )
Answer:
आगमाः

10. ज्ञानमास्ते च भार: _________ वै विना ।
(कर्म, करणं, क्रियां)
Answer:
क्रियां

11. जाह्नवी – चन्द्रभागा-जलैः _________ ।
(लालितम्, भावितम्, पावितम् )
Answer:
पावितम्

12. शुभ्रहमाद्रि – हासप्रभा _________ ।
(पूरितम्, पूजितम्, अन्वितम्)
Answer:
पूजितम्

13. विद्युदुत्पादने त्रैल _________ ।
(अन्वेषणे, संशोधने, निष्पादने)
Answer:
संशोधने

14. _________ चिकित्सालयस्थापनैः ।
(शल्यशोधं, छिद्रजालं, रोगजालं)
Answer:
रोगजालं

15. _________ वियन्मण्डले स्थापयत् भारतं भूतले भाति ।
(वाणभट्टं, सूर्यभट्ट, आर्यभट्टं)
Answer:
आर्यभट्ट

16. _________ भूमिगर्भेऽणुशक्तिं किरत् ।
(पोखरण, आजमेर, चान्दिपुरम् )
Answer:
पोखरण

17. _________ अणुं सन्ततं प्रेरयत् भूतले भारतं भाति ।
(मैत्रिकार्येषु, प्रीतिकार्येषु, शान्तिकार्येषु)
Answer:
शान्तिकार्येषु

18. राष्ट्रभाषायुतं _________ भारतम् ।
(तावकं, मामकं, मे)
Answer:
मामकं

19. कालिदासेन _________ संद्योतितम् ।
(व्यासेन, रत्नाकरेण, भासेन )
Answer:
भासेन

20. हस्तिगुम्फामजन्तामलौरां _________ ।
(दधत्, लसत्, मण्डितम्)
Answer:
दधत्

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

21. निष्क्रियं पार्थमाश्वासयद् _________ ।
( सीतया, गीतया, कृपया )
Answer:
गीतया

22. स्वीयरूपेण तं च व्याघात् _________ ।
(कर्मठम्, पुण्यजम्, पापिष्ठम् )
Answer:
कर्मठम्

23. मृत्युपाशं च _________ सहर्षं धरन् ।
(शिरे, कण्ठे, हृदि )
Answer:
कण्ठे

24. येन _________ सदा शिक्ष्यते प्रेर्यते ।
( समाजं, राष्ट्र, विश्वं)
Answer:
विश्वं

25. गायतिक्रान्तदर्शी _________ चयः ।
( मुनीनां, कवीनां, मतीनां )
Answer:
कवीनां

26. मोदतां मे सदा _________ भारतम् ।
(पावनं, कुशलं, विशुद्धं)
Answer:
पावनं

(ख) अतिसंक्षेपेण उत्तरं लिखत-

1. किम्भूतं विश्ववन्धुत्वम् उद्घोषयत् भारतं भाति ?
Answer:
ପାବନମ୍ ।

2. पावनं किम् उद्घोषयत् भारतं भाति ?
Answer:
ବିଶ୍ଵବନ୍ଧୁତ୍ଵମ୍ ।

3. किम्भूतैः चरित्रैः जगत् पावयत् भारतं भाति ?
Answer:
ବିଶ୍ବବନ୍ଦ୍ୟଃ ।

4. अनारतं भारतं कुत्र भाति ?
Answer:
ଭୂତଳେ ।

5. मोक्षस्य मार्गं भणन्ति ?
Answer:
ଆଗମ ।

6. आगमाः मोक्षस्य किं भणन्ति ?
Answer:
ମାର୍ଗମ୍ ।

7. क्रियां विना किं भारः आस्ते ?
Answer:
ଜ୍ଞାନମ୍।

8. कां विना ज्ञानं भारः आस्ते ?
Answer:
କ୍ରିୟାମ୍ ।

9. जाह्नवी – चन्द्रभागा-जलैः किं पावितम् ?
Answer:
ଭାରତମ୍।

10. भानुजा – नर्मदा – वीचिभिः किं लालितम् ?
Answer:
ଭାରତମ୍।

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

11. किम् अर्वुदारावलीश्रेणीसंपूजितम् ?
Answer:
ଭାରତମ୍।

12. चिकित्सालयस्थापनै: किम् उन्मूलयत् भारतं भाति?
Answer:
ରୋଗଜାଲମ୍ ।

13. औषधोत्पादनै: किम् उन्मूलयत् भारतं भाति ?
Answer:
ରୋଗଜାଲମ୍ ।

14. शल्यशोधै: किम् उन्मूलयत् भारतं भाति ?
Answer:
ରୋଗଜାଲମ୍ ।

15. वियन्मण्डले किं स्थापयत् भूतले भारतं भाति ?
Answer:
ଆର୍ଯ୍ୟଭଟ୍ଟମ୍ ।

16. पोखरण्-भूमिगर्भे किं किरत् भूतले भारतं भाति ?
Answer:
ଅଣୁଶକ୍ତିମ୍ ।

17. कीदृक्कार्येषु अणुं सन्ततं प्रेज्यत् भूतले भारतं भाति ?
Answer:
ଶାନ୍ତକାର୍ଯ୍ୟଷୁ ।

18. सेतुबन्धं कुत्र वर्तते ?
Answer:
ରାମେଶ୍ଵରେ

19. सूर्यमन्दिर कुत्र वर्तते ?
Answer:
କୋଣାର୍ଥେ ।

20. ‘ओनम्’ पर्व कुत्र पालयते ?
Answer:
କେରଳେ ।

21. श्रीहरि: निष्क्रियं पार्थं कया आश्वासयत् भारतं भाति ?
Answer:
ଗୀତୟା ।

22. कुत्र श्रीहरि: निष्क्रियं पार्थम् आश्वासयत् ?
Answer:
କୁରୁକ୍ଷେତ୍ରମଧେ ।

23. ‘क: मृत्युपाशं कण्ठे सहर्षं धरन् अमरत्वं गतः ?
Answer:
ଭକ୍ତସିଂହଃ ।

24. किम्भूतं भारतं सदा मोदताम् ?
Answer:
ପାବିନମ୍ ।

GROUP – B

(ग) संक्षेपेण (षड्भि: वाक्यै:/ पज्चविंशत्यापदै:) उत्तरं लिखत –

1. भारतं काभि: नदीभि: भाति ?
Answer:
ଅସ୍ମାକଂ ଭାରତଂ ନଦୀମାତୃକାୟା ଦେଶଃ । ଭାରତେ ବହବ ସୁଜଳା ସରିତଃ ଶୋଭନ୍ତେ । ତାସୁ ସୁବିଖ୍ୟାତା ଜହୁ କନ୍ୟା ଗଙ୍ଗା ପବିତ୍ରା ପାପନାଶିନୀ ଚ ଭବତି । ତଥୈବ ଯମୁନା, ନର୍ମଦା, ସିନ୍ଧୁ, ଗୋଦାବରୀ, କାବେରୀ, ତୁଙ୍ଗଭଦ୍ରା, ଚନ୍ଦ୍ରଭାଗା, ବିପାଶା ଇତ୍ୟାଦୟ ପବିତ୍ରା ନଦ୍ୟ ବହନ୍ତି । ଏତାଭି ନଦୀଭି ଅସ୍ମାକଂ ଭାରତଂ ସଦା ଭାତି । ପୂଜାବିଧୌ ସଂଳ୍ପକାଳେ ନଦୀନାମ୍ ଆବାହନଂ ଭବତି ।

2. चिकित्साविज्ञाने भारतस्य उन्नतिं वर्णयत ।
Answer:
ପ୍ରାଚୀନକାଳାଦାରଭ୍ୟ ଆଧୁନିକକାଳଂ ଯାବତ୍ ଚିକିତ୍ସାକ୍ଷେତ୍ରେ ଅସ୍ମାକଂ ଭାରତଂ ସମୁନ୍ନତମ୍ । ଚରକସୁଶ୍ରୁତୟୋଃ ଚିକିତ୍ସାପଦ୍ଧତିଃ ଅତୀବ ଆଦରଣୀୟଃ ସ୍ପୃହଣୀୟଂ ଚ ଭବତି । ଚିକିତ୍ସାଳୟାନାଂ ସ୍ଥାପନେନ, ଔଷଧାନାମ୍ ଅନୁସନ୍ଧାନେନ ଉତ୍ପାଦନେନ ଚ, ସୁଶ୍ରୁତକାଳାଦାରଭ୍ୟ ଅଦ୍ୟପର୍ଯ୍ୟନ୍ତ ଶଲ୍ୟପଦ୍ଧତିମାଧମେନ ଚିକିତ୍ସୟା ବା ଉପଚାରେଣ, ନବୀନପଦ୍ଧତ୍ୟା ଚିକିସୟା ରୋଗନିରାକରଣେନ ଚ ପୃଥ‌ିବ୍ୟା ପ୍ରଶଂସନୀୟଂ ଭାରତଂ ସର୍ବଦୈବ ଶୋଭତେ ।

3. भारतं विविधभाषाभिः समृद्धम् – इत्यस्य उदाहरणं प्रदत्त ?
Answer:
ଭାଷାମାଂ ବୈଭବେନ ଭାରତଂ ସୁସମୃଦ୍ଧ ଭବତି । ଭାରତେ ବିବିଧା ଭାଷାଃ ଦୃଶ୍ୟନ୍ତେ ଯଥା – ସଂସ୍କୃତଂ, ପ୍ରାକୃତଂ, ତାମିଲଂ, ତେଲୁଗୁ, କନ୍ନଙ୍‌, କୈରଳୀ, ବାଙ୍ଗଲାମ୍, ଔତ୍କଳୀ ବାଚମ୍ ଅନ୍ୟା ଚ ତାଂ ତାଂ ବ୍ରୁବଦ୍ ବର୍ଧତେ ରାଷ୍ଟ୍ରଭାଷାଯୁତଂ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି । ଭାଷାଭିଭିକସମୃଦ୍ଧି ଭାରତସ୍ୟ ପରିଲକ୍ଷ୍ୟତେ ।

4. अस्माकं देशः केन केन रामायणेन शेभते ?
Answer:
ଅସ୍ଵାକଂ ଦେଶ ବିବିଧନ ରାମାୟଣେନ ଶୋଭତେ । ବାଲ୍ମୀକି ବିରଚିତଂ ମୂଳସଂସ୍କୃତରାମାୟଣମ୍, କନ୍ନଡ଼ଭାଷାରଚିତଂ ପମ୍ପରାମାୟଣମ୍, ତାମିଲ୍‌ ଭାଷା ରଚିତଂ କମ୍ବରାମାୟଣମ୍, ଉତ୍କଳଭାଷାରଚିତଂ ଦାଣ୍ଡିରାମାୟଣମ୍, ବଙ୍ଗଭାଷାରଚିତଂ କୃତ୍ତିବାସରାମାୟଣମ୍ ଚ ତେରାଂ ପବିତ୍ରଧାରାଭିଂ ସର୍ବାନ୍‌ ଶ୍ରାବୟନ୍ତି ପବିତ୍ରୟନ୍ତି ଚ । ଏବଂଭୂତଃ ଦେଶଃ ଅସ୍ମାକଂ ଭାରତଂ ଶାଶ୍ଵତଂ ଶୋଭିତେ ।

5. भारतस्य नृत्यकलां सूचयत ?
Answer:
ଭାରତେଷୁ ବିଭିନ୍ନ ନୃତ୍ୟପରମ୍ପରା ସୂଚିତା । ପ୍ରମୁଖତୟା ଉତ୍କଳପ୍ରଦେଶେ ଓଡ଼ିଶୀନୃତ୍ୟ, ବନବାସିନାମ୍ ଆଦିବାସିନାଂ ଲୋକନୃତ୍ୟ, ଛଉନୃତ୍ୟ, ମଣିପୁରେ ମଣିପୁରୀନୃତ୍ୟ, ଉତ୍ତରଭାରତେ ପ୍ରଚଳିତଂ କଥକନୃତ୍ୟ, ଗୁର୍ଜରପ୍ରଦେଶେ ଗର୍ବାନୃତ୍ୟ, ଡାଣ୍ଡିଆନୃତ୍ୟ ଚ, ଆନ୍ଧ୍ରପ୍ରଦେଶେ କୂଚିପୂଡ଼ିନୃତ୍ୟ, ପଞ୍ଜାବପ୍ରଦେଶେ ଗିନୃତ୍ୟ ଭାଙ୍ଗଡ଼ାନୃତ୍ୟ ଚ, କେରଳପ୍ରଦେଶେ କଥକଳୀନୃତ୍ୟ, ତାମିନାଡ଼ୁପ୍ରଦେଶ ଭାରତନାଟ୍ୟମ୍ ଇତି ନୃତ୍ୟାନାଂ ପ୍ରଚଳନଂ ଦୃଶ୍ୟତେ । ଅନେନ ନୃତ୍ୟେନ ମମ ଭାରତମ୍ ଅନାରତଂ ଶୋଭତେ ।

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

6. भक्तसिंहः कथम् अमरत्वं गतः ?
Answer:
ଏକଦା ପରିଧୀନସ୍ୟ ଭାରତସ୍ୟ ସ୍ଵତନ୍ତଡାନିମିତ୍ତମ୍ ଆମ୍ବକଳିଂ ପ୍ରଦାୟ ଭଗତସିଂହ ଇତି ଦେଶଭକ୍ତଃ ଯୁବକଃ ଅମରତ୍ଵ ପ୍ରାପ୍ତବାନ୍ । ମାତୃଭୂମେ ବିପଦଜାଲିଂ ଉଚ୍ଛେଦୟନ୍ ଭକ୍ତସିଂହଃ ଇତି ଯୁବତଃ ସ୍ଵଜୀବନସ୍ୟ ବିନିମୟେନ ଅପି ଆଙ୍ଗଲଶାସକାନାଂ ବିତାଡ଼ନାର୍ଥେ ଯତ୍ନ କୃତବାନ୍ ! ମୃତ୍ୟୁପାଶଂ କଣ୍ଠେ ସହର୍ଷ ଧରନ୍ ଭକ୍ତସିଂହଃ ଅତ୍ର ଅମରତ୍ୱ ଗତଃ । ଏତାଦୃଶଃ ସୁପୁତ୍ର ଅସ୍ଵାକଂ ଦେଶସ୍ୟ ଗର୍ବ ଗୌରବଂ ଚ ଭବତି ।

7. विश्वं भारतेन कथं शिक्ष्यते ?
Answer:
ଅସ୍ମାକଂ ଦେଶସ୍ୟ ଅଭୂତପୂର୍ବ ସ୍ବାଧୀନତାସଂଗ୍ରାମିଂ ପ୍ରଦର୍ଶିତମ୍ । ରକ୍ତପାତଂ ବିନା କୁତ୍ରାପି କଶ୍ଚିତ୍ ଦେଶଃ ବିଶ୍ୱେଽସ୍ମିନ୍ ସ୍ଵାଧୀନତାଂ ନ ପ୍ରାୟୋତି । ପରନ୍ତୁ ଭାରତମେବ ପୃଥ‌ିବ୍ୟାମ୍ ଏତାଦୃଶଂ ଭୂଖଣ୍ଡ ଯତ୍ର ରକ୍ତପାତଂ ବିନା ଶସ୍ତ୍ରପାତଂ ବିନା ମନ୍ଦସ୍ମିତା ସଂକ୍ରାନ୍ତି ନାମ ସ୍ବାଧୀନତା ଆୟାତି । ଯେନ ବିଶ୍ଵ ସଦା ଶିଷ୍ୟତେ ପ୍ରେର୍ଯ୍ୟତେ ଚ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

8. भारते के के प्रसिद्धा: उत्सवा: सन्ति ?
Answer:
ଉତ୍ସବପ୍ରିୟା ଖଳୁ ମାନବାଃ । ଅସ୍ମାକଂ ଦେଶ ବିବିଧା ଉତ୍ସବାଃ ଋତୁ ଅନୁସାରଂ ମାସାନୁସାରଂ ଚ ପରିପାଳିତଂ ଭବତି । ବସନ୍ତେ ହୋଲୀଉତ୍ସବ, ଦୋଳୋତ୍ସବାଃ ବା, ଆଶ୍ୱିନମାସେ ଦଶହରାପର୍ବ, କାର୍ତ୍ତିକମାସେ ଦୀପାବଲ୍ୟୁସନଃ, ଆଶ୍ୱିନମାସେ ଲକ୍ଷ୍ମୀପୂଜନାର୍ଥୀ କୋଜାଗରୀ ଉତ୍ସବ, ଦକ୍ଷିଣଭାରତେ ପୋଙ୍ଗଲ୍ ଇତି ଉତ୍ସବାଃ, ଶ୍ରାବଣମାସେ ଶ୍ରାବଣ୍ୟତ୍ସବ, ରକ୍ଷାବନ୍ଧନୋତ୍ସଦଃ ଚ, ପଞ୍ଜାବ-ହରିୟାଣାଦି ରାଜ୍ୟେଷ୍ଠ ଲୋହଡ଼ୀ ଇତି ଉତ୍ସବାଃ, ରମଜାନ୍‌ମାସେ ‘ଇଦ୍’ ଇତି ଉତ୍ସବ, କେରଳରାଜ୍ୟ ‘ଓଣମ୍’ ଇତି ଉତ୍ସରଃ ପରିପାଲ୍ୟନ୍ତେ । ଏତାଦୃଶମ୍ ଉତ୍ସବମୁଖରଂ ମମ ଭାରତଂ ଭୂତଳେ ଭାତି ।

GROUP – C

(घ) प्रायशः चत्वारिंशतापदै: अष्टाभि: वाक्यै: वा उत्तरं लिखत –

1. कथं विश्वम् एकं कुटुम्वं समालोकयत् भारतं भाति ?
Answer:
ଭାରତସ୍ୟ ମହତ୍ତ୍ଵମ୍ ଅଖଣ୍ଡତ୍ଵ ବିଭୂତଂ ଚ ବିଶ୍ଵେଽସ୍ମିନ୍ ରାଜତେ । ସମସ୍ତେ ଧରଣୀତଳେ ଭାରତଂ ହି ଏକମାତ୍ର ଭୂଖଣ୍ଡ ଯତ୍ ସମଗ୍ରସ୍ୟ ବିଶ୍ବସ୍ୟ ହେତୋ ଚିନ୍ତା ପ୍ରକଟିତା । ତସ୍ମାତ୍ କାରଣାତ୍ ବିଶ୍ଵବନ୍ଧୁତ୍ଵମ୍ ଉଦ୍‌ଘୋଷୟତି, ଜନସମୁଦାୟସ୍ୟ କୃତେ ରାମାଦୀନାଂ ସଚ୍ଚରିତ୍ରାଣା ବର୍ଣନଂ କୃତ୍ୱା ଧରିତ୍ରୀ ପବିତ୍ରୀ କରୋତି । ‘ବସୁଧୈବ କୁଟୁମ୍ବକମ୍’’ ଇତ୍ୟେବ ବାର୍ତ୍ତା ବିଶ୍ୱେଽସ୍ମିନ୍ ପ୍ରଚାରିତା ପ୍ରସାରିତା ଚ ଭବତି । ସମଗ୍ର ବିଶ୍ୱମ୍ ଏବଂ କୁଟୁମିଂ ସମାଲୋକୟତ୍ବ ଅସ୍ମାକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

2. भिन्नमपि भारतं केन रूपेण अभिन्नम् ?
Answer:
‘‘ବିବିଧତା ମଧ୍ୟ ଏକତା’’ – ଅସ୍ମାକଂ ଦେଶସ୍ୟ ଲକ୍ଷ୍ୟ ଭବତି । ବିଭିନ୍ନତ୍ୱ ଅପି ଭାରତସ୍ୟ ଅଭିନ୍ନତଂ ବର୍ଣ୍ଣିତଂ ଭବତି । ଅସ୍ମାକଂ ଭାରତଭୂଖଣ୍ଡସ୍ୟ ଜନଃ ପରସ୍ପରଂ ବେଶଃ, ଭୂଷଣି, ଖାଦ୍ୟ, ପାନୀୟୈ, ପୂଜାପଦ୍ଧତିଭଃ, କ୍ରୀଡ଼ାଭ୍ୟାସସ୍ୟ ବୈବିଧେନ ଉତ୍ସବାନାଂ ଭେଦଃ ଜୀବନୋପାୟୈ ଚ ଭିନ୍ନା ଦୃଶ୍ୟନ୍ତେ । କିନ୍ତୁ ତେଷାମ୍ ଅନ୍ତଃକରଣେଷୁ ଭାରତୀୟତା ଏବ ବିରାଜତେ । ଏବଂ ବିବିଧେଷୁ ସହ ଅପି ଅଭିନଂ ସନ୍‌ ତନ୍ନାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

3. भारते के पर्वता: शेभन्ते ?
Answer:
‘ଭୂଭୃତ୍’ ଇତି ପର୍ବତସ୍ୟ ପର୍ଯ୍ୟାୟ ଶବ୍ଦ । ଅନେନ କାୟତେ ଯତ୍ ପର୍ବତଃ ଅସ୍ମାକଂ ଦେଶସ୍ୟ ପ୍ରାକୃତିକୀଶୋଭା ଭବନ୍ତି । ପର୍ବତଃ ଭାରତସ୍ୟ ମହିମମଣ୍ଡିତାଂ ଶୋଭାଂ ପବିତ୍ରତାଂ ଚ ପ୍ରକଟୟନ୍ତି । ବିନ୍ଧ୍ୟ, ସହ୍ୟପର୍ବତଃ, ନୀଳପର୍ବତଃ ଇତ୍ୟାଦିଭିଃ ପର୍ବତଃ ବିଭୂଷିତଂ, ଶୁଭ୍ରହାସଦୀପ୍ରେନ ହିମାଳୟେନ ପୁନଃ ଅର୍ବୁଦାରାବଳୀପର୍ବତମାଳାଭି ଅନାରତଂ ପୂଜିତଂ ମେ ଭାରତମ୍ ଅନାରତଂ ଭୂତଳେ ଭାତି । ମହାକବିକାଳିଦାସେନ କୁମାରସମ୍ଭବମ୍ ମହାକାବ୍ୟ ବର୍ଣ୍ଣିତଂ ଯତ୍ – ‘ଅସ୍ତୁଭରସ୍ୟା ଦିଶି ଦେବତାତ୍ମା ହିମାଳୟ ନାମ ନଗାଧୁରାଜଃ ପୂର୍ବାପରତୋୟ ନିଧ୍ଵଗାହ୍ୟ ସ୍ଥିତାପୃଥ୍ବ୍ୟାମ୍ ଇବ ମାନଦଣ୍ଡ ।’’ ଇତି ।

4. भारतं कस्मिन् कस्मिन् विषये समर्थम् ?
Answer:
ନ କେବଳଂ ପ୍ରାଚୀନକାଳେ ଅପିତୁ ଆଧୁନିକକାଳେ ଭାରତସ୍ୟ ସାମର୍ଥ୍ୟ ବିଶ୍ୱେଽସ୍ମିନ୍ ରାଜତେ । ବିଦ୍ୟୁତଃ ଉତ୍ପାଦନେ, ମୃତ୍ତିକାଧଃସ୍ଥିତସ୍ୟ ତୈଳସ୍ୟ ଉତ୍ତୋଳନେ ବିଶୋଧନେ ଚ, ପ୍ରାକୃତିକବାଷ୍ପରୂପେଣ ଅଙ୍ଗାରରୂପେଣ ଚ ଇନ୍ଧନସ୍ୟ ଅନୁସନ୍ଧାନେ, ଲୌହସ୍ୟ ଉତ୍ପାଦନେ ପୁନଃ ଯନ୍ତ୍ରଣା ବିନିର୍ମାଣେ ଚ ଦେଶସ୍ୟାସ୍ୟା ଉନ୍ନତତରାଂ କୌଶଲଂ ସାମର୍ଥ୍ୟ ବା ଅସ୍ତି ଇତି ବିଶ୍ବବାସିନଃ ଜାନନ୍ତି । ଏତସ୍ମିନ୍ କ୍ଷେତ୍ରେ ଭାରତସ୍ୟ ସାମର୍ଥ୍ୟ ସର୍ବେ ଜାନନ୍ତି ଏବ । ଏତାଦୃଶଂ ସମୁଜ୍ଜ୍ବଳଂ ମେ ଭାରତଂ ଭୂତଳେ ଅନାରତଂ ରାଜତେ ।

5. अणुशक्तिप्रयोगप्रभृतिषु भारतस्य नैपुण्यं प्रदर्शयत ।
Answer:
ଅତ୍ୟାଧୁନିକେ,ବିଜ୍ଞାନକ୍ଷେତ୍ରେ ଅପି ଅସ୍ମାକଂ ଭାରତଂ ଯାଦୃଶଂ ବୈଶିଷ୍ଟ୍ୟ ପ୍ରକଟୟତି ତଦେବ ବିଶ୍ଵେଽସ୍ମିନ୍ ଆଦୃତଃ ଭବତି । ଖଗୋଳ ବିଦ୍ୟାସୁ ନିପୁଣସ୍ୟ ଆର୍ଯ୍ୟଭଟ୍ଟସ୍ୟ ନାମାନୁସାରଂ ନାମିତଃ ଆର୍ଯ୍ୟଭଟ୍ଟ ଇତି ସ୍ଵାନକୌଶଳନିର୍ମିତଃ ଉପଗ୍ରହ ଭାରତୀୟୈ ବୈଜ୍ଞାନିକୈ ମହାକାଶେ ସ୍ଥାପିତଃ । ରାଜସ୍ଥାନମ୍ ଇତି ପ୍ରଦେଶସ୍ୟ ‘ପୋଖରାନ୍’ ଇତି ମରୁପ୍ରାନ୍ତରେ ଭୂଗର୍ଭମଧେ ଅସ୍ମାକମ୍ ଅଣୁବୈଜ୍ଞାନିକଃ ଅଣୁଶକ୍ତିବିସ୍ଫୋରଣଂ କୃତମ୍ । ସଫଳତୟା ଅଣୁଶକ୍ତିପ୍ରୟୋଗପରୀକ୍ଷା କୃତା । ମାନବସମାଜସ୍ୟ କଲ୍ୟାଣାୟ ଅଣୁଖଣ୍ଡିଂ ସନ୍ତତଂ ନିୟୋଜୟତ୍ ମେ ଭାରତମ୍ ଅନାରତଂ ମହୀତଳେ ରାଜତେ ।

6. के कवयः भारतस्य गोरवम् अवर्द्धयन् ?
Answer:
ଅସ୍ମାକଂ ଦେଶସ୍ୟ ଭାରତୀବୈଭବଂ ବିଶ୍ୱେଽସ୍ମିନ୍ ରାଜତେ । କବୟଃ ସ୍ବରଚିତଂ କାବ୍ୟମ୍ ଅତ୍ର ଶ୍ରାବୟନ୍ତି । କବୟଃ କ୍ରାନ୍ତଦର୍ଶନଃ ଭବନ୍ତି । ପୁରାଣମୁନଃ ଆଦିକବି ବାଲ୍ମୀକି ରାମାୟଣକାବ୍ୟ, ବେଦବ୍ୟାସ୍ୱ ମହାଭାରତଂ ଚ, ରତ୍ନାକରଶ୍ଚ ‘ହରବିଜୟମ୍’ ନାମ ମହାକାବ୍ୟ ବିରଚ୍ୟ ଭାରତବର୍ଷ ବିଭୂଷିତଂ କୃତବର୍ଷେ । ତେଷା ରଚନାବଳୀମ୍ ଉପଜୀବ୍ୟ କାବ୍ୟରୂପେଣ ଗୃହୀତ୍ଵା ଅନ୍ୟ କବୟଃ ସ୍ଵକାବ୍ୟ ବିରଚିତମ୍ । କବିକୁଳଗୁରୁ କାଳିଦାସ, ହାସସ୍ଵରୂପଃ ଭାସ୍ୱ, ସର୍ବତନ୍ତ୍ରସ୍ବତନ୍ତ୍ର ବିଳାପଃ କାଦମ୍ବରୀକାବ୍ୟରସପରିବେଷକଃ ବାଣଭଟ୍ଟ ଚ ଚେଷ୍ଟା କାବ୍ୟାଲୋକେନ ଦେଶଂ ସଂଦ୍ୟୋତିତଂ ସୌନ୍ଦର୍ଯ୍ୟାନ୍ବିତଂ ଚ କୃତବନ୍ତଃ । ଏତେଷା କାବ୍ୟସମୃଦ୍ଧା ମେ ଭାରତଂ ଭୂତଳେ ନିରନ୍ତରଂ ଶୋଭତେ ।

7. द्वारकाप्रभृतितीर्थानां नामानि लिखत ।
Answer:
ତୀର୍ଥକ୍ଷେତ୍ର ଭାରତମ୍ । ଅସ୍ଵାକଂ ଦେଶେ ପବିତ୍ରସ୍ଥାନାନି ତୀର୍ଥରୂପାଣି ବିରାଜନ୍ତେ । ଆକୁମାରୀହିମାଚଳପର୍ଯ୍ୟନ୍ତ ସମଗ୍ରଦେଶ ତୀର୍ଥକ୍ଷେତ୍ରାଣି ରାଜନ୍ତେ । ଗୁଜୁରାଟପ୍ରଦେଶସ୍ଥିତା ଦ୍ୱାରକା, ତାମିଲନାଡ଼ୁପ୍ରଦେଶସ୍ୟ ରାମେଶ୍ଵରସ୍ୟ ସବିଧେ ସାଗରମଧ୍ୟେ ନିର୍ମତଃ ସେତୁବନ୍ଧଃ ଉତ୍କଳପ୍ରଦେଶସ୍ୟ ପୁରୀସ୍ଥିତ ଭ।ଗଦତଃ ଜଗନ୍ନାଥସ୍ୟ ବଦରିକାକ୍ଷେତ୍ରମ୍ । ଆନ୍ଧ୍ରପ୍ରଦେଶସ୍ୟ ତିରୁପତୌ ତିରୁମଲାକ୍ଷେତ୍ରେ ପ୍ରସିଦ୍ଧ ବେଙ୍କଟେଶ ମନ୍ଦିରମ୍, ଉତ୍ତରପ୍ରଦେଶସ୍ୟ ମଥୁରାୟାଂ ଶ୍ରୀକୃଷ୍ଣଜନ୍ମଭୂମି, ସୁପୀସମ୍ପ୍ରଦାୟସ୍ୟ ଖ୍ରୀଜ୍ୟମଇନୁଦ୍ଦିସ୍ତି ଦର୍ଘଶୋଭିତମ୍ ଆଜମେରକ୍ଷେତ୍ରମ୍ ତତଂ ଚ ସ୍ଥିତଂ ପାବନଂ ପୁଷ୍କରତୀର୍ଥମ୍, ପଞ୍ଜାବପ୍ରଦେଶସ୍ୟ ଅମୃତସରଃ ଇତି ସ୍ଥାନେ ଶିଖଧର୍ମପୀଠମ୍ । ଏବମ୍ ଅନେକୈ ସୁପ୍ରସିଦ୍ଧି କ୍ଷେତ୍ରେ ତୀର୍ଥେ ଚ ବିରାଜିତଂ ଭାରତମ୍ ଅନାରତଂ ଭାତି ।

8. भारतस्य विविधेषु क्षेत्रेषु पालितान् उत्सवान् परिचाययत ।
Answer:
ଉତ୍ସବପ୍ରିୟା ଖଳୁ ମାନବାଃ । ଧର୍ମପ୍ରାଣସ୍ୟ ଭାରତବର୍ଷସ୍ୟ ବିଭିନ୍ନେଷୁ ସ୍ଥାନେଷୁ ସୁପ୍ରସିଦ୍ଧା ମନେଜ୍ଞା ଉତ୍ସବାଃ ପରିପାଲ୍ୟନ୍ତେ । ବସନ୍ତକାଳେ ଭଗବତଃ ଶ୍ରୀକୃଷ୍ଣସ୍ୟ ଦୋଳୋତ୍ସବେନ ସହ ସମଗ୍ର ଭାରତେ ପରିପାଳିତଃ ହୋଲିକୋତ୍ସବାଃ, ମହିଷମର୍ଦ୍ଦିନୀଦୁର୍ଗତିନାଶିନୀ ଦୁର୍ଗା ଆଶ୍ୱିନମାସେ ଶୁକ୍ଳପକ୍ଷେ ସଂପୂଜିତା ଭବତି । ଜନାସ୍ତୁ ଆଶ୍ୱିନମାସସ୍ୟ ଶୁକ୍ଳଦଶମାଂ ତୀର୍ଥେ ଦଶହରା ଇତି ଉତ୍ସବଂ ସାନନ୍ଦ ସଗୌରବଂ ପାଳୟନ୍ତି । କାର୍ତ୍ତିକମାସେ ଚ ଦୀପାଲୋକସଜିତାଂ ଦୀପାବଳୀ ଇତି ଉତ୍ସବଂ ସମଗ୍ର ଦଶହରା ଇତି ଉତ୍ସବଂ ସାନନ୍ଦ ସଗୌରବଂ ପାଳୟନ୍ତି ।

କାର୍ତ୍ତିକମାସେ ଚ ଦୀପାଲୋକସଜିତାଂ ଦୀପାବଳୀ ଇତି ଉତ୍ସବଂ ସମଗ୍ର ତାମିଲନାଡ଼ୁ ପ୍ରଦେଶସ୍ୟ ନବବର୍ଷେତ୍ସବରୂପେଣ, ଆନ୍ଧ୍ରପ୍ରଦେଶେ ଦକ୍ଷିଣଭାରତେ ଚ ପୋଙ୍ଗଲ୍‌ଇତି ଉତ୍ସବ ପରିପାଲ୍ୟତେ । ଶ୍ରାବଣପୂର୍ଣ୍ଣିମାୟାଂ ଶ୍ରାବଣ୍ୟସମଃ, ରକ୍ଷାବନ୍ଧାନତ୍ସତଃ ବଳଭଦ୍ରଜନ୍ମୋତ୍ସବ ବା ପରିପାଲ୍ୟତେ । ଦକ୍ଷିଣ-ପୂର୍ବ-ଭାରତୀୟା ଚ ଯଥା ଅଭ୍ୟୁ ସବଂ ପାଳୟନ୍ତ ତଥୈବ ପଞ୍ଜାବ-ହରିୟାଣାଦିରାଜ୍ୟେଷ୍ଠ ଲୋହଡ଼ୀ ଇତି ଉତ୍ସବଂ, ରମଜାନ୍‌ମାସେ ‘ଇଦ୍‌’ ଇତି ଉତ୍ସବଂ, କେରଳରାଜ୍ୟ ଜନାଃ ‘ଓଲ୍‌ମ୍’ ଇତି ଉତ୍ସବଂ ପାଳୟନ୍ତି । ଅତଃ ଏତାଦୃଶମ୍ ଉତ୍ସବମୁଖରଂ ସମଭରାତମ୍ ଅନାରତଂ ବିଶ୍ଵେଽସ୍ମିନ୍ ବିରାଜତେ ।

ଶ୍ଳୋକ – ୧

विश्वबन्धुत्वमुद्घोषयत्पावनं
विश्ववन्द्येश्चरित्रैर्जगत्पावयत् ।
विश्वमेकं कुटुम्बं समालोकयद्
भूतले भाति मेनारतं भारतम् ॥ १ ॥

ବିଶ୍ଵବନ୍ଧୁତ୍ୱମୁଦ୍ଘେ।ଷୟତ୍ପାବନଂ
ବିଶ୍ୱବନ୍ଦ୍ୟଶ୍ଚ ରିତ୍ରୈର୍ଜଗତ୍ପାବୟତ୍ ।
ବିଶ୍ଵମେକଂ କୁଟୁମ୍ବୀ ସମାଲୋକୟଦ୍
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ୧ ॥

ଅନ୍ବୟ – ପାବନଂ ବିଶ୍ଵବନ୍ଧୁତ୍ଵମୁଦ୍‌ଷୟତ୍, ବିଶ୍ୱବନ୍ଦ୍ୟ ଚରିତଃ ଜଗତ୍ ପାବୟତ୍, ବିଶ୍ବମ୍ ଏବଂ କୁଟୁମ୍ବ ସମାଲୋକୟତ୍, ମେ ଭାରତଂ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ପାବନମ୍ = ପବିତ୍ର । ଉଦ୍‌ଘୋଷୟତ୍ = ଘୋଷଣା କରୁଥିବା । ଚରିତିଃ = ଚରିତ୍ରଦ୍ଵାରା । ଜଗତ୍ = ସଂସାର । ବିଶ୍ୱମ୍ = ପୃଥ‌ିବୀକୁ । କୁଟୁମ୍ବମ୍ = ପରିବାର । ସମାଲୋକୟତ୍ = ଦେଖୁଥ‌ିବା । ମେ = ମୋର । ଭୂତଳେ = ପୃଥ‌ିବୀରେ । ଅନାରତମ୍ = ଅବିରତ । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ପବିତ୍ର ବିଶ୍ୱବନ୍ଧୁତ୍ଵକୁ ଉଦ୍‌ଘୋଷଣ କରି, ବିଶ୍ଵବନ୍ଦିତ ଚରିତ୍ରଦ୍ଵାରା ସଂସାରକୁ ପବିତ୍ରିତ କରି ବିଶ୍ବକୁ ଏକ ପରିବାରଭାବେ ପ୍ରଦର୍ଶିତ କରି ମୋର ଭାରତ ଦେଶ ପୃଥ‌ିବୀ ପୃଷ୍ଠରେ ନିରନ୍ତର ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ସଂସ୍କୃତସାହିତ୍ୟସ୍ୟ ପ୍ରଥଯଶା ଲବ୍ଧପ୍ରତିଷ୍ଠ କବି ରମାକାନ୍ତଶୁକ୍ଳ ବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ କାବ୍ୟଭାଗାତ୍ ଆନୀତଃ । ଅତ୍ର ଅସ୍ଥିନ୍ ପଦ୍ୟ ଭାରତସ୍ୟ ଶୋଭା କୀଦୃଶୀ ଭବତି ତଦେବ ବର୍ଣ୍ଣିତମ୍ ।

ବିଶ୍ୱେଽସ୍ମିନ୍ ଭାରତଦେଶସ୍ୟ ସ୍ଥାନଂ ତୁ ସ୍ଵତନ୍ତ୍ରମେବ । କବି ଅମ୍ଳୀକଂ ଦେଶସ୍ୟ ପାବିଦ୍ର୍ୟ ବନ୍ଧୁତ୍ୱ ଚ ପ୍ରଦର୍ଶିତମ୍ । ‘ବସୁଧୈବ କୁଟୁମ୍ବକମ୍’ ଇତ୍ୟେବ ଭାବନା ଭାରତୀୟାନାଂ ଭବତି । ଅମ୍ଳୀକଂ ଦେଶଃ ପବିତ୍ର ବିଶ୍ୱବନ୍ଧୁତ୍ଵମ୍ ଉଦ୍‌ଘୋଷୟତି, ଦେଶେଽସ୍ମିନ ମହାପୁରୁଷଚରତ୍ୟେ ଇୟଂ ଭୂମି ପବିତ୍ରିତଂ ଭବତି । ଅସ୍ମାକଂ ଦେଶସ୍ୟ ମହତୀ ସଂସ୍କୃତିଃ କୁଟୁମ୍ବୀଭାବନା ଚ ସମଗ୍ର ବିଶ୍ଵ ବିରାଜତେ । ଅନେନ ମାଧ୍ୟମେନ ମମ ଭାରତଦେଶଃ ବିଶ୍ଵ ରାଜତେ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ସଂସ୍କୃତ ସାହିତ୍ୟର ପ୍ରଥଯଶା ଲବ୍ଧପ୍ରତିଷ୍ଠ କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ‘ଭାତି ମେ ଭାରତମ୍’ କାବ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଆଲୋଚିତ ପଦ୍ୟରେ ଭାରତର ଶୋଭା କିଭଳି ତାହା ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ସମଗ୍ର ବିଶ୍ଵରେ ଭାରତର ସ୍ଥାନ ସ୍ଵତନ୍ତ୍ର । ଏହି ଦେଶରେ ବିଶ୍ଵବନ୍ଧୁତ୍ଵର ଭାବନା ଅତ୍ୟନ୍ତ ପବିତ୍ର । ଏଠାରେ ଥ‌ିବା ମହାପୁରୁଷମାନଙ୍କର ପବିତ୍ର ଚରିତ୍ର ଏ ଭାରତଭୂଇଁକୁ ଅତୀବ ପବିତ୍ରିତ କରିଛି । ‘ବସୁଧୈବ କୁଟୁମ୍ବକମ୍’ – ଏହି ଭାବନାରେ ଉଦ୍‌ବୁଦ୍ଧ ସମସ୍ତ ଭାରତୀୟ ବିଶ୍ବକୁ ଏକ କୁଟୁମ୍ବରୂପେ ଅବଧାରଣ କରିଛନ୍ତି । ଭାରତର ଏଭଳି ମହାନ୍ ସଂସ୍କୃତି ସମଗ୍ର ବିଶ୍ବରେ ପ୍ରତିଷ୍ଠିତ ହୋଇଛି । ଏଥ‌ିପାଇଁ ମୋର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ଵରେ ନିରନ୍ତର ଶୋଭା ପାଉଛି । ସତରେ କବିଙ୍କର ଏତାଦୃଶ ଭାବନା ଏକାନ୍ତ ଯଥାର୍ଥ ଅଟେ ।

ବ୍ୟାକରଣ:

ସନ୍ଧି ବିଚ୍ଛେଦ – ବିଶ୍ୱବନ୍ଧୁତ୍ବ ମୁଦ୍‌ଷୟପ୍ଲାବନମ୍ = ବିଶ୍ଵବନ୍ଧୁତ୍ଵମ୍ + ଉଦ୍‌ଘୋଷୟତ୍ + ପାବନମ୍ । ବିଶ୍ବବନ୍ଦୋଶ୍ଚରିତୈର୍ଜଗତ୍ପାବୟତ୍ = ବିଶ୍ଵବନ୍ଦ୍ୟ + ଚରିତିଃ + ଜଗତ୍ + ପାବୟତ୍ । ବିଶ୍ଵମେକମ୍ = ବିଶ୍ୱମ୍ + ଏକମ୍ । ମେଽନାରତମ୍ = ମେ + ଅନାରତମ୍ ।

ସମାସ – ବିଶ୍ୱବନ୍ଧୁତ୍ଵମ୍ = ବିଶ୍ୱସ୍ୟ ବନ୍ଧୁତ୍ଵମ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ବିଶ୍ଵବନ୍ଧ୍ୟା = ବିଶ୍ଵ ବନ୍ଦୀ, ତସ୍ମି (ସପ୍ତମୀ ତପୁରୁଷ) । ଭୂତଳେ = ଭୁନଃ ତଳମ୍, ତସ୍ମିନ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ)

ସକାରଣବିଭକ୍ତି – ପାବନମ୍ = କର୍ମଣି ୨ୟା । ବିଶ୍ୱବନ୍ଦ୍ୟ, ଚରିତଃ = କରଣେ ୩ୟା । ବିଶ୍ୱ, ଏବଂ, କୁଟୁମ୍ବମ୍ = କର୍ମଣି ୨ୟା । ମେ = ସମ୍ବନ୍ଧ ୬ଷ୍ଠୀ । ଭୂତଳେ = ଅଧ୍ଵରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ପାବନମ୍ = ପୂ + ଣିଚ୍ + ଲୁଟ୍ । ବନ୍ଧୁତ୍ଵମ୍ = ବନ୍ଧୁ+ ତ୍ଵ ।

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

ଶ୍ଳୋକ – ୨

वेशभूषाशनोपासनापद्धति –
क्रीडनामोद-संस्कार-वृत्त्यादिषु ।
यद्धि भिन्नं सदप्यस्त्यभिन्नं सदा
भूतले भातिं तन्मामकं भारतम् ॥ २ ॥

ବେଶଭୂଷାଶନେ।ପାସନାପଦ୍ଧତି –
କ୍ରୀଡ଼ନାମୋଦ – ସଂସ୍କାର – ବୃତ୍ୟାଦିଷ୍ଣୁ ।
ଯଦ୍ଧି ଭିନଂ ସଦଶ୍ୟସ୍ତ୍ୟଭିନଂ ସଦା
ଭୂତଳେ ଭାତି ତନ୍ମାମକଂ ଭାରତମ୍ ॥ ୨ ॥

ବେଶଭୂଷା – ଅଶନ – ଉପାସନାପଦ୍ଧତି-କ୍ରୀଡ଼ନ-ଆମୋଦ-ସଂସ୍କାର-ବୃତ୍ୟାଦିଷ୍ଣୁ ଯଦ୍ ହି ଭିନ୍ନ ସତ୍ ସଦା ଅପି ଅଭିନ୍ନମ୍ ଅସ୍ତି । ତତ୍ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଅଶନ = ଭୋଜନ । କ୍ରୀଡ଼ନ = ଖେଳକୁଦ । ଆମୋଦ = ଆନନ୍ଦ । ଯଦ୍ = ଯାହା । ହି = ଯେହେତୁ । ଅପି = ମଧ୍ୟ । ମାମକଂ = ମୋର । ଭୂତଳେ = ପୃଥ‌ିବୀପୃଷ୍ଠରେ । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ବେଶଭୂଷା-ଭୋଜନ-ଉପାସନ।ପଦ୍ଧତି-କ୍ରୀଡ଼।-ଆମୋଦପ୍ରମୋଦ-ସଂସ୍କାର ବ୍ୟବହାର ମାନଙ୍କରେ ଯାହା ହିଁ ଭିନ୍ନ ହୋଇ ମଧ୍ୟ ସର୍ବଦା ଅଭିନ୍ନ ଅଟେ, ତାହା ମୋର ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର ଅସ୍ମାକଂ ଭାରତଦେଶସ୍ୟ ଶୋଭା ବର୍ଣିତମସ୍ତି ।

ଅସ୍ମାକଂ ଭାରତଦେଶେ ବିଭିନ୍ନପ୍ରାନ୍ତେଷୁ ବେଶଭୂଷା ଭିନ୍ନା ଅସ୍ଥି । ଅସ୍ମାକଂ ଭୋଜନଂ ତୁ ଭିନ୍ନମେବ ଭବତି । ସର୍ବେଷା ଜନାନାମ୍ ଉପାସନାପଦ୍ଧତିଃ ଭିନ୍ନମସ୍ତା । କ୍ରୀଡ଼ା-ଆମୋଦଃ ପ୍ରମୋଦଣ୍ଟ ଭିନ୍ନମସ୍ତା । ସଂସ୍କାରମପି ଭିନ୍ନମସ୍ତା । ଏତେଷୁ ବିଭିନ୍ନେଷୁ ବୃଦ୍ଧିଶୁ ଯଦ୍ୟପି ଭିନ୍ନତା ପରିଲକ୍ଷ୍ୟତେ ତଥାପି ସର୍ବବ ଅଭିନ୍ନତାମ୍ ଅସ୍ଥି । ଏତତ୍ ତୁ ଅସ୍ମାକଂ ଦେଶସ୍ୟ ମହତ୍ତ୍ଵମ୍ । ଏତାଦୃଶଂ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ନାମ ବିଭାତି । କବି ଯଥା ଅତ୍ର ଅସ୍ମାକଂ ଦେଶସ୍ୟ ମହନୀୟତାଂ ବର୍ଣ୍ଣୟତି ତଦେବ ଅତୀବ ସ୍ପୃହଣୀୟଃ ଗ୍ରହଣୀୟ ଭବତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଉପରୋକ୍ତ ଶ୍ଳୋକଟି ପଠିତ ‘ସଂସ୍କୃତପ୍ରଭା’ ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭାରତର ସଂସ୍କୃତି ଓ ପରମ୍ପରା ପ୍ରସଙ୍ଗ ବର୍ଣିତ ହୋଇଛି ।

ଆମ ଦେଶ ଭାରତବର୍ଷରେ ବିଭିନ୍ନ ପ୍ରାନ୍ତରେ ଲୋକମାନଙ୍କର ବେଶଭୂଷା ଭିନ୍ନ ଅଟେ । ସେମାନଙ୍କର ପୋଷାକ ପରିଚ୍ଛଦ ମଧ୍ୟ ଭିନ୍ନ ଅଟେ । ଆମ୍ଭମାନଙ୍କର ଖାଦ୍ୟପାନୀୟ ମଧ୍ୟ ସ୍ଥଳବିଶେଷରେ ଭିନ୍ନ ଅଟେ । ପ୍ରତ୍ୟେକ ଲୋକର ପୂଜା ବା ଉପାସନା ପଦ୍ଧତି ସ୍ଵତନ୍ତ୍ର ଅଟେ । ବିଭିନ୍ନ ସଂପ୍ରଦାୟର ଲୋକେ ସ୍ଵ ସ୍ବ ଶୈଳୀରେ ଉପାସନା କରିଥା’ନ୍ତି । କ୍ରୀଡ଼ା, ଆମୋଦପ୍ରମୋଦ, ସଂସ୍କାରାଦି କାର୍ଯ୍ୟରେ ଆମ ଦେଶର ଲୋକମାନଙ୍କର ରୁଚିବୋଧ ଏକାନ୍ତ ଗ୍ରହଣୀୟ ଅଟେ । ଏସବୁ କାର୍ଯ୍ୟକଳାପରେ ଏତେସବୁ ଭିନ୍ନତା ସତ୍ତ୍ବେ ମଧ୍ଯ ଅଭିନ୍ନତା ହିଁ ରହିଛି । ବସ୍ତୁତଃ ଏହିଭଳି ଭାବେ ଆମର ଭାରତଦେଶ ଏତେସବୁ ବିବିଧତା ସତ୍ତ୍ବେ ମଧ୍ଯ ଏକରୂପେ ପ୍ରତୀତ ହେଉଛି । ଏହିଭଳି ଆମର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ବରେ ଶୋଭାପାଉଅଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ବେଶଭୂଷାଶନୋପାସନାପଦ୍ଧତି = ବେଶଭୂଷା + ଅଶନ + ଉପାସନାପଦ୍ଧତିଃ । କ୍ରୀଡ଼ନାମୋଦ = କ୍ରୀଡ଼ନ + ଆମୋଦ । ବୃତ୍ୟାଦିଷୁ = ବୃତ୍ତି + ଆଦିଶୁ । ଯଦି ଯଦ୍ + ହି । ସଦସ୍ୟସ୍ତ୍ୟଭିନ୍ନମ୍ = ସତ୍ + ଅପି + ଅସ୍ତି + ଅଭିନ୍ନମ୍ । ତନ୍ମାମକମ୍ = ତତ୍ + ମାମକମ୍ ।

ସମାସ – ଉପାସନାପଦ୍ଧତିଃ = ଉପାସନାୟା ପଦ୍ଧତିଃ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ଅଭିନ୍ନମ୍ = ନ ଭିନ୍ନମ୍ (ନୡ ତତ୍‌ପୁରୁଷ) । ।

ସକାରଣବିଭକ୍ତି – ବୃତ୍ୟାଦିଷ୍ଣୁ = ଅଧିକରଣେ ୭ମୀ ଭାରତମ୍ = କର୍ଭରି ୧ ମା । ଭୂତଳେ = ଅଧିକରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ ଅଶନମ୍ = ଅଶ୍ + ଲୁଟ୍ । ଆମୋଦଃ = ଆ+ ମୁଦ୍ + ଘଞ୍ଚ । ସଂସ୍କାରଃ = ସମ୍ + କୃ + ଘଞ୍ଚ । ଭିନ୍ନମ୍ = ଭିଦ୍ + କ୍ତ ।

ଶ୍ଳୋକ – ୩

दर्शनज्ञानचारित्र्य सम्मेलनं
यत्र मोक्षस्य मार्गं भणन्त्यागमाः ।
ज्ञानमास्ते च भारः क्रियां वै विना
भूतले भाति तन्मामकं भारतम् ॥ ३ ॥

ଦର୍ଶନଜ୍ଞ।ନଚାରିତ୍ତ୍ର୍ୟସମ୍ମଳନଂ
ଯତ୍ର ମୋକ୍ଷସ୍ୟ ମାର୍ଗ ଭଣତ୍ୟାଗମା ।
ଜ୍ଞାନମାସ୍ତେ ଚ ଭାରଃ କ୍ରିୟାଂ ବୈ ବିନା
ଭୂତଳେ ଭାତି ତନ୍ମାମକଂ ଭାରତମ୍ ॥ ୩ ॥

ଅନ୍ବୟ – ଯତ୍ର ଆଗମଃ ଦର୍ଶନ-ଜ୍ଞାନ-ଚାରିଦ୍ର୍ୟ-ସମେଳନଂ ମୋକ୍ଷସ୍ୟ ମାର୍ଗ ଭଣନ୍ତି (ଯତ୍ର) ଜ୍ଞାନଂ ବୈ କ୍ରିୟା ବିନା ଭାରଃ ଚ ଆସ୍ତେ । ତତ୍ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଯତ୍ର = ଯେଉଁଠାରେ । ଆଗମା = ବେଦାଦିଶାସ୍ତ୍ରମାନ । ସମ୍ମଳନମ୍ = ଏକତ୍ରିତ ହୋଇ । ମୋକ୍ଷସ୍ୟ = ମୁକ୍ତିର । ମାର୍ଗମ୍ = ପଥକୁ । ଭଣନ୍ତି = କୁହନ୍ତି । ଭୂତଳେ = ପୃଥ‌ିବୀପୃଷ୍ଠରେ । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ଯେଉଁଠି ବେଦାଦିଶାସ୍ତ୍ର-ଦର୍ଶନ-ଜ୍ଞାନ ଏବଂ ଚାରିଦ୍ର୍ୟ ମିଳିତଭାବେ ମୁକ୍ତିର ପଥକୁ କହନ୍ତି (ଯେଉଁଠି) କ୍ରିୟା

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର କରଂ ବେଦାଦିଶାସ୍ତ୍ରଜ୍ଞାନଂ ମୋକ୍ଷମେବ କଥୟନ୍ତି ତଦେବ ବର୍ଣ୍ଣିତମ୍ ।

ଅସ୍ମାକଂ ଭାରତଦେଶେ ବେଦୋପନିଷଦାଦିଗ୍ରନ୍ଥା ମୁକ୍ତେ ମାର୍ଗମେବ କଥୟନ୍ତି । ସର୍ବେ ଦର୍ଶନଶାସ୍ତ୍ର ଅପି ମୋକ୍ଷବିଷୟମ୍ ଆଲୋଚୟନ୍ତି । ଜ୍ଞାନଂ ଚାରିଦ୍ର୍ୟ ଚ ମୋକ୍ଷାର୍ଥୀ ଭବତି । ଅତ୍ର କ୍ରିୟାଂ ବିନା ଜ୍ଞାନଂ ଭାରଃ ଭବତି । ଏତାଦୃଶଂ ମାମକଂ ଭାରତମ୍ ଅସ୍ମିନ୍ ଭୂତଳେ ଶୋଭତେ । ଅତ୍ର ଯଥା କବି ଭାରତଦେଶସ୍ୟ ଗ୍ରନ୍ଥଶାସ୍ତ୍ରାମିଂ ଜ୍ଞାନଚରିତ୍ରାଣଂ ଚ ବିଶ୍ଳେଷଣଂ କୃତମ୍ ତଦେବ ଅତୀବ ଶିକ୍ଷଣୀୟଂ ଗ୍ରହଣୀୟଂ ଚ ଭବତି । ଯେ ତାବତ୍ ମୁକ୍ତିକାମୀଜନା ସନ୍ତ ତେ ଅସ୍ମାକଂ ଦେଶସ୍ୟ ଶାସ୍ତ୍ର ଜ୍ଞାନଂ ଚରିତ୍ର ଚ ଅବଶ୍ୟ ସ୍ଵୀକରିଷ୍ୟନ୍ତ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଉଦ୍ଧୃତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ସଂଗୃହୀତ । ଏଠାରେ ଭାରତୀୟ ଶାସ୍ତ୍ର, ଜ୍ଞାନ ଓ ଚାରିଦ୍ର୍ୟର ମହତ୍ତ୍ବ ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଭାରତୀୟ ଶାସ୍ତ୍ରୀୟ ପରମ୍ପରା ଅତୀବ ପ୍ରାଚୀନ । ଏଠାରେ ବେଦ ଉପନିଷଦାଦି ଗ୍ରନ୍ଥ ବିଭିନ୍ନ ଦର୍ଶନଶାସ୍ତ୍ର, ଜ୍ଞାନ ତଥା ବିବିଧ ଚରିତ୍ର ମିଳିତଭାବେ ମୁକ୍ତିର ମାର୍ଗକୁ କହିଥା’ନ୍ତି । ମନୁଷ୍ୟ ଜୀବନରେ ମୁକ୍ତି ବା ମୋକ୍ଷ ହିଁ ସର୍ବଶ୍ରେଷ୍ଠ ପୁରୁଷାର୍ଥ । ଏଣୁ ଏସବୁ ଶାସ୍ତ୍ରଗ୍ରନ୍ଥମାନ ମୁକ୍ତିର ପଥ ନିର୍ଦ୍ଦେଶ କରନ୍ତି । ଏଠାରେ କ୍ରିୟା ବିନା ଜ୍ଞାନ ଭାରସଦୃଶ ହୋଇଥାଏ । ଶାସ୍ତ୍ରୀୟ ଜ୍ଞାନ ଅପେକ୍ଷା ଭାରତୀୟମାନେ ବ୍ୟାବହାରିକ କ୍ରିୟା ଉପରେ ଅଧିକ ଗୁରୁତ୍ଵ ଦେଇଥା’ନ୍ତି । ଏହିଭଳି ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ଵରେ ଶୋଭିତ ହେଉଛି । ଏଣୁ ଆମେ ଏ ଦେଶର ଅସ୍ଵାସୀ ଭାବେ ଗର୍ବ କରିବା ଉଚିତ ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ଭଣନ୍ତ୍ୟାଗମାଃ = ଭଣନ୍ତି + ଆଗମା । ଜ୍ଞାନମାସ୍ତେ = ଜ୍ଞାନମ୍ + ଆସ୍ତେ । ତନ୍ମାମକମ୍ = ତତ୍ + ମାମକମ୍ ।
ସମାସ – ଭୂତଳେ = ଭୁନଃ ତଳମ୍, ତସ୍ମିନ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) ।
ସକାରଣବିଭକ୍ତି – ଆଗମା = କଉଁରି ୧ ମା । ମୋକ୍ଷସ୍ୟ = ସମ୍ବନ୍ଧ ୬ଷ୍ଠୀ । ମାର୍ଗମ୍ = କର୍ମଣି ୨ୟା । ଜ୍ଞାନମ୍ = କଉଁରି ୧ ମା । କ୍ରିୟାମ୍ = ବିନା ଯୋଗେ ୨ୟା । ଭାରଃ = କଉଁରି ୧ ମା । ଭୂତଳେ = ଅଧ୍ଵରଣେ ୭ମୀ
ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଜ୍ଞାନମ୍ = ଜ୍ଞା + ଲ୍ୟୁଟ୍ । ଆଗମା = ଆ + ଗମ୍ + ଅଚ୍ + ଜସ୍ । ଭାରଃ = ଭୂ + ଘଞ୍ଚ୍ ।

ଶ୍ଳୋକ – ୪

जाह्नवी – चन्द्रभागा-जलैः पावितं
भानुजा-नर्मदा-वीचिभिर्लालितम् ।
तुङ्गभद्रा – विपाशादिभिर्भावितं
भूतले भाति मेनारतं भारतम् ॥ ४ ॥

ଜାହ୍ନବୀ-ଚନ୍ଦ୍ରଭାଗା -ଜୟଃ ପାବିତଂ
ଭାନୁଜା-ନର୍ମଦା -ବୀଚିଭିର୍ଲାଳିତମ୍ ।
ତୁଙ୍ଗଭଦ୍ରା-ବପାଶ।ଦିଭିର୍ଭ।ବିତଂ
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ४ ॥

ଅନ୍ୱୟ – ଜାହ୍ନବୀ – ଚନ୍ଦ୍ରଭାଗା -ଜଳି ପାବିତମ୍ବ, ଭାନୁଜା – ନର୍ମଦା – ବିଚିଭି ଲାଳିତମ୍, ତୁଙ୍ଗଭଦ୍ରା – ବିପାଶାଦିଭିଃ ଭାବିତମ୍, ମେ ଭାରତଂ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଜାହ୍ନବୀ = ଗଙ୍ଗା । ଜକୈଃ = ଜଳଦ୍ଵାରା । ପାବିତମ୍ = ପବିତ୍ରିତ । ଭାନୁଜା = ଯମୁନା । ବିଚିଭିଃ = ତରଙ୍ଗଦ୍ଵାରା । ଲାଳିତମ୍ = ଆନ୍ଦୋଳିତ । ଭାବିତମ୍ = ଆଦ୍ରୀଭୂତ । ମେ = ମୋର । ଅନାରତମ୍ = ଅବିରତ । ଭାତି = ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋୟାଂ ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର ଭାରତେ ପ୍ରବାହିତାନାଂ ନଦୀନାଂ ବର୍ଧନଂ ଭବତି ।

ପୂତଂ ପବିତଂ ଚ ଅସ୍ଵାକଂ ଭାରତବର୍ଷମ୍ । ଗଙ୍ଗା-ଚନ୍ଦ୍ରଭାଗା ଜଳେ ନିତାଂ ପବିତ୍ରିତମ୍ । ଯମୁନା-ନର୍ମଦା ଜଳତରଙ୍ଗି ତରଙ୍ଗାୟିତମ୍ ଆନ୍ଦୋଳିତଂ ଚ । ତୁଙ୍ଗଭଦ୍ରା-ବିପାଶାଦିଭିଂ ଜ ଭାବିତମ୍ । ଏତାଦୃଶଂ ମମ ଦେଶଂ ଭାରତମ୍ ଅସ୍ମିନ୍ ବିଶ୍ଵେ ଅବିରତଂ ବିଶୋଭିତମ୍ । ଅସ୍ମାକଂ ଭାରତଦେଶେ ପ୍ରବାହିତନଦୀନାଂ ବର୍ଣ୍ଣନଂ ଯଥା କବିନା କ୍ରିୟତେ, ତଦେବ ଅତୀବ ବର୍ଣ୍ଣନୀୟଂ ଭବତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଏହି ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭାରତଭୂମିରେ ପ୍ରବାହିତ ନଦୀମାନଙ୍କର ବର୍ଣ୍ଣନା କରାଯାଇଚ୍ଛା

ଆମ ଦେଶର ମାଟି ଅତ୍ୟନ୍ତ ପବିତ୍ର କାରଣ ଏଠାରେ ଗଙ୍ଗା ଓ ଚନ୍ଦ୍ରଭାଗାର ଜଳ ପ୍ରବାହିତ ହେଉଅଛି । ଭାରତୀୟମାନେ ଗଙ୍ଗାକୁ ମାତା ରୂପରେ ପୂଜା କରନ୍ତି । ଏହାର ପବିତ୍ର ଜଳ ସମସ୍ତ ପୁଣ୍ୟକାର୍ଯ୍ୟରେ ବ୍ୟବହୃତ ହୋଇଥାଏ । ଯମୁନା, ନର୍ମଦାର ତରଙ୍ଗରେ ତରଙ୍ଗାୟିତ ଏ ଭାରତଭୂମି । ତୁଙ୍ଗଭଦ୍ରା, ବିପାଶାଦି ନଦୀମାନଙ୍କଦ୍ୱାରା ଭାରତର ମାଟି ଆର୍ଦ୍ର ହୋଇଛି । ଏହିଭଳି ମୋର ଭାରତଦେଶ ପୃଥ୍ବୀପୃଷ୍ଠରେ ନିରନ୍ତର ଶୋଭାପାଉଅଛା

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ବୀଚିଭିର୍ଲାଳିତମ୍ = ବୀଚିରଃ + ଲାଳିତମ୍ । ବିପାଶାଦିଭିର୍ଭାବିତମ୍ = ବିପାଶା + ଆଦିଭିଂ + ଭାବିତମ୍ । ମେଽନାରତମ୍ = ମେ + ଅନାରତମ୍ ।
ସମାସ – ଭୂତଳେ = ଭୁବ ତଳମ୍, ତସ୍ମିନ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷଃ) ।
ସକାରଣବିଭକ୍ତି – ଜାହ୍ନବୀ, ଚନ୍ଦ୍ରଭାଗା, ଭାନୁଜା, ନର୍ମଦା, ତୁଙ୍ଗଭଦ୍ରା = କଉଁରି ୧ମା । ଜଳେ, ବିଚିଭି, ବିପାଶାଦିଭିଂ = କରଣେ ୩ୟା । ମେ = ସମ୍ବନ୍ଧ ୬ଷ୍ଠୀ । ଭୂତଳେ = ଅଧିକରଣେ ୭ମୀ ।
ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ପାବିତମ୍ = ପୂ + ଣିଚ୍ + କ୍ତ । ଭାବିତମ୍ = ଭୂ + ଣିଚ୍ + କ୍ତ ।

ଶ୍ଳୋକ – ୫

विन्ध्य – सह्याद्रि- नीलाद्रिमालान्वितं
शुभ्रहमाद्रि – हासप्रभापूरितम् ।
अर्बुदारावलीश्रेणी – सम्पूजितं
भूतले भाति मेनारतं भारतम् ॥ ५ ॥

ବିନ୍ଧ୍ୟ-ସହହ୍ୟାଦ୍ରି-ନୀଳାଦ୍ରିମାଳାନ୍ୱିତଂ
ଶୁଭ୍ରହୈମାଦ୍ରି -ହାସପ୍ରଭାପୂରିତମ୍ ।
ଅର୍ବୁଦାରାବଳୀଶ୍ରେଣୀ – ସମ୍ପୂଜିତଂ
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ || ୫ ||

ଅନ୍ୱୟ – ବିନ୍ଧ୍ୟ – ସହ୍ୟାଦ୍ରି-ନୀଳାଦ୍ରିମାଳାନ୍ନିତମ୍, ଶୁଭ୍ରହିମାଦ୍ରି-ହାସପ୍ରଭାପୂରିତମ୍, ଅର୍ବୁଦାରାବଳୀ ଶ୍ରେଣୀ – ସମ୍ପୂଜିତମ୍, ମେ ଭାରତଂ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଅଦ୍ରି = ପର୍ବତ । ଶୁଭ୍ର = ସ୍ଵଚ୍ଛ । ଶ୍ରେଣୀ = ସମୂହ । ମେ = ମୋର । ଅନାରତମ୍ = ନିରନ୍ତର । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ବିନ୍ଧ୍ୟ-ସହ୍ୟାଦ୍ରି-ନୀଳାଦ୍ରି ପର୍ବତମାଳାଦ୍ୱାରା ପରିପୂରିତ, ଧବଳ ହିମାଳୟ ପର୍ବତ ହାସପ୍ରଭାରେ ପରିପୂର୍ଣ,

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର ଭାରତଭୂମୌ ଅବସ୍ଥିତାନାଂ ପର୍ବତାନାଂ ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତମସ୍ତି ।

ପର୍ବତଃ ଭୂମିଭାଗଂ ଧାରୟତ୍ତି । ଅସ୍ମାକଂ ଭାରତସ୍ୟ ସର୍ବତ୍ରୈବ ପ୍ରାକୃତିକ ସଂପଦ ସ୍ବରୂପ ପର୍ବତଃ ବିରାଜନ୍ତେ । ଅତ୍ର ବିନ୍ଧ୍ୟ – ସହ୍ୟାଦ୍ରି-ନୀଳାଦ୍ରିମାଳାନ୍ଵିତମ, ଶୁଭ୍ରହିମାଦ୍ରି-ହାସପ୍ରଭାପୂରିତମ୍, ଅର୍ବୁଦାରାବଳୀଶ୍ରେଣୀ-ସଂପୂଜିତମ୍ । ଅନେନ ପ୍ରକାରେଣ ବିଭିନ୍ନେ ପର୍ବତଃ ଅସ୍ଵାକଂ ଦେଶ ଭାରତଂ ସୁଶୋଭିତଂ ସୁସମୃଦ୍ଧ ଚ ଭବତି । ଏତାଦୃଶଂ ମମ ଭାରତଂ ଭୂତଳେ ନିରନ୍ତରଂ ଶୋଭତେ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଏହି ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆସିଅଛି । ଏଠାରେ ଭାରତ ଭୂଖଣ୍ଡରେ ଅବସ୍ଥିତ ପର୍ବତମାନଙ୍କ ବିଷୟରେ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ପର୍ବତମାନେ ଭାରତଭୂଖଣ୍ଡର ଧାରକ ତଥା ପ୍ରାକୃତିକ ସୌନ୍ଦର୍ଯ୍ୟ ଓ ବୈଭବର ପ୍ରତୀକ ରୂପେ ଦଣ୍ଡାୟମାନ । ଏହି ଭାରତଦେଶରେ ବିନ୍ଧ୍ୟ ସହ୍ୟାଦ୍ରି ଓ ନୀଳାଦ୍ରି ମାଳାରୂପରେ ଶୋଭାପାଉଛନ୍ତି । ସ୍ବଚ୍ଛ ହିମାଳୟ ପର୍ବତ ସତେ ଅବା ହାସପ୍ରଭାରେ ପରିପୂରିତ ହୋଇଛି । ଅର୍ବୁଦ ଓ ଆରାବଳୀ ପର୍ବତଶ୍ରେଣୀ ସମ୍ୟକ୍ ପୂଜିତ ହୋଇଅଛନ୍ତି । ଏହିଭଳି ଅନେକ ପର୍ବତମାନଙ୍କଦ୍ଵାରା ବିଶୋଭିତ ମୋର ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ନିରନ୍ତର ଶୋଭାପାଉଅଛି । ଭାରତର ପ୍ରାକୃତିକ ସୌନ୍ଦର୍ଯ୍ୟ ଅତ୍ୟନ୍ତ ମନୋହାରୀ ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ସହ୍ୟାଦ୍ରିଃ = ସହ୍ୟ + ଅନ୍ତଃ । ନିଳାଦ୍ରିମାଳାନ୍ଵିତମ୍ = ନୀଳ + ଅଦ୍ରିମାଳା + ଅନ୍ବିତମ୍ । ହୈମ + ଅନ୍ତଃ । ଅର୍ବୁଦାରାବଳୀଶ୍ରେଣୀ ମେଽନାରତମ୍ = ମେ + ଅନାରତମ୍ ।

ସମାସ – ସହ୍ୟାଦ୍ରିଃ = ସହ୍ୟ + ନାମ ଅନ୍ତଃ (ମଧ୍ୟପଦଲୋପୀ କର୍ମଧାରୟ) । ଶୁଭ୍ରହୈମାଦ୍ରି = ହୈମ ଅର୍ଦ୍ର ଯସ୍ୟ ଡଃ, ହୈମାଦ୍ରି (ବହୁବ୍ରୀହିଃ), ଶୁଭ୍ର ହୈମାଦ୍ରି (କର୍ମଧାରୟ) । ହାସପ୍ରଭାପୂରିତମ୍ = ହାସ ପ୍ରଭା ଯସ୍ୟ ଡଃ (ବହୁବ୍ରୀହିଃ), ହାସପ୍ରଭୟା ପୂରିତମ୍ (୩ୟା ତତ୍ତ୍ଵପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ବିନ୍ଧ୍ୟ, ସହ୍ୟାଦ୍ରି, ଭାରତମ୍ = କଉଁରି ୧ ମା । ମେ = ସମ୍ବନ୍ଧ ୬ଷ୍ଠୀ । ଭୂତଳେ = ଅଧ୍ଵରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ହାସ୍ୟ = ହସ୍ + ଘଞ୍ଚ୍ । ପୂରିତମ୍ = ପୂର୍ + ଣିଚ୍ + କ୍ତ । ସମ୍‌ପୂଜିତମ୍ = ସମ୍ + ପୂଜ୍ + ଣିବ + କ୍ତ ।

ଶ୍ଳୋକ – ୬

विद्युदुत्पादने तैलसंशोधने
इन्धनान्वेषणे लौहनिष्पादने ।
यन्त्रनिर्माणकार्ये समर्थं च सद्
भूतले भाति मेनारतं भारतम् ॥ ६ ॥

ବିଦ୍ୟୁତୁତ୍ପାଦନେ ତୈଳସଂଶୋଧରେ
ଇନ୍ଧନାନ୍ଵେଷଣେ ଲୌହନିଷ୍ପାଦନେ ।
ଯନ୍ତନିର୍ମାଣକାର୍ଯ୍ୟେ ସମର୍ଥଂ ଚ ସଦ୍‌
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ୬ ॥

ଅନ୍ୱୟ – ବିଦ୍ୟୁତ୍‌ ଉତ୍ପାଦଳନ ତୈଳସଂ ଶୋଧନେ ଇନ୍ଧନ ଅନ୍ୱେଷଣେ ଲୌହନିଷ୍ପ।ଦନେ ସଦ୍ ମେ ଭାରତଂ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ବିଦ୍ୟୁତ୍ = ବିଜୁଳି । ଇନ୍ଧନ = ଜାଳେଣି । ଅନ୍ଵେଷଣେ = ଅନୁସନ୍ଧାନ କାର୍ଯ୍ୟରେ । ସଦ୍ = ହୋଇ । ମେ = ମୋର । ଅନାରତମ୍ = ନିରନ୍ତର । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ବିଦ୍ୟୁତ୍ ଉତ୍ପାଦନରେ, ତୈଳଶୋଧନ କାର୍ଯ୍ୟରେ, ଇନ୍ଧନସାମଗ୍ରୀ ଅନୁସନ୍ଧାନରେ, ଲୌହ ନିଷ୍ପନ୍ନ କରିବାରେ ଓ ଯନ୍ତ୍ର ନିର୍ମାଣ କାର୍ଯ୍ୟରେ ସମର୍ଥ ହୋଇ ମୋର ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ନିରନ୍ତର ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଯନ୍ତ୍ରବିଜ୍ଞାନେ ଭାରତସ୍ୟ ସାମର୍ଥ୍ୟ କଥ୍ୟ ବିଦ୍ୟତେ ତଦେବ ଅତ୍ର ବର୍ଣ୍ଣିତମ୍ ।

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

ଅସ୍ମାକଂ ଭାରତଦେଶଃ ବିବିଧଯାନ୍ତ୍ରିକ ତଥା ବୈଷୟିକ ବିଦ୍ୟାୟାଂ ପାରଙ୍ଗମଃ ଭବତି । ବିଦ୍ୟୁତ୍-ଉତ୍ପାଦନେ, ତୈଳ ସଂଶୋଧନେ, ଇନ୍ଧନାନ୍ଵେଷଣେ, ଲୌହନିଷ୍ପାଦନେ, ଯନ୍ତ୍ରନିର୍ମାଣକାର୍ଯ୍ୟ ଚ ସଦା ସମର୍ଥ ଭବତି । ଏତେଷୁ କାର୍ଯ୍ୟଷୁ ଭାରତୀୟ ଯାନ୍ତ୍ରିକବୈଜ୍ଞାନିକାନାଂ ଭୂମିକା ଅତୀବ ଗୁରୁତ୍ଵପୂର୍ଣ୍ଣା ଆସୀତ୍ । ଅସ୍ମିନ୍ କାର୍ଯ୍ୟ ଭାରତସ୍ୟ ସ୍ୱତନ୍ତ୍ରତା ବିଦ୍ୟତେ । ଏତେନ ଅସ୍ମାକଂ ଭାରତବର୍ଷ ଭୂତଳେ ଅବିରତଂ ଭାତି । ବିଶ୍ଵେଽସ୍ମିନ୍ ଭାରତସ୍ୟ ପ୍ରଗତିଃ ସମୃଦ୍ଧି ଚ ପରିଲକ୍ଷିତଂ ଭବତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ସଂଗୃହୀତ । ଏଠାରେ ଭାରତବର୍ଷ କିଭଳି ଭାବେ ବିଭିନ୍ନ ଯାନ୍ତ୍ରିକ ତଥା ବୈଷୟିକ ବିଦ୍ୟାରେ ପାରଙ୍ଗମ ବା ସିଦ୍ଧହସ୍ତ ତାହା ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ମନୁଷ୍ୟମାନଙ୍କର ନିତ୍ୟ ବ୍ୟବହାର୍ଯ୍ୟ ରୂପେ ବିଜୁଳି, ଇନ୍ଧନ, ଲୌହ, ଯନ୍ତ୍ରାଂଶ ଆଦି ବ୍ୟବହୃତ ହୋଇଥାଏ । ଏସବୁ ବ୍ୟତିରେକେ ମାନବଜୀବନ ଗତିଶୀଳ ହେବା କଷ୍ଟକର । ଏଣୁ ଏସବୁ କ୍ଷେତ୍ରରେ ଭାରତର ସୁମହତ୍ତ୍ବ ରହିଛି । ବିଦ୍ୟୁତ୍ ଉତ୍ପାଦନରେ, ତୈଳ ସଂଶୋଧନ କାର୍ଯ୍ୟରେ, ବିବିଧପ୍ରକାର ଇନ୍ଧନ ଯଥା – ଡିଜେଲ୍, ପେଟ୍ରୋଲ୍, କିରାସିନି ଆଦିର ଅନ୍ଵେଷଣରେ ପୁନଶ୍ଚ ଲୌହନିଷ୍ପାଦନ ଓ ବିବିଧ ଯନ୍ତ୍ର ନିର୍ମାଣ କାର୍ଯ୍ୟରେ ଭାରତୀୟ ବୈଜ୍ଞାନିକମାନେ ସଦା ଚେଷ୍ଟିତ । ଏହାଦ୍ଵାରା ଆମର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ଵରେ ଶୋଭିତ ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ବିଦ୍ୟୁଦୂତ୍ପାଦନେ = ବିଦ୍ୟୁତ୍ + ଉତ୍ପାଦନେ । ଇନ୍ଧନାନ୍ଵେଷଣେ = ଇନ୍ଧନ + ଅନ୍ଵେଷଣେ । ମେଽନାରତମ୍ ମେ + ଅନାରତମ୍ ।

ସମାସ – ତୈଳସଂଶୋଧନେ = ତୈଳାନାଂ ସଂଶୋଧନମ୍, ତସ୍ମିନ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ଲୌହନିଷ୍ପାଦନେ = ଲୌହସ୍ୟ ନିଷ୍ପାଦନମ୍, ତସ୍ମିନ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ବିଦ୍ଯୁଦୁତ୍ପାଦନେ, ତୈଳସଂଶୋଧନେ, ଇନ୍ଧନାନ୍ଵେଷଣେ, ଲୌହନିଷ୍ପାଦନେ, ଯନ୍ତ୍ରନିର୍ମାଣକାର୍ଯ୍ୟ ଅଧ୍ୟାକରଣେ ୭ମୀ । ଭୂତଳେ = ସ୍ଥାନାଧ୍ଵରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ସଂଶୋଧନମ୍ = ସମ୍ – ଶୁଧ୍ + ଲ୍ୟୁଟ୍ । ନିର୍ମାଣମ୍ = ନିର୍ – ମା! + ଲ୍ୟୁଟ୍ । ଭାତି = ଭା + ସ୍କ୍ରିନ୍ ।

ଶ୍ଳୋକ – ୭

रोगजालं चिकित्सालयस्थापनै-
रोषधोत्पादनैः शल्यशोधैस्तथा ।
नूतनाभिश्चिकित्साभिरून्मूलयद्
भूतले भाति मेऽनारतं भारतम् ॥ ७

ରେ।ଗାଜ।ଳଂ ଚିକିତ୍ସାଳୟସ୍ଥାପନୈ-
ରୋଷଧୋତ୍ପାଦନଃ ଶଲ୍ୟଶୋଧୈସ୍ଥଥା ।
ନୂତନାଭିଶ୍ଚିକିତ୍ସାଭିରୂନୂ ଳୟଦ୍
ଭୂତଳେ ଭାତି ମେଽନାଇତଂ ଭାରତମ୍ || ୭ ||

ଅନ୍ୱୟ – ଚିକିତ୍ସାଳୟସ୍ଥାପନୌଃ ଔଷଧୋପ୍।ଦନୈଃ ଶଲ୍ୟ ଭୂତଳେ ମେ ଭାରତମ୍ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଚିକିତ୍ସାଳୟ ସ୍ଥାପନଦ୍ଵାରା । ଔଷଧୋତ୍ପାଦନଃ = ଔଷଧ ଉତ୍ପାଦନଦ୍ଵାରା । = ଶଲ୍ୟଶୋଧୈ = କଣ୍ଟାପ୍ରଭୃତି ବାହାର କରିବାଦ୍ଵାରା । ରୋଗଜାଳମ୍ = ବିଭିନ୍ନ ରୋଗକୁ । ଉନ୍ମୁଳୟତ୍ = ଦୂରକରି । ମେ = ମୋର । ଅନାରତମ୍ = ଅବିରତ ବା ନିରନ୍ତର । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ଡାକ୍ତରଖାନା ସ୍ଥାପନଦ୍ୱାରା, ଔଷଧ ଉତ୍ପାଦନଦ୍ଵାରା, ଶଲ୍ୟ (କଣ୍ଟା ବା ଶର) ଶୋଧନ ବା ବାହାର କରିବାଦ୍ଵାରା ତଥା ନୂତନ ଚିକିତ୍ସାଦ୍ଵାରା ରୋଗସମୂହକୁ ନିରାକରଣ କରି ପୃଥ‌ିବୀ ପୃଷ୍ଠରେ ମୋର ଭାରତଦେଶ ନିରନ୍ତର ଶେ।ଭାପାଉଛା

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଚିକିତ୍ସାକ୍ଷେତ୍ରେ ଭାରତସ୍ୟ ଅବଦାନମ୍ ଉଲ୍ଲେଖ୍ୟ ଭବତି ।

ଭାରତୀୟ ଚିକିତ୍ସାବିଜ୍ଞାନ ପରମ୍ପରା ଅତୀବ ପ୍ରାଚୀନା ଭବତି । ଚରକସୁଶ୍ରୁତାଦୟ ଚିକିତ୍ସାବିଜ୍ଞାନେ କ୍ରାନ୍ତିକାରୀ ପରିବର୍ତ୍ତନଂ ପ୍ରଦର୍ଶୟନ୍ତି । ଚିକିତ୍ସାଳୟସ୍ଥାପନଃ, ଔଷଧୋତ୍ପାଦନଃ, ଶଲ୍ୟଶୋଧୈ ତଥା ନୂତନାଭି ଚିକିତ୍ସାଭି ରୋଗଜାଳଂ ଉନ୍ମୁଳୟତ୍ ଭୂତଳେ ମେ ଭାରତମ୍ ଅନାରତଂ ଭାତି । ସୁନମ୍ୟଚିକିତ୍ସାୟା ଜନନୀ ଭାରତଭୂମି ଇତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଉଦ୍ଧୃତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭାରତୀୟ ଚିକିତ୍ସା ତଥା ରୋଗନିଦାନ ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଆମ ଭାରତବର୍ଷର ଚିକିତ୍ସା ପରମ୍ପରା ଅତୀବ ପ୍ରାଚୀନ । ଚରକ ଓ ସୁଶ୍ରୁତ ପ୍ରଭୃତି ଚିକିତ୍ସାବିଜ୍ଞାନୀ ଚିକିତ୍ସାର ବିଭିନ୍ନ ପ୍ରସଙ୍ଗ ସ୍ୱରଚିତ ଗ୍ରନ୍ଥମାନଙ୍କରେ ବର୍ଣ୍ଣନ କରିଅଛନ୍ତି । ରୋଗର ନିଦାନ ତଥା ଚିକିତ୍ସାବ୍ୟବସ୍ଥା ନିମନ୍ତେ ବିଭିନ୍ନ ସ୍ଥାନମାନଙ୍କରେ ଚିକିତ୍ସାଳୟ ବା ଡାକ୍ତରଖାନାମାନଙ୍କର ସ୍ଥାପନ, ବିଭିନ୍ନ ଔଷଧଗୁଡ଼ିକର ଉତ୍ପାଦନ, କଣ୍ଟା ବା ଶର ଫୋଡ଼ି ହୋଇଥିଲେ ସେସବୁର ଶୋଧନ ତଥା ନୂତନ ନୂତନ ଚିକିତ୍ସା ପଦ୍ଧତିଦ୍ଵାରା ସମସ୍ତ ପ୍ରକାର ରୋଗସମୂହର ଉନ୍ମଳନ କରି ଆମ ଦେଶ ଚିକିତ୍ସାକ୍ଷେତ୍ରରେ କ୍ରାନ୍ତିକାରୀ ସଫଳତା ଲାଭ କରିଛି । ଏହିଭଳି ମୋର ଭାରତଦେଶ ବିଶ୍ବରେ ଶୋଭାପାଉଅଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ଚିକିତ୍ସାଳୟସ୍ଥାପନୈରୋଷଧୋତ୍ପାଦନୈଃ = ଚିକିତ୍ସାଳୟସ୍ଥାପନଃ + ଔଷଧ + ଉତ୍ପାଦନଃ । ଶଲ୍ୟଶୋଧୈ + ତଥା । ନୂତନାଭିଶ୍ଚିକିତ୍ସାଭିରୁନୁ ଳୟଦ୍ = ନୂତନାଭଃ + ଚିକିତ୍ସାଭଃ + ଉନ୍ମ ଳୟଦ୍ । ମେଽନାରତମ୍ : = ମେ + ଅନାରତମ୍ ।

ସମାସ – ଚିକିତ୍ସାଳୟଃ = ଚିକିତ୍ସାୟୋଃ ଆଳୟ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ଶଲ୍ୟଶୋଧୈ = ଶଲ୍ୟାନାଂ ଶୋଧଃ, ତୈଃ (ଷଷ୍ଠୀ ତତ୍ତ୍ଵପୁରୁଷଃ) ।

ସକାରଣବିଭକ୍ତି – ଚିକିତ୍ସାଳୟସ୍ଥାପନଃ, ଔଷଧୋତ୍ପାଦନଃ, ଶଲ୍ୟଶୋଧୈ, ନୂତନାଭିଂ, ଚିକିତ୍ସାଭି = କରଣେ ୩ୟା । ରୋଗଜାଳମ୍ = କର୍ମଣି ୨ୟା । ଭୂତଳେ = ଅଧ୍ଵରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ସ୍ଥାପନମ୍ = ସ୍ଥାପ୍ + ଲୁଟ୍ ।

ଶ୍ଳୋକ – ୮

आर्यभट्ट वियन्मण्डले स्थापयत्
पोखरण – भूमिगर्भेऽणुशक्तिं किरत् ।
शान्तिकार्येष्वणं प्रेरयत्सन्ततं
भूतले भाति मेऽनारतं भारतम् ॥ ८ ॥

ଅ।ର୍ଯ୍ୟଭଙ୍ଗ ବିୟନ୍କଣ୍ଡଳେ ସ୍ଥ।ପୟତ୍‌
ପୋଖରଣ୍ -ଭୂମିଗର୍ଭେଽଣୁଶଣ୍ଡିଂ କିରତ୍
ଶାନ୍ତିକାର୍ଯ୍ୟୋଷ୍ଠଭ୍ୟୁ ପ୍ରେରୟସ୍ଥନ୍ତତଂ
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ।। ୮ ।।

ଅନ୍ବୟ – ବିୟନ୍ମଣ୍ଡଳେ ଆର୍ଯ୍ୟଭଙ୍ଗ ସ୍ଥାପୟତ୍, ପୋଖରଣ୍-ଭୂମିଗର୍ଭେ ଅଣୁଶକ୍ତି କିରତ୍ୱ, ଶାନ୍ତକାର୍ଯ୍ୟଷୁ ଅଣଂ ପ୍ରେରୟତ୍ ସନ୍ତତଂ ମେ ଭାରତମ୍ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ବିୟନ୍ମଣ୍ଡଳେ = ଆକାଶ ମଣ୍ଡଳରେ। ଆର୍ଯ୍ୟଭଟ୍ଟମ୍ = ଆର୍ଯ୍ୟଭଟ୍ଟକୁ । ସ୍ଥାପୟତ୍ ସଂସ୍ଥାପନ କରି । ଭୂଗର୍ଭରେ । କିରତ୍ = ବିକିରଣ କରି । ପ୍ରେରୟତ୍ = ପ୍ରେରଣ କରି । ସନ୍ତତମ୍ = ସର୍ବଦା । ମେ = ଅନାରତମ୍ = ନିରନ୍ତର । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ଆକାଶ ମଣ୍ଡଳରେ ଆର୍ଯ୍ୟଭଟ୍ଟକୁ ସ୍ଥାପନ କରି, ପୋଖରଣ୍ ଭୂମିଗର୍ଭରେ ଅଣୁଶକ୍ତିକୁ ବିକିରଣ କରି, ଶାନ୍ତକାର୍ଯ୍ୟରେ ଅଣୁକୁ ପ୍ରେରିତ କରି ସଦୈବ ମୋର ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ନିରନ୍ତର ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ଥିତ୍ଵ ଶ୍ଳୋକେ ଅଣୁଶକ୍ତିକ୍ଷେତ୍ରେ ଭାରତସ୍ୟ ସଫଳତାଂ ପ୍ରଗତଂ ଚ ବର୍ଣ୍ଣିତଂ ଭବତି ।

ଅଣୁବିଜ୍ଞାନୀରୂପେଣ ଆର୍ଯ୍ୟଭଟ୍ଟ ସୁପରିଚିତଃ । ତସ୍ୟ ସିଦ୍ଧାନ୍ତ ତୁ ବିଜ୍ଞାନସମ୍ମତଃ । ଆକାଶମଣ୍ଡଳେ ଆର୍ଯ୍ୟଭଙ ଉପଗ୍ରହ ସ୍ଥାପୟତ୍, ପୋଖରଣ୍ୟଭୂମିଗର୍ଭେ ଅଣୁଶଭିଂ ବିକିରତ୍, ଶାନ୍ତକାର୍ଯ୍ୟଷୁ ଅଭ୍ୟୁ ପ୍ରେରୟତ୍‌ ସନ୍ତତଂ ମମ ଦେଶଂ ଭାରତଂ ନିରନ୍ତରଂ ଭୂତଳେ ଭାତି । ପ୍ରାଚୀନଭାରତେ ଭୂଗର୍ଭଶାସ୍ତ୍ରମ୍ ଅତୀବ ବିଖ୍ୟାତମାସୀତ୍ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଏହି ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆସିଅଛି । ଏଠାରେ ଭାରତୀୟ ଅଣୁଶକ୍ତିର ସଫଳ ପ୍ରୟୋଗ ପ୍ରସଙ୍ଗ ବର୍ଣିତ ହୋଇଛି ।

ପ୍ରାଚୀନ ଭାରତୀୟ ଭୂଗର୍ଭଶାସ୍ତ୍ର ବିଶ୍ୱବିଖ୍ୟାତ । ଭାରତର ଅଣୁବିଜ୍ଞାନୀ ଆର୍ଯ୍ୟଭଟ୍ଟ ସ୍ୱକୀୟ ସିଦ୍ଧାନ୍ତ ନିମନ୍ତେ ସର୍ବଜନବିଦିତ ଆର୍ଯ୍ୟଭଟ୍ଟଙ୍କର ଭୂବ୍ୟାସ ଓ ପୃଥ‌ିବୀର ଆକୃତିବିଷୟକ ମତବାଦ ସମସ୍ତ ବୈଜ୍ଞାନିକମାନଙ୍କଦ୍ୱାରା ଆଦୃତ । ସେହି ଆର୍ଯ୍ୟଭଟ୍ଟଙ୍କ ନାମାନୁସାରେ ନାମିତ ଆର୍ଯ୍ୟଭଟ୍ଟ ନାମକ ଉପଗ୍ରହକୁ ଆକାଶମଣ୍ଡଳରେ ସ୍ଥାପନ କରି ପୋଖରଣ ଭୂମିଗର୍ଭରେ ଅଣୁଶକ୍ତିକୁ ବିକିରଣ କରି ଶାନ୍ତିକାର୍ଯ୍ୟରେ ଅଣୁଶକ୍ତିକୁ ପ୍ରୟୋଗ କରି ସଦୈବ ଆମର ଶାନ୍ତିକାମୀ ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ନିରନ୍ତର ଶୋଭାପାଉଅଛି।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ବିୟନ୍ମଣ୍ଡଳେ = ବିୟତ୍ + ମଣ୍ଡଳେ । ଶାନ୍ତକାର୍ଯ୍ୟଶ୍ୱମ୍ = ଶାନ୍ତକାର୍ଯ୍ୟଷୁ + ଅଣୁମ୍ । ପ୍ରେରୟସନ୍ତତମ୍ : ପ୍ରେରୟତ୍ + ସନ୍ତତମ୍ ।

ସମାସ – ଭୂମିଗର୍ଭେ = ଭୂମେ ଗର୍ଭମ୍, ତସ୍ମିନ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ଶାନ୍ତିକାର୍ଯ୍ୟଶ୍ରୁ = ଶାନ୍ତ୍ କାର୍ଯ୍ୟମ୍, ତେଷୁ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ଆର୍ଯ୍ୟଭଟ୍ଟମ୍, ଅଣୁଶକ୍ତିମ୍, ଅଣୁମ୍ = କର୍ମଣି ୨ୟା । ବିୟନ୍ମଣ୍ଡଳେ, ଭୂମିଗର୍ଭ, ଶାନ୍ତକାର୍ଯ୍ୟଷୁ, ଭୂତଳେ = ଅଧ୍ଵରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଶକ୍ତିମ୍ = ଶକ୍ + ସ୍କ୍ରିନ୍ । ଶାନ୍ତି = ଶମ୍ + ସ୍କ୍ରିନ୍ ।

ଶ୍ଳୋକ – ୯

संस्कृतं प्राकृर्त तामिलं तेलुगुं
कन्नडं कैरलीं वाड्नलामौत्कलीम् ।
वाचमन्यां च तां तां ब्रुवद्वर्धते
राष्ट्रभाषायुतं मामकं भारतम् ॥

ସଂସ୍କୃତଂ ପ୍ରାକୃତଂ ତାମିଲଂ ତେଲୁଗୁଂ
କନ୍ନଡ଼ କୈରଳୀ ବାଙ୍ଗଲାମୌତ୍କଳୀମ୍ ।
ବାଚମସ୍ୟା ଚ ତାଂ ତାଂ ବ୍ରୁବଦ୍‌ବର୍ଧତେ
ରାଷ୍ଠ୍ରଭାଷାୟୁତଂ ମାମକଂ ଭ।ରତମ୍ ॥ ୯ ॥

ଅନ୍ବୟ – ସଂସ୍କୃତଂ ପ୍ରାକୃତଂ ତାମିଲଂ ତେଲୁଗୁ କନ୍ନଡ଼ କୈରଳୀ ବାଙ୍ଗଲାମ୍ ଔତ୍କଳୀ ଚ ତାଂ ତାମ୍ ଅନ୍ୟା ବା ନଂ ବ୍ରୁବତ୍ ରାଷ୍ଟ୍ରଭାଷାୟୁତଂ ମାମକଂ ଭାରତଂ ବର୍ଧତେ ।

ଶବ୍ଦାର୍ଥ – ଔତ୍କଳୀ = ଉତ୍କଳୀୟ ଭାଷା । ବାଚଂ = ଭାଷାକୁ । ବ୍ରୁବତ୍ = କହି କହି । ରାଷ୍ଟ୍ରଭାଷାୟୁଙ = ରାଷ୍ଟ୍ରଭାଷାସମୂହ ବା ସହିତ । ମାମକଂ = ମୋର । ବର୍ଧତେ = ବଢ଼ିଛି ବା ସମୃଦ୍ଧ ହୋଇଛି ।

ଅନୁବାଦ – ସଂସ୍କୃତ, ପ୍ରାକୃତ, ତାମିଲ୍, ତେଲୁଗୁ, କନ୍ନଡ଼, କେରଳୀ, ବଙ୍ଗଳା ଓ ଓଡ଼ିଆ ସେହି ସେହି ଅନ୍ୟ ଭାଷାକୁ କହି ମଧ୍ୟ ରାଷ୍ଟ୍ରଭାଷା ସମ୍ମିଳିତ ମୋର ଭାରତର ସମୃଦ୍ଧି ହୋଇଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍କିନ୍ ପଦ୍ୟ ଅସ୍ମାକଂ ଭାରତଦେଶେ କଳଂ ବିବିଧା ଭାଷା ପ୍ରଚଳିତଃ ତଦେବ ବର୍ଣ୍ଣିତମ୍ ।

ଅସ୍ମାକଂ ଭାରତଦେଶଃ ବହୁଭାଷାଭାଷୀ ରାଷ୍ଟ୍ର ଭବତି । ଅଗ୍ନିନ୍ ଦେଶେ ଜନାଃ ଅନେକାନାଂ ପ୍ରାନ୍ତୀୟଭାଷାନାଂ ବ୍ୟବହାରଂ କୁର୍ବନ୍ତି । ଯଥା – ସଂସ୍କୃତଂ – ପ୍ରାକୃତଂ – ତାମିଲଂ – ତେଲୁଗୁ – କନ୍ନଡିଂ – କୈରଳୀ – ବାଙ୍ଗଲାମ୍ – ଔତ୍କଳୀ ଚ । ଅନେକେଷୁ ଭାଷାସୁ କଥୟତଃ ଜନାଃ ଅତ୍ର ରାଷ୍ଟ୍ରଭାଷାୟୁତଂ ମାମକଂ ଭାରତଂ ବର୍ଧତେ । ଅତଃ ଭାଷାମାଧ୍ଯମେନ ବିବିଧତା ମଧ୍ୟ ଏକତା ପରିଲକ୍ଷ୍ୟତେ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଏହି ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ସଂଗୃହୀତ । ଏଠାରେ ଭାରତରେ ବିଭିନ୍ନ ପ୍ରାନ୍ତରେ ଭିନ୍ନ ଭିନ୍ନ ଭାଷା କଥନ ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଆମ ଭାରତ ହେଉଛି ବହୁ ଭାଷାଭାଷୀର ଦେଶ । ଏଠାରେ ବିଭିନ୍ନ ରାଜ୍ୟର ଲୋକମାନେ ବିଭିନ୍ନ ପ୍ରାନ୍ତୀୟ ଭାଷାର ବ୍ୟବହାର କରିଥା’ନ୍ତି । ସଂସ୍କୃତ, ପ୍ରାକୃତ, ତାମିଲ, ତେଲୁଗୁ, କନ୍ନଡ଼, କେରଳୀ, ‘ବଙ୍ଗଳା ଓ ଓଡ଼ିଆ ଆଦି ବିଭିନ୍ନ ଅନ୍ୟ ଭାଷାମାନ କହିଥା’ନ୍ତି । ଏହି ଅନେକ ଭାଷାକଥନ ମାଧ୍ୟମରେ ରାଷ୍ଟ୍ରଭାଷା ସମ୍ମିଳିତ ହୋଇଥିବା ମୋର ଭାରତଦେଶ ସଦା ସମୃଦ୍ଧି କରିଅଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ବାଙ୍ଗଲାମୌତ୍କଳୀମ୍ = ବାଙ୍ଗଲାମ୍ + ଔତ୍କଳୀମ୍ । ବାଚମନ୍ୟାମ୍ = ବାଚମ୍ + ଅନ୍ୟାମ୍ । ବୃବଦ୍‌ବର୍ଧତେ ବୃବତ୍ + ବର୍ଧତେ ।

ସମାସ – ରାଷ୍ଟ୍ରଭାଷାୟୁତମ୍ = ରାଷ୍ଟ୍ରସ୍ୟ ଭାଷା (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ତାଭି ୟୁତମ୍ (ତୃତୀୟା ତତ୍‌ପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ବାଚମ୍, ଧନ୍ୟାମ୍, ତାମ୍ = କର୍ମଣି ୨ୟା ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ବୁବତ୍ = ବ୍ଲୁ + ଶତୃ ।

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

ଶ୍ଳୋକ – ୧୦

व्यासवाल्मीकिरत्नाकरैरुज्वल
स्वादुकादम्बरीपानलुब्धं सदा ।
कालिदासेन भासेन संद्योतितं
भूतले भाति मेऽनारतं भारतम् ॥ १० ॥

ବ୍ୟାସବାଲ୍ମୀକିରନାକରୋଗରୁଜ୍ଜଳଂ
ସ୍ଵାଦୁକାଦମ୍ବରୀପାନଲୁକ୍କ ସଦା ।
କାଳିଦାସେନ ଭାସେନ ସଂଦ୍ୟୋତିତଂ
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ୧୦ ॥

ଅନ୍ୱୟ – ସଦା ସ୍ଵାଦୁକାଦମ୍ବରୀପାନ ଲୁବ୍ଧ ବବ୍ୟାସବାଲ୍ମୀକିରନାକରେ ଉଜ୍ଜ୍ଵଳ, କାଳିଦାସେନ ଭାସେନ ସଂଦ୍ୟୋତିତଂ ମେ ଭାରତଂ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ବ୍ୟାସବାଲ୍ମୀକିରନାକରଃ = ବ୍ୟାସ-ବାଲ୍ମୀକି-ରନାକରଙ୍କଦ୍ଵାରା । ଉଜ୍ଜ୍ବଳମ୍ = ସମୁଜ୍ଜ୍ବଳ । ସଦା = ସର୍ବଦା । କାଦମ୍ବରୀ = କାବ୍ୟରସ ।

ଅନୁବାଦ – ସର୍ବଦା ଆସ୍ବାଦନୀୟ କାଦମ୍ବରୀ କାବ୍ୟରସ ପାନରେ ଲୁବ୍ଧ ବ୍ୟାସ-ବାଲ୍ମୀକି-ରନାକରଙ୍କଦ୍ଵାରା ସମୁଜ୍ଜ୍ଵଳିତ, କାଳିଦାସ-ଭାସଙ୍କଦ୍ୱାରା ସମ୍ୟକ୍ ପ୍ରକାଶିତ (କାବ୍ୟକାଦମ୍ବରୀ) ମୋର ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ଅବିରତ ବା ନିରନ୍ତର ଶେ।ଭ।ପାଉଛା

ସଂସ୍କୃତ ବରାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଭାରତୀୟ କାବ୍ୟ ତଥା କବି ପରମ୍ପରା ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତମ୍ ।

ସଦୈବ କାବ୍ୟରସ କାଦମ୍ବରୀ ଆସ୍ବାଦେନ ଲୁଚଂ କବୟଃ କାବ୍ୟ ବିରଚୟନ୍ତି । ବ୍ୟାସ-ବାଲ୍ମୀକି-ରନ୍‌କରଃ ପ୍ରଭୃତି କବିଭିଂ ସ୍ବରଚିତଂ କାବ୍ୟମାଧ୍ଯମେନ କାବ୍ୟରସଂ ସମ୍ମୁଜ୍ଜ୍ବଳଂ କୃତମ୍ । କାବ୍ୟପରମ୍ପରାୟାଂ ରାମାୟଣଂ ମହାଭାରତଂ ହରବିଜୟଂ ଚ ସୁପ୍ରସିଦ୍ଧ ଭବତି । ସଂସ୍କୃତକାବ୍ୟଧାରାଂ କାଳିଦାସେନ ଭାସେନ ଚ ସଂଦ୍ୟୋତିତମ୍ । ଏତାଦୃଶଂ କାବ୍ୟରସେନ ରସାନ୍ବିତଂ ସନ୍ ଅସ୍ମାକଂ ଭାରତଦେଶଃ ବିଶ୍ଵେଽସ୍ମିନ୍ ଅବିରତଂ ଶୋଭିତେ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭାରତୀୟ କାବ୍ୟ ତଥା କବିପରମ୍ପରା ସଂପର୍କରେ ଉଲ୍ଲେଖ କରାଯାଇଛି ।

‘ଅପାରେ କାବ୍ଯସଂସାରେ କବିରେବପ୍ରଜାପତିଃ’ – ଅର୍ଥାତ୍ କାବ୍ୟସଂସାରରେ କବି ହେଉଛି ସ୍ରଷ୍ଟା ବା ବ୍ରହ୍ମା । ସଂସ୍କୃତ କାବ୍ୟସାହିତ୍ୟରେ ଯେଉଁ କେତେଜଣ କବି ନିଜର କାବ୍ୟକାଦମ୍ବରୀକୁ ଆସ୍ବାଦ କରାଇଅଛନ୍ତି ସେମାନେ ହେଲେ – ଆଦିକାବ୍ୟ ରାମାୟଣର ରଚୟିତା ବାଲ୍ମୀକି, ମହାଭାରତର ରଚୟିତା ବ୍ୟାସ ଏବଂ ହରବିଜୟ ମହାକାବ୍ୟର ପ୍ରଣେତା ରନ୍‌କର ଇତ୍ୟାଦି । ଏହା ବ୍ୟତିରେକ କାବ୍ୟ ତଥା ନାଟ୍ୟ ସାହିତ୍ୟରେ ଯେଉଁମାନେ ପ୍ରସିଦ୍ଧି ଅର୍ଜନ କରିଅଛନ୍ତି ସେମାନେ ହେଲେ ମହାକବି କାଳିଦାସ ଓ ଭାସ । ଏମାନେ କାବ୍ୟ ଜଗତିକୁ ସମ୍ୟକ୍ ପ୍ରକାଶିତ କରିଛନ୍ତି । ଏହିଭଳି ବିଭିନ୍ନ କବିମାନଙ୍କଦ୍ୱାରା କାବ୍ୟରସ କାଦମ୍ବରୀ ସହୃଦୟର ହୃଦୟକୁ ଆହ୍ଲାଦିତ କରିଛି । ଏହାଦ୍ଵାରା ଆମର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ବରେ ନିରନ୍ତର ଶୋଭାପାଉଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ବ୍ୟାସବାଲ୍ମୀକିରନାକରେରୁଜ୍ଜ୍ବଳମ୍ = ବ୍ୟାସବାଲ୍ମୀକିରନାକରଃ + ଉଜ୍ଜ୍ୱଳମ୍ । ମେଽନାରତମ୍ = ମେ + ଅନାରତମ୍ ।

ସମାସ – ବ୍ୟାସବାଲ୍ମୀକିରନାକରଃ =ବ୍ୟାସ ଚ ବାଲ୍ମୀକି ଚ ରନ୍ଧାକରଃ ଚ, ଡଃ (ଦ୍ବନ୍ଦ୍ବ) । ସଂଦ୍ୟାତିତମ୍ ସମ୍ୟକ୍ ଦ୍ୟୋତିତମ୍ (ପ୍ରାଦି ତତ୍‌ପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – କାଳିଦାସେନ, ଭାସେନ = ଅନୁସ୍ତେ କଉଁରି ୩ୟା । ଭୂତଳେ

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଲୁବ୍ଧମ୍ = ଲୁଭ୍ + କ୍ତ ।

ଶ୍ଳୋକ – ୧୧

मूलरामायणं पम्परामायणं
कम्बरामायणं दाण्डिरामायणम् ।
कृत्तिवासादिरामायणं श्रावयद्
भूतले भाति मेडनारतं भारतम् ॥ ११ ॥

ମୂଳରାମାୟଣଂ ପମ୍ପରାମାୟଣଂ
କମ୍ବରାମାୟଣଂ ଦାଣ୍ଡିରାମାୟଣମ୍ ।
କୃତ୍ତିବାସାଦିରାମାୟଣଂ ଶ୍ରାବୟଦ୍
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ || ୧୧ ||

ଅନ୍ୱୟ – ମୂଳରାମାୟଣଂ ପମ୍ପରାମାୟଣଂ କମ୍ବରାମାୟଣଂ କୃଭିବା ସାଦି ରାମାୟଣଂ ଶ୍ର।ବୟତ୍‌ ମେ ଭାରତମ୍ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ମୂଳରାମାୟଣମ୍ = ବାଲ୍ମୀକିକୃତ ସଂସ୍କୃତ ରାମାୟଣ । ଶ୍ରାବୟତ୍ = ଶୁଣାଇ । ମେ = ମୋର । ଭୂତଳେ = ପୃଥ‌ିବୀପୃଷ୍ଠରେ । ଅନାରତମ୍ = ନିରନ୍ତର ।

ଅନୁବାଦ – ମୂଳରାମାୟଣ, ପମ୍ପରାମାୟଣ, କମ୍ବରାମାୟଣ, ଦାଣ୍ଡିରାମାୟଣ, କୃଭିବାସାଦି ରାମାୟଣକୁ ଗାନ କରାଇ ବା ଶୁଣ।ଇ ମୋର ଭାରତଦେଶ ପୃଥିବୀପୃଷ୍ଠରେ ନିରନ୍ତର ଶୋଭାପାଉଛ ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର କରଂ ବିବିଧରାମାୟଣଂ ଗ୍ରନ୍ଥ ଶ୍ରାବୟତ୍ ଅସ୍ମାକଂ ଭାରତଦେଶଃ ଭୂତଳେ ଭାତି ତଦେବ ବର୍ଣ୍ଣିତମ୍ ।

ପ୍ରାଚୀନ ଭାରତୀୟ ସଂସ୍କୃତିଃ । ଅତ୍ର ମହର୍ଷିବାଲ୍ମୀକିନା ରାମାୟଣଗ୍ରନ୍ଥ ବିରଚିତମ୍ । ଲୌକିକସଂସ୍କୃତସାହିତ୍ୟ ଆଦିକାବ୍ୟରୂପେଣ ଖ୍ୟାତଃ ରାମାୟଣମ୍ । ରାମସ୍ୟ ଅୟନଂ ରାମାୟଣମ୍ । ଅସ୍ୟ ଗ୍ରନ୍ଥାଧାରେଣ ବିଭିନ୍ନାନି ରାମାୟଣାନି ରଚିତାନି ସନ୍ତି । ମୂଳରାମାୟଣଂ, ପମ୍ପରାମାୟଣଂ, କମ୍ବରାମାୟଣ, ଦାଣ୍ଡିରାମାୟଣଂ, କୃତ୍ତିବାସାଦିରାମାୟଣଂ ଶ୍ରାବୟତ୍ ମେ ଭାରତଂ ଭାତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ରାମାୟଣ ଗ୍ରନ୍ଥର ବିବିଧତା ଏବଂ ତଦ୍ବାରା ଭାରତର ଶୋଭା ବିଷୟ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଭାରତୀୟ ସଂସ୍କୃତି ଅତୀବ ପ୍ରାଚୀନା । ଲୌକିକ ସଂସ୍କୃତ ସାହିତ୍ୟରେ ଆଦିକାବ୍ୟ ରୂପେ ଖ୍ୟାତ ରାମାୟଣ ମହାକାବ୍ୟର ରଚୟିତା ହେଉଛନ୍ତି ମୂହର୍ଷି ବାଲ୍ମୀକି । ତାଙ୍କର ଅମରଲେଖନୀ ନିଃସୃତ କାଳଜୟୀ କୃତି ରାମାୟଣରେ ରାମଙ୍କର ଅୟନ ବା ମାର୍ଗ ବିଷୟ ଉଲ୍ଲେଖ କରାଯାଇଅଛି । ପରବର୍ତ୍ତୀ କାଳରେ ଏହି ଗ୍ରନ୍ଥର ଆଧାରରେ ଅର୍ଥାତ୍ ମୂଳରାମାୟଣର ଅନୁକରଣରେ ପମ୍ପରାମାୟଣ, କମ୍ବରାମାୟଣ, ଦାଣ୍ଡିରାମାୟଣ, କୃଭିବାସାଦି ରାମାୟଣର ରଚନା କରାଯାଇଅଛି । ଏସବୁ ରାମାୟଣ ଗ୍ରନ୍ଥଦ୍ୱାରା ଆମର ଭ।ରତଦେଶ ବିଶ୍ଵରେ ଚିରପରିଚିତ ହେ।ଇପାରିଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – କୃତ୍ତିବାସାଦିରାମାୟଣମ୍ = କୃତ୍ତିବାସ + ଆଦିରାମାୟଣମ୍ । ମେଽନାରତମ୍ = ମେ + ଅନାରତମ୍ ।

ସମାସ – ମୂଳରାମାୟଣମ୍ = ମୂଳ ରାମାୟଣମ୍ (କର୍ମଧାରୟ) । ପମ୍ପରାମାୟଣମ୍ = ପମ୍ପ ନାମ ରାମାୟଣମ୍ (ମଧ୍ୟପଦଲୋପୀ କର୍ମଧାରୟ) । ଦାଣ୍ଡିରାମାୟଣମ୍ = ଦାଣ୍ଡି ନାମ ରାମାୟଣମ୍ (ମଧ୍ୟପଦଲୋପୀ କର୍ମଧାରୟ) ।

ସକାରଣବିଭକ୍ତି – ଭୂତଳେ = ଅଧିକରଣେ ୭ମୀ । ମେ = ସମ୍ବନ୍ଧ ୬ଷ୍ଠୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଭାତି = ଭା + ସ୍କ୍ରିନ୍ ।

ଶ୍ଳୋକ – ୧୨

द्वारकां सेतुबन्धं पुरीं बदरिकां
तिरुपतिं मधुपुरीं चाजमेरं दधत् ।
पुष्करामृतसरस्तीर्थराजैर्युतं
भूतले भाति मेडनारतं भारतम् ॥। १२ ॥

ଦ୍ଵାରକାଂ ସେତୁବନ୍ଧ ପୁରୀ ବଦରିକାଂ
ତିରୁପଣ୍ଟିଂ ମଧୁପୁରୀ ଚାଜମେରଂ ଦଧତ୍ ।
ପୁଷ୍କରାମୃତସରସ୍ତୀର୍ଥ ରାଜୈର୍ୟୁତଂ
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ୧୨ ॥

ଅନ୍ୱୟ – ଦ୍ବାରକାଂ ସେତୁବନ୍ଧ ପୁରୀ ବଦରିକାଂ ତିରୁପତଂ ମଧୁପୁରୀମ୍ ଆଜମେରଂ ଚ ଦଧତ୍, ପୁଷ୍କର-ଅମୃତସରଃ-ତୀର୍ଥରାଜ୍ୟୈତଂ ମେ ଭାରତମ୍ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଦଧତ୍ = ଧାରଣ କରି । ତୀର୍ଥରାଜଃ = ଶ୍ରେଷ୍ଠ ତୀର୍ଥଦ୍ବାରା । ୟୁତମ୍ = ସମ୍ମିଳିତ ହୋଇ ।

ଅନୁବାଦ – ଦ୍ଵାରକା, ସେତୁବନ୍ଧ, ପୁରୀ, ବଦରିକା, ତିରୁପତି, ମଧୁପୁରୀ ଓ ଆଜମେର୍ ପ୍ରଭୃତିକୁ ଧାରଣ କରି;

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଭାରତେ ବିଦ୍ୟମାନାନାଂ ତୀର୍ଥକ୍ଷେତ୍ରାଣଂ ବର୍ଣ୍ଣନଂ ଭବତି ।

ପୁଣ୍ୟଭୂମି ଭାରତଃ । ଅତ୍ର ବିଭିନ୍ନାନି ତୀର୍ଥକ୍ଷେତ୍ରାଣି ବିରାଜିତାନି ସନ୍ତି । ଦ୍ଵାରକାଂ-ସେତୁବନ୍ଧ -ପୁରୀ-ବଦରିକା-ତିରୁପତି-ମଧୁପୁରୀ – ଆଜମେରଂ ଚ ଧାରୟତ୍,ସନ୍ ଦେଶାତ୍ମାକଂ ବିରାଜତେ । ପୁନରପି ଅମ୍ଳୀକଂ ଦେଶେ ପୁଷ୍କର-ଅମୃତସରଃ-ଇତ୍ୟାଦୟଃ ତୀର୍ଥରାଜଃ ସମ୍ମିଳିତଃ ସନ୍‌ ଅସ୍ମାକଂ ଦେଶଃ ଭାରତଃ ବିଶ୍ଵଽସ୍ମିନ୍ ନିରନ୍ତରଂ ଶୋଭିତେ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ସମାନୀତ । ଏଠାରେ ଭାରତବର୍ଷରେ ଶୋଭାପାଉଥିବା ତୀର୍ଥରାଜିର ବର୍ଣ୍ଣନ କରାଯାଇଅଛି ।

ପୁଣ୍ୟଭୂମିର ଦେଶ ଭାରତ । ଏହାର ମାଟି-ପାଣି-ପବନ ସବୁକିଛି ପବିତ୍ର । ଗଙ୍ଗାର ପବିତ୍ର ଜଳଧାରା ଏଠି ପ୍ରବାହିତ । ବିଭିନ୍ନ ତୀର୍ଥକ୍ଷେତ୍ରଦ୍ଵାରା ପରିବେଷ୍ଟିତ ଏ ଦେଶ । ଶଙ୍କରାଚାର୍ଯ୍ୟଙ୍କଦ୍ବାରା ଧର୍ମପ୍ରଚାର ନିମନ୍ତେ ଚାରିଧାମର ପ୍ରତିଷ୍ଠା ଏ ଦେଶର ପବିତ୍ରତାକୁ ସୂଚିତ କରେ । ଦ୍ଵାରକା, ସେତୁବନ୍ଧ, ପୁରୀ, ବଦରିକା, ତିରୁପତି, ମଧୁପୁରୀ, ଆଜମେର୍ ପ୍ରଭୃତି ପବିତ୍ର ତୀର୍ଥକ୍ଷେତ୍ର ଭାରତର ଐକ୍ୟ ଓ ସଂସ୍କୃତିକୁ ପରିରକ୍ଷିତ କରୁଛି । ପୁଷ୍କର-ଅମୃତସର ପ୍ରଭୃତି ସୁପ୍ରସିଦ୍ଧ ତୀର୍ଥସମୂହଦ୍ଵାରା ସମ୍ମଳିତ ଭାରତଦେଶ ବିଶ୍ବରେ ନିରନ୍ତର ଶୋଭା ପାଉଛି । ଏଭଳି ତୀର୍ଥଭୂମି ବା ପୁଣ୍ୟଭୂମିରେ ଆମେ ଜନ୍ମ ହୋଇଥିବାରୁ ଧନ୍ୟ ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ଚାଜମେରମ୍ = ଚ + ଆଜମେରମ୍ । ପୁଷ୍କରାମୃତସରସ୍ତୀର୍ଥରାଜୈର୍ୟୁତମ୍ = ପୁଷ୍କର + ଅମୃତସରଃ + ତୀର୍ଥରାଜଃ + ୟୁତମ୍ ।

ସମାସ – ମଧୁପୁରୀମ୍ = ମଧୁ ନାମ ପୁରୀମ୍ (ମଧ୍ୟପଦଲୋପୀ କର୍ମଧାରୟ) । ତୀର୍ଥରାଜଃ = ତୀର୍ଥେଷୁ ରାଜଃ, ତୈ (ସପ୍ତମୀ ତତ୍‌ପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ଦ୍ବାରକା, ସେତୁବନ୍ଧ, ପୁରାଂ, ବଦରିକା, ତିରୁପଣ୍ଟିଂ, ମଧୁପୁରୀ, ଆଜମେରମ୍ = କର୍ମଣି ୨ୟା । ତୀର୍ଥରାଜୈ = କରଣେ ୩ୟା । ଭୂତଳେ = ଅଧ୍ଵରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଦଧତ୍ = ଧା + ଶତୃ ।

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

ଶ୍ଳୋକ – ୧୩

हस्तिगुम्फामजन्तामलौरां दधत्
खर्जुराहो – गया – सारनाथैर्लसंत् ।
ताज-कोणार्क-विष्णुध्वजैर्मण्डितं
भूतले भाति मेऽनारतं भारतम् ॥ १३ ॥

ହସ୍ତୀଗୁମ୍ଫାମଜନ୍ତାମଲୌରାଂ ଦଧତ୍
ଖର୍ଜୁରାହୋ-ଗୟା-ସାରନାଥୈର୍ଲସତ୍ ।
ତାଜ କୋଣାର୍କ ବିଷ୍ଣୁଧ୍ୱଜୈର୍ମଣ୍ଡିତଂ
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ୧୩ ॥

ଅନ୍ୱୟ – ହସ୍ତିଗୁମ୍ଫାଂ ଅଜନ୍ତ୍ରାଂ ଏଲୋରାଂ ଦଧତ୍ ଖର୍ଜୁରାହଃ ଗୟା ସାରନାଥୈଃ ଲସତ୍ ତାଜ କୋଣାର୍କ ବିଷ୍ଣୁଧ୍ୱଜୈଃ ମଣ୍ଡିତଂ (ସନ୍) ମେ ଭାରତମ୍ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଦଧତ୍ = ଧାରଣ କରି । ଲସତ୍ = ଉଲ୍ଲସିତ ହୋଇ । ମଣ୍ଡିତମ୍ = ଶୋଭିତ ହୋଇ । ଅନାରତମ୍ = ନିରନ୍ତର ।

ଅନୁବାଦ – ହସ୍ତୀଗୁମ୍ଫା, ଅଜନ୍ତା, ଏଲୋରାକୁ ଧାରଣ କରି, ଖର୍ଜୁରାହ, ଗୟା, ସାରନାଥଦ୍ୱାରା ବିଳସିତ; ତାଜମହଲ, କୋଣାର୍କ, ବିଷ୍ଣୁସ୍ତମ୍ଭଦ୍ଵାରା ବିଶୋଭିତ ମୋର ଭାରତଦେଶ ବିଶ୍ବରେ ନିରନ୍ତର ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଭାରତୀୟ ସ୍ଥାପତ୍ୟକଳାୟାଂ ଉତ୍କର୍ଷତାଂ ବର୍ଣ୍ଣିତଂ ଭବତି ।

କଳାୟା ଦେଶଃ ଭାରତଃ । ଅତ୍ର ସ୍ଥାପତ୍ୟକଳାୟା ମହତ୍ତ୍ୱ ସର୍ବେଷାମ୍ ଆକର୍ଷୟତି । ଅସ୍ୟ ଦେଶସ୍ୟ ସୁମହାନ୍ କୀର୍ତ୍ତିରାଜି ବିଶ୍ୱେଽସ୍ମିନ୍ ବିଖ୍ୟାତଃ ଆସୀତ୍ । ଅଜନ୍ତା-ଏଲୋରା, ହସ୍ତୀଗୁମ୍ଫା ଦଧୂତ୍, ଖର୍ଜୁରାହଃ, ଗୟା, ସାରନାର୍ଥେ ଭଗବାନ୍ ବୁଦ୍ଧସ୍ୟ ମହିମା ଉଲ୍ଲସିତା ଅଭବତ୍ । ତାଜ-କୋଣାର୍କ-ଦେହଲୀସ୍ଥ ବିଷ୍ଣୁସ୍ତତଃ ଇତ୍ୟାଦିଭିଂ ବିମଣ୍ଡିତମ୍ ଅସ୍ମାକଂ ଭାରତଦେଶଃ ଅନାରତଂ ଭୂତଳେ ଭାତି । ଭାରତୀୟ ସ୍ଥାପତ୍ୟକଳାୟା ଉତ୍କର୍ଷତାମ୍ ଆବହତି ଇତ୍ୟେପି କ୍ଷେତ୍ରମ୍ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଉଦ୍ଧୃତ ଶ୍ଳୋକଟି ମହାମହିମ କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକରେ ସ୍ଥାନିତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭାରତୀୟ ସ୍ଥାପତ୍ୟକଳାର ଉତ୍କର୍ଷତା ବର୍ଣ୍ଣିତ ହୋଇଛି ।

କଳାର ଦେଶ ଭାରତ । ଏଠାରେ ବିଭିନ୍ନ ସ୍ଥାପତ୍ୟକଳାର ନମୁନା ଉପଲବ୍ଧ ହୁଏ । ଅଜନ୍ତା ଓ ଏଲୋରାର ହାତୀଗୁମ୍ଫା, ଖର୍ଜୁରାହ, ଗୟା ଓ ସାରନାଥଠାରେ ଭଗବାନ୍ ବୁଦ୍ଧଙ୍କର କୀର୍ତ୍ତିରାଜି, ତାଜମହଲ, କୋଣାର୍କକ୍ଷେତ୍ର, ଦିଲ୍ଲୀର ବିଷ୍ଣୁସ୍ତମ୍ଭ ଆଦିଦ୍ୱାରା ଆମର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ଵରେ ଶୋଭାପାଉଛି । ଭାରତୀୟ ସ୍ଥପତି ବିଶ୍ଵରେ ଚିରପରିଚିତ ହୋଇଛନ୍ତି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ସ।ରନାଥୌର୍ଲସତ୍‌ = ସାରନାର୍ଥେ + ଲସତ୍ । ବିଷ୍ଣୁଧ୍ଵଜୈର୍ମଣ୍ଡିତମ୍ = ବିଷ୍ଣୁଧ୍ଵଜଃ + ମଣ୍ଡିତମ୍ । ମେଽନାରତମ୍ = ମେ + ଅନାରତମ୍ ।

ସମାସ – ହସ୍ତୀଗୁମ୍ଫାମ୍ = ହସ୍ତୀନାଂ ଗୁମ୍ଫା, ତାମ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ବିଷ୍ଣୁଧ୍ଵ = ବିଷୋ ଧ୍ଵଜଃ, ତୋ (ଷଷ୍ଠୀ ତପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ହସ୍ତୀଗୁମ୍ଫା, ଏଲୋରାମ୍ = କର୍ମଣି ୨ୟା । ସାରନାଥେ, ବିଷ୍ଣୁଧ୍ଵ = କରଣେ ୩ୟା । ଭୂତଳେ = ଅଧ୍ଵରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଦଧତ୍ = ଧା + ଶତୃ । ଲସତ୍ = ଲସ୍ + ଶତୃ ।

ଶ୍ଳୋକ – ୧୪

होलिका-दशहरा-पर्वकोजागरी-
पोड़ल-श्रावणी-दीपमालामयम् ।
लोहडीदौणमाद्युत्सवै: ‘पूरितं
भूतले भाति मेडनारतं भारतम् ॥ १४ ॥

ହେ।ଲିକା ଦଶହରା ପର୍ବକୋଜ।ଗରୀ
ପୋଙ୍ଗଲ-ଶ୍ରାବଣୀ-ଦୀପମାଳାମୟମ୍ ।
ଲୋହଡୀଦୌଣମାଦ୍ୟୁତ୍ସୟଃ ପୂରିତଂ
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ୧୪ ॥

ଅନ୍ୱୟ – ହୋଲିକା ଦଶହରା ଉଜାଗରୀପର୍ବକ ପୋଙ୍ଗଲ ଶ୍ରାବଣୀ ଦୀପମାଳାମୟଂ ଲୋହଡ଼ୀ ଦୌଣମ୍ ଆଦି ଉତ୍ସବୈ ପୂରିତଂ ମେ ଭାରତମ୍ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଉଜାଗରୀପର୍ବକ = ଉଜାଗର ପର୍ବ (ଜାଗର ବା ଶିବରାତ୍ରି) । ଦୀପମାଳାମୟମ୍ = ଦୀପାବଳୀ ଉତ୍ସବ। ପୂରିତମ୍ = ପୂର୍ଣ୍ଣ ।

ଅନୁବାଦ – ହୋଲି, ଦଶହରା, ଉଜାଗର ଉତ୍ସବ (ଶିବରାତ୍ରି), ପୋଙ୍ଗଲ, ଶ୍ରାବଣ ପୂର୍ଣ୍ଣିମା (ରକ୍ଷାବନ୍ଧନ), ଦୀପାବଳୀ ଉତ୍ସବ, ଲୋହଡ଼ୀ, ଦୌଣୋତ୍ସବ ଆଦି ବିଭିନ୍ନ ଉତ୍ସବଦ୍ବାରା ପରିପୂର୍ଣ୍ଣ ମୋର ଭାରତଦେଶ ବିଶ୍ବରେ ନିରନ୍ତର ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ରମାକାନ୍ତଶୁକ୍ଳବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅତ୍ର ଅସ୍ଥିନ୍ ଶ୍ଳୋକେ ଭାରତୀୟ ଉତ୍ସବାନାଂ ବିଷୟେ ବର୍ଣ୍ଣିତଂ ଭବତି ।

ଉତ୍ସବପ୍ରିୟା ଖଳୁ ମାନବାଃ । ଅସ୍ମାକଂ ଦେଶସ୍ୟ ବିଭିନ୍ନେଷୁ ପ୍ରାନ୍ତେଷୁ ବିବିଧା ଉତ୍ସବାଃ ପରିପାଳିତଂ ଭବନ୍ତି । ହୋଲିକା – ସନ୍ ମେ ଭାରତଂ ଅନାରତଂ ଭୂତଳେ ଭାତି । ଅସ୍ମାକଂ ଦେଶସ୍ୟ ସଂସ୍କୃତିଃ ସୁମହତୀ ଭବତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ସଂଗୃହୀତ । ଏଥିରେ ଭାରତୀୟ ଉତ୍ସବଗୁଡ଼ିକର ବିଷୟରେ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଭାରତୀୟ ସଂସ୍କୃତି ମହାନ୍ । ଏଠାରେ ପ୍ରତ୍ୟେକ ମନୁଷ୍ୟ ଅତ୍ୟନ୍ତ ଉତ୍ସବପ୍ରିୟ । ଭାରତର ବିଭିନ୍ନ ପ୍ରାନ୍ତରେ ଭିନ୍ନ ଭିନ୍ନ ଭାଷାଭାଷୀ ଲୋକମାନେ ଅନେକପ୍ରକାର ଉତ୍ସବ ପାଳନ କରିଥା’ନ୍ତି । ସେସବୁ ଉତ୍ସବମାନଙ୍କ ମଧ୍ୟରେ ହୋଲି, ଦଶହରା, ଜାଗର ଅମାବାସ୍ୟା ବା ଶିବରାତ୍ରି, ପୋଙ୍ଗଲ, ଶ୍ରାବଣ ପୂର୍ଣ୍ଣିମା ବା ରକ୍ଷାବନ୍ଧନ, ଦୀପାବଳୀ, ଲୋହଡ଼ୀ, ଦୌଣାଦି ବହୁବିଧ ଉତ୍ସବ ରହିଛି । ଏହିସବୁ ଉତ୍ସବଗୁଡ଼ିକରେ ବିମଣ୍ଡିତ ଆମର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ବରେ ଶୋଭାପାଉଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ପର୍ବକୋଜାଗରୀ = ପର୍ବକ + ଉଜାଗରୀ । ଦୌଣମାଦ୍ୟୁତ୍ସବୈ = ଦୌଣମ୍ + ଆଦି + ଉତ୍ସବାଃ ।

ସମାସ – ଦୀପମାଳା = ଦୀପାନାଂ ମାଳା (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ଉତ୍ସବୈ = କରଣେ ୩ୟା । ଭୂତଳେ = ଅଧୂକରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ପୂରିତମ୍ = ପୂର୍ + ଣିଚ୍ + କ୍ତ ।

ଶ୍ଳୋକ – ୧୫

ओडिशीं मणिपुरीं कथक -गर्बादिकं
कूचिपूडिं च गिद्दां छऊं भङ्गडाम् ।
कथकलीं डाण्डियां भरतनाट्यं दधद्
भूतले भाति मे नारतं भारतम् ॥ १५ ॥

ଓଡ଼ିଶୀ ମଣିପୁରୀ କଥକ-ଗର୍ବାଦିକଂ
କୂଚିପୂରିଂ ଚ ଗିଦ୍‌ଦାଂ ଛଉଂ ଭଙ୍ଗଡ଼ାମ୍ ।
କଥକଲୀ ଡାଣ୍ଡିୟାଂ ଭରତନାଟ୍ୟମ୍ ଦଧଦ୍
ଭୂତଳେ ଭାତି ମେଽନାରତଂ ଭାରତମ୍ ॥ ୧୫ ॥

ଅନ୍ବୟ – ଓଡ଼ିଶୀ, ମଣିପୁରୀ, କଥକଂ, ଗର୍ବାଦିକ, କୂଚିପୂରିଂ, ଗି, ଛଉଂ, ଭଙ୍ଗଡ଼ା, କଥକଳ୍କୀ, ଡାଣ୍ଡିୟାଂ ଭାରତନାଟ୍ୟ ବ ଦଧତ୍ ମେ ଭାରତମ୍ ଭୂତଳେ ଅନାରତଂ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଦଧଦ୍ = ଧାରଣ କରି । ମେ = ମୋର ।

ଅନୁବାଦ – ଓଡ଼ିଶୀ, ମଣିପୁରୀ, କଥକ, ଗରବା, କୁଚିପୁଡ଼ି, ଗି, ଛଉ, ଭାଙ୍ଗଡ଼ା, କଥକଲୀ, ଡାଣ୍ଡିୟା ଓ ଭାରତନାଟ୍ୟମ୍ ଆଦି ବିଭିନ୍ନ ନୃତ୍ୟକୁ ଧାରଣ କରି ମୋର ଭାରତ ବିଶ୍ବରେ ଅବିରତ ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ରମାକାନ୍ତଶୁକ୍ଳ ବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଭାରତୀୟନୃତ୍ୟକଳାୟା ମହତ୍ତ୍ୱ ବଣ୍ଡିତଂ ଭବତି ।

ଭାରତସ୍ୟ ବିଭିନ୍ନ ପ୍ରାନ୍ତେଷୁ ବିବିଧ୍ୱଂ ନୃତ୍ୟ ପରିଦୃଶ୍ୟତେ । ନୃତ୍ୟ ଜୀବନସ୍ୟ କଷ୍ଟ ଦୂରିକରୋତି । ଭାରତଦେଶେ ବିବିଧାନି ନୃତ୍ୟାନି ପରିବେଶିତାନି ଭବନ୍ତି । ଓଡ଼ିଶୀ, ମଣିପୁରୀ, କଥକଂ, ଗବାଂ, କୂଚିପୂଡ଼ି, ଗିଦ୍‌, ଛଉଂ, ଭଙ୍ଗଡ଼ା, କଥକଲୀ, ଡାଣ୍ଡିୟା, ଭାରତନାଟ୍ୟ ଚ ଧାରୟନ୍ ମମ ଭାରତଦେଶଃ ବିଶ୍ଵେଽସ୍ମିନ୍ ବିରାଜତେ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ରମାକାନ୍ତ ଶୁକ୍ଳ ବିଗତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭାରତୀୟ ନୃତ୍ୟକଳାର ମହନୀୟତା ପ୍ରକଟିତ ହୋଇଛି ।

ସଂସ୍କୃତିସଂପନ୍ନ ଦେଶ ଭାରତ । ଏହାର କଳାସଙ୍ଗୀତ ତଥା ନୃତ୍ୟ ବିଶ୍ଵବିଶ୍ରୁତ । ଭାରତର ବିଭିନ୍ନ ପ୍ରାନ୍ତରେ ଭିନ୍ନ ଭିନ୍ନ ପ୍ରକାରର ନୃତ୍ୟ ପରିଲକ୍ଷିତ ହୁଏ । ଓଡ଼ିଶୀ, ମଣିପୁରୀ, କଥକ, ଗରବା, କୁଚିପୁଡ଼ି, ଗି, ଛଉ, ଭାଙ୍ଗଡ଼ା, କଥକଲୀ, ଡାଣ୍ଡିୟା ଓ ଭାରତନାଟ୍ୟମ୍ ଆଦି ବିଭିନ୍ନ ନୃତ୍ୟ ପରିବେଶିତ ହେଉଛି । ଏସବୁ ନୃତ୍ୟ ପାରମ୍ପରିକ ବେଶପୋଷାକ, ଉପରେ ଆଧାରିତ । ଏହିସବୁ ନୃତ୍ୟଦ୍ଵାରା ସମଗ୍ର ବିଶ୍ଵରେ ଭାରତଦେଶ ବିଶୋଭିତ ହୋଇଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ଗର୍ବାଦିକମ୍ = ଗର୍ବା + ଆଦିକମ୍ । ଦଧଦ୍‌ଭୂତଳେ = ଦଧତ୍ + ଭୂତଳେ ।

ସକାରଣବିଭକ୍ତି – ଓଡ଼ିଶୀ, ମଣିପୁରୀ, କୂଚିପୂଡ଼ି, ଗିର୍ଦ୍ଦା, ଛଉଂ, ଭଙ୍ଗଡ଼ାମ୍, କଥକଳ୍କୀ, ଡାଣ୍ଡିୟାଂ, ଭରତନାଟ୍ୟମ୍ କର୍ମଣି ୨ୟା ଭୂତଳେ = ଅଧିକରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଦଧତ୍ = ଧା + ଶତୃ ।

ଶ୍ଳୋକ – ୧୬

यत्कुरुक्षेत्रमध्ये स्वयं श्रीहरि –
निष्क्रियं पार्थमाश्वासयद् गीतया ।
स्वीयरूपेण तं च व्वधात्कर्मठ
भूतले भाति तन्मामकं भारतम् ॥ १६ ॥

ଯତକୁରୁକ୍ଷେତ୍ରମଧ୍ୟେ ସ୍ୱୟଂ ଶ୍ରୀହରି –
ନିଷ୍କ୍ରିୟଂ ପାର୍ଥମାଶ୍ଵାସୟଦ୍ ଗୀତୟା ।
ସ୍ୱୀୟରୂପେଣ ତଂ ଚ ବ୍ୟଧାତ୍ମକର୍ମଠ
ଭୂତଳେ ଭାତି ତନ୍ମାମକଂ ଭାରତମ୍ ॥ ୧୬ ॥

ଅନ୍ୱୟ – ଯତ୍‌କୁରୁକ୍ଷେତ୍ରମଧ୍ୟ ସ୍ବୟଂ ଶ୍ରୀହରିଂ ଗୀତୟା ନିଷ୍କ୍ରିୟଂ ପାର୍ଥମ୍ ଆଶ୍ୱାସୟତ୍ ସ୍ୱୀୟରୂପେଣ ତଂ କର୍ମଠ ବ୍ୟଧାତ୍ ଚ, ତତ୍ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ସ୍ୱୟମ୍ = ନିଜେ । ଶ୍ରୀହରି = ଶ୍ରୀକୃଷ୍ଣ । ଗୀତୟା = ଗୀତାଦ୍ଵାରା । ପାର୍ଥମ୍ = ଅର୍ଜୁନକୁ । ସ୍ବୀୟରୂପେଣ ନିଜ ସ୍ଵରୂପଦ୍ବାରା । ବ୍ୟଧାତ୍ = ବିଧାନ କରି । ମାମକମ୍ = ମୋର । ଭାତି = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ଯେଉଁ କୁରୁକ୍ଷେତ୍ର ମଧ୍ୟରେ ନିଜେ ଶ୍ରୀକୃଷ୍ଣ ଗୀତାଦ୍ୱାରା ନିଷ୍କ୍ରିୟ ଅର୍ଜୁନକୁ ଆଶ୍ୱାସନା ଦେଇଛନ୍ତି ଓ ସ୍ୱ ସ୍ୱରୂପଦ୍ୱାରା ତାକୁ କର୍ମଠ ବିଧାନ କରିଛନ୍ତି ମୋର ସେହି ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ଶୋଭାପାଉଛି ।

CHSE Odisha Class 11 Sanskrit Solutions Poem 2 भाति मे भारतम्

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ କବିରମାକାନ୍ତଶୁକ୍ଳ ବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ଥିନ୍ ଶ୍ଳୋକେ ଭାରତଦେଶେ ଗୀତାଗାସ୍ୟ ମହତ୍ତ୍ୱ ବଣ୍ଡିତଂ ଭବତି ।

ବ୍ୟାସବିରଚିତଂ ମହାଭାରତମ୍ । ଅସ୍ୟ ଗ୍ରନ୍ଥସ୍ୟ ଭୀଷ୍ମପର୍ବାନ୍ତର୍ଗତଂ ଭବତି ଶ୍ରୀମଦ୍‌ଭଗବଦ୍‌ଗୀତା । କୁରୁକ୍ଷେତ୍ରସ୍ୟ ସମରାଙ୍ଗଣସ୍ୟ ଗାଥା ଭଗବାନ୍ ଶ୍ରୀକୃଷ୍ଣସ୍ୟ ଅର୍ଜୁନଂ ପ୍ରତି ଉପଦେଶସାରଂ ଭବତି ଗୀତା ବିଶାଦୀ ନିଷ୍କ୍ରିୟମ୍ ଅର୍ଜୁନଂ ପ୍ରତି ଭଗବତଃ ଶ୍ରୀକୃଷ୍ଣସ୍ୟ ଆଶ୍ୱାସନାବାଣୀ ଗୀତା । ଭଗବାନ୍ ସ୍ଵୟଂ ଗୀତୟା ନିଷ୍କ୍ରିୟଂ ପାର୍ଥମ୍ ଆଶ୍ୱାସୟତ୍ । ସ୍ବୀୟରୂପେଣ ତଂ କର୍ମତଂ କୃତମ୍ । ଅନେନ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଉଦ୍ଧୃତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର ମହାମହିମ କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ସମାନୀତ । ଏଠାରେ ଭାରତରେ ଗୀତାଗାନର ମହତ୍ତ୍ବ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

‘ରୁଚୁ ନ ରୁଚୁ ଖାଅ ପିତା, ବୁଝ ନ ବୁଝ ପଢ଼ ଗୀତା’ – ଏହି ଲୋକୋକ୍ତି ଆଧାରରେ ସମଗ୍ର ଭାରତରେ ଗୀତାଗାନର ମହତ୍ତ୍ବ ରହିଛି । କୁରୁକ୍ଷେତ୍ରର ସମରାଙ୍ଗଣରେ ବିଷାଦିତ ଅର୍ଜୁନଙ୍କ ପ୍ରତି ଭଗବାନ୍ ଶ୍ରୀକୃଷ୍ଣଙ୍କର ଉପଦେଶବାଣୀ ଗୀତା ରୂପେ ଖ୍ୟାତ । ମହାଭାରତର କୁରୁକ୍ଷେତ୍ରରେ ଭଗବାନ୍ ଶ୍ରୀକୃଷ୍ଣ ନିଜେ ନିଷ୍କ୍ରିୟ ଅର୍ଜୁନଙ୍କୁ ଗୀତାଦ୍ଵାରା ଆଶ୍ଵାସନାବାଣୀ ଶୁଣାଇଛନ୍ତି । ନିଜସ୍ଵରୂପ ପ୍ରଦର୍ଶନଦ୍ଵାରା ଅର୍ଜୁନଙ୍କୁ କର୍ମଠ କରିଅଛନ୍ତି । ଏହିଭଳି ସୁମହାନ୍ ଗୀତାଗାନରେ ମୋର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ଵରେ ଶେ।ଭିତ ହେଉଅଚ୍ଛା

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ଶ୍ରୀହରିନିଷ୍କ୍ରିୟମ୍ = ଶ୍ରୀହରଃ + ନିଷ୍କ୍ରିୟମ୍ । ପ୍ରାର୍ଥମାଶ୍ଵାସୟଦ୍ = ପାର୍ଥମ୍ + ଆଶ୍ୱାସୟଦ୍ । ତନ୍ମାମକମ୍ ତତ୍ + ମାମକମ୍ ।
ସମାସ – ସ୍ବୀୟରୂପେଣ = ସ୍ଵୟଂ ରୂପମ୍, ତେନ (କର୍ମଧାରୟ) ।

ସକାରଣବିଭକ୍ତି – କୁରୁକ୍ଷେତ୍ରମଧେ = ଅଧ୍ଵରଣେ ୭ମୀ । ଶ୍ରୀହରିଃ = କଉଁରି ୧ ମା । ପାର୍ଥୀ, ତମ୍ = କର୍ମଣି ୨ୟା । ଗୀତୟା, ସ୍ୱୀୟରୂପେଣ = କରଣେ ୩ୟା ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଭାତି = ଭା + ସ୍କ୍ରିନ୍ ।

ଶ୍ଳୋକ – ୧୭

मातृभूमेर्विपज्जालमुच्छेदयन्
मृत्युपाशं च कण्ठे सहर्षं धरन् ।
भक्तसिंहोऽस्ति यत्रामरत्वं गुतो
भूतले भाति तन्मामकं भारतम् ॥ १७ ॥

ମାତୃଭୂମେର୍ବିପଜାଲମୁଚ୍ଛେଦୟନ୍
ମୃତ୍ୟୁପାଶଂ ଚ କଣ୍ଠ ସହର୍ଷ ଧରନ୍ ।
ଭକ୍ତସିଂହୋ ଽସ୍ତୀ ଯତ୍ରାମରତ୍ଵ ଗତୋ
ଭୂତଳେ ଭାତି ତନ୍ମାମକଂ ଭାରତମ୍ ।। ୧୭ ।।

ଅନ୍ୱୟ – ମାତୃଭୂମେ ବିପଜାଲମ୍ ଉଚ୍ଛେଦୟନ୍ କଣ୍ଠ ସହର୍ଷୀ ମୃତ୍ୟୁପାଶଂ ଧରନ୍ ଚ ଯତ୍ର ଭକ୍ତସିଂହଃ ଅମରତ୍ୱ ଗତଃ ଅସ୍ତି ତତ୍ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ମାତୃଭୂମେ = ମାତୃଭୂମିର । ବିପଜାଲମ୍ = ବିପଦର ଜାଲକୁ । କଣ୍ଠେ = ଗଳାରେ । ସହର୍ଷମ୍ = ଆନନ୍ଦର ସହିତ । ମୃତ୍ୟୁପାଶମ୍ = ମରଣର ଫାଶକୁ । ଧରନ୍ = ଧାରଣ କରି । ଅସ୍ତି = ଅଛି ।

ଅନୁବାଦ – ମାତୃଭୂମିର ବିପଦଜାଲକୁ ଉଚ୍ଛେଦ କରି ଓ ଗଳାରେ ଆନନ୍ଦର ସହ ମୃତ୍ୟୁଫାଶକୁ ଧାରଣ କରି ଯେଉଁଠି ଭକ୍ତସିଂହ ଅମରତ୍ୱକୁ ପ୍ରାପ୍ତ ହୋଇଛନ୍ତି ସେହି ମୋର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ବରେ ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ କବି ରମାକାନ୍ତଶୁକ୍ଳା ବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଭକ୍ତସିଂହସ୍ୟ ବଳିଦାନ ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତଂ ଭବତି ।

ଭାରତୋଽୟଂ ବୀରାମାଂ ଦେଶଃ । ଅସ୍ଟିନ୍ ଦେଶେ ବହବାଃ ସୁରୟଃ ସ୍ଵସ୍ୟ ବଳିଦାନଂ ପ୍ରଦତ୍ତବନ୍ତଃ । ଅସ୍ମିନ୍ ଦେଶେ ଭକ୍ତସିଂହଃ ମାତୃଭୂମେ ବିପଜ୍ଜାଲମ୍ ଉଚ୍ଛେଦୟନ୍ କଣ୍ଠ ସହର୍ଷୀ ମୃତ୍ୟୁପାଶଂ ଧାରୟନ୍ ଚ ଅମରତଂ ଗତଃ । ତତ୍ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭକ୍ତସିଂହର ବଳିଦାନ ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଭାରତୀୟ ବୀରମାନେ ଏ ଦେଶରେ ନିଜର ପ୍ରାଣବଳି ଦେଇଛନ୍ତି । ସେମାନଙ୍କ ମଧ୍ୟରେ ଭକ୍ତସିଂହ ଅନ୍ୟତମ । ସେ ମାତୃଭୂମିର ବିପଦ ଜାଲକୁ ଉଚ୍ଛେଦ କରି ଗଳାରେ ଆନନ୍ଦ ସହକାରେ ମୃତ୍ୟୁପାଶକୁ ଧାରଣ କରି ଅମରତ୍ୱକୁ ପ୍ରାପ୍ତ ହୋଇଅଛନ୍ତି । ଏହିଭଳି ବୀରମାନଙ୍କର ବଳିଦାନଦ୍ଵାରା ମୋର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ଵରେ ଶୋଭାପାଉଅଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ମାତୃଭୂମେର୍ବିପଦଜ୍ଜାଲମୁଚ୍ଛେଦୟନ୍ = ମାତୃଭୂମଃ + ବିପତ୍ + ଜାଲମ୍ + ଉଚ୍ଛେଦୟନ୍ । ଭକ୍ତସିଂହୋଽସ୍ତି = ଭକ୍ତସିଂହଃ + ଅସ୍ତି । ତନ୍ମାମକମ୍ = ତତ୍ + ମାମକମ୍ ।

ସମାସ – ମୃତ୍ୟୁପାଶମ୍ = ମୃତ୍ୟୁ ପାଶମ୍, ତମ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ସହର୍ଷମ୍ = ହର୍ଷେଣ ସହ ବର୍ତ୍ତମାନମ୍ ( ସହାର୍ଥ ବହୁବ୍ରୀହିଃ) ।

ସକାରଣବିଭକ୍ତି – ମାତୃଭୂମେଃ = ସମ୍ବନ୍ଧ ୬ଷ୍ଠୀ । ବିପଜାଲଂ, ମୃତ୍ୟୁପାଶମ୍ = କର୍ମଣି ୨ୟା । କଣ୍ଠ, ଭୂତଳେ = ଅଧିକରଣେ ୭ମୀ । ଭକ୍ତସିଂହଃ = କଉଁରି ୧ମା।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଧରନ୍ = ଧର୍ + ଶତୃ । ଗତଃ = ଗମ୍ + କ୍ତ ।

ଶ୍ଳୋକ – ୧୮

रक्तपातं विना शस्त्रपातं विना
यत्र संक्रान्तिरायाति मन्दस्मिता ।
येन विश्वं सदा शिक्ष्यते प्रेर्यते
भूतले भाति तन्मामकं भारतम् ॥ १८ ॥

ରକ୍ତପାତଂ ବିନା ଶସ୍ତ୍ରପାତଂ ବିନା
ଯତ୍ର ସଂକ୍ରାନ୍ତିରାୟାତି ମନ୍ଦସ୍ମିତା ।
ଯେନ ବିଶ୍ଵ ସଦା ଶିଷ୍ୟତେ ପ୍ରେର୍ଯ୍ୟତେ
ଭୂତଳେ ଭାତି ତନ୍ମାମକଂ ଭାରତମ୍ ।। ୧୮ ।।

ଅନ୍ବୟ – ଯତ୍ର ରକ୍ତପାତଂ ବିନା ଶସ୍ତ୍ରପାତଂ ବିନା ମନ୍ଦସ୍ମିତା ସଂକ୍ରାନ୍ତି ଆୟାତି, ଯେନ ସଦା ବିଶ୍ଵ ଶିଷ୍ୟତେ ପ୍ରେର୍ଯ୍ୟତେ ଚ ତତ୍‌ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଯତ୍ର = ଯେଉଁଠାରେ । ମନ୍ଦସ୍ମିତା = ଈଷତ୍ ମନ୍ଦହାସ୍ୟଯୁକ୍ତା । ସଂକ୍ରାନ୍ତଃ = ପରିବର୍ତ୍ତନ । ଶିକ୍ଷ୍ୟତେ ଶିକ୍ଷାଲାଭ କରେ । ପ୍ରେର୍ଯ୍ୟତେ = ପ୍ରେରିତ ହୋଇଥାଏ ।

ଅନୁବାଦ – ଯେଉଁଠି ବିନା ରକ୍ତପାତରେ ବିନା ଶସ୍ତ୍ରପାତରେ ଈଷତ୍ ହାସ୍ୟଯୁକ୍ତା ପରିବର୍ତ୍ତନ ଆସିଛି, ବିଶ୍ବ ଶିକ୍ଷାଲାଭ କରିଛି ଓ ପ୍ରେରିତ ହୋଇଛି, ସେହି ମୋର ଭାରତଦେଶ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ କବି ରମାକାନ୍ତଶୁକ୍ଳା ବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ସଂଗୃହୀତଃ । ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଭାରତୀୟଶାନ୍ତଃ କଥ୍ୟ ସକଳଂ ବିଶ୍ୱ ଶିକ୍ଷୟତି ତଦେବ ବର୍ଣ୍ଣିତମ୍ ।

ଭାରତଃ ଶାନ୍ତିପ୍ରିୟ ଦେଶଃ । ସର୍ବେ ଜନଃ ଅତ୍ର ଶାନ୍ତିକାମିନଃ ଭବନ୍ତି । ଅସ୍ଥିନ୍ ଦେଶେ ରକ୍ତପାତଂ ବିନା ଶସ୍ତ୍ରପାତଂ ବିନା ମନ୍ଦସ୍ମିତା ସଂକ୍ରାନ୍ତଃ ଆୟାତି, ଯେନ ବିଶ୍ଵ ସଦା ଶିଷ୍ୟତେ ପ୍ରେର୍ଯ୍ୟତେ ଚ । ମାମକଂ ତତ୍ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଏହି ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ଭାରତୀୟ ଶାନ୍ତି କିପରି ସମଗ୍ର ବିଶ୍ବ ପାଇଁ ଶିକ୍ଷଣୀୟ ହୋଇଛି ତାହା ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଭାରତ ଅତ୍ୟନ୍ତ ଶାନ୍ତିପ୍ରିୟ ଦେଶ ଅଟେ । ଏହି ଦେଶର ଲୋକମାନେ ସମଗ୍ର ବିଶ୍ବରେ ଶାନ୍ତିର ବାର୍ତ୍ତା ପ୍ରଚାର କରିଥା’ନ୍ତି । ଏଠାରେ ହିଂସାର ସ୍ଥାନ ନାହିଁ, କେହି କାହା ଉପରେ ଶସ୍ତ୍ରର ପ୍ରୟୋଗ କରେନାହିଁ । ଅହିଂସାର ମନ୍ତ୍ର ଉଚ୍ଚାରଣ କରି ଏ ଦେଶ ସ୍ଵାଧୀନତା ଲାଭକରିଛି । ବିନା ରକ୍ତପାତରେ ବିନା ଶସ୍ତ୍ରପାତରେ ଏ ଦେଶରେ ହାସ୍ୟଯୁକ୍ତା ସଂକ୍ରାନ୍ତି ଅର୍ଥାତ୍ ପରିବର୍ତ୍ତନ ସମ୍ଭବ ହୋଇଛି, ଯଦ୍ବାରା ସମଗ୍ର ବିଶ୍ବ ସର୍ବଦା ଶିକ୍ଷାଲାଭ କରିଛି ଓ ପ୍ରେରିତ ହୋଇଛି । ଏହିଭଳି ଶାନ୍ତିକାମୀ ମୋର ଦେଶ ସମଗ୍ର ବିଶ୍ବରେ ନିରନ୍ତର ଶୋଭାପାଉଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ସଂକ୍ରାନ୍ତିରାୟାତି = ସଂକ୍ରାନ୍ତଃ + ଆୟାତି ।

ସମାସ – ରକ୍ତପାତମ୍ = ରକ୍ତସ୍ୟ ପାତମ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । ଶସ୍ତ୍ରପାତମ୍ = ଶସ୍ତ୍ରସ୍ୟ ପାତମ୍ (ଷଷ୍ଠୀ ତତ୍‌ପୁରୁଷ) । = ମନଃ ସ୍ମିତଃ ଯସ୍ୟ ଡଃ (ବହୁବ୍ରୀହିଃ) ।

ସକାରଣବିଭକ୍ତି – ରକ୍ତପାତଂ, ଶସ୍ତ୍ରପାତମ୍ = ବିନା ଯୋଗେ ୨ୟା । ମନ୍ଦସ୍ମିତା, ସଂକ୍ରାନ୍ତି = କର୍ଭରି ୧ମା । ଯେନ = ଅନୁସ୍ତେ କର୍ଭରି ୩ୟା । ବିଶ୍ବମ୍ = ଉକ୍ତ କର୍ମଣି ୧ ମା । ଭୂତଳେ = ଅଧିକରଣେ ୭ମୀ ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ପାତମ୍ = ପତ୍ + କ୍ତ ।

ଶ୍ଳୋକ – ୧୯

यस्य कीर्त्तिं प्रतिष्ठां च शोभां मुदा
गायतिक्रान्तदर्शी कविनां चयः ।
यस्य वाणी- विहारोऽतुलो राजते
भूतले भाति तन्मामकं भारतम् ॥ १९॥

ଯସ୍ଯ କୀର୍ତ୍ତି ପ୍ରତିଷ୍ଠା ଚ ଶୋଭାଂ ମୁଦା
ଗାୟତି କ୍ରାନ୍ତଦର୍ଶୀ କବିନାଂ ଚୟଃ ।
ଯସ୍ୟ ବାଣୀ-ବିହାରୋଽତୁଳୋ ରାଜତେ
ଭୂତଳେ ଭାତି ତନ୍ମାମକଂ ଭାରତମ୍ ॥ ୧୯ ॥

ଅନ୍ବୟ – ଯସ୍ୟ କୀର୍ଡିଂ ପ୍ରତିଷ୍ଠା ଶୋଭାଂ ଚ କ୍ରାନ୍ତଦର୍ଶୀ କବିନାଂ ଚୟଃ ମୁଦା ଗାୟତି, ଯସ୍ୟ ବାଣୀବିହାରଃ ଅତୁନଃ ରାଜତେ, ତତ୍ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି ।

ଶବ୍ଦାର୍ଥ – ଯସ୍ୟ = ଯାହାର । କୀର୍ତିମ୍ = ଯଶକୁ । କ୍ରାନ୍ତଦର୍ଶୀ = ସର୍ବଦ୍ରଷ୍ଟା । ଚୟଃ = ସମୂହ । ମୁଦା = ଆନନ୍ଦ । ଅତୁନଃ = ଅତୁଳନୀୟ । ରାଜତେ = ଶୋଭାପାଉଛି ।

ଅନୁବାଦ – ଯାହାର କୀର୍ତ୍ତି, ପ୍ରତିଷ୍ଠା ଓ ଶୋଭା ସର୍ବଦ୍ରଷ୍ଟା କବିମାନେ ସମୂହ ଆନନ୍ଦରେ ଗାନ କରନ୍ତି, ଯାହାର ବାଣୀ ବିହାର ଅତୁଳନୀୟ ଭାବେ ଶୋଭାପାଉଛି ସେହି ମୋର ଭାରତ ପୃଥ‌ିବୀପୃଷ୍ଠରେ ଶୋଭାପାଉଛି ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ କବି ରମାକାନ୍ତଶୁକ୍ଳ ବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ମିନ୍ ଶ୍ଳୋକେ ଭାରତସ୍ୟ କୀର୍ତ୍ତି ପ୍ରତିଷ୍ଠା ଚ କଥ୍ୟ କବୟଂ ଗାୟନ୍ତ ତଦେବ ବର୍ଣ୍ଣିତମ୍ ।

ବିଶ୍ୱେଽସ୍ମିନ୍‌ ଭାରତସ୍ୟ କୀଭିଂ ପ୍ରତିଷ୍ଠା ଶୋଭାଂ ଚ ପରିଲକ୍ଷ୍ୟତେ । କ୍ରାନ୍ତଦର୍ଶୀ କବିନାଂ ସମୂହଃ ସଦୈବ ଅସ୍ୟ ଦେଶସ୍ୟ ଯଶୋଗାନଂ କୁର୍ବନ୍ତି । ଯସ୍ୟ ବାଣୀ ବିହାରଃ ଚ ଅତୁଳ ରାଜତେ ତତ୍ ମାମକଂ ଭାରତଂ ଭୂତଳେ ଭାତି । ଅତ୍ର ଯଥା ଭାରତସ୍ୟ ମହତ୍ତ୍ୱ କବିନା ବର୍ଣିତଂ ତଦେବ ସ୍ପୃହଣୀୟଂ ଗ୍ରହଣୀୟଂ ଚ ଭବତି ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଉଦ୍ଧୃତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକର କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ କବିମାନଙ୍କଦ୍ୱାରା କିପରି ଭାରତର ଯଶ ଓ ପ୍ରତିଷ୍ଠା ଗାନ କରାଯାଇଛି ତାହା ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ଭାରତର ମହାନତା ବିଶ୍ବରେ ସଦା ପରିଚିତ । ଏହାର କୀର୍ତ୍ତି, ପ୍ରତିଷ୍ଠା ଓ ଶୋଭାକୁ କ୍ରାନ୍ତଦର୍ଶୀ କବିମାନେ ଅତି ସୁନ୍ଦରଭାବେ ଗାନ କରିଅଛନ୍ତି । ଏହାର ବାଣୀ ବିହାର ଅତୁଳନୀୟ ଭାବେ ଶୋଭାପାଉଛି । ଏହିଭଳି ଯଶୋଯୁକ୍ତ ପ୍ରତିଷ୍ଠିତ ଶୋଭାୟମାନ ଆମର ଭାରତଦେଶ ସମଗ୍ର ବିଶ୍ଵରେ ବିରାଜିତ ହୋଇଅଛି ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ବିହାରୋଳୋ ରାଜତେ = ବିହାରଃ + ଅତୁଳ + ରାଜତେ । ତନ୍ମାମକମ୍ = ତତ୍ + ମାମକମ୍ ।

ସମାସ – ଅତୁଳ = ନ ତୁନଃ (ନଞ୍ଜ୍ ତତ୍‌ପୁରୁଷ) ।

ସକାରଣବିଭକ୍ତି – ଯସ୍ୟ, କବିନାମ୍ = ସମ୍ବନ୍ଧ ୬ଷ୍ଠୀ । କୀର୍ତ୍ତି, ପ୍ରତିଷ୍ଠା, ଶୋଭାମ୍ = କର୍ମଣି ୨ୟା ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଚୟ = ଚି + ଅଚ୍ ।

ଶ୍ଳୋକ – ୨୦

शोषितो नात्र कश्चिद् भवेत् केनचित्
व्याधिना पीडितो नो भवेत् कश्चन ।
नात्र कोऽपि, व्रजेद्दीनतां हीनतां
मोदतां मे सदा पावनं भारतम् ॥ २० ॥

ଶୋଷିତୋ ନାତ୍ର କଶ୍ଚିଦ୍ ଭବେତ୍ କେନଚିତ୍
ବ୍ୟାଧୁନା ପୀଡ଼ିତୋ ନୋ ଭବେତ୍ କଷ୍ଟନ ।
ନାତ୍ର କୋଽପି, ବ୍ରଜେଦ୍‌ଦୀନତାଂ ହୀନତାଂ
ମୋଦତାଂ ମେ ସଦା ପାବନଂ ଭାରତମ୍ ॥ ୨୦ ॥

ଅନ୍ୱୟ – ଅତ୍ର କଣ୍ଠିତ୍ ଶୋଷିତଃ ନ ଭବେତ୍, କେନଚିତ୍ ବ୍ୟାଧନା ପୀଡ଼ିତଃ ନ ଭବେତ୍ । ଅତ୍ର ଅପି କଣ୍ଟନ ଦୀନତାଂ ହୀନତାଂ ନ ବ୍ରଜେତ୍ । ସଦା ମେ ପାବନଂ ଭାରତଂ ମୋଦତାମ୍ ।

ଶବ୍ଦାର୍ଥ – ଅତ୍ର = ଏଠାରେ । କଶ୍ଚିତ୍ = କେହିହେଲେ । ଭବେତ୍ = ହେବା ଉଚିତ । ନ = ନୁହେଁ। ବ୍ୟାଧନା = ରୋଗଦ୍ଵାରା । ଅପି = ମଧ୍ୟ । ଦୀନତାମ୍ = ଦୈନ୍ୟକୁ । ହୀନତାମ୍ = ହୀନପଣକୁ । ବ୍ରଜେତ୍ = ଯିବା ଉଚିତ । ସଦା = ମୋଦତାମ୍ = ଆନନ୍ଦିତ । ପାବନମ୍ = ପବିତ୍ର ।

ଅନୁବାଦ – ଏଠାରେ କେହିହେଲେ ଶୋଷିତ ହେବା ଅନୁଚିତ । କେହିହେଲେ ରୋଗଦ୍ଵାରା ପୀଡ଼ିତ ହେବା ଉଚିତ ନୁହେଁ । କେବେହେଲେ କେହି ମଧ୍ୟ ଦୈନ୍ୟତା ଓ ହୀନତାକୁ ଯିବା ଉଚିତ ନୁହେଁ । ସର୍ବଦା ମୋର ପବିତ୍ର ଭାରତଦେଶ ଆନନ୍ଦମୟ ହେଉ ।

ସଂସ୍କୃତ ବ୍ୟାଖ୍ୟା – ଶ୍ଳୋକୋଽୟଂ ପଠିତଃ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ କବି ରମାକାନ୍ତଶୁକ୍ଳ ବିରଚିତଃ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟାତ୍ ଆନୀତଃ । ଅସ୍ଥିନ୍ ଶ୍ଳୋକେ ପବିତ୍ର ଭାରତସ୍ୟ ବନ୍ଦନଂ ଭବତି ।

ଅସ୍ମାକଂ ଦେଶଃ ଭାରତଃ ବିଶ୍ଵଽସ୍ମିନ୍ ସଦା ରାଜତେ । ଅତ୍ର କଣ୍ଠିତ୍ ଶୋଷିତଃ ନ ଭବେତ୍, କଣ୍ଟନ କେନଚିତ୍ ବ୍ୟାଧୁନା ପୀଡି଼ତଃ ନ ଭବେତ୍ । ଅତ୍ର କୋଽପି ଦୀନତାଂ ହୀନତାଂ ନ ବ୍ରଜେତ୍ । ସଦୈବ ମମ ପବିତ୍ର ଭାରତଂ ମୋଦତାଂ ଭବତୁ । ଏତାଦୃଶଂ ଭାରତଂ ବୟଂ କାମୟେ ।

ଓଡ଼ିଆ ବ୍ୟାଖ୍ୟା – ଶଂସିତ ଶ୍ଳୋକଟି ପଠିତ ସଂସ୍କୃତପ୍ରଭା ପୁସ୍ତକସ୍ଥ ମହାମହିମ କବି ରମାକାନ୍ତ ଶୁକ୍ଳ ବିରଚିତ ‘ଭାତି ମେ ଭାରତମ୍’ ପଦ୍ୟରୁ ଆନୀତ । ଏଠାରେ ପବିତ୍ର ଭାରତର ଆନନ୍ଦମୟତା ପ୍ରସଙ୍ଗ ବର୍ଣ୍ଣିତ ହୋଇଛି ।

ସତରେ ଭାରତ ଏକ ପବିତ୍ର ଆନନ୍ଦମୟ ଦେଶ ଅଟେ । ସମଗ୍ର ବିଶ୍ବରେ ଏ ଦେଶ ଆଦର୍ଶଭୂତ ଅଟେ । ଏ ଦେଶରେ କେହି ଶୋଷିତ ନାହିଁ, କେହିହେଲେ କେବେ ରୋଗରେ ପୀଡ଼ିତ ହୁଏ ନାହିଁ । ଏ ଦେଶରେ କେହି ଦୀନହୀନ ଦୁଃଖୀରକି ଦରିଦ୍ର ନାହିଁ । ସଦାସର୍ବଦା ଆନନ୍ଦରେ ଏ ଦେଶରେ ସମସ୍ତେ ବସବାସ କରିଥା’ନ୍ତି । କାରଣ ଏ ଦେଶ ହେଉଛି ପବିତ୍ର ଆନନ୍ଦମୟ ଦେଶ । ଏଭଳି ଦେଶବାସୀ ହୋଇ ଆମେ ଧନ୍ୟ ।

ବ୍ୟାକରଣ:

ସନ୍ଧିବିଚ୍ଛେଦ – ନାତ୍ର = ନ + ଅତ୍ର । କଶ୍ଚିଦ୍‌ଭବେତ୍ = କଃ + ଚିତ୍ + ଭବେତ୍ । କଣ୍ଟନ = କଃ + ଚନ । କୋଽପି = କଃ + ଅପି । ବ୍ରଜେଦ୍‌ଦୀନତାମ୍ = ବ୍ରଜେତ୍ + ଦୀନତାମ୍ ।

ସକାରଣବିଭକ୍ତି – କଃ = କଉଁରି ୧ ମା । ବ୍ୟାଧୁନା = କରଣେ ୩ୟା । ଦୀନତାଂ, ହୀନତାମ୍ = କର୍ମଣି ୨ୟା ।

ପ୍ରକୃତିପ୍ରତ୍ୟୟ – ଶୋଷିତଃ = ଶୋଷ୍ + ଣିଚ୍ + କ୍ତ । ପୀଡ଼ିତଃ = ପୀଡ୍ + କ୍ତ । ପାବନମ୍ = ପୂ + ଲ୍ୟୁଟ୍ ।

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(a)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(a)

Question 1.
\(\lim _{x \rightarrow 3}\)(x + 4)
Solution:
Clearly, if we take x very close to 3, x + 4 will go very close to 7.
Now let us use ε – δ technique to confirm the result.
Given ε > 0, we seek for δ > 0 depending on ε such that
|x – 3| < δ ⇒ |(x + 4) – 7|< ε
Now |(x + 4) – 7| < ε
if |x – 3| < ε
∴ We can choose ε = 8
Hence for given ε > 0, there exist 8 = ε > 0
such that |x – 3| < δ ⇒ |(x + 4) – 7| < ε
∴ \(\lim _{x \rightarrow 3}\)(x + 4) = 7

Question 2.
\(\lim _{x \rightarrow 1}\)(4x – 1)
Solution:
By taking very close to 1 we have 4x- 1 tends to 3.
Let us use ε – δ technique to confirm the result.
Given ε > 0. We shall find δ > 0 depending on ε such that
|x – 1| < 5 ⇒ |(4x – 1) – 3| < ε
Now |(4x – 1 ) – 3| < ε
if |4x – 1| < ε i.e.|x – 1| < \(\frac{\varepsilon}{4}\)
Let us choose δ = \(\frac{\varepsilon}{4}\)
∴ For given ε > 0 there exists δ = \(\frac{\varepsilon}{4}\) > 0
such that |x – 1| < δ
⇒ |(4x – 1) – 3| < ε
∴ \(\lim _{x \rightarrow 1}\)(4x – 1) = 3

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(a)

Question 3.
\(\lim _{x \rightarrow 1}\)(√x + 3)
Solution:
As x → 1 we see √x + 3 → 4
We will confirm the result using ε – δ technique
Let ε > 0, we will choose δ > 4
such that |x – 1| < 8 ⇒ |√x + 3 – 4| < ε
Now |√x + 3 – 4| = |√x – 1|
\(=\frac{|x-1|}{|\sqrt{x}+1|}\)
But |√x + 1| > 1
⇒ \(\frac{1}{|\sqrt{x}+1|}\) < 1
⇒ \(\frac{|x-1|}{|\sqrt{x}+1|}<\frac{\delta}{1}\)
∴ (√x + 3) – 4 < \(\frac{\delta}{1}\)
We can take δ < ε i.e. δ = min {1, ε}
∴ |x – 1| < δ ⇒ |(√x + 3) – 4| < ε
for given ε > 0 and (δ = ε)
⇒ \(\lim _{x \rightarrow 1}\)(√x + 3) = 4

Question 4.
\(\lim _{x \rightarrow 0}\) (x2 + 3)
Solution:
As x → 0 we observe that x3 + 3 → 3
Let us use ε – δ technique to confirm the result.
Let ε > 0, we seek for a δ > 0 such that
|x – 0| < ε ⇒ |x2 + 3 – 3| < ε
Let |x| < 8
Now |x2 + 3 – 3| < ε
We have |x|2 < ε ⇒| x| < √ε
(∴ |x| and ε are positive.)
∴ we can choose δ = √ε
∴ We have for given δ > 0 there exists
δ = √ε > 0 such that |x| < δ ⇒ |x2 + 3 – 3| < ε
∴ \(\lim _{x \rightarrow 0}\) (x2 + 3) = 3

Question 5.
\(\lim _{x \rightarrow 0}\) 7
Solution:
If x → 0 we observe that 7 → 7.
Let us use e- 8 technique to confirm the limit.
Let f(x) = 7
Given ε > 0, we will choose a δ > 0
such that |x – 0| < δ ⇒ |f(x) – 7| < ε
Now |f(x) – 7| < ε
If f(x) ∈ (7 – ε . 7 + ε)
But for every x, f(x) = 7
⇒ for|x| < δ also f(x) = 7 ∈ (7 – ε . 7 + ε)
∴ Choosing ε = δ we have
|x| < δ ⇒ |f(x) – 7| < ε
∴ \(\lim _{x \rightarrow 0}\) (7) = 7

Question 6.
\(\lim _{x \rightarrow 1} \frac{(x-1)^3}{(x-1)^3}\)
Solution:
We guess the limit is 1.
Let us confirm using ε – δ technique.
Let ε > 0, f(x) = \(\frac{(x-1)^3}{(x-1)^3}\)
We will choose a δ > 0 such that
|x – 1| < δ ⇒ |f(x) – 1)| < ε
Now |f(x) – 1| < ε
if 1 – ε < f(x) < 1 + ε
∴ We will choose a δ > 0 such that
x ∈ (1 – δ, 1 + δ) – { 1 }
⇒ f(x) ∈ ( 1 – ε, 1 + ε)
As f(x) = for x ≠ 1
We have f(x) ∈ (1 – ε. 1 + ε) for all x ∈ (1 – δ, 1 + δ) – [1]
∴ We can choose δ = ε
for given ε > 0, there exists δ = ε
s.t. |x – 1| < δ ⇒ |f(x) – 1| < ε
∴ \(\lim _{x \rightarrow 1}\) f(x) = 1

Question 7.
\(\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}\)
Solution:
If we take x very close to 3 (≠ 3)
we have \(\frac{x^3-9}{x-3}\)
= \(\frac{(x-3)\left(x^2+3 x+3^2\right)}{2}\) → 27
Let ε > 0 and x ≠ 3
Now |\(\frac{x^3-9}{x-3}\) – 27| = |x2 + 3x +9 – 27|
=|x2 – 9 + 3(x – 3)| ≤ |x2 – 9| + 3|x – 3|
= |x – 3| [|x + 3| + 3] ≤ |x – 3| [|x + 6| < |x – 3| [|x – 3 + 9|]]
If |x – 3| < δ and δ < 1 then |x – 3| [x – 3 + 9| < δ {1 + 9} = 10 δ
Let δ = min {1, \(\frac{\varepsilon}{10}\)}
∴ For given ε > 0 we have a δ = min {1, \(\frac{\varepsilon}{10}\)} >0 such that
|x – 3| < δ ⇒ |\(\frac{x^3-9}{x-3}\) – 27|
∴ \(\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}\) = 27

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(a)

Question 8.
\(\lim _{x \rightarrow 1} \frac{3 x+2}{2 x+3}\)
Solution:
we observe that as x → 1, \(\frac{x+2}{2 x+3}\) → 1
To establish this let ε > 0,
we seek a δ > 0,
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(a)

Question 9.
\(\lim _{x \rightarrow 0}|x|\)
Solution:
We see that when x → 0,|x| → 0
Let us establish this using ε – δ technique.
Let ε > 0 we seek a δ > 0 depending on
ε s.t.|x – 0| < ε ⇒ ||x| – 0| < ε
Now ||xl – 0| = ||x|| = |x| < δ
By choosing ε = δ we have |x| < ε ⇒ ||x| – 0| < ε
∴ \(\lim _{x \rightarrow 0}|x|\) = 0

Question 10.
\(\lim _{x \rightarrow 2}(|x|+3)\)
Solution:
We see that as x → 2, |x| + 3 → 5
Let ε > 0 we were searching for a, δ > 0
such that |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
Now ||x|| + 3 – 5| = ||x| – 2| < |x – 2| < δ
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ \(\lim _{x \rightarrow 2}(|x|+3)\) = 5

CHSE Odisha Class 11 English Writing Description

Odisha State Board CHSE Odisha Class 11 Invitation to English 3 Solutions Writing Description Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Writing Description

Description

Descriptions are of two types: factual (or realistic) and impressionistic. For a factual or realistic description, you will have to go looking for descriptions of objects or processes. An impressionistic description, on the other hand, maybe factual in places but it is chiefly concerned with recording the impression produced by the describer of a person, place or object. Above all, there is the description of a person.
Some common words used to describe a person
Words that go with hair: long, wavy, curly, brown, dark
eyes : pale, blue, black, flashing
nose : long, high, fleshy
lips : full, thin
shoulder : broad, drooping
age : mid-thirties
voice : commanding
Important points in connection with the description of the person :
A person’s height (tall/short), lean or fat, age, physical appearance, the shape of the face with its prominent features* eyes (bright/dull/dark), hair, forehead, dress and nature.

CHSE Odisha Class 11 English Writing Description

Activity 5

Can you replace them with more favourable descriptions

Read passage C again. Note that there are some points of description which are not very favourable to the person being described. Can you replace them with more favourable descriptions? Reorganise the passage, starting with the sentence: “She was fair and her face was round.”
Answer:
She was fair and her face was round. The prominent feature of her face was its sparkle. She was tall and full of youth. She was lean. She had thick hair and her forehead was not that broad. Her nose was long and her cheeks chubby. She was elegantly dressed. She was brimming with confidence.

Activity 6

You met the people in the picture at a party. Describe them to your friend in a letter. (Have a close look at the picture first.)

You met the people in the picture at a party. Describe them to your friend in a letter

Answer:

Satyanagar, Plot No. 121
1st January 20

Dear Sarika,
Yesterday evening I had been to a great party. There was plenty to eat, and numerous games to play but the highlight of it all was the music and dance with which the party closed. Besides the music, we had a dance competition in which the best couple was chosen. The dance went on for an hour. Mr and Mrs Das were adjudged the best couple. They were a perfect match for each other. Mr Das was dressed in a black suit, white shirt and a black bow. He was tall, dark and handsome and he held himself elegantly as he held his wife.

His jet black hair, thick moustache and long sharp nose showed forth a man of character. Mrs Das just a shade shorter than her husband was a cute-looking lady. Her hair was tied in a chignon, her eyebrows were shaped like thin orange pieces. She too had a sharp nose, and a face as clear as glass. She was fair, slim and tall. She sported a short sleeveless gown and a plain slip-on. She also held a dupatta in her hand. She looked really chick and beautiful. They deserved the prize. Well, the party was over at around 4.00 a.m. in the morning. I really enjoyed the party even though it was exhausting.

Please write back.
Yours sincerely,
Rita.

CHSE Odisha Class 11 English Writing Description

Activity 7

Describe the person in the picture below as interestingly as you can.
Answer:
He wore a long full-sleeved magenta gown and long loose magenta trousers to go along with it. The gown opened onto a large star-shaped neck which was shining black with confetti pasted onto it. His face he had put on a mask with his face farded with cosmetic paint. He had a huge bulbous nose painted in red, large lips painted again in red. He wore no shoes but grey socks which looked extremely funny. In his hand, he held a conical hat which had feathers on its top.

Activity 8

Here is a short dialogue between two friends. They are talking about a mutual friend whose name is Prabhakar. Read the dialogue carefully and write a short description of Prabhakar.
Asaf : You remember Prabhakar? He has become a doctor. He is doing very well, in fact.
Krishna : Prabhakar? The name is familiar, but I don’t remember who you are talking about. What did he look like?
Asaf : He was that short chap with a shining pink face. Always dressed in white. He had long hair, like a girl’s. We used to call him Prabhavati, and how he used to blush then!
Answer:
Prabhakar is a short chap with a shining pink face. He sported long hair like a girl and was always clad in white.

Activity 9

(a) A stranger visited your house during your father’s absence. You received him and talked to him. When your father returned, he wanted to know if any visitor had come while he was away. Describe the visitor to your father so that he can know who you are referring to.
(b) Your mother is looking for a bride for your elder brother. You have seen a girl who, in your opinion, will be ideal for your brother. Describe her to your mother.
Answer:
(a) Dad, this man was around six feet high and darkly complexioned. He had peculiar hairstyle which was parted right in the middle like a girl’s. He also wore ear¬rings in both ears and four stone rings of different colours on his right hand. Two of his toes on the right also had rings in them. He was carrying a shoulder bag and putting on a white dhoti as well as white kameez. There was a long tilak on his forehead. He spoke in the Sambalpuri dialect and he is perhaps in his thirties. I hope you recognise him.

(b) She is very fair and has eyes shaped like petals of a rose, eyes-lashes thin and long like leaves of the touch-me-not. Her hair is thick and black, falling down even below the knees. Her face is spotlessly clean and has a soft look. She is twenty-four but looks like she is in her teens. There is a black birthmark on her chin which adds to the beauty of her face. Her sharp nose and jaw looked as if it was sculpted like a Grecian statue. Her height is about five-feet and she is slim. On the whole, she looked like a model.

CHSE Odisha Class 11 English Writing Description

Activity 10

Here is a picture of a rhinoceros. Write a short description of the animal for a friend who has not seen a rhinoceros.

Here is a picture of a rhinoceros. Write a short description of the animal for a friend who has not seen a rhinoceros.
Answer:
The very look of it is ferocious. It is a huge mammal, 4 feet high and perhaps 5 feet long, that is bigger than a cow and smaller than an adult elephant. Just like a cow, it has a long face with the snout protruding forward, underneath which is a large mouth. The lower jaw looks like a concave plateau. Its most remarkable features are two 1 horns protruding right out from the snout. Moreover, it has three toes on each of its legs. Its skin is so thick that they look like shielding pad. Besides this, it has a short tail. It is found mostly in Asia and Africa and it is a herbivore.

Activity 11

Here is the description of a particular dog. All the details are present, but not in order. Rewrite the description. Begin with the general appearance and size, then describe the features of the animal, which you find most striking.
(a) He has huge paws, with joined fingers and retractable claws.
(b) Achilles isn’t an ordinary dog.
(c) But the most incredible characteristic is his face, which looks sad and solemn.
(d) Firstly, he is larger than any dog I have ever seen, and he is more like a wolf.
(e) It seems as if he can almost speak if he is given the chance.
Answer:
Achilles isn’t an ordinary dog. He is larger than any normal dog and looks more like a wolf. He has huge paws, with jointed fingers and retractable claws. But the most incredible characteristic is his face, which looks solemn and sad. It seems as if he can speak, if he is given a chance.

Activity 12

Write short descriptions of the following animals. A few questions are given to help you organise your descriptions.

(a) A giraffe How tall is it? What makes it look so tall? Where is it found? What does it eat and how? How does it fight other giraffes and enemies? What kind of sound does it make?
(b) A tiger Where is it found? How tall/long/heavy is it? What is its colour? What is its food? What are man-eaters? How long does a tiger live (life span)? How do tiger cubs look?

Answer:
(a) A giraffe :
Excepting the now extinct Dinosaurs, the giraffe is the tallest mammal found originally on the African continent. Its most characteristic feature is its long neck that protrudes out angularly from its body and is usually about 4 feet to 6 feet long. Besides, just like the leopard, it has black spots spread all over its body which is white-skinned. It is due to this that it was formerly called a came lopard. The giraffe is herbivorous and lives on the leaves of plants which it can easily reach due to its height. It has a very long stride and therefore it is difficult for any preying animal to catch it.

(b) A tiger :
The tiger is a savage and cruel animal. We say, “As cruel as a tiger”. It is really a great cat. It is a large, strong, and fine-looking animal. Its hair is yellow, marked with black stripes. It is shaped like a cat, with a long tail, round head, and thickly padded feet. It has sharp claws and strong teeth. The tiger is an Indian animal. There are many in the jungles of Bengal. Like cats, tigers hunt at night. They kill big animals, like deer, cows, sheep and goats. They attack men, too. Some become “man-eaters”. They like men’s flesh best to eat. Tigers are feared by farmers. They come at night to steal their sheep and cows. Its average lifespan is 17 years and its cubs look like domestic cats with shining black eyes.

CHSE Odisha Class 11 English Writing Description

Activity 13

Rewrite passage (c), using your own words as far as possible. Divide your description into two paragraphs.

A telephone comes with a bell which can ring, a microphone which converts human speech into electrical signals, an earphone which converts incoming electrical signals back into speech, and a dial which is used to send electrical pulses along the line to an automatic exchange.
Answer:
When we want to make a call, we must lift the handset and then dial the number we want to call. Immediately after we dial the number an automatic selector connects us to an outgoing junction cable that is linked to the exchange we want. A ringing bell indicates the fact that a telephone has come. Then the operation starts. The exchange first connects our phone to automatic selector equipment which in turn connects us to an outgoing cable linked to the exchange that is connected to the number we rang. Finally, this exchange connects us to the phone we are interested in.

Activity 14

Add a short paragraph to passage (d). The hints below will help you in writing the paragraph.

Butter is a rich food made from the cream of milk. It is usually eaten as a spread on bread, but cooks may use butter for frying and making cakes and pastries. Butter contains about 80 per cent fat, the remainder being water, salt and protein. Butter is made from cream, by churning the cream so that the fat [ is separated out. For many centuries, farmers have made butter from cream by churning it by hand in a wooden vat. Nowadays, however, butter is made by machines. First, the milk is whirled in a centrifuge to separate the cream. The l cream is then pasteurised by heating it and then cooling it quickly. This action kills germs in the cream and prevents the butter from going rancid quickly. The pasteurised cream is then churned in huge revolving drums, which separate fat from the liquid in the cream. When the liquid, called the buttermilk, is ‘ drained away, the resulting mass of butter is then cut into pieces and packed.
Hints : How does butter feel when you touch it? Is it tasty to eat? / Is it expensive? j How is butter used in India? etc.
Answer:
Butter is hard but smooth to the touch because it is kept mostly in the refrigerator. Of course, it is tasty but expensive. A mere hundred grams cost thirteen rupees. It is mostly used in cakes and as a spread on bread sandwiches. Sometimes it is used to fry almonds and cashew nuts. However, its use is limited and common people seldom buy it.

Activity 15

Here is a conversation in which an uncle describes a saw to his nephew. Read through the conversation and write a paragraph describing a saw.

Boy : What is a saw, uncle?
Uncle: It’s something we use to cut a piece of wood into two.
Boy : You mean it’s a sort of axe, uncle?
Uncle: No, not an axe, This one has a thinner blade and a short ring-like handle of wood.
Boy : Oh, I know what it is. It’s like a sword.
Uncle: Not really. A saw has one edge sharp. The other edge does not cut.
Boy : Like a big knife?
Uncle: Partly, but the sharp edge does not cut like a knife. There are teeth on the sharp edge. When you press the blade against the wood and move it forward and backwards, like the bow of a violin, the wood gets cut along that line. There are big saws, too, which two people hold at either end to make cuts along the whole length of a log.
Answer:
A saw is an instrument made of either iron or steel with curved teeth on one side or both. The blade of a saw is thin but strong. There is a ring-like wooden handle fitted to one end of the instrument held during sawing wood or wood planks. When it is pressed against the wood, the wood gets cut along that line. Nowadays, the saw has developed a lot from its crude form to a sophisticated one.

CHSE Odisha Class 11 English Writing Description

Activity 16

Write a short description of the following objects :
(a) A football
Hints:
(i) Size, shape and colour
(ii) Is it smooth, rough or soft to the touch? Is it hard? Is it light or heavy?
(iii) How does it smell?
(iv) Does it bounce? How high?

(b) A ripe mango
Hints:
(i) Size, shape, colour and smell –
(ii) How does it feel to the touch?
(iii) What happens when you press it hard?
(iv) How does it taste?

(c) A pressure cooker
Hints:
(i) What is it? (Definition)
(ii) What does it look like? (parts, size, make, etc.)

Answer:
(a) A football :
A football is a spherical object made of hexagonal leather pieces of alternately white and black colour or plain grey. It has a bladder inside which is inflated by air to give it a round shape. The leather is smooth and soft in the evening but gradually becomes rough because of wear and tear as a result of frequent use. It is light when inflated and bounces up to a height of 15 feet to 20 feet depending on how much it has inflated and how hard it is hit.

(b) A ripe mango :
It is a tropical fruit which consists of a hard kernel, a central core around which is a fleshy pulp. It is yellowish-red in colour and in ovalish in shape. The mango smells sweet and is soft and smooth to the touch; when pressed hard the outer pulp along with the juice comes out. It has a very sweet taste to it.

(c) A pressure cooker :
A pressure cooker is a vessel in which food is cooked in steam under pressure. It consists of a very strong vessel, made of aluminium alloy with a lid that fits tightly on the top. The lid can be sealed onto the vessel by means of a rubber ring. At the centre of the lid, there is a vent or hole through which steam can escape. The food to be cooked is placed in a smaller vessel inside the cooker and a little water is poured into the outer vessel.

Water boils in the vessel and steam begins to escape through the vent. Then we stop the steam by placing a weight on the vent. Steam pressure inside increases and the temperature rises. So the food gets cooked at a higher temperature. This takes only one-third of the time taken by the ordinary method.

CHSE Odisha Class 11 English Writing Description

Activity 17

Describe the following objects :
(a) A bicycle
(b) A teapot
(c) A fountain pen
(d) A gold necklace

Answer:
(a) A bicycle :
A bicycle is the cheapest and simplest form of transport on wheels. It consists of the main frame and a secondary frame both joined together and triangular in shape. The main frame has a head tube in its front. The handle of the bicycle protrudes out from the upper end of the head tube while the fork protrudes out from the lower end of the head tube. An inner bolt holds both the handle and the fork in place. At the lower end of the fork is the front axle which holds the wheel.

At the opposite end of the upper end of the head, the tube is a tube that protrudes out of the hollow of the main frame. This has a nut-bolt arrangement to hold the seat. Similarly, the peak of the triangular main frame has a hole and axle arrangement to which the crank is connected. The secondary frame has a seat stay which serves as a support for the weight on the seat. The upper end of the seat stay is joined to the main frame while the lower end forks out into two legs which hold the rear wheel in place.

The wheels consist of a central spoke holder from which spokes radiate out into the rim of the wheel where it is screwed. The spokes keep the rim in shape and support it. Besides this, the wheel has an inflatable tube and an outer tyre. The tube has a valve through which air is pumped into it. This valve emerges on to the outer side of the rim through a hole in it. The crank is held by the main frame while the rear frame holds a sprocket wheel. A chain extends from the crank and is wound around the sprocket wheel. The chain is fitted onto them and locked.

The crank further has two pedals joined to it. When force is applied to the pedals, the crank turns and this chain transmits this force applied to the sprocket wheel which is attached to the rear wheel, thereby moving it. Consequently, the cycle moves. To facilitate proper control of the bike, there are brakes. Brake levers are attached to the handle and have brake brackets with rubber on them which are fitted close to the rear and front wheels. Besides this mudguards are provided for both wheels. Finally, a bell and a stand complete the bike. The stand serves as a prop to keep the bicycle standing.

(b) A teapot:
A teapot is a vessel used to brew tea. It can be of various shapes and sizes but most often it is cylindrical in shape with a hollow inside. It is open at one end and this top end has a lid which can be closed or opened as required. The lid is attached to the body of the teapot. The teapot also has a snout with an opening in it from which brewed tea is poured. In teapots of other kinds, instead of a snout, there is a long neck with a mouth at one from which the tea is to be poured. The teapot can be made of various materials like clay, bone china, wood, steel or copper.

(c) A fountain pen:
A fountain pen has two parts. A hollow cylindrical lower part two inches in diameter and a nib holder that is screwed onto it. Besides this, it has a cap with which to cover the nib and protect the ink from spreading. The lower cylindrical half of the pen is filled with ink. This ink passes to the nib which has a sharp pointed end to which it drips. It is with this pointed end of the nib that one writes.

(d) A gold necklace :
Gold is the most precious of all metals. Its bright yellow colour is very pretty. It takes fine polish. Gold is used to make many ornaments out of which a necklace is one. The goldsmith artistically makes it. It is of different sizes and designs. Each gold necklace has a beauty of its own. It is studded with rare stones and diamonds. This necklace is made by hand as well as by machine.

CHSE Odisha Class 11 English Writing Description

Activity 18

Your father has bought the item in the picture for you. Write a letter to your sister describing what it is, how it looks and what you are going to do with it.

Your father has bought the item in the picture for you. Write a letter to your sister describing what it is, how it looks and what you are going to do with it.

Answer:

Shanti Vihar
7 August 20

Dear sister,
Dad presented me with a very useful gift. I really needed it. Well, don’t hazard any guesses because you might think of the wrong thing. It is a wall clock. The clock is a fairly big one. It has a huge round dial which is fixed onto a round plastic case. This in turn is encrusted into a squarish plastic body. The glass on the dial case is spotlessly clean. One significant feature of the watch is its radium-coated hands.

These shine in the darkness and allow me to know the time even though the lights are off. Besides this, the clock has an alarm system. This serves the purpose of waking up a lazy boy like me. Nowadays I use the alarm to wake up at 5.00 a.m. to go out jogging. That is a thoughtful present from dad, isn’t it? I am writing to him separately to thank him but do tell him how useful it is to me. Thanks for sending cakes and biscuits through your classmate Suneeta.

Your loving brother,
Sushil.

Activity 19

Read through the following paragraph and answer the following questions.

Coal mining-digging coal out of the earth – is a very big industry. Some coal is mined on the surface, but most of it has to be mined deep underground. Both forms of mining are now highly mechanised. On or near the surface, coal is mined by the open-cast method. Huge power shovels first strip off the earth’s overburden above the coal seam. Then the coal is broken up by explosives and shovelled into trucks.

Underground mining is more complicated, more expensive, and more dangerous. Shafts are sunk down into the earth and tunnels are struck outwards from the shafts to the coal seams. Then a machine, called a continuous miner, rips coal from the mine face and loads it onto a conveyor belt, which carries the coal up.

Now answer the following questions :

(i) What is the paragraph about?
(iii) The sentences below give us a simple description of the process of surface ( mining, but they are not in order. Rewrite them in the proper order and use the connectives “first,” “second”, “third” and “finally”. “The coal is thus broken up. Explosives are detonated. The earth above the coal seam is removed. It is loaded into the trucks”.
(iii) There are certain steps involved in underground mining. Write down the steps in proper order. The first one is given to you as an example.
a. Shafts are sunk down in the ground.
b. _______________________
c. Coal is __________________
d. The coal is _______________
e. Then it is ________________

Answer:
(i) The paragraph defines coal mining and enumerates the two kinds of mining. The paragraph also describes the open-cast method of mining.
(ii) Firstly the earth above the coal seam is removed. Secondly, explosives are detonated. Thirdly, the coal is thus broken up. Finally, it is loaded into trucks.
(iii) (b) Tunnels are struck outward from the shafts to the coal seam.
(c) Coal is then ripped from the mine face by a machine called a continuous miner.
(d) The coal is broken up by explosives and shovelled into trucks.
(e) Then it is loaded onto a conveyor belt which carries the coal up.

CHSE Odisha Class 11 English Writing Description

Activity 20

Here is a description of an experiment demonstrating the process of photosynthesis in green leaves. Read it carefully and note the steps in the experiment.

Two leaves are removed from a de-starched plant. The upper side of one and the lower side of the other are greased with vaseline. The stalk of ‘ each leaf is dipped in water and the leaves are left in the light for four hours so that photosynthesis takes place. Most of the vaseline is wiped off and the leaves are placed in a solution of potassium iodide. The leaf greased on the upper side develops a blue colour, showing that starch has formed by photosynthesis from carbon dioxide, which entered through the leaf pores which are mainly on the underside. No colour develops in the other leaf in which vaseline blocked the pores.

Have you understood the steps involved in the experiment? Can you now help your younger sister conduct this experiment? For this, you may have to give her instructions and let her do the experiment. Give her instructions step by step. You may proceed like this :

1. Take two leaves from a de-starched plant.
2. Grease one leaf on the upper side.
3. _______________________
Continue the instructions till the experiment is over.
Answer:
1. Take two leaves from a de-starched plant
2. Grease one leaf on the upper-side
3. Grease the other on the lower side
4. Dip both leaves in water
5. Then leave it under light for four hours so that photosynthesis takes place.
6. After this wipe off vaseline from the leaves and place it in a solution of potassium iodide.
7. You will now notice that the leaf greased on the upper side develops a blue colour.
8. This shows that starch has formed photosynthesis from carbon dioxide which entered through the leaf pores on the underside of the leaf.
9. No colour develops on the leaf in which vaseline blocked the pores. Structural items used in the passages. We use technical writing while describing the processes etc. These technical writings are commonly impersonal and formal. In this type, the action referred to is more important than the doer of that action. Hence, we express this importance using active voice. Try to fill in the blanks.

1. Coal is mined ___________.
2. Both forms ________ are now highly mechanised.
3. Then the coal is broken up ________ and shovelled
4. Shafts are sunk ________ and tunnels are stuck ________.
5. Two leaves are removed ___________.
6. ________ are greased with vaseline
7. Stalks are dipped __________.
8. Leaves are left ____________.

Answer:
1. Coal is mined on the surface. but most of it has to be mined deep underground.
2. Both forms of mining are now highly mechanised.
3. Then the coal is broken up by explosives and shovelled into trucks.
4. Shafts are sunk down into the earth and tunnels are stuck outwards from the shafts to the coal seams.
5. Two leaves are removed from a de-starched plant.
6. The upper side of one and the lower side of the other are greased with vaseline.
7. Stalks are dipped in water.
8. Leaves are left in light for four hours.

Activity 21

Here is a set of instructions for an experiment on transpiration in plants. Rewrite the description in the passive voice. Select a potted plant and water it sufficiently before the experiment. Cover the soil surface by means of oil paper and check the ordinary evaporation of water. Put the pot on the workbench of the laboratory and cover it with a bell jar. Allow the experimental set-up to continue for one hour. Observe that drops of water stick to the inner wall of the bell jar.
Hint: A plotted plant is selected and it is watered sufficiently before the experiment. (Continue)
Answer:
A potted plant is selected and watered sufficiently before the experiment. Then its soil surface is covered by an oil paper to check ordinary evaporation of water. After this, the pot is put on the workbench of the laboratory and covered with a bell- jar. This experimental setup is allowed to continue for an hour. It is observed that drops of water stick to the inner wall of the bell jar.

Activity 22

The following sentences are from a passage, which tells us about the ideal temperature necessary for the growth of plants. But the sentences are not in order. Put them in order.
1. At lower temperatures the activity of enzymes is reduced; therefore, the growth is also retarded.
2. Most plants grow well between 20-30 degrees centigrade, which may be called the optimum temperature range.
3. The effect of temperature on growth may be indirectly related to the activity of enzymes.
4. But some plants grow well at temperatures lower than 20° C, while other plants grow best at temperatures higher than 30° C.
5. At higher temperatures, the activity of the enzymes in the plant is considerably increased, leading to a kind of ‘exhaustion’ of the plant. Beyond 40° C, the enzymes themselves are destroyed.
Answer:
Most plants grow well between 20°-30° centigrade, which may be called the optimum temperature range. But some plants grow well at temperatures lower than 20° C while other plants grow best at temperatures higher than 30° C. At higher temperatures, the activity of enzymes in the plant is considerably increased, leading to a kind of ‘exhaustion’ of the plant. Beyond 40° C, the enzymes themselves are destroyed. At lower temperatures the activity of enzymes is reduced; therefore, growth is also retarded. Thus the effect of temperature on growth may be indirectly related to the activity of enzymes.

CHSE Odisha Class 11 English Writing Description

Activity 23

Given below is a diagram which describes how water mixed with solid substances or impurities is distilled. Write a description of the process of distillation.

Given below is a diagram which describes how water mixed with solid substances or impurities is distilled

Answer:
The water mixed with solid substances or impurities is first put into a round-bottomed flask. This flask is then placed on a tripod stand. A rectangular glass tube is then put into the flask through the hole in the cork covering the flask. The tube must reach down to the depth of the water level in the flask. The other hand of the tube must be kept under an empty glass beaker. After this, the flask is to be heated by a glass flame. As the flask is heated, it gradually reaches boiling point and water starts turning into water vapour.

This steam passes through the rectangular tube. As it passes through the tube, the water vapour condenses and droplets of water start falling into the glass beaker. They quicken the process of condensation and the rectangular tube can be attached to a condenser tube through which cold water passes. As all the water in the flask evaporates, the impurities or the solid substance will remain behind in the flask and pure water will be deposited in the beaker.

(a) In ironing a shirt, you first press the cuffs and the sleeves. You then press the collar, inside and outside. After that you ……………….
Answer:
While ironing a shirt, first we press the cuffs and then the sleeves. We then press the collar on both sides with the iron. Then we can press the front part and then | the back ……………….

CHSE Odisha Class 11 English Writing Description

Activity 24

Describe the following simple processes.
(a) how to make a glass of lassi
(b) how to make a booking for a berth in a reserved compartment (on a train)
(c) how to clean and polish your shoes
(d) how to cook rice
(e) how to send a letter by registered post

Answer:
(a) how to make a glass of lassi:
1. First, take the required amount of curd and milk.
2. Pour them into the jar of the mixer.
3. Add ice cubes and sugar (to your own taste) to it.
4. Then churn and grind it in the mixer using the whipper till the mixture of curd, milk sugar and ice, foam is.
5. Pour it back into a glass.
6. Top ft with garnished coconut, cream, ground cashew nut, bournvita powder or cocoa powder and dried grapes.
7. The lassi is now ready.

(b) how to make a booking for a berth in a reserved compartment (on a train):
1. Procure a reservation slip/form from the reservation counter at the railway station or city booking centres.
2. Fill in the form giving details of the train you want to travel, the class you want to travel to, the date of your journey, your name, age, sex and preference for lower, middle or upper berth.
3. Then give this reservation slip/form to the reservation clerk.
4. The clerk will then check on his computer to find out whether a berth is available on the train and on the particular date you asked for.
5. After finding the availability if it is available, he will print the details on the ticket and pass it to you asking you the fee for it.
6. If the clerk finds that no berth is available, he will tell you what other options are available and you can fill out a new reservation slip with the options available and thereby start the whole process again.

(c) how to clean and polish your shoes:
1. First bring a cherry blossom or a polishing cream (white or black).
2. Use a soft polishing brush for cleaning the dust and dirt.
3. Apply the cherry cream on the brush.
4. Polish the shoes slowly and continuously for some time so as to give them a shining colour.
5. Then apply the cream for a better glaze on the shoes.

(d) how to cook rice:
1. Clean the rice off stones, chaff and burnt rice.
2. Then clean it with water.
3. After this take water that is twice the volume of rice you have taken and set it to boil on the stove in a pot or vessel.
4. When you notice the water boiling, pour the rice into it.
5. Keep it over the fire till the grains of rice become soft.
6. Then drain the water from the pot, so that the cooked rice is left behind ready for consumption.

(e) how to send a letter by registered post:
1. Procure an envelope of the size required by you from the stationery shop.
2. Put the letter inside it and seal it with gum or cello tape.
3. Then write the name and address of the person you want to send it to on the right-hand side of the envelope. Add your name and address to the envelope in the left bottom corner.
4. Take it to the post office and hand it to the registration clerk. He will weigh it and tell you how much stamp it requires. Buy the required amount of stamp from him and paste it on the envelope.
5. Then hand it back to him. He will enter it in a registration journal, write the registration number and date on the envelope and put his Initials on it. He will then hand you a receipt for the letter he received from you.
6. The registration work is done.

Read the following description of a hill station.

(a) Ootacamund, or Ooty (as it is popularly known), which nestles in the Nilgiri Hills, lies on the borders of Kerala, Tamil Nadu and Karnataka. Tourists from both home and abroad flock to this beautiful little hill station for a holiday. The most prominent attraction for them is the Botanical Garden, which was established in 1847. A variety of exotic and ornamental plants adorn this garden. The chief attraction of the garden is a fossil tree trunk which is 20 million years old. A small lake runs through the garden. The government organises in this garden a flower festival in May every year.

(i) What is Ootacamund’s other name? Where is it situated?
(ii) What is its main attraction?
(iii) Where is the lake?
Answer:
(i) Ootacamund’s other name is Ooty. It is situated in the Nilgiri Hills which . “ lies on the borders of Kerala, Tamilnadu and Karnataka.
(ii) The most prominent attraction is the Botanical Garden, which was established in 1847.
(iii) The lake runs through the Botanical Garden of Ooty.

Can you draw up an outline of the passage above and how the description progresses?
Answer:
Paragraph 1: Popular name and location of Ooty – a tourist spot.
Paragraph 2: The Botanical Garden – the most prominent tourist spot.
Paragraph 3: The lake and the flower festival
Now read another description of a place of tourist interest in India.

(b) Junagadh is an ancient city in Gujarat. It is situated among the shadows of Mount Gimar. The name “Junagadh”- Juna (old) and Gadh (fort)- literally means “old fort”. On the outskirts of the city, there is a dark basalt rock. It stands on the way to Mount Gimar. The rock holds the inscriptions of three mighty dynasties. They include the Maurya and Gupta dynasties. The inscriptions are in Sanskrit.

Notice some keywords and phrases used in the descriptions.
existence: Ooty lies on the borders of Kerala, Tamil Nadu and Karnataka,
location: on the outskirts of the city there is a dark basalt rock.

CHSE Odisha Class 11 English Writing Description

Activity 25

Describe the following places, highlighting their size, location and type. Also, mention the interesting or outstanding features of each place.
(a) Your home town or village
(b) An important place you have visited
(c) Your college

Answer:
(a) My Village :
My village, Mahendragiri is situated in the Gajapati district. It is one-hundred and eighty km away from the silk city, of Berhampur. To reach my village one has to take a bus from Berhampur and after the Tapta Pani Ghat take the route leading to Ramgiri Udayagiri. This place was in the news recently because of communal clashes. Mahendragiri is just 60 km away from R. Udaygiri. The village is situated at the foot of the Mahendra Hills and hence it is called Mahendragiri.

The whole village consists of a cluster of huts, asbestos roof houses and a few concrete buildings. It has only two sahis namely the Nuasahi and the Puranasahi. These sahis are situated one after the other. When one approaches Mahendragiri from R. Udayagiri, one will first see the Nuasahi and after that the Puranasahi. Each ship has rows of houses facing each other. Thus in Nuasahi, we have two rows of houses facing each other and in Puranasahi too there are two rows of houses facing each other.

There are only 200 families living in the whole village. The village has only one main road, the state highway which comes from Berhampur goes past R. Udayagiri to our village Mahendragiri and then continues upto Parlakhemundi, the district headquarters of Gajapati district, which is just 20 km from our village. Nuasahi which is in the south of Berhampur is surrounded by a huge mango grove and tamarind trees.

Puranasahi which is on the north of Parlakhemundi is bordered by cashew-nut plantations. Beyond the mango grove and the cashew, plantation lie the hills. On the top of a hill is a Shiva temple. It can be reached after climbing 480 steps. The temple is a very ancient one. It is now almost in ruins because of a lack of maintenance. Nevertheless, one can see the crude Shiva Lingam in the inner sanctuary always covered with fresh flowers.

The view from the temple courtyard is thrilling. One can see the streams flowing down on the rear of the hill. The sahis looked like tiny rows of toy houses. T.V. antennas look like minute clothes hangers and the mango grove and casuarina trees look like beds of cauliflowers. 3 km away from the village, on the road to Parlakhemundi is our marketplace. It does not have any permanent shops but only rows of rectangular concrete platforms on which businessmen put up their shops.

The market meets on three days of the week, Monday, Wednesday and Friday. It is a very colourful market where one can get everything necessary. Beyond the market is a thick jungle. On the hills, one can also see patches of barren land. This is because the farmers of our village practice shifting cultivation. Some of the plants on the hill are seen as half burnt. Some are already yellow with the flowers of the mustard.

(b) An important place I have visited :
One of the most important and unforgettable places that I have ever visited is the Taj Mahal at Agra. I have not seen anything else that surpasses the beauty of this marble mausoleum. Built by Shah Jahan, as a tomb for his wife and as an enduring symbol of his love, the Taj Mahal is true “an elegy in stone.” It has a gateway of red stone with verses from the Quran inscribed on it.

The gateway leads to a garden with three pathways. Besides that, there are fountains and pillars that lead to the marble platform at whose four comers are four towers or minarets. In the middle is the main dome with two smaller domes flanking it on either side. The red and white marble walls are decorated with stones of various colours encrusted in them. Their insides too are covered by flowers wrought in stone and lace work of green foliage.

The hall of death has a verse inscribed on it. Words cannot describe the splendour of the Tajmahal in the moonlight. It glitters and appears radiant as a bride. Moreover, the large rooms, cool ambience and solitariness about it, give it a sober air whereby one becomes reverential and meditative.

(c) My college :
My college is situated on the National Highway No.5 between Cuttack and Bhubaneswar on the outskirts of our village Kamalpur when proceeding towards Cuttack. It is a red-brick, single-storeyed straight-line building that faces the East. It is constructed only on an acre of land which is marked by a similar red-brick boundary wall that has only one gate which opens onto the highway. The gate has a gravel path which leads to the portico of the college building.

Immediately after the portico is steps which lead to the principal’s office. On the right of the principal’s office is the staff common-room, while to its left is the Administrative Office of the college. All three are housed in single rooms. Beyond the staff common- room and the administrative office on both sides are the stores, the right one storing sports equipment while the left one has stationary. Following the sports store on the right are the classrooms. There are three classrooms in all for 1st Year Arts students.

Similarly on the left beyond the stationeries store are three classrooms for 2nd year Arts students. Beyond the boundary wall, on every side are paddy fields. It is interesting that the college itself has been constructed on an erstwhile paddy field donated by a farmer whose son is one of the teachers in the college. Thus in the rainy season the earth there does not absorb water and as a result, we often have 2-3 ft. of standing water in the College compound. One can even fish inside the college during rainy reason.

CHSE Odisha Class 11 English Writing Description

Activity 26

Describe the place shown in the picture below.

Describe the place shown in the picture below.

Answer:
It was a wonderful and picturesque sight just like a picture postcard. A waterfall nestling among the mountains peopled by the evergreen pines standing tall on the mountains. The water cascaded down the fall spraying itself into the air looking like tiny globules of diamond and then crashing into the rock below where it turns into white foam and then cascades down the mountain forming divergent streams that end up in a rivulet.

Activity 27

Describe the following people of our country and the places they live in.
(a) The Kashmiris
(b) The Sikhs
(c) The Santals

Here are some helpful points for (a)
(i) Live in the valley of Kashmir, fair complexion, tall, long noses, about 5 million people
(ii) Very cold winters – snow, frozen lakes, poorly heated mud houses, individual firepots. Woollen clothes, long gowns and rubber shoes
(iii) Food: meat, fish and rice; fruit (apples, pears, peaches, cherries, etc.); Drink: a lot of tea with or without milk
(iv) handicrafts: carpets, silk, wood carvings, etc.
(v) well-known tourist spots: Shalimar Gardens, Gulmarg, Dal lake, etc.

Answer:
(a) The Kashmiris:
The Kashmiris are a fair complexioned people most of who live in the valley of Kashmir which lies in the north of India. They usually have tall and long noses. They experience very cold winters because of snow, frozen lakes and badly heated mud houses. To get rid of the cold, they wear woollen clothes, long gowns and rubber shoes. However, most of the population is poor and hence they live in poorly heated mud houses but each of them has separate firepots. The majority of the Kashmiris are Muslims. As a result, most men wear caps while women are veiled. They wear purdah. This is of course more common among the orthodox folk.

The Kashmiris are mostly non-vegetarian people eating meat, fish and rice. They also consume fruits like apples, pears, peaches, cherries etc. and drink a lot of tea. Their main occupation is handicrafts. Whole families including young children are engaged in weaving carpets, silk clothes, wood carvings etc. all of which are exquisitely done. The main revenue, however, comes from tourism. Kashmir which is considered earth’s paradise has many famous tourist spots like the Dal Lake, Shalimar Gardens and the Gulmarg. Terrorism has however decreased tourist trade in recent times.

(b) The Sikhs :
The Sikhs are the residents of Punjab but over the years they have migrated to almost all the states of India and to many countries abroad. Sikhism began as a socio-religious movement which was more interested in fighting evils but in its process of evolution, it was forced by circumstances to become a militant sect. It was Guru Gobind Singh who transformed the Sikhs into a militant sect and created Khalsa. The Sikh people are easily distinguishable by the turban they put on. Every Sikh is bound by the laws of his religion to never have his hair cut.

Men, therefore, tie their hair in a plait, bind it on the head and wear a turban upon it. Besides this, all Sikhs who are part of the Khalsa are armed with a Kirpan and put on a steel bangle called Kada. Most men are dressed in long Kurtas that reach down to the knees and pyjamas. Women are dressed in salwar kameez. The Sikhs are very hard-working and industrious people. They mainly cultivate wheat, rice, maize, gram and pulses. They produce the largest amount of wheat in India.

Unlike other States, the Sikhs use all mechanised equipment for agriculture and adopt the latest techniques and methods of production. Besides this, they are engaged in several industries like bicycle parts, auto parts, sports and leather goods, hosiery, knitwear, footwear, nuts and bolts, textiles etc. Most Sikhs eat roti or parathas along with Makhan, and dal and drink large glasses of milk. They celebrate the birthdays of their Guru by offering prayer and distributing sweets. The important tourist centres and places of worship in Punjab are the Golden Temple at Amritsar, the Durgiana Mandir, the Anandpur Sahib and the Jalianawala Bagh.

(c) The Santhals :
The Santhals or Santals are an indigenous aboriginal tribe inhabiting Bihar and some parts of Orissa. They lend their name to the Santhal Pargana district of Bihar, which is known after them. They are short dark-skinned people having broad noses, thick lips, coarse and curly hair and very prominent cheekbones. Their main occupation is cultivating and cattle breeding. Most of them are uneducated and illiterate and rarely mix with mainstream society. They are also good hunters.

They are animistic in their beliefs and enjoy sexual liberty practising polyandry and polygamy. Their dialect is called Santali. They live in mud houses short in height. But their villages are extremely clean. The Santals have an elaborate tribal structure, with 12 exogamous clans. More over, each village has its cadre of village officials the head of whom is the chief.

CHSE Odisha Class 11 English Writing Description

Activity 28

Describe how the Money Order which you send reaches the addressee.
Answer:
The money order form is first filled up and the money to be sent as well as the amount of exchange is given to the postal clerk who gives a receipt in return. It is understood that a customer has commissioned money to be sent to a customer in another place for paying a sum for the service. The postal clerk thus sends this form to the post office where the addressee is to be found. This is taken as a direction by that post office to pay the sum to the addressee, which is done by a postal peon.

Activity 29

Write about 150 words on each of the following :
(a) A peacock
(b) A cat
(c) An elephant

Answer:
(a) A Peacock :
The male peacock is a beautiful bird. Its neck is covered with lovely blue feathers. Its body is green and blue. Its glory is its long tail. It can open its tail like a great fan. The colours are blue and green and gold. The bird is very proud of its tail. It opens it, and struts about to be admired. The peacock stands for pride. We say, “as proud as a peacock”. The lady peacock is a plain brown bird. She has no tail like her husband. He has all the beauty.

In India, peacocks are sacred birds. The peacock is called the mount of Saraswati, the goddess of learning. So it is very wrong to kill a peacock. But peacocks are great thieves. They do great damage to growing crops. Some people say that peacocks kill snakes. Some say they can smell the coming rain. Then they give harsh screams.

(b) A Cat :
People keep cats as pets. Cats are pretty animals, covered with soft fur. They are of different colours. Some are black, some white, some grey, and some brown. Kittens, or young cats, are very playful. They will play for hours with little balls, fallen leaves, or bits of string. The chief use of cats is to catch mice and rats. Like their big cousins, lions and tigers, cats can see in the dark. They hunt for mice at night. Mice are a great pest in a house. A cat will soon kil them, or drive them away.

Cats have been tamed for thousands of years. They were kept as pets in ancient Egypt. Cats are very different from dogs. Dogs love persons, but cats love places. A dog will follow his master anywhere. But a cat loves the comfort of the house and stays at home. Their love is what we call “cupboard love”.

(c) An elephant :
The elephant is the largest of all animals. It is a strange animal to look at. It has thick legs, huge sides and back, large ears, small eyes, a short tail, and great white tusks. Its long nose, or trunk, is the strangest thing about it. It uses its trunk like a hand. It picks things up with its trunk and puts them into its mouth. It sucks up water with its trunk and squirts it into its mouth for drinking.

Elephants are very strong. And they are very clever. So tame elephants are very useful. They are trained to draw heavy loads. They are taught to carry logs of wood on their tusks and pile them up in perfect order. They are used, too, in hunting tigers in the jungle. In old days they were used in battle. And Indian Rajas ride on elephants in state processions.

CHSE Odisha Class 11 English Writing Description

Activity 30

Write short paragraphs on :
(a) A refrigerator
(b) A screwdriver
(c) A motorcycle

Answer:
(a) A refrigerator :
A refrigerator is a common household electrically powered equipment/appliance that is used to chill or freeze food items for preservation. It consists of an outer metal cabinet or box, rectangular in shape and an inner polyurethane foam lining (pdf) to ensure zero gaps, insulation and provide no space for insects to breed. Its size ranges from 165 litres to 310 litres. Its inside can be cooled to temperatures as low as – 16°C.

The cooling is affected by a thermostat which controls the temperature inside the refrigerator as well as of the freezer compartment. The cabinet contains a separate freezer compartment in which ice can be formed and food kept frozen. The freezer also has ice trays with which ice cubes are made. Below it is the chill tray which is used to store soft drinks as well as milk jackets for quick cooling. Besides, the cabinet may have adjustable shelves, which are found mostly in domestic refrigerators.

In which vessels of different sizes can be accommodated to store cooked food, jellies, pies etc. Right below is the San crisper which is a compartment for storing leafy vegetables and fresh fruits. The door inside also has a dairy compartment for cheese and butter, removable egg racks and adjustable bottle racks. Today frost-free refrigerators are available.

(b) A screwdriver:
A screwdriver is a common tool used for turning screws. It consists of a metal rod that is fixed in a wooden, plastic or rubber groove that has been moulded into a handle grip. The rod is chiselled in the front to facilitate its getting into the groove of the screw. The rod varies in length and diameter.

(c) A motorcycle :
Motor cycle is one of the most popular means of conveyance. Now different brands are manufactured by different companies. Each of them has a distinctive feature. The motorcycle consists of various parts, such as a handle, brake, fuel tank, silencer pipe, engine (two-stroke/four-stroke) carburettor, clutch lever, speedometer, two tyres, indicator (front and rear), and battery compartment. These parts are systematically set in the bike’s comfortable seat, headlight, shock absorber and so on.

Four-stroke motor-bike is superior to a two-stroke one because the former has smooth pick-up. Besides, it doesn’t produce defeaning sound. On the other hand, the motorbike has a two-stroke engine that doesn’t have that smooth pick-up. It produces sound. The fuel consumption of a four-stroke engine is better than that of a two-stroke engine. The former is economical. Replacement of engine oil at the scheduled time is of great importance. Now wherever we notice, we see varieties of wonderful bikes playing on the road.

CHSE Odisha Class 11 English Writing Narration

Odisha State Board CHSE Odisha Class 11 Invitation to English 3 Solutions Writing Narration Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Writing Narration

There are different ways of developing a paragraph depending on the topic and the purpose of writing. We shall study some of those ways now and in a subsequent chapter, we will develop our ideas into a longer piece of writing in the same light. We will also learn to mix different ways of developing our ideas into one or more paragraphs.

The common ways of developing paragraphs are :
(i) Narration – To tell an event, incident., or experience in chronological order.
(ii) Description -To describe a person, animal, object, place or process.
(iii)Exposition – To explain an idea, instrument, or problem.
(iv)Argumentation or persuasion -To argue for or against a view, in order to influence the reader’s opinion.

Narration

Read the following paragraph and see how the sentences have been arranged.

On Sunday morning, I get up at six in the morning. After a quick wash, I get into my jogging rig and go for a run. By 6.30 I am on the road. I run half an hour. I return home and have a leisurely bath, a luxury. I cannot afford it on weekdays. The bath is over, and I get ready quickly. What have you done? You have described your activities on a Sunday morning, in the order in which they take place.

You start with what you do first, then go on to what you do next, and so on and you come to your last activity. This brings you to the most important thing about a narrative paragraph. In a narrative paragraph, the events or happenings are arranged in chronological order, that is, in order of time.

CHSE Odisha Class 11 English Writing Narration

A. Narrative Passage

A. Account of Events with Sequence of Action:
It was late in the evening. The bride’s place was richly illuminated and decorated. There was music and dancing all around. Around half past 10, the bridegroom’s procession arrived and there was a flurry of excitement. The bridegroom was sitting erect on a young horse. The bride’s parents came out to receive the bridegroom and the bride was brought out too, her face covered with a pink and gold veil. She was barely 18. But, suddenly, without any provocation, the horse broke away from his master’s grip and ran away taking the bridegroom with him.

The bridegroom shouted for help and clung on to the horse for dear life. Soon his turban fell off, revealing his sparse white hair flying in the breeze. “He was too old to marry”, everyone concluded. He couldn’t be less than 70. The young bride stood aghast. Turning to her parents she cried aloud, “He’s old ! He’s a doddering old man ! I won’t marry him”. Tearing off her finery she stormed back into the house. And the old man was taken to hospital with multiple fractures.

B. A Different point of view:
At last, we reached the bride’s place. It was well lighted and there was music and dancing all around. The atmosphere was exciting and I liked it. The bride’s parents came out to receive the bridegroom and then the bride was brought too, her face covered with a pink and gold veil. I stood on the tips of my hooves to get a better view of her. Suddenly the veil on her face blew up in the breeze and I had a glimpse of her face. I was shocked.

She was only a child! She couldn’t be more than 18. And she was about to marry a man old enough to be her grandfather! “Something is seriously wrong”, I thought. “This marriage must be stopped! I then looked sly at my master. He was looking the other way. Without losing any time I broke away from his hold and ran, taking the bridegroom with me. When I reached the street I threw him on the ground and bolted away.

CHSE Odisha Class 11 English Writing Narration

Activity 1

Read the accounts in passages A and B again. Then narrate the incident from the point of view of :
(i) the bridegroom
(ii) the bride’s father
(iii) the bride
Answer:
(i) The bridegroom:
It was late in the evening when I reached the bride’s place which was richly decorated and illuminated. It was a treat to watch. The bride’s parents then came out to receive me and my friends. She was there too, with her face covered with a pink and gold veil. I wanted to take a good look at her as I sat on my horse. Fortunately, a breeze blew her veil away and I got a full glimpse of her face. she was young and beautiful! My heart beat fast in excitement. I couldn’t wait to get married.

Suddenly without warning my horse broke away from my grip and galloped away as if frightened. The sudden jerk loosened my turban and it fell off my head. Alas ! my white hairs were now visible! I could hear shouts of amazement from people gathered for the marriage. They were calling me an old man, fit to be a grandad. All my hopes of marriage were dashed. Suddenly, I was thrown off the horse and consciousness. When I regained it, I found myself in the hospital. I was told that the girl refused to marry me.

(ii) The bride’s father:
It was the happiest day of my life. My one and only daughter were getting married. I spent money lavishly on decorating my house. The bridegroom had not arrived till now. It was almost ten o’clock. Fortunately, however, they arrived half an hour later. I went out to receive him along with my wife and daughter. He sat astride on a horse. He had put on a coffee-colored sherwani suit. A gold brocaded yellow silk turban wrapped his head.

He looked handsome. I was proud of him – a man of many means. We now faced to face and I was asking him to get down from the horse so that the rituals to welcome him could be started, but he was inattentive. I followed his eyes. He was looking at my daughter. “Natural”, I thought. Suddenly, the horse raised its forelegs high up in the air, broke its master’s grip, and galloped away toward the street. I thought I could hold its reins but it was too fast for me.

My would-be son-in-law was now crying out loudly for help but in vain. Being afraid, I ran behind the horse. At that moment my would-be son-in-law’s turban fell off and I was aghast by what I saw. There was only a little white hair clinging to his bald head! O God, he could not.be less than seventy! “Marry my daughter”, would he ?” I said to myself. I then looked back to see my daughter rushing into the house with tears rolling down her cheeks.

(iii) The bride:
The moment of my marriage had arrived. My parents took me out with them to receive the bridegroom. As usual, this was an arranged marriage and I had not even had so much as a glimpse of my fiance before. So when we neared him sitting astride on a house, I tried to look at him through my veil but was not satisfied with the hazy figure that I perceived. Luckily, a breeze lifted off my veil and I had a fair look at him. He looked like a chivalrous knight sitting on a horse. He looked smart and handsome.

Suddenly, however, his horse neighed loudly, raised its forelegs high up into the air, and bolted away. This movement disrobed his head and his turban fell off. What I saw turned me speechless for a moment. He was completely bald, except for a few strands of impeccably white hair. I was shocked. I couldn’t believe my eyes. I had been deceived! Screaming, I ran into the house. “I couldn’t marry this old man, no not at all”. I thought to myself. Tears filled my eyes and I was disconsolate. It was indeed providential that the horse had bolted and the turban had fallen off.

CHSE Odisha Class 11 English Writing Narration

Activity 2

One day you saw a woman faint at a bus stop. Narrate what happened; describe the sequence of events and the reaction of the people at the bus stop.
Answer:
It was a blazing summer afternoon. I could feel the heat almost scalding my cheeks. I was perspiring like a faucet as I reached the bus stop from the college. It was mid-noon then. The streets were deserted and desolate. The tar on the road was melting. There was no one at the bus stop except for a lady in a synthetic saree. “How the hell can she put on a synthetic saree on this hot noon ?” I thought she was holding a bag of vegetables – “A housewife, returning home after marketing44, I surmised.

My lips were dry and so I crossed the road to have sugarcane juice. There were four to five people at the crushing trolley. As I ordered a glass of juice and started sipping it all the while facing the bus stop to see if the bus is arriving. Suddenly I saw the woman standing across the road, at the stop, fell down unconscious. For seconds I did not realize that she had fainted.

We all did but none moved to help – they were men and she was a woman. She was old enough to be my mother. I ran to her; saw that she was faint and her lips were trembling, her mouth was completely dry. I ran back, took an ice block from the juice seller and a glass of water, and rushed back. Meanwhile, all my fellow drinkers had gathered around her. They asked me to sprinkle water on her face which I did. But then I also poured water into her mouth and massaged her head with the ice block.

Within a few minutes, she regained consciousness and blushed with confusion as she saw the crowd around her. Holding her hand I lifted her up and handed her another glass of water. She drank that and her cheeks regained color. I helped her pick up the vegetables that had fallen all around her, hailed an auto at her request, and saw her off.

CHSE Odisha Class 11 English Writing Narration

Activity 3

Write short accounts of the following imaginary incidents :
(i) You saw a man trying to steal someone’s wallet (purse) and caught hold of him.
(ii) A road accident you witnessed.
Answer:
(i) It was at the Bhubaneswar railway station that this happened. I was standing in the queue, nonchalantly like the rest of the others, waiting for my turn at the booking counter. It was around 10 o’clock in the morning. The heat and sweat were making everyone restless. To add to this, persons in the queue were having unnecessary arguments with those trying to bypass the queue and go to the counter.

I was fidgeting with my watch, noting how much time each person’s booking was taking and calculating roughly how much time it would take for my turn to come. Just then a man, handsomely dressed arrived and went straight to the counter. He then asked something to the clerk and stood there. Watching his demeanor, I did not like to ask him to move from the I place and take his place in the queue. Instead, I was admiring his mustache jeans. It looked smart on him.

Suddenly I saw him reach for the wallet in the back pocket of the person, first in the queue. He picked it up and turned back. Spontaneously I shouted out, [ “you thief’ and embraced him in a hug. He threw it on the floor and feigned ignorance about it. However, everyone had seen him throw it and so he was caught red-handed. Meanwhile, hearing the commotion the railway police arrived and arrested him.

(ii) It was a Sunday morning, 7.00 a.m. I was off on my bicycle to IRC village to buy vegetables from the Gandhi Market. I was on the road that runs parallel to the National Highway in Jaydev Vihar. All of a sudden a mini truck sped past me, raising a lot of dust. It immediately turned left to go onto the highway taking the mud track road. Just before it reached the highway, there was a loud thud. The four wheels of the truck had fallen into the ditch that had been recently dug to make a drain along the highway.

The driver did not know this and had taken that track to reach the highway. He was unfortunate. The four wheels were deep in the ditch and the truck’s back had been thrown up into the air with the rear tires hanging and wheeling. I rushed to the spot. The driver had escaped unhurt but the cleaner was lying unconscious in the driver’s cabin. We quickly brought him out through the open door of the truck. Pieces of glass had pierced his cheeks and forehead. He was bleeding profusely. The driver hailed a taxi and took him immediately to the hospital.

CHSE Odisha Class 11 English Writing Narration

Activity 4

The Prime Minister is to visit Bhubaneswar next week and the following is the tour program. Write a short account of the planned tour, using the points below.
10.0 am : arrival by a special Air Force plane.
10.10 a.m.: reception by the Chief Minister at the airport.
10.30 a.m.: meeting at the State Secretariat; discussion with the Chief Minister
11.30 a.m.: meeting with party workers.
12.30 p.m.: lays the foundation of the Software Technology Park.
1.0 p.m. : meeting the press.
2.0 p.m. : lunch at Raj Bhavan.
3.0 p.m. : return to Delhi.
Answer:
The Prime Minister arrives at the Biju Pattanaik Airport at 10.00 a.m. sharp by the special Air Force plane. He is to be received there by the Chief Minister and other Cabinet Colleagues. After this, he heads straight for the Secretariat where he discusses relief measures granted by the World Bank and other funding agencies for the cyclone-affected area. After this, at 11.30 p.m.

he reaches the B.J.B. party office to meet party workers. There he discusses organizational elections. At 12.30 p.m. he reaches the site for the Software Technology Park and lays its foundation stone. Following this, he attends a press conference organized by the BBSR Press Club. At 2.00 p.m. he has lunch with the Governor at the Raj Bhavan. At 3.00 he once again boards the plane to leave for Delhi.