BSE Odisha 10th Class Physical Science Solutions Chapter 4 କାର୍ବନ ଓ ଏହାର ଯୌଗିକ

Odisha State Board BSE Odisha 10th Class Physical Science Solutions Chapter 4 କାର୍ବନ ଓ ଏହାର ଯୌଗିକ Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Physical Science Solutions Chapter 4 କାର୍ବନ ଓ ଏହାର ଯୌଗିକ

Question 1.
ଇଥେନ୍‌ର ଆଣବିକ ସଙ୍କେତ ହେଉଛି C2H6। ଏଥରେ କେତୋଟି ସହସଂଯୋଗ୍ୟ ବନ୍ଧ ରହିଛି ?
(a) 6
(b) 7
(c) 8
Answer:
(b) 7 [C-H ବନ୍ଧସଂଖ୍ୟା = 6 ଓ C–C ବନ୍ଧ ସଂଖ୍ୟା = 1]

Question 2.
ବ୍ୟୁଟାନୋନ୍‌ରେ ଥ‌ିବା ସକ୍ରିୟ ଗ୍ରୁପଟି କ’ଣ ?
(a) କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍
(b) ଆଲତିହାଇଡ୍‌
(c) କିଟୋନ
(d) ଆଲକହଲ
Answer:
(c) କିଟୋନ

Question 3.
ରୋଷେଇ କଲାବେଳେ ଯଦି ରନ୍ଧାପାତ୍ରର ତଳପଟର ବହିର୍ଭାଗ କଳା ପଡ଼ିଯାଏ, ଏହାର ଅର୍ଥ :
(a) ଖାଦ୍ୟ ସଂପୂର୍ଣ୍ଣ ରୂପେ ପ୍ରସ୍ତୁତ ହୋଇ ନାହିଁ
(b) ଜାଳେଣି ସଂପୂର୍ଣ୍ଣ ରୂପେ ଜଳୁନାହିଁ
(C) ଜାଳେଣିଟି ଆର୍ଦ୍ର ଅଛି
(d) ଜାଳେଣି ସଂପୂର୍ଣ୍ଣ ଭାବେ ଜଳୁଛି ।
Answer:
(b) ଜାଳେଣି ସଂପୂର୍ଣ୍ଣ ରୂପେ ଜଳୁନାହିଁ

Question 4.
ସହସଂଯୋଜ୍ୟ ବନ୍ଧର ଗଠନ CH3CI ର ଉଦାହରଣ ଦେଇ ବୁଝାଅ ।
Answer:
(i) CH3CI ଯୌଗିକ କାର୍ବନ, ହାଇଡ୍ରୋଜେନ ଓ କ୍ଲୋରିନ୍‌ ମୌଳିକମାନଙ୍କୁ ନେଇ ଗଠିତ ।

(ii) କାର୍ବନ, ହାଇଡ୍ରୋଜେନ୍‌ ଓ କ୍ଲୋରିନ୍‌ର ପରମାଣୁ କ୍ରମାଙ୍କ ଯଥାକ୍ରମେ 6, 1 ଏବଂ 17 ।
କାର୍ବନର ଇଲେକ୍‌ଟ୍ରନିକ୍‌ ସଂରଚନା K(2) L(4) ।
ହାଇଡ୍ରୋଜେନର ଇଲେକ୍‌ଟ୍ରନିକ ସଂରଚନା K (1) |
କ୍ଲୋରିନ୍‌ର ଇଲେକ୍‌ଟ୍ରନିକ୍ ସଂରଚନା K(2), L(8), M(7) ।

(iii) ଅକ୍ଲେଟ୍ ପାଇଁ କାର୍ବନ 4ଟି ଇଲେକ୍‌ଟ୍ରନ, ହାଇଡ୍ରୋଜେନ୍ 1 ଟି, ଇଲେକ୍‌ଟ୍ରନ ଏବଂ କ୍ଲୋରିନ୍ 1 ଟି ଇଲେକ୍‌ଟ୍ରନ୍

(iv) କାର୍ବନ ତା’ର 4ଟି ଇଲେକ୍‌ଟ୍ରନକୁ 3ଟି ହାଇଡ୍ରୋଜେନ ସହ (ପ୍ରତ୍ୟେକ ଗୋଟିଏ ଲେଖାଏଁ) ଏବଂ ଗୋଟିଏ କ୍ଲୋରିନ୍ ସହ ଭାଗ କରିଥାଏ । ତେଣୁ କାର୍ବନ ତା’ର 4ଟି ଇଲେକ୍‌ଟ୍ରନରୁ 3ଟି ଇଲେକ୍‌ଟ୍ରନ 3ଟି ହାଇଡ୍ରୋଜେନ ସହ ଏବଂ ଗୋଟିଏ ଇଲେକ୍‌ଟ୍ରନ କ୍ଲୋରିନ ସହ ସହଭାଜନ କରି ସହସଂଯୋଜ୍ୟ ବନ୍ଧ ଗଠନ କରେ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-1

(v) ଫଳରେ କାର୍ବନ ନିକଟତମ ନିଷ୍କ୍ରିୟ ଗ୍ୟାସ୍ Ne ର ସଂରଚନା ଧାରଣ କରେ । ହାଇଡ୍ରୋଜେନ ନିକଟତମ ନିଷ୍କ୍ରିୟ ଗ୍ୟାସ୍ He ର ସଂରଚନା ଏବଂ କ୍ଲୋରିନ୍ ନିକଟତମ ନିଷ୍କ୍ରିୟ ଗ୍ୟାସ୍ Arର ସଂରଚନା ଧାରଣ କରି C – H ଏବଂ C – Cl ସହସଂଯୋଜ୍ୟ ବନ୍ଧ ଗଠନ କରେ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 4 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 5.
ତଳେ ଦିଆଯାଇଥ‌ିବା ଅଣୁଗୁଡ଼ିକର ଇଲେକ୍‌ଟ୍ରନ ଡଟ୍ ସଂରଚନାର ଚିତ୍ର ଦିଅ ।
(a) ଇଥାନୋଇକ୍ ଏସିଡ୍ (CH3COOR)
(b) H2S
(c) ପ୍ରେ|ପେନ୍
(d) F2
Answer:
(a)
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-2 BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-3

Question 6.
ହୋମୋଲଗସ୍ ଶ୍ରେଣୀ କ’ଣ ଉଦାହରଣ ସହ ବୁଝାଅ ।
Answer:
(i) ଯେଉଁ ଶ୍ରେଣୀରେ ଥ‌ିବା ଜୈବ ଯୌଗିକଗୁଡ଼ିକର ସମାନ ସଂରଚନା, ସମାନ ସକ୍ରିୟ ଓ ସମାନ ରାସାୟନିକ ଧର୍ମ ଥାଏ । ସେହି ଶ୍ରେଣୀକୁ ହୋମୋଲଗସ୍ ଶ୍ରେଣୀ କୁହାଯାଏ ।

(ii) ଏହି ଶ୍ରେଣୀରେ ଥ‌ିବା ପ୍ରତ୍ୟେକ ଯୌଗିକ ତା’ର ପୂର୍ବବର୍ତୀ ଓ ପରବର୍ତ୍ତୀ ଯୌଗିକଠାରୁ – CH2 ଗ୍ରୁପ୍ ଦ୍ବାରା ଭିନ୍ନ ହୋଇଥାଏ ।

ଉଦାହରଣ :
CH3OH (ମିଥାନଲ୍) → CH3OH
C2H5OH (ଇଥାନଲ୍) → CH3CH2OH
C3H7OH (ପ୍ରୋପାନଲ୍) → CH3CH2CH2OH
C4H9OH (ବ୍ୟୁଟାନଲ୍) → CH3CH2CH2CH2OH
ଇତ୍ୟାଦି ହୋମୋଲଗସ୍ ଶ୍ରେଣୀ ଗଠନ କରନ୍ତି ।

Question 7.
ଇଥାନଲ୍ ଓ ଇଥାନୋଇକ୍ ଏସିଡ୍ ମଧ୍ୟରେ ଭୌତିକ ଓ ରାସାୟନିକ ଧର୍ମରେ ପ୍ରଭେଦ ଲେଖ ।
Answer:
ଭୌତିକ ଧର୍ମ
ଇଥାନଲ୍:

  • ଇଥାନଲ୍ ଅମ୍ଳ କିମ୍ବା କ୍ଷାର ନୁହେଁ, ଏହା ପ୍ରଶମିତ । ଅଟେ । ଉଭୟ ଲିଗ୍‌ସ୍ କାଗଜର କୌଣସି ରଙ୍ଗର ପରିବର୍ତ୍ତନ କରେ ନାହିଁ ।
  • ଏହା ଅନେକ ଜୈବ ଯୌଗିକ ପାଇଁ ଏକ ଉତ୍କୃଷ୍ଟ
  • ଇଥାନଲ୍‌ର ଗଳନାଙ୍କ 156 K ।
  • ସ୍ଫୁଟନାଙ୍କ 351 K
  • ଥଣ୍ଡା ଜଳବାୟୁରେ ଅର୍ଥାତ୍ ଶୀତଋତୁରେ ଏହା ଘନୀଭୂତ | ହୁଏ ନାହିଁ ।
  • ଇଥାନଲ୍‌ର ସ୍ବାଦ ପୋଡ଼ିଗଲା ଭଳି ଲାଗେ ।
  • ଏହା ତୀବ୍ର ଗନ୍ଧଯୁକ୍ତ ନୁହେଁ |

ଇଥାନୋଇକ୍ ଏସିଡ୍ :

  • ଇଥାନୋଇକ୍ ଏସିଡ୍ ନୀଳ ଲିଟ୍‌ସ୍‌କୁ ନାଲି କରିଥାଏ ଅର୍ଥାତ ଏହା ଅମ୍ଳୀୟ ।
  • ଏହା ଉତ୍ତମ ଦ୍ରାବକ ନୁହେଁ । ଦ୍ରାବକ ।
  • ଇଥାନୋଇକ୍ ଏସିଡ୍‌ର ଗଳନାଙ୍କ 290 K ।
  • ଏହାର ସ୍ଫୁଟନାଙ୍କ 391 K ।
  • ଥଣ୍ଡା ବାୟୁରେ ଏହା ଘନୀଭୂତ ହୁଏ ।
  • ଏହାର ସ୍ବାଦ ଖଟା ।
  • ଏହା ତୀବ୍ର ଗନ୍ଧଯୁକ୍ତ ଅଟେ ।

ରାସାୟନିକ ଧର୍ମ :
ଇଥାନଲ୍:
(i) ଇଥାନଲ୍ ସୋଡ଼ିୟମ ସହ ପ୍ରତିକ୍ରିୟା କରି ହାଇଡ୍ରୋଜେନ୍ ଗ୍ୟାସ୍ ନିର୍ଗତ କରେ ।
2Na + 2CH3 CH2OH → 2CH3CHO – Na + H2

(ii) ଇଥାନଲ୍‌କୁ ଅଧିକ ପରିମାଣ ଗାଢ଼ ସଲ୍‌ଫ୍ୟୁରିକ୍ ଏସିଡ୍ ସହିତ 443Kରେ ଉତ୍ତପ୍ତକଲେ ଇଥାନଲ ଅଣୁରୁ ଗୋଟିଏ ଜଳ ଅଣୁ ବାହାରି ଏଥୁନ ସୃଷ୍ଟି ହୁଏ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-4

(iii) ଇଥାନଲ ସୋଡ଼ିୟମ କାର୍ବୋନେଟ୍ ବା ସୋଡ଼ିୟମ୍ ବାଇକାର୍ବୋନେଟ୍ ସହ ପ୍ରତିକ୍ରିୟା କରେ ନାହିଁ ।

ଇଥାନୋଇକ୍ ଏସିଡ୍:
(i) ଇଥାନୋଇକ୍ ଏସିଡ୍ ସୋଡ଼ିୟମ ହାଇଡ୍ରକ୍‌ସାଇଡ଼ ସହ ପ୍ରତିକ୍ରିୟା କରି ଲବଣ ଓ ଜଳ ସୃଷ୍ଟିକରେ ।
NaOH + CH3COOH → CH3COONa + H2O

(ii) ଏସିଟିକ୍ ଏସିଡ଼୍କୁ ଫସ୍‌ଫରସ୍ ପେଣ୍ଟକ୍‌ସାଇଡ ସହିତ ଉତ୍ତପ୍ତ କଲେ ଦୁଇଟି ଏସିଡ୍ ଅଣୁରୁ ଗୋଟିଏ ଜଳ ଅଣୁ ଅପସାରିତ ହୁଏ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-5

(iii) ଏହା ସୋଡ଼ିୟମ କାର୍ବୋନେଟ୍ ଓ ସୋଡ଼ିୟମ୍‌ କାର୍ବୋନେଟ୍ ସହ ପ୍ରତିକ୍ରିୟା କରେ ।
2CH3COOH + Na2CO3 → CH3COONa + H2O + CO2
CH3COOH + NaHCO3 → CH3COONa + H2 + CO2

BSE Odisha 10th Class Physical Science Solutions Chapter 4 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 8.
ଜଳରେ ସାବୁନ ମିଶାଇଲେ ମିସେଲ କାହିଁକ ସୃଷ୍ଟି ହୁଏ ବୁଝାଅ । ଅନ୍ୟ ଦ୍ରାବକ ଯଥା : ଇଥାନଲରେ ମଧ୍ୟ ମିସେଲ୍ ସୃଷ୍ଟି ହେବ କି ?
Answer:
(i) ସାବୁନର ଅଣୁ ଗୁଡ଼ିକ ଦୀର୍ଘ-ଶୃଙ୍ଖଳ କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍‌ର ସୋଡ଼ିୟମ୍ କିମ୍ବା ପୋଟାସିୟମ୍ ଲବଣ । ସାବୁନ ଅଣୁର ଆୟନିକ ପ୍ରାନ୍ତଟି ଜଳରେ ଦ୍ରବୀଭୂତ ହେଉଥ‌ିବା ବେଳେ କାର୍ବନ ଶୃଙ୍ଖଳଟି ତେଲରେ ଦ୍ରବୀଭୂତ ହୁଏ । ତଦନୁଯାୟୀ ସାବୁନ୍ ଅଣୁଗୁଡ଼ିକ ଏକ ସଂରଚନା ସୃଷ୍ଟି କରିଥାନ୍ତି । ଏହାକୁ ମିସେଲ୍ କୁହାଯାଏ ।

(ii) ଏହି ଅଣୁର ଗୋଟିଏ ପ୍ରାନ୍ତ ତେଲର ଛୋଟବିନ୍ଦୁ ଆଡ଼କୁ ରହୁଥ‌ିବା ବେଳେ ଆୟନିକ୍ ପ୍ରାନ୍ତଟି ବାହାରକୁ ମୁହଁ କରିଥାଏ । ଏହା ଜଳରେ ଏକ ଅପଦ୍ରବ ସୃଷ୍ଟିକରେ । କିନ୍ତୁ ସାବୁନ ଇଥାନଲ୍‌ରେ ଦ୍ରବଣୀୟ, ତେଣୁ ମିସେଲ୍ ସୃଷ୍ଟି ହୁଏ ନାହିଁ ।

Question 9.
କାର୍ବନ ଏବଂ ଏହାର ଯୌଗିକକୁ ଅଧିକାଂଶ ପ୍ରୟୋଗରେ ଇନ୍ଧନ ରୂପେ ବ୍ୟବହାର କରାଯାଏ କାହିଁକି ?
Answer:
(i) କାର୍ବନର ଯେକୌଣସି ରୂପ ଅକ୍‌ସିଜେନ୍‌ରେ ଜଳିଲେ କାର୍ବନ ଡାଇଅକ୍‌ସାଇଡ୍ ସୃଷ୍ଟି ହେବା ସହିତ ତାପ ଓ ଆଲୋକ ଉତ୍ପନ୍ନ ହୁଏ ।

(ii) ଅଧ୍ୟାକାଂଶ କାର୍ବନ ଯୌଗିକ ମଧ୍ୟ ଦହନ ପ୍ରତିକ୍ରିୟାରେ ବହୁପରିମାଣର ତାପଶକ୍ତି ଓ ଆଲୋକ ଶକ୍ତି ସୃଷ୍ଟି କରିଥାନ୍ତି । ଏକକ ବସ୍ତୁତ୍ଵ ବିଶିଷ୍ଟ କାର୍ବନ ଯୌଗିକରୁ ଅନ୍ୟ ତୁଳନାରେ ଅଧିକ ଶକ୍ତି ମିଳେ ।

Question 10.
ଖରଜଳ ସହିତ ସାବୁନ କିପରି ପ୍ରତିକ୍ରିୟା କରେ ?
Answer:
ଖରଜଳରେ କ୍ୟାଲସିୟମ୍ ଓ ମ୍ୟାଗ୍ନେସିୟମ୍‌ର ଲବଣ ଥାଏ । ସାବୁନ ଅଣୁ ସହିତ କ୍ୟାଲସିୟମ୍ ଓ ମ୍ୟାଗ୍ନେସିୟମ୍ ଲବଣ ପ୍ରତିକ୍ରିୟାକରି ଧଳା ଅଦ୍ରବଣୀୟ ପଦାର୍ଥ (କ୍ୟାଲସିୟମ୍ ଓ ମ୍ୟାଗ୍ନେସିୟମ୍ ଲବଣ) ସୃଷ୍ଟିକରେ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-6

Question 11.
ସାବୁନକୁ ଲାଲ ଏବଂ ନୀଳ ଲିଟମସ୍ କାଗଜରେ ପରୀକ୍ଷା କଲେ କି ପରିବର୍ତ୍ତନ ଦେଖାଯିବ ?
Answer:
ସାବୁନ୍ କ୍ଷାରୀୟ ଅଟେ । ଏହା ଲାଲ୍ ଲିଟ୍‌ସ୍କୁ ନୀଳ କରେ; କିନ୍ତୁ ନୀଳ ଲିଟମସ୍‌ର କୌଣସି ପରିବର୍ତ୍ତନ କରେ ନାହିଁ ।

Question 12.
ହାଇଡ୍ରୋଜେନୀକରଣ କ’ଣ ? ଶିଳ୍ପରେ ଏହାର ପ୍ରୟୋଗ କ’ଣ ?
Answer:
ଉତ୍ପ୍ରେରକ ଉପସ୍ଥିତିରେ ଅପୃକ୍ତ ହାଇଡ୍ରୋକାର୍ବନ ସହ ହାଇଡ୍ରୋଜେନର ଯୋଗକୁ ହାଇଡ୍ରୋଜେନୀକରଣ କହନ୍ତି ।

ଶିଳ୍ପରେ ପ୍ରୟୋଗ :
ବନସ୍ପତି ତେଲଗୁଡ଼ିକରେ ସାଧାରଣତଃ ଅପୃକ୍ତ କାର୍ବନ ଶୃଙ୍ଖଳ ନିକେଲ୍‌ ଉତ୍ପ୍ରେରକ ଉପସ୍ଥିତିରେ ହାଇଡ୍ରୋଜେନୀକରଣ କରାଯାଇ ପୃକ୍ତ କାର୍ବନ ଚେନ୍ ଗଠନ କରେ, ଯାହାକି ପଶୁ ଚର୍ବି ବା ଘିଅ ଅଟେ । ତେଣୁ କଳକାରଖାନାରେ ହାଇଡ୍ରୋଜେନୀକରଣ ପଦ୍ଧତିରେ ବନସ୍ପତି ତେଲ ବା ଘିଅ ପ୍ରସ୍ତୁତ କରାଯାଏ ।

Question 13.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ହାଇଡ୍ରୋକାର୍ବନଗୁଡ଼ିକ ମଧ୍ୟରୁ କେଉଁଗୁଡ଼ିକ ଯୋଗ ପ୍ରତିକ୍ରିୟା ଦେଇଥା’ନ୍ତି ? C2H6, C3H8, C3H6, C2H2, ଏବଂ CH4,
Answer:
C3H6 ଏବଂ C2H2,

Question 14.
ଲହୁଣୀ ଏବଂ ରୋଷେଇ ପାଇଁ ବ୍ୟବହୃତ ତେଲ ମଧ୍ଯରେ ରାସାୟନିକ ଧର୍ମରେ ପ୍ରଭେଦ ନିର୍ଣ୍ଣୟ କରିବାକୁ ଏକ ପରୀକ୍ଷା ଦର୍ଶାଅ ।
Answer:
ଲହୁଣି ପୃକ୍ତ (Saturated) ହାଇଡ୍ରୋକାର୍ବନ ଅଟେ । ବ୍ରୋମିନ୍ ଲହୁଣିରେ ମିଶାଇଲେ ବ୍ରୋମିନ୍‌ର ବାଦାମୀ ରଙ୍ଗରେ କୌଣସି ପରିବର୍ତ୍ତନ ହୁଏନାହିଁ । କାରଣ ବ୍ରୋମିନ ପୃକ୍ତ ହାଇଡ୍ରୋକାର୍ବନ ସହ ପ୍ରତିକ୍ରିୟା କରେ ନାହିଁ କିନ୍ତୁ ରୋଷେଇ ପାଇଁ ବ୍ୟବହୃତ ତେଲ ଅପୃକ୍ତ (unsturated) ହାଇଡ୍ରୋକାର୍ବନ ଅଟେ । ବ୍ରୋମିନ୍‌ର କିଛି ବୁନ୍ଦା ମିଶାଇଲେ ବାଦାମୀ ରଙ୍ଗ ବର୍ଣ୍ଣହୀନ ହୁଏ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 4 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 15.
ସାବୁନ କିପରି ସଫାକରେ ବୁଝାଅ ।
(i) ସାବୁନ ଅଣୁରେ ଥ‌ିବା ଦୁଇଟି ପ୍ରାନ୍ତର ଧର୍ମ ଭିନ୍ନ । ଗୋଟିଏ ହେଉଛି ଜଳାସକ୍ତ ଅନ୍ୟ ପ୍ରାନ୍ତଟିଏ ଜଳାତଙ୍କୀ ଜଳାସକ୍ର (Hydrophobic) ଏବଂ ଅନ୍ୟ ପ୍ରାନ୍ତ୍ରଟିଏ ଜଳାତଙ୍କ1 (Hydrophobic) |

(ii)
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-7 ହୋଇଥାଏ ।

(iii)
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-8
(iv) ଜଳର ପୃଷ୍ଠରେ ସାବୁନ ରହିଲେ ଜଳାତଙ୍କୀ ଲାଞ୍ଜ ଜଳରେ ଦ୍ରବୀଭୂତ ହେବ ନାହିଁ ଏବଂ ସାବୁନଟିର ଆୟନିକ ପ୍ରାନ୍ତ ସହ ଜଳରେ ଏବଂ ହାଇଡ୍ରୋକାର୍ବନ ‘ଲାଞ୍ଜ’ ଜଳ ବାହାରକୁ ବଢ଼ିଆସି ଜଳର ପୃଷ୍ଠ ନିକଟରେ ଶ୍ରେଣୀବଦ୍ଧ ଭାବେ ସଜାଇ ହୋଇ ରହେ ।

(v) ଜଳଭିତରେ ଏହି ଅଣୁଗୁଡ଼ିକର ଏକ ଅନନ୍ୟ ଅଭିବିନ୍ୟାସ ରହିଛି ଯାହା ହାଇଡ୍ରୋକାର୍ବନ ଅଂଶଗୁଡ଼ିକୁ ଜଳଠାରୁ ଅଲଗା ରଖେ ।

(vi) ଅଣୁ ପେନ୍ଥାମାନ ସୃଷ୍ଟି କରିବା ଦ୍ଵାରା ଏହା ସଂପାଦିତ ହୋଇଥାଏ ।

(vii) ଜଳାତଙ୍କୀ ଲାଞ୍ଜଗୁଡ଼ିକ ଅଣୁପେନ୍ଥାର ଭିତର ପାର୍ଶ୍ବରେ ଏବଂ ଆୟନିକ ପ୍ରାନ୍ତ ଗୁଡ଼ିକ ପେନ୍ଥାର ଉପରି ଭାଗରେ ରହିଥାଏ ।

(vi) ଏହି ଗଠନକୁ ମିସେଲ୍ କୁହାଯାଏ ।

(ix) ମିସେଲ୍‌ର କେନ୍ଦ୍ରରେ ତୈଳାକ୍ତ ମଇଳା ସଂଗୃହୀତ ହୁଏ ।

(x) ମିସେଲ୍ ଦ୍ରବଣରେ କଲଏଡ୍ ରୂପରେ ରହେ ଏବଂ ଆୟନ-ଆୟନ ବିକର୍ଷଣ ଯୋଗୁଁ ଏକତ୍ରିତ ହୋଇ ଅବକ୍ଷେପିତ ହୁଏ ନାହିଁ ।

(xi) ତେଣୁ ଲୁଗାକୁ ଧୋଇଦେଲେ ମିସେଲ୍‌ରେ ରହିଥିବା ମଇଳା ମଧ୍ୟ ସହଜରେ ଜଳ ସହିତ ପଦାକୁ ବାହାରି ଆସେ ।

ବିଶେଷ ଦ୍ରଷ୍ଟବ୍ୟ :
ଏହି ଅଧ୍ୟାୟରେ (*) ତାରକା ଚିହ୍ନିତ ତଥ୍ୟ, ପ୍ରଶ୍ନ ଓ ବାକ୍ସ ଅନ୍ତର୍ଗତ ବିଷୟବସ୍ତୁ ସମ୍ପର୍କିତ ତଥ୍ୟ ପରୀକ୍ଷାରେ ଆସିବ ନାହିଁ ।

ପ୍ରଶ୍ନବଳୀ ଓ ଉତ୍ତର:

Question 1.
କାର୍ବନ ଡାଇଅକ୍‌ସାଇଡ୍ (ସଙ୍କେତ CO2)ର ଇଲେକ୍‌ଟ୍ରନ ଡଟ୍ ସଂରଚନା କ’ଣ ହେବ ?
Answer:
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-9

Question 2.
ଆଠଟି ସଲ୍ଫର ପରମାଣୁରେ ଗଠିତ ଏକ ସଲ୍ଫର ଅଣୁର ଇଲେକ୍‌ଟ୍ରନ ଡଟ୍ ସଂରଚନା କ’ଣ ହେବ ?
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-10

Question 3.
ପେଣ୍ଟେନ ପାଇଁ ତୁମେ କେତୋଟି ସଂରଚନାତ୍ମକ ଆଇସୋମର୍ ଚିତ୍ର କରିପାରିବ ?
Answer:
ପେଣ୍ଟେନ୍ ପାଇଁ ତିନୋଟି ସଂରଚନାତ୍ମକ ଆଇସୋମ୍‌ର ଚିତ୍ର ଅଙ୍କନ କରାଯାଇଛି –
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-11

Question 4.
ଆମ ଚାରିପଟେ ଦେଖୁଥ‌ିବା କାର୍ବନ ଯୌଗିକର ବହୁଳତା ପାଇଁ କାର୍ବନର କେଉଁ ଦୁଇଟି ପ୍ରକୃତି ଦାୟୀ ?
Answer:
(i) କାଟିନେସନ୍ : କାର୍ବନର କ୍ଷୁଦ୍ର ଆକାର ଓ ଉଚ୍ଚ ବିଦ୍ୟୁତ୍ ଋଣାତ୍ମକତା ଗୁଣ ଯୋଗୁଁ ଏହା ଅନ୍ୟ କାର୍ବନ ପରମାଣୁଗୁଡ଼ିକ ସହିତ ସ୍ଥାୟୀ ବନ୍ଧ ଗଠନ କରିପାରେ । ଯୌଗିକଗୁଡ଼ିକର ସଂରଚନା ସଳଖ ଚେନ, ଶାଖା ଚେନ, କିମ୍ବା ବକ୍ରାକାର ଚେନ ହୋଇପାରେ ।

(ii) କାର୍ବନର ଚତୁଃସଂଯୋଗ୍ୟତା : କାର୍ବନ ଯୌଗିକର କକ୍ଷରେ 4ଟି ଇଲେକ୍‌ଟ୍ରନ୍‌ ଥିବାରୁ ଏହି ଇଲେକ୍‌ଟ୍ରଗୁଡ଼ିକ ଅନ୍ୟ 4ଟି କାର୍ବନ ପରମାଣୁ ସହ କିମ୍ବା ହାଇଡ୍ରୋଜେନ୍, ଅକ୍‌ସିଜେନ୍, ନାଇଟ୍ରୋଜେନ୍, ସଲ୍‌ଫର୍‌ ପରି ଅନ୍ୟ ମୌଳିକଗୁଡ଼ିକ ସହ ସହଭାଜନ କରିପାରେ । ତେଣୁ କାର୍ବନ ଯୌଗିକର ସଂଖ୍ୟା ବୃଦ୍ଧିପାଏ

Question 5.
ସାଇକ୍ଳୋପେଣ୍ଟେନର ସଙ୍କେତ ଏବଂ ଇଲେକ୍‌ଟ୍ରନ ଡଟ୍ ସଂରଚନା କ’ଣ ହେବ ?
Answer:
ଫଂକେତ:
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-12

Question 6.
ନିମ୍ନଲିଖୂତ ଯୌଗିକଗୁଡ଼ିକ ପାଇଁ ସଂରଚନା ଚିତ୍ର ଦିଅ ।
(i) ଇଥାନୋଇକ୍ ଏସିଡ଼୍
(ii) ବ୍ରୋମୋପେଣ୍ଟନ୍‌
(iii) ବ୍ୟୁଟାନୋନ୍
(iv) ହେକ୍ସାନାଲ
Answer:
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-13

Question 7.
ନିମ୍ନଲିଖୂ ଯୌଗିକଗୁଡ଼ିକୁ କିପରି ନାମକରଣ କରିବ ?
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-14
Answer:
(i) ବ୍ରୋମୋଇଥେନ୍
(ii) ମିଥାନଲ୍
(iii) ହେକ୍‌ସାଇନ୍

Question 8.
ଇଥାନଲ୍‌ରୁ ଇଥାନୋଇକ୍ ଏସିଡ୍କୁ ରୂପାନ୍ତର କାହିଁକି ‘କ ଜାରଣ ପ୍ରତିକ୍ରିୟା ?
Answer:
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-15
ହାଇଡ୍ରୋଜେନ ପରମାଣୁ ସଂଖ୍ୟାର ହ୍ରାସ ଘଟିଛି, ତେଣୁ ଏହା ଏକ ଜାରଣ ପ୍ରତିକ୍ରିୟା ।

Question 9.
ଝଳାଇ ପାଇଁ ଅକ୍‌ସିଜେନ୍ ଓ ଇଥାଇନ୍‌ର ଏକ ମିଶ୍ରଣ ଜଳାଯାଏ I ଇଥାଇନ୍ ଓ ବାୟୁର ମିଶ୍ରଣ ବ୍ୟବହାର କରାଯାଏ ନାହିଁ କାହିଁକି କହିପାରିବ ?
Answer:
ବାୟୁରେ ଯବକ୍ଷାରଜାନ ସହିତ ନିଷ୍କ୍ରିୟ ଗ୍ୟାସ୍ ଅଛି, ଯାହାକି ଇଥାଇନ୍ ଜଳିବା ପାଇଁ ଯେଉଁ ଅମ୍ଳଜାନ ଆବଶ୍ୟକ ସେଗୁଡ଼ିକ ପ୍ରତିରୋଧ କରିଥାନ୍ତି । ତେଣୁ ବାୟୁ ଓ ଇଥାଇନ୍‌ର ମିଶ୍ରଣ ବ୍ୟବହାର କରାଯାଏ ନାହିଁ ।

Question 10.
ପରୀକ୍ଷା କରି ଆଲ୍‌କହଲ ଏବଂ କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍ ମଧ୍ଯରେ ପାର୍ଥକ୍ୟ କିପରି ଜାଣିବ ?
Answer:
କ୍ଷାରୀୟ ପୋଟାସିୟମ ପରମାଙ୍ଗାନେଟ୍ ଟେଷ୍ଟ : ଆଲ୍‌କହଲ ଓ କାର୍ବୋସିଲିକ୍ ଭିନ୍ନ ଭିନ୍ନ ପରୀକ୍ଷାନଳୀ ଦୁଇଟିରେ ନିଅ ଏବଂ କିଛି ବୁନ୍ଦା କ୍ଷାରୀୟ ପୋଟାସିୟମ୍ ପରମାଙ୍ଗାନେଟ୍ ଦ୍ରବଣ ପକାଅ ଓ ଗରମ୍ କର ପୋଟାସିୟମ୍ ପରମାଙ୍ଗାନେଟ୍‌ର ଗୋଲାପୀ ରଙ୍ଗ ରହିଲେ ତାହା ଆଲ୍‌କହଲ, ଅନ୍ୟଥା ତାହା କାର୍ବୋକସିଲିକ୍ ଏସିଡ୍ ।

Question 11.
କାରଣ କ’ଣ ?
Answer:
NaHCO ଦ୍ଵାରା ପରୀକ୍ଷା – ଦୁଇଟି ଭିନ୍ନ ଭିନ୍ନ ପରୀକ୍ଷାନଳୀରେ ଆଲ୍‌କହଲ ଓ କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍ ନେଇ କିଛି NaHCO3 ର ଗାଢ଼ ଦ୍ରବଣ ମିଶାଅ । ଯେଉଁ ପରୀକ୍ଷାନଳୀରେ ଗାଢ଼ ଧୂଆଁଳିଆ CO2 ଗ୍ୟାସ ବାହାରିବ ସେହି ପରୀକ୍ଷାନଳୀରେ କାର୍ବୋକସିଲିକ୍ ଏସିଡ୍ ଅଛି ।

BSE Odisha 10th Class Physical Science Solutions Chapter 4 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 12.
ଡିଟରଜେଣ୍ଟ ବ୍ୟବହାର କରି ଜଳ ଖର କି ନୁହେଁ ଜାଣିବାକୁ ତୁମେ ସମର୍ଥ ହୋଇପାରିବ କି ?
Answer:
ନାଁ, କାରଣ ଖରଜଳ ଓ ମୃଦୁଜଳ ଉଭୟରେ ଡିଟରଜେଣ୍ଟ ସମାନ ଭାବରେ କାର୍ଯ୍ୟ କରେ ।

Question 13.
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-16
ଦେଇସାରି ସେମାନେ ଲୁଗାକୁ ପଥର ଉପରେ ବାଡ଼େଇ ଥା’ନ୍ତି କିମ୍ବା ଏକ ଦଣ୍ଡରେ ବାଡ଼େଇଥା’ନ୍ତି ବ୍ରସ୍‌ରେ ଘଷିଥା’ନ୍ତି କିମ୍ବା ୱାସିଂମେସିନ୍‌ରେ ମନ୍ଥନ (agitate) କରିଥା’ନ୍ତି । ସଫାଲୁଗା ପାଇବା ପାଇଁ ମନ୍ଥନ କାହିଁକି ଆବଶ୍ୟକ ?
Answer:
ମଇଳା ସହିତ ପ୍ରତିକ୍ରିୟା କରି ସାବୁନ ଅଣୁଗୁଡ଼ିକ ମିସେଲ୍‌ ସୃଷ୍ଟି କରନ୍ତି ଓ ଯେଉଁ ଅପଦ୍ରବ ସୃଷ୍ଟିହୁଏ ପାଣିରେ କିଛି ଧୋଇ ହୋଇଯାଏ ଆଉ କିଛି ଲୁଗାର ଉପର ସ୍ତରରେ ହାଲୁକା ଭାବେ ଲାଗିରହେ । ତେଣୁ ସାବୁନ ଦେଇ ସାରି ସେମାନେ ଲୁଗାକୁ ପଥର ଉପରେ ବାଡ଼େଇଥାନ୍ତି କିମ୍ବା ଏକ ଦଣ୍ଡରେ ବାଡ଼େଇଥାନ୍ତି ବା ମନ୍ଥନ କରିଥାନ୍ତି |

କାର୍ଯ୍ୟାବଳୀ (Activity):

କାର୍ଯ୍ୟାବଳୀ – 1 (Activity – 1):
(i) ସକାଳୁ ତୁମେ ବ୍ୟବହାର କରିଥିବା କିମ୍ବା ଖାଇଥିବା ଦଶଟି ଜିନିଷର ଏକ ତାଲିକା କର ।
ଏହି ତାଲିକା ସହ ତୁମ ସହପାଠୀମାନେ କରିଥିବା ତାଲିକାକୁ ଏକାଠି କର ଏବଂ ତା’ପରେ ଦିଆଯାଇଥ‌ିବା ସାରଣୀ ଭିତରେ ସେହି ଦ୍ରବ୍ୟଗୁଡ଼ିକୁ ତାଲିକାଭୁକ୍ତ କର । ଯେଉଁ ଦ୍ରବ୍ୟ ଏକରୁ ଅଧିକ ପଦାର୍ଥରୁ ପ୍ରସ୍ତୁତ ହେଉଛି, ସେଗୁଡ଼ିକୁ ସଂପୃକ୍ତ ସ୍ତମ୍ଭରେ ରଖ ।
ଡ :
ସକାଳୁ ବ୍ୟବହାର କରିଥିବା କିମ୍ବା ଖାଇଥିବା ଦଶଟି ଜିନିଷ ହେଲା
ବ୍ୟବହାର କରିଥିବା ଜିନିଷ: ସାବୁନ୍, ତେଲ, କାଗଜ, ଜୋତା ପଲିସ୍, ପ୍ଲାଷ୍ଟିକ ମର, ବହି, ଖବରକାଗଜ, ଷ୍ଟିଲ୍ ମର୍ ।
ଖାଇଥ‌ିବା ଜିନିଷ – କ୍ଷୀର, ଚା, ଅଣ୍ଡା, ପାଉଁରୁଟି, ଭାତ, ଡାଲି, ଔଷଧ ।

(ii)
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-17

ପର୍ଯ୍ୟବେକ୍ଷଣ:
କେବଳ ଜଳ ରନ୍ଧନପାତ୍ର, (ଯେଉଁଗୁଡ଼ିକ ଧାତୁ, କାଚ କିମ୍ବା ମାଟିରେ ତିଆରି) ଛାଡ଼ି ପ୍ରାୟ ଅନ୍ୟ ସମସ୍ତ ଜିନିଷରେ କାର୍ବନ ଅଛି ।

ସିଦ୍ଧାନ୍ତ:
ଆମେ ଦୈନନ୍ଦିନ ଜୀବନରେ ବ୍ୟବହାର କରୁଥିବା ଅଧିକାଂଶ ଜିନିଷ କାର୍ବନରେ ତିଆରି ।

କାର୍ଯ୍ୟାବଳୀ – 2 (Activity – 2):

ତଳେ ଦିଆଯାଇଥ‌ିବା ଯୋଡ଼ିଗୁଡ଼ିକ ପାଇଁ ସଙ୍କେତଗୁଡ଼ିକ ମଧ୍ୟରେ ଏବଂ ଆଣବିକ ବସ୍ତୁତ୍ଵ ମଧ୍ୟରେ ପ୍ରଭେଦ କଳନା କର ।
(a) CH3 OH ଏବଂ C2H5 OH
(b) C2H5 OH ଏବଂ C3 H7OH
(c) C3H7 OH ଏବଂ C4H9 OH
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-18
ଏହି ତିନୋଟିରେ କିଛି ସାଦୃଶ୍ୟ ଅଛି କି ?
ଉ :
ସମସ୍ତଙ୍କର ସକ୍ରିୟ ଗ୍ରୁପ୍ ଏକ ।

ଗୋଟିଏ ଶ୍ରେଣୀ (Family) ପାଇବା ପାଇଁ ଏହି ଆଲକହଲଗୁଡ଼ିକୁ କାର୍ଟୁନ ପରମାଣୁର ବର୍ଦ୍ଧିତ କ୍ରମରେ ସଜାଅ ।
ଊ :
CH3OH, C2H5OH, C3H8OH, C4H9OH

ସାରଣୀର ଅନ୍ୟ ସକ୍ରିୟଗ୍ରୁପ୍ ପାଇଁ ଚାରୋଟି କାର୍ବନ ପର୍ଯ୍ୟନ୍ତ ଯୌଗିକଗୁଡ଼ିକର ହୋମୋଲଗସ୍ ଶ୍ରେଣୀ ଲେଖ ।
ଊ :
କ୍ଲୋରୋ ହୋମୋଲୋଗସ୍ ଶ୍ରେଣୀ : CH3CI, C2H5CI, C3H7Cl, C4H9Cl
ଆଲ୍‌ହାଇଡ଼ ହୋମୋଲଗସ୍ ଶ୍ରେଣୀ : CH3CHO, C2H5CHO, C3H7CHO
କିଟୋନ୍ ହୋମୋଲଗସ୍ ଶ୍ରେଣୀ : C2H6CO, C3H8CO
କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍ ହୋମୋଲଗସ୍ ଶ୍ରେଣୀ : CH3COOH, C2H5COOH, C3H7COOH

କାର୍ଯ୍ୟାବଳୀ -3 (Activity-3)

ଆବଶ୍ୟକ ଉପକରଣ :
ଗୋଟିଏ ଚାମଚ, କିଛି କାର୍ବନ ଯୌଗିକ (ଗନ୍ଧକର୍ପୂର, କର୍ପୂର, ଆଲ୍‌କହଲ) ।

ପରୀକ୍ଷଣ :
ଗୋଟିଏ ଚାମଚରେ କେତୋଟି କାର୍ବନ ଯୌଗିକ (ଗନ୍ଧକର୍ପୂର, କର୍ପୂର, ଆଲକହଲ) ଗୋଟିଏ ପରେ ଗୋଟିଏ ନିଅ ଏବଂ ଜଳାଅ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ :
(a) ଅଗ୍ନିଶିଖା ଉପରେ ଗୋଟିଏ ଧାତବ ଥାଳି ରଖ । ଆଲ୍‌କହଲ ଭଳି ପୃକ୍ତ ହାଇଡ୍ରୋକାର୍ବନ ଜଳିଲେ ସ୍ୱଚ୍ଛ ଶିଖା ଦେଇଥାଏ ।
(b) ଗନ୍ଧକର୍ପୂର ଭଳି ଅପୃକ୍ତ ହାଇଡ୍ରୋକାର୍ବନ ଜଳିଲେ କଳାଧୂଆଁ ସହିତ ହଳଦିଆ ରଙ୍ଗର ଶିଖା ଦେବ ଏବଂ ଧାତବ ଥାଳିରେ କଳାକଣିକା ଜମିଯିବ ।

ସିଦ୍ଧାନ୍ତ:
ପୃକ୍ତ ହାଇଡ୍ରୋକାର୍ବନ ଜଳିଲେ ସ୍ବଚ୍ଛ ଶିଖା ଦିଏ । ଅପୃକ୍ତ ହାଇଡ୍ରୋକାର୍ବନ ଜଳିଲେ କଳାଧୂଆଁ ଓ ହଳଦିଆ ରଙ୍ଗର ଶିଖା ଦେବ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 4 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -4 (Activity-4)

ପରୀକ୍ଷଣ:
ଗୋଟିଏ ବୁନ୍‌ସେନ୍ ବର୍ଣ୍ଣର ଜଳାଅ ଏବଂ ବିଭିନ୍ନ ପ୍ରକାର ଶିଖା ବା ଧୂଆଁର ଉପସ୍ଥିତି ପାଇବା ପାଇଁ ଏହାର ନିମ୍ନଅଂଶରେ ଥ‌ିବା ବାୟୁଛିଦ୍ର (Airhole)କୁ ନିୟନ୍ତ୍ରଣ କର ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
(a) କଳାଧୂଆଁ ଓ ହଳଦିଆ ଅଗ୍ନିଶିଖା ମିଳିବ ଯେତେବେଳେ ବର୍ଣ୍ଣରର ନିମ୍ନ ଅଂଶରେ ଥ‌ିବା ବାୟୁଛିଦ୍ରକୁ ନିୟନ୍ତ୍ରଣ କରିବା ଫଳରେ ଅମ୍ଳଜାନର ପରିମାଣ ହ୍ରାସପାଇବ ।
(b) ବର୍ଣ୍ଣରର ନିମ୍ନ ଅଂଶରେ ଥ‌ିବା ବାୟୁଛିଦ୍ର ସମ୍ପୂର୍ଣ୍ଣ ନିୟନ୍ତ୍ରଣମୁକ୍ତ ରହିଲେ ପ୍ରଚୁର ପରିମାଣର ଅମ୍ଳଜାନ ପ୍ରବେଶ କରିବ ଏବଂ ସମ୍ପୂର୍ଣ ଜାରଣ ହେବା ଫଳରେ ନୀଳ ଅଗ୍ନି ଶିଖା ଦେଖାଯିବ ।

ସିଦ୍ଧାନ୍ତ:
ଯଥେଷ୍ଟ ପରିମାଣର ଅକ୍‌ସିଜେନ୍‌ରେ ଜଳିଲେ ସ୍ବଚ୍ଛ ନୀଳଶିଖା ମିଳିବ ଏବଂ ସୀମିତ ବାୟୁ (ଅକ୍‌ସିଜେନ)ରେ

କାର୍ଯ୍ୟାବଳୀ -5 (Activity-5)

ଆବଶ୍ୟକ ଉପକରଣ:
ଗୋଟିଏ ପରୀକ୍ଷାନଳୀ, 3 ମିଲି ଇଥାନଲ୍ , ଏକ ଜଳ ଉଷ୍ମକ (water bath), 5% କ୍ଷାରୀୟ ପୋଟାସିୟମ ପରମାଙ୍ଗାନେଟ୍ ଦ୍ରବଣ ।

ପରୀକ୍ଷଣ:
ଗୋଟିଏ ପରୀକ୍ଷାନଳୀରେ ପ୍ରାୟ 3 ମିଲି ଇଥାନଲ୍ ନିଅ ଏବଂ ଏହାକୁ ଧୀରେ ଧୀରେ ଉଷୁମ କର । 5% କ୍ଷାରୀୟ ପୋଟାସିୟମ୍ ପରମାଙ୍ଗାନେଟ୍ ଦ୍ରବଣକୁ ବୁନ୍ଦା ବୁନ୍ଦା କରି ପକାଅ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
(a) 5 % କ୍ଷାରୀୟ ପୋଟାସିୟମ୍ ପରମାଙ୍ଗାନେଟ୍ ଦ୍ରବଣକୁ ବୁନ୍ଦା ବୁନ୍ଦା କରି ପକାଇଲେ ଗୋଲାପୀ ରଙ୍ଗ କ୍ରମଶଃ ଅଦୃଶ୍ୟ ହୋଇଯିବ ।
(b) ଯେତେବେଳେ ଅଧିକ KMnO4 ମିଶାଗଲା ସେତେବେଳେ ଇଥାନଲ୍ ଜାରଣ ହୋଇ ଇଥାନୋଇକ୍ ଅମ୍ଳରେ ପରିଣତ ହେଲା ଏବଂ ଅଧ୍ଵ KMnO4 ରଙ୍ଗ ସେହିଭଳି ରହିଲା ଓ ଅଦୃଶ୍ୟ ହେଲା ନାହିଁ ।

ସିଦ୍ଧାନ୍ତ:
କ୍ଷାରୀୟ KMnO4 ଦ୍ରବଣ ଇଥାନଲ୍‌କୁ ଜାରଣ କରେ ଏବଂ ନିଜେ ବିଜାରିତ ହୁଏ ।

କାର୍ଯ୍ୟାବଳୀ -6 (Activity-6)
ପରୀକ୍ଷଣ:
ଦୁଇଟି ଚାଉଳଦାନା ଆକାରର ଖଣ୍ଡେ ଛୋଟ ସୋଡ଼ିୟମକୁ ଇଥାନଲ (ବିଶୁଦ୍ଧ ଆଲକହଲ) ମଧ୍ୟରେ

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ସୋଡ଼ିୟମ୍ ଧାତୁ ଅତ୍ୟନ୍ତ ପ୍ରତିକ୍ରିୟାଶୀଳ ଅଟେ । ବିଶୁଦ୍ଧ ଆଲ୍‌କହଲ ସହିତ ରାସାୟନିକ ପ୍ରତିକ୍ରିୟା କରି ସୋଡ଼ିୟମ୍‌ ଇଥକ୍‌ସାଇଡ ଓ ହାଇଡ୍ରୋଜେନ୍ ଗ୍ୟାସ୍ ଦେଇଥାଏ ।

2C2H5OH + 2Na → 2C2H5ONa + H2
ଗୋଟିଏ ଜଳୁଥ‌ିବା କାଠିକୁ ଗ୍ୟାସ୍ ପାଖକୁ ଆଣିଲେ ପପ୍ ଶବ୍ଦ କରି ଜଳିଉଠିବ ।

ଇଥାନୋଇକ୍ ଏସିଡ୍‌ର ସାଧା (Properties of Ethanoic Acid):
(i) ଇଥାନୋଇକ୍ ଏସିଡ୍କୁ ସାଧାରଣତଃ ଏସିଟିକ୍ ଏସିଡ୍ (acetic acid) କହନ୍ତି ଏବଂ ଏହା କାର୍ବୋକ୍‌ଲିକ୍

(ii) ଜଳରେ ଏସିଟିକ୍ ଏସିଡ୍‌ର 5-8% ଦ୍ରବଣକୁ ଭିନେଗାର କୁହାଯାଏ ଏବଂ ଏହାକୁ ଆଚାର ସଂରକ୍ଷଣ ପାଇଁ ବ୍ୟବହାର କରାଯାଏ ।

(iii)ବିଶୁଦ୍ଧ ଇଥାନୋଇକ୍ ଏସିଡ୍‌ର ଗଳନାଙ୍କ ହେଉଛି 290 K ।

(iv) ଏସିଟିକ୍ ଏସିଡ୍‌ର ଅନ୍ୟ ନାମ ଗ୍ଲାସିଆଲ୍ (Glacial) ଏସିଟିକ୍ ଏସିଡ୍ । କାରଣ ଏହା ଥଣ୍ଡାଜଳ ଓ ବାୟୁରେ ଅଧିକାଂଶ ସମୟରେ ଘନୀଭୂତ ହୋଇଯାଏ ।

(v) କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍ ଏକ ଦୁର୍ବଳ ଏସିଡ୍ ।

ଏଷ୍ଟରୀକରଣ ପ୍ରତିକ୍ରିୟା:
(i) ଗୋଟିଏ ଏସିଡ୍ ଏବଂ ଗୋଟିଏ ଆଲ୍‌କହଲର ରାସାୟନିକ ପ୍ରତିକ୍ରିୟାଦ୍ୱାରା ଏଷ୍ଟର (Ester) ସୃଷ୍ଟି ହୁଏ । ଇଥାନୋଇକ୍ ଏସିଡ୍ ବିଶୁଦ୍ଧ ଆଲକହଲ ସହିତ ପ୍ରତିକ୍ରିୟା କରି ଏଷ୍ଟର ଦେଇଥାଏ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-19

(ii) ଏଷ୍ଟରଗୁଡ଼ିକ ସୁଗନ୍ଧଯୁକ୍ତ ପଦାର୍ଥ ଚ

(iii) ଏହାକୁ ଅତରରେ ବ୍ୟବହାର କରାଯାଏ ।

(iv) ଏହାକୁ ଖାଦ୍ୟ ଓ ପାନୀୟ ସୁଗନ୍ଧକାରୀ ଦ୍ରବ୍ୟ ରୂପେ ମଧ୍ୟ ବ୍ୟବହାର କରାଯାଏ ।

ସାବୁନୀକରଣ :
ସୋଡ଼ିୟମ୍ ଲବଣରେ ପରିଣତ ହୁଏ । ଏହି ପ୍ରତିକ୍ରିୟାକୁ ସାବୁନୀକରଣ କୁହାଯାଏ କାରଣ ଏହାକୁ ସାବୁନ୍ ପ୍ରସ୍ତୁତି କରିବା ପାଇଁ ବ୍ୟବହାର କରାଯାଏ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-20

(ii) କ୍ଷାରକସହ ପ୍ରତିକ୍ରିୟା: ଇଥାନୋଇକ୍ ଏସିଡ୍ + ସୋଡ଼ିୟମ୍‌ ହାଇଡ୍ରକ୍‌ସାଇଡ୍ → ଲବଣ + ଜଳ
NaOH + CH3COOH → CH3COONa + H2O

(iii) କାର୍ବୋନେଟ୍ ଏବଂ ହାଇଡ୍ରୋଜେନ କାର୍ବୋନେଟ୍ ସହ ପ୍ରତିକ୍ରିୟା:
କାର୍ବୋନେଟ୍ ଏବଂ ହାଇଡ୍ରୋଜେନ କାର୍ବୋନେଟ୍ ସହ ଇଥାନୋଇକ୍ ଏସିଡ୍ ପ୍ରତିକ୍ରିୟା କରି ଲବଣ, ଡାଇଅକ୍‌ସାଇଡ୍ ଏବଂ’ ଜଳ ଦେଇଥାଏ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-21

କାର୍ଯ୍ୟାବଳୀ -7 (Activity-7)
ଆବଶ୍ୟକ ଉପକରଣ: ଦୁଇଟି ପରୀକ୍ଷାନଳୀ, 10 ମି.ଲି. ଜଳ, ରୋଷେଇରେ ବ୍ୟବହୃତ ତେଲ ।

ପରୀକ୍ଷଣ:

  • ଦୁଇଟି ପରୀକ୍ଷାନଳୀରେ ପ୍ରାୟ 10 ମି.ଲି. ଜଳ ନିଅ, ପ୍ରତ୍ୟେକ ପରୀକ୍ଷାନଳୀରେ ଗୋଟିଏ ବୁନ୍ଦା ଲେଖାଏଁ ତେଲ ପକାଅ ଓ ପରୀକ୍ଷାନଳୀକୁ A ଓ B ରୂପେ ଚିହ୍ନଟ କର ।
  • ପରୀକ୍ଷାନଳୀ B ରେ ଅଳ୍ପ କେତେ ବୁନ୍ଦା ସାବୁନ ଦ୍ରବଣ ପକାଅ ।
  • ଦୁଇଟି ଯାକ ପରୀକ୍ଷାନଳୀକୁ ଖୁବ୍ ଜୋର୍‌ରେ ଏକା ସମୟରେ ହଲାଅ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ପରୀକ୍ଷାନଳୀ B ରେ ତେଲ ଓ ଜଳ ସାବୁନର ଗୋଟିଏ ସ୍ତର ଦେଖାଯାଉଛି । କାରଣ ତେଲ ସାବୁନ୍ ଦ୍ରବଣରେ ଦ୍ରବୀଭୂତ ହୁଏ ଫଳରେ ସଫା କରିବାରେ ସାହାଯ୍ୟ କରେ । ପରୀକ୍ଷାନଳୀ A ଜଳ ଓ ରୋଷେଇ ରେ ବ୍ୟବହୃତ ତେଲ ଅଛି । ଜୋର୍‌ରେ ହଲାଇଲା ପରେ କିଛି ସମୟ ସ୍ଥିର ହେବାକୁ ଦିଅ । 2ଟି ସ୍ତର ହୋଇ ରହିଥ‌ିବା ଦେଖାଯିବ କାରଣ ତେଲ ଓ ଜଳ ପରସ୍ପର ଅଦ୍ରବଣୀୟ ।

ସିଦ୍ଧାନ୍ତ:
ସାବୁନ ଦ୍ରବଣରେ ତେଲ ଦ୍ରବଣୀୟ, ଜଳ ମଧ୍ଯରେ ତେଲ ଅଦ୍ରବଣୀୟ ।

ସାଚୁନ୍‌ ଓ ଡିଟରଜେଣ୍ଠ (Soaps and Detergents):
ସାବୁନ୍ :

  • ସାବୁନର ଅଣୁଗୁଡ଼ିକ ଦୀର୍ଘ-ଶୃଙ୍ଖଳ କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍‌ର ସୋଡ଼ିୟମ କିମ୍ବା ପୋଟାସିୟମ ଲବଣ ।
  • BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-22
  • ପ୍ରାକୃତିକ ତୈଳ ଓ ଚର୍ବିରୁ ସାବୁନ ପ୍ରସ୍ତୁତ କରାଯାଏ ।
  • ଜୈବ ନିମ୍ବୀକରଣ ଯୋଗ୍ୟ ।
  • ଅମ୍ଳୀୟ ମାଧ୍ୟମରେ ବ୍ୟବହାର କରାଯାଇ ପାରିବନାହିଁ ।

ଡିଟରଜେଣ୍ଠ:

  • ଦୀର୍ଘଶୃଙ୍ଖଳ ବିଶିଷ୍ଟ କାର୍ବୋକ୍‌ଲିକ୍ ଏସିଡ୍‌ର ଏମୋନିୟମ୍ କିମ୍ବା ସଲ୍‌ଫୋନେଟ୍ ଲବଣ ।
  • BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-23
  • ପେଟ୍ରୋଲିୟମ୍ର ଉତ୍ପାଦଗୁଡ଼ିକରୁ ଡିଟରଜେଣ୍ଟ୍ ପ୍ରସ୍ତୁତ କରାଯାଏ ।
  • ଜୈବନିମ୍ନୀକରଣ ଅଯୋଗ୍ୟ ।
  • ଅମ୍ଳୀୟ ମାଧ୍ୟମରେ ବ୍ୟବହାର କରାଯାଇ ପାରିବ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 4 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -8 (Activity-8)
ଆବଶ୍ୟକ ଉପକରଣ:
ଦୁଇଟି ପରୀକ୍ଷାନଳୀ, 20 ମି.ଲି. ପାତିତ ଜଳ କିମ୍ବା ବର୍ଷାଜଳ, ଅନ୍ୟଟିରେ 10 ମି.ଲି.

ପରୀକ୍ଷଣ :
ଦୁଇଟି ଭିନ୍ନ ଭିନ୍ନ ପରୀକ୍ଷାନଳୀରେ ପାତିତ ଜଳ ଓ ଖରଜଳ ନିଅ । ପ୍ରତ୍ୟେକ ପରୀକ୍ଷାନଳୀରେ ଦୁଇବୁନ୍ଦା ସାବୁନ ଦ୍ରବଣ ପକାଅ । ପରୀକ୍ଷାନଳୀ ଦୁଇଟିକୁ ଖୁବ୍ ଜୋର୍‌ରେ ଏକା ସମୟ ପାଇଁ ହଲାଅ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଯେଉଁ ପରୀକ୍ଷାନଳୀରେ ପାତିତ ଜଳ |ବର୍ଷାଜଳ ଅଛି ସେହି ପରୀକ୍ଷାନଳୀରେ ଅଧ୍ଵକ ଫେଣ ସୃଷ୍ଟି ହୋଇଛି । ଯେଉଁ ପରୀକ୍ଷାନଳୀରେ ଖରଜଳ ଅଛି ସେଥ‌ିରେ ଦହିଭଳି ଧଳା ଅବକ୍ଷେପ ଦେଖାଯାଉଛି ।

ସିଦ୍ଧାନ୍ତ:
ମୃଦୁଜଳରେ ସାବୁନ ଫେଣ ସୃଷ୍ଟିକରେ, ଖରଜଳରେ ସାବୁନ୍ ଦହିଭଳି ଧଳା ଅବକ୍ଷେପ ସୃଷ୍ଟିକରେ ।

କାର୍ଯ୍ୟାବଳୀ -9 (Activity-9)
ଆବଶ୍ୟକ ଉପକରଣ:
2ଟି ପରୀକ୍ଷାନଳୀ, 20 ମି.ଲି. ଖର ଜଳ, ସାବୁନଦ୍ରବଣ, ଡିଟରଜେଣ୍ଟ ଦ୍ରବଣ ।

ପରୀକ୍ଷଣ:
ପ୍ରତ୍ୟେକ ପରୀକ୍ଷାନଳୀରେ 10 ମିଲି ଲି. ଖରଜଳ ନିଅ । ଗୋଟିଏ ପରୀକ୍ଷାନଳୀରେ ପାଞ୍ଚବୁନ୍ଦା ସାବୁନ ଦ୍ରବଣ ଓ ଅନ୍ୟ ଏକ ପରୀକ୍ଷାନଳୀରେ ପାଞ୍ଚ ବୁନ୍ଦା ଡିଟରଜେଣ୍ଟ୍ ଦ୍ରବଣ ପକାଅ । ଉଭୟ ପରୀକ୍ଷାନଳୀକୁ ନିର୍ଦ୍ଦିଷ୍ଟ ସମୟ ପାଇଁ ହଲାଅ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ :
ଦୁଇଟି ଯାକ ପରୀକ୍ଷାନଳୀରେ ସମାନ ପରିମାଣର ଫେଣ ରହୁନାହିଁ । ସାବୁନ ଦ୍ରବଣ ଥିବା
BSE Odisha Class 10 Physical Science Solutions Chapter 4 img-24
ପରୀକ୍ଷାନଳୀରେ ଅଧ୍ବକ ପରିମାଣରେ ଫେଣ ସୃଷ୍ଟି ହୋଇଛି ।

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 4 The One and only Houdini

Odisha State Board CHSE Odisha Class 11 Invitation to English 2 Solutions Non-Detailed Chapter 4 The One and only Houdini Textbook Exercise Questions and Answers.

CHSE Odisha 11th Class English Solutions Non-Detailed Chapter 4 The One and only Houdini

CHSE Odisha Class 11 English The One and only Houdini Text Book Questions and Answers

Unit – I

Gist :
The writer takes us back to the swimming pool at New York’s Shelton Hotel. In response to Houdini’s declaration, a box is ready. After he enters inside, it will be tightly closed. The pool now becomes the scene of an incredible performance. Driven by curiosity, people are gathering here. The reporter finds a telephone and an alarm bell with batteries inside the box for eventualities. According to Collins, he is going without air for even an hour. Houdini is as good as words. In the mean time, Dr. MacConnell emerges on the scene. He has already checked Houdini’s blood pressure and pulse. Everything is fine. The reporter is interested to know how long he can live without air. To the doctor, it is a difficult question.

He comes to know of Houdini’s knowledge of breathing easily and saving the oxygen. His condition is stable, yet he is fifty-two. The doctor is sure of Houdini’s ability to cope with danger. Now Houdini clad in a black swimming suit appears, smiling and determined. He expresses his thanks to the ladies and gentlemen for coming to see his performance. He is going to explode the myth that a man can live only three minutes without air. Now we find Houdini in the box. Several men are engaged in sealing it tightly. Then the box was submerged in the swimming pool of Hotel Shelton. Joseph Rinn, the official time keeper is in charge of counting every minute of Houdini’s stay in the box. The reporter is optimistic.

Glossary :
edge : the sharp side of something (କୌଣସି ଜିନିଷର ତୀକ୍ଷ୍ଣ ପାର୍ଶ୍ୱ)
sealed: closed tightly (ଜୋରରେ ବନ୍ଦ)
soldered: firmly fixed with solder (solder is easily melted metal) (ସୋଲ୍ଡର ସହିତ ଦୃଢ଼ ଭାବରେ ସ୍ଥିର କରାୟାଇଛି)
trick : cheating, a fraud (ଠକେଇ, ଜାଲିଆତି)
emergencies : sudden happenings which make it necessary to act without delay (ଜରୁରୀକାଳୀନ ପରିସ୍ଥିତି)
And days : Houdini is as good as his words (ଏବଂ ଦିନଗୁଡିକ)
Concerned : anxious (t) (ଚିନ୍ତିତ)
signal (v): to give a sign (ଏକ ଚିହ୍ନ ଦେବାକୁ)
build (n): general shape or size of a person’s body (ଜଣେ ବ୍ୟକ୍ତିର ଶରୀରର ସାଧାରଣ ଆକାର ବା ଆକାର)
He himself: Housini is firm in his determination (ସେ ନିଜେ)
motioning: making a sign (ଏକ ଚିହ୍ନ ତିଆରି କରିବା)
will : desire (ଇଚ୍ଛା)
rocked: moved backwards and forwards (ପଛକୁ ଏବଂ ଆଗକୁ ଗତି କଲା)
admire : praise (ପ୍ରଶଂସା)

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 4 The One and only Houdini

Unit – II

Gist:
Five minutes have passed ever since he was in a sealed casket submerged in the swimming pool. In the meantime, the reporter probes his past. His real name is Ehrich Weiss. His native place is Appleton, Wisconsin. He was born in 1874. His father was a Jewish priest. Houdini, renowned for his remarkable magic tricks, has an astonishing control over his body. He excels in the art of putting his shoulders out of position. When he was working for a locksmith, Ehrich knew technique of opening handcuffs with small, sharp pointed instruments. He also took interest in reading about Robert Houdini, the renowned French magician. At last he relinquished his job and became a professional magician under the name of Harry Houdini.

The reporter gives a picture of Houdini’s married life. Bessie is his wife. She was a magician too. Houdini is known for his sensational escape acts. Bessie has stood by her husband through thick and thin. Rinn informs that twenty minutes have passed. Tension mounts. Houdini has a tension-ridden career. His underwater box escape is a case in point. Handcuffed and chained, Harry Houdini is placed in a wooden box which firmly shut with nails. The box is then left underwater. Houdini accepts the challenge in his characteristic tricky manner. He swims to the surface safe and sound in a few minutes. This also demonstrates his skill in using tricks. Rinn, the official time-keeper keeps on counting time. Twenty five minutes have gone. In Houdini, the reporter finds an unusual person. He admires Houdini’s incredible control of chest muscles. He excels others in the act of moving around inside the jacket at the time of relaxation. The reporter is lavish in his praise, ‘There’s only one Houdini.’

Thirty minutes have trickled by. The reporter sees one of the swimmers standing on the box lose his balance. Other swimmers are no better. The box has moved up quickly above the water level. The reporter is worried. He enquires of Mr. Collins, about his condition. Collins talks to Houdini and says that he is fine. Rinn announces that thirty minutes have passed since Houdini’s adventure. The reporter brims with confidence. Houdini’s rigorous training for three weeks seems to be rewarded. The reporter thinks of Houdini’s march to a spectacular success. Forty minutes have elapsed. Meanwhile, the reporter refers eloquently to Houdini’s wonderful stomach muscles, that have resisted the fists of big, strong men. Forty-five minutes have already gone.

Glossary:
amazing : astonishing ବିସ୍ମୟକର)
dislocate : put out of position (ପଦରୁ ବାହାର)
straitjacket(ଷ୍ଟ୍ରେଟ୍ ଜାକେଟ୍) : kind of garment once used to prevent madman from struggling
rabbi: Jewish priest (ଇହୁଦୀ ପୁରୋହିତ)
locksmith: one who makes locks (ଯିଏ ତାଲା ତିଆରି କରେ)
a bit picks : a little small, sharp-pointed instruments (ଟିକିଏ ବାଛି ନେଲେ)
quit: give up (ତ୍ୟାଗ କରିବା)
idol : somebody greatly admired or loved (ଆଦର୍ଶ)
catch sight of: see (ଦେଖିବା)
career : way of making a living; profession (ବୃଭି)
bobbed up : moved up quickly (ଦ୍ରୁତ ଗତିରେ ଉପରକୁ ଉଠିଲା)
paying off : rewarding (ପୁରସ୍କାର ପ୍ରଦାନକାରୀ)
sources : places from which information comes or is got (ସୂତ୍ର)
claims : says, declares (କହୁଛି, ଘୋଷଣା କରୁଛି)
punch (v) : strike hard with the fist (ମୁଠା ସାହାୟ୍ୟରେ ଜୋରରେ ପ୍ରହାର କରନ୍ତୁ)
tensed : stiffened or hardened (ଦୃଢ଼ୀଭୂତ)
withstand : resist (ପ୍ରତିରୋଧ କରିବା)

Unit – III

Gist :
Houdini’s well-being concern the people. Everybody seems to be nervous. The doctors seem especially panicky. The reporter enquires of Dr. MaConnell about Houdini’s present condition. The doctor is really not aware of anything about him. Fifty minutes have gone. Other reporters are furiously taking track of Houdini’s adventurous mission. Time does not wait for anybody. Rinn announces – “Fifty minutes !” A group of grim-looking doctors seem to arguing with James Collins, Houdini’s assistant. Collins pleads his helplessness. The doctor advises him to be sensible and pull Houdini out of the box when an hour is up. Collins is adamant. The doctor insists him on responding to his advise. Collins contacts Houdini and apprising him of the doctors’ pressure on him to bring him out of the casket and an hour has passed. He gives them good news. Houdini is doing well. But the reporter’s anxiety continues.

Now one hour and thirteen minutes are up. Collins learns that the box has developed a leak and only a little trickle of water is entering. Houdini tells him that he faces no danger. The reporter promptly responds: “Houdini has nerves of steel.’’ This section comes to a terrific end. In spite of spending one hour and thirty-one minutes, underwater without air, Houdini emerges victorious. His secret lies in making very few body movements while in the box and taking short breaths. Houdini knows no panic. He has trained to reconcile himself to any situation. We see him in the reporter’s eyes: ‘You ’re a fantastic man, Mr. Harry Houdini!’

Glossary:
on edge : nervous (ସ୍ନାୟବିକ)
exhaustion : tiredness (ଅବସାଦ)
superstitious: full of superstition (ଅନ୍ଧବିଶ୍ୱାସରେ ପୂର୍ଣ୍ଣ)
grim : (here) anxious (ଉତ୍କଣ୍ଠିତ)
sensible : intelligent (ବିଚକ୍ଷଣ)
urging : persuading (ପ୍ରରୋଚନା)
sprung a leak : appeared to have a leak (ଲିକ୍ ହୋଇଥିବା ଜଣାପଡ଼ିଥିଲା)
trickle : thin flow (ପତଳା ପ୍ରବାହ)
dizzy : feeling as if everything is whirling (ସବୁ ଘୂରିବା ଭଳି)
panic (v) : a sudden fright (ହଠାତ୍ ଭୟ)
calm : quiet (ଶାନ୍ତ)
fantastic : wonderful (ଅଦ୍ଭୁତ)

Think it out :

Question 1.
What was the feat that Houdini took up in Hotel Shelton?
Answer:
The feat that Houdini took up in Hotel Shelton was to remain in a sealed casket or coffin submerged in a swimming pool. He was to stay inside the tightly closed box for an hour and that too without air. According to Mr. Collins, his assistant, Houdini was serious about being buried alive.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 4 The One and only Houdini

Question 2.
What were Houdini’s purposes behind doing this miraculous act?
Answer:
Houdini’s purposes behind doing this miraculous act were to disprove science and thereby, to show that determination can shake mountains. It is well-known that man can only live only three minutes without air, but he is going to stay without air for over an hour. Houdini is committed and according to Collins: ‘Houdini does what he says.’

Question 3.
Discuss how he prepared to do the act.
Answer:
Houdini undertook a rigorous training for three weeks at a stretch. He practised the habit of holding his breath in the course of his underwater escapes. He trained long and hard with great patience before doing the act. He developed a mastery over his body control. Determination, hard work and calm acceptance of any challenge were his forte.

Question 4.
Focus on the role of the reporter during Houdini’s performance.
Answer:
From beginning to the end, the reporter had a keen watch on Houdini’s, spectacular performance in New York’s Hotel Shelton. He is the first to communicate Houdini’s ordeal in the swimming pool of New York’s Hotel Shelton to the public. He always keeps on touch with Dr. McConnell, Collins, Houdini’s assistant and Joseph Rinn, the official time-keeper. He probes Houdini’s past. As minutes trickle by, the reporter’s reaction fluctuates from anxiety to confidence. Houdini’s incredible feat of remaining in a sealed box, submerged in the swimming pool of the Hotel Shelton, overwhelms him. His admiration of Houdini’s knows no bound.

Question 5.
Describe some of Houdini’s previous achievements.
Answer:
Houdini was an Austrian-Hungarian born. American stunt performer, noted for his sensational escape acts from handcuffs and jails in the length and breadth of the country and Europe. Once he was handcuffed to the bars in a jail cell. While kissing him for good luck, his wife Bessie passed a small lock tool from her mouth to his. Several minutes later Houdini took his jailers by surprise. His handcuff was no more. He walked freely into their office. His underwater escape is another achievement. Besides, Houdini became a professional magician under the name of Harry Houdini.

Question 6.
Why does the reporter say to Houdini, You’re a fantastic man?
Answer:
The reporter here is a keen observer of Houdini’s miraculous act of remaining in a tightly-closed box, submerged in the swimming pool of New York’s Hotel Shelton for one hour and thirty minutes. Houdini comes out with flying colours, putting an end to everyone’s fear, tension and anxiety. That he is able to survive so long in such a situation bears the stamp of his determination. Houdini states that he had trained himself to remain calm in the face of any situation. These facts make the reporter say to Houdini, ‘You’re a fantastic man.’

Question 7.
Bring out the important aspects of Houdini’s character.
Answer:
Harry Houdini is a many-faceted character. He is a magician. He is also known for his sensational escape acts. He is an incredible stuntman. His act of remaining in a sealed box, submerged in the swimming pool of New York’s Hotel Shelton for one hour and a half is indeed unprecedented. He is the epitome of commitment. He does not understand the language of fear. To maintain calmness in any situation is his forte. We see him through the doctor’s eyes: ‘Houdini’s an amazing man!’ The reporter aptly remarks, ‘You’re a fantastic man, Mr. Harry Houdini.’ Houdini’s life shows that an ordinary human being can achieve anything through determination and hard work.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 4 The One and only Houdini

CHSE Odisha Class 11 English The One and only Houdini Important Questions and Answers

Question 1.
Read through the extract and answer the questions that follow.

Reporter Thanks, Mr. Collins. It’s hard to believe that a man can live without air for over an hour. Oh, here come the doctors. Dr. McConnell, have you seen Houdini?
Dr. MaConnell: Yes, I just checked his blood pressure and pulse rate. They’re both normal.
Reporter: How last without long do you think Houdini will air?
Dr. MaConnell: That’s a difficult question. When they seal the box, there will be enough oxygen inside for the average man to take fifty breaths. Houdini says he knows how to breathe easily and save the oxygen. He is in good condition, but still he’s fifty two years old. I’d guess he might be able to get by without air for fifteen minutes.
Reporter: And after that?
Dr. MaConnell: After that Houdini had better telephone for help or ring his alarm bell – if he still has the strength.
Reporter: Are you doctors worried?
Dr. MaConnell: Well, we’re concerned. Houdini’s no fool. He isn’t trying to kill himself. But there’s always the danger that he’ll pass out before he can signal for help.
Reporter: There’s Houdini now! He’s wearing a black swimming suit. For a man his age, his build is great. He’s smiling. He seems sure of himself. Now he’s motioning for silence!
Houdini: Ladies and gentlemen, thank you for coming. As you know, it has often been written that a man can live only three minutes without air. I’m going to prove that that is wrong. If I die, it will be by the will of God and my own foolishness.
Reporter: Houdini’s in the box now. Several men are putting the iron cover in place. This is it! A man is soldering the cover. The box will be lowered into the shallow end of this pool. Then a team of swimmers will stand on the box to keep it level beneath the surface of the water. So far, the only person I’ve seen who doesn’t look nervous is Houdini. Well, they’re lowering the box. It’s under! Joseph Rinn, the official
timekeeper, has started the clock. Mr. Rinn, what are your plans?

Questions :
(i) Throw light on the conversation between the reporter and Dr. McConnell. What does it reveal?
(ii) What does the reporter tell us about Houdini’s imperial ordeal?

Answers :
(i) The reporter asks Dr. McConnell if he has checked Houdini. The doctor replies that his blood pressure and pulse rate are Both normal. The reporter asks the doctor how long he thinks Houdini will breathe. McConnell in his reply, states that the average man breaths fifty times in a sealed box, but Houdini is an exception. He knows the technique of breathing easily, but the doctor guessing he could remain without air for fifteen minutes. After that Houdini would be advised to seek help. The conversation between the two reveals their concern for Houdini.
(ii) The reporter apprises all of Houdini’s presence inside the box. After sealing it firmly, several men will put the box into the shallow end of the swimming pool. After that a team of swimmers will stand on it to maintain it level below the surface of the water. The reporter says that he has never seen such a person as Houdini, for he never looks nervous, despite this great ordeal. We also learn that Joseph Rinn, the official time-keeper has a look at the watch.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 4 The One and only Houdini

Question 2.
Read through the extract and answer the questions that follow.
Reporter: Well, Houdini is going past the limit suggested by Dr. Mc Connell! I wonder how his wife Bessie, is taking this. I haven’t been able to catch sight of her. She’s also a magician. After she married Houdini, they both had an act that played in many theatres. She dropped out after Houdini became famous for his escapes from handcuffs and jails throughout the country and in Europe. She still helped him, though. Once, after Houdini had been searched and handcuffed to the bars in a jail cell, she kissed him for good luck. It was more than good luck she offered. She also passed a lock pick from her mouth to his. Several minutes later, Houdini surprised his jailers by walking into their office.
Rinn: Twenty minutes!
Reporter: The tension here is growing. But tension has followed Houdini through most of his career. Take his underwater box escape, for example. He is handcuffed and chained. Then he’s put in a wooden box and it’s nailed shut. After that the box is dropped underwater. Minutes later Houdini swims to the surface. In that act, though, he uses tricks. He has lock picks and other special tools hidden on him or in the box. But there’s always the risk that the picks or tools will fail. Most people wouldn’t try Houdini’s escape tricks even if they knew how to do them.

Questions :
(i) Describe the part played by Bessie in Houdini’s life.
(ii) Describe of Houdini’s feat of underwater escape.

Answers :
(i) Houdini is widely known for his sensational escapes. Bessie, a magician, plays an important part in her husband’s life. Both had showed magic in many theatres. Houdini’s life is a saga of escape from handcuffs and jails in the length and breadth of the country and in Europe. Bessie stood by him through thick and thin. Once she kissed her husband for good luck. He was in a prison cell. In the course of offering her good luck, Bessie passed a small lock tool from her mouth to his. Several minutes later, the jailers, to their astonishment, saw Houdini walk into their office.
(ii) Houdini, handcuffed and chained, was put in a wooden box which was tightly sealed with nails. Then it was dropped underwater. After some minutes, Houdini swims to the surface safe and sound by using tricks, though he possesses lock picks and other special instruments not visible to others. This underwater box escape is one of Houdini’s wonderful feats.

Question 3.
Read through the extract and answer the questions that follow.
Reporter: Well, one can only wonder how he’s doing in the box. He still has a long time to go. Of course, he is unusual. Did you know that he can tie and untie rope with either foot? That won’t help him now, but it’s another example of his body control. He also has great control of his chest muscles. Before he’s bound into a straitjacket, he fills his chest with air. Later, when he relaxes, he can move around inside the jacket. Perhaps it sounds easy. But others have tried the same trick and failed. There’s only one Houdini!
Rinn: Thirty minutes!
Reporter: Oh – oh! There’s trouble! One of the swimmers standing on the box has lost his balance. He’s failing into the water! Now the other swimmers are falling, too. The box has bobbed up above the water level. I wonder if Houdini caused the box to move. Mr. Collins, is Houdini all right?
Collins: Quiet, sir, I’m calling Houdini. Harry, can you hear me?
Houdini: Yes.
Collins: The swimmers fell off the box. They’re getting back on now. That’s what caused the movement of the box. Are you all right?
Houdini: Fine.
Reporter: Mr. Collins, Houdini’s voice sounded faint. Is he well?
Collins: That’s what he said. I believe him.
Rinn: Thirty-five minutes!
Reporter: Houdini’s training seems to be paying off. According to sources around here, he has been training for three weeks. Also, he probably learned a lot about holding his breath while doing his underwater escapes. Anyway, it’s no secret that Houdini always trains long and hard before he does anything for the public. He wants to be sure that he can do anything he claims he will do. And right now he seems on his way to an amazing success.
Rinn: Forty minutes!
Reporter: When I was talking before about Houdini’s chest muscles, I failed to mention his stomach. Quite often, he lets local strongmen punch him in the stomach as hard as they can. Of course, his stomach muscles are tensed, but they have to be almost as hard as steel to withstand the fists of big, strong men.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 4 The One and only Houdini

Questions :
(i) Give an account of Houdini’s chest muscles and stomach muscles.
(ii) What picture of Houdini do you get in the extract?

Answers :
(i) The reporter presents a very clear picture of Houdini’s chest and stomach muscles. He eloquently refers to Houdini’s great control of his chest muscles. Before being bound into a straitjacket, he fill his chest with air. But, in times of relaxation, he has the ability to move around inside the jacket. The reporter then throws light on Houdini’s stomach muscles. Very often, he allows local strong men punch his Stomach in the hardest possible manner. His stomach muscles tensed, but they have to be as hard as steel to resist the fists of big, strong men.
(ii) We see Houdini through the reporter’s eyes, “Of course, he is unusual.” He has the ability to tie and untie rope with either foot. He is remarkable for his great control over his chest muscles. Before being bound into a straightjacket, he pumps air into his chest. In his moments of relaxation, he is capable of moving around inside the jacket. But others have tried the same trick but in vain. The reporter aptly remarks, ‘There ‘s only one Houdini.’ Houdini resorts to rigorous training before he does anything for the public.

Question 4.
Read through the extract and answer the questions that follow.
Collins: All right. Houdini, can you hear me? You have just passed an hour. The doctors are urging me to pull you out. What do you say?
Houdini I’ll let you know when I’m ready.
Rinn: One hour and one minute!
Collins: Well, you heard him, gentlemen. He seems to be doing well.
Reporter: Rinn’s going to be calling out each minute from here on. I’ll get back to him. Meanwhile, I’d like to hear from one of the doctors. Is this a world’s record, Doctor?
Doctor: As far as I know, it is. If I weren’t here, I don’t think I would believe it. Houdini’s an amazing man!
Reporter: He certainly is, Doctor, and thank you. Well, people are whispering to one another around the pool. When Houdini passed the hour mark, some of the crowd seemed to relax somewhat. Still Houdini is far from safe. I’m going to try to pick up the time.
Rinn: One hour and thirteen minutes!
Reporter: Did you hear that? What a man. Wait! Collins has a call from Houdini.
Collins: Had enough, sir?
Houdini: No. The box has sprung a leak. But there’s no danger. Only a slight trickle of water is coming in.
Collins: Are you sure?
Houdini: Yes.
Reporter: Houdini has nerves of steel. He’s under there with no air, and the water is leaking in on him. Yet he’s going to stay.
Rinn: One hour and twenty-five minutes.
Reporter: I must say, I wish he would come up. I’m probably more nervous than he is. Even Collins seems to be showing some concern.
Rinn: One hour and thirty minutes.
Reporter: This may be it! Collins has a call.
Houdini: Jim, get me up.
Reporter: They’re raising the box. Now they’re opening it. I wish they’d hurry.
Rinn: One hour and thirty-one minutes.
Reporter: Dr. McConnell has Houdini’s arms. He’s checking the magician’s pulse and blood pressure. How is he, Doctor?
Dr. McConnell: His pulse and blood pressure are very low. The man’s suffering from exhaustion.
Houdini: Nonsense! I feel a little dizzy, that’s all! If I do an hour or so of exercise. I’ll feel fine.
Reporter: Mr. Houdini, congratulations! How did you do it? Was it a trick?
Houdini: It was no trick. I took a series of deep breaths before the cover was soldered on. I made very few body movements while in the box and took short breaths. That’s all. I certainly hope that trapped coal miners and deep-sea divers take a lesson from this.
Reporter: How can they take a lesson from your performance. Mr. Houdini? You’re in perfect physical condition.
Houdini: They must learn not to panic. If you panic, your body needs more air. I have trained myself to remain calm in all situations. Everyone should do the same.

Questions :
(i) Describe Collin’s contact with Houdini.
(ii) What happens after Houdini comes out of the box?

Answers :
(i) The doctor and the reporter are concerned about Houdini, because there has been no sign from the man in the box yet. They insist on Collins, Houdini’s assistant, to talk to him. Collins informs him of the passage of an hour and the doctors pressure on him to pull him (Houdini) out. Collins waits for his reply. He learns that he is doing well and shares this information with them. Collins gets a call from Houdini concerning the leak in the box and a slight trickle of water coming in.
(ii) After Houdini comes out of the iron box in a triumphant fashion, the reporter congratulated him on remaining for one hour and thirty-one minutes in a sealed box and that too submerged in a swimming pool. He asks Houdini if he had resorted to a trick for his success. Houdini says that there was no trick. He had taken a series of deep breaths before the cover of the box was firmly fixed with solder. During his stay inside the box, he made few body movements and took short breaths. Above all, he has trained himself to remain calm in the face of any situation. In the reporter’s view, Houdini is a fantastic man.

Introducing the Author :
Dr. Robert Lado is one of the founders of contrastive linguistics which as a sub-disciple of applied linguistics served the purpose of improving language teaching material. He is a prolific writer. His works include over 60 books and many articles that deal with various topics ranging from linguistics to language testing and cross-cultural understanding.

About the Story :
‘The One and Only Houdini’, as the title signifies, throws light on Houdini’s miraculous act. Houdini’s second variation on Buried Alive was an endurance test designed to expose mystical Egyptian performer Rahman Bey, who claimed to use supernatural powers to remain in a sealed casket for an hour. Houdini bettered Bey on August 5, 1926, by remaining in a sealed casket, or coffin, submerged in the swimming pool of New York’s Hotel Shelton for one hour and a half. Houdini claimed he did not use any trickery or supernatural powers to accomplish this feat, just controlled breathing.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 4 The One and only Houdini

Summary :
It was August 5, 1926. All eyes were set on the swimming pool at New York’s Hotel Shelton. James Collins, Houdini’s assistant, Dr. Mc Connell and Joseph Rinn, the official time-keeper were present. People were gathering gradually at the pool. They were placing the iron box near the pool’s edge. It comprised a telephone and an alarm bell with batteries.

Harry Houdini clad in a black swimming suit appeared there. He was fifty-two, yet he was in good physical condition. Before getting inside the box, Houdini thanked everyone for coming to see his performance. He was going to disprove the theory that a man could live only three minutes without air. Houdini was inside the box now. Then it was tightly sealed. A team of swimmers would stand on the box before it was submerged in the swimming pool. The swimmers were instructed to keep the box level beneath the surface of the water as a result of which Houdini will not be moved backwards and forwards. The countdown started.

Houdini who was keen on going without air for an hour made him an object of attention. His success would come out in all newspapers all over the world. The reporter probed Houdini’s past. His real name was Ehrich Weiss, who was born in Appleton, Wisconsin. He was interested in magic. He became a professional magician under the name of Harry Houdini. He was married. Houdini was widely known for his sensational escape acts. Bessie had played a great role in her husband’s life. Minutes trickled by. Nobody knows what Houdini was doing inside the box. He had undertaken this ordeal after undergoing rigorous training for three weeks. Houdini possessed strong chest muscles. His tough stomach muscles resisted the fists of big, strong men.

With the passing of minutes, fear and anxiety gripped the people. There was no response from Houdini. The doctor in particular advised Collins to pull him out of the box when an hour was up. After interacting with Houdini, Collins declared that he seems to have smooth sailing. Doctor’s anxiety gave way to admiration. Houdini was an amazing man! Collins got a call from him. The box had developed a leak and only a slight trickle of water was entering. Everybody was understandably concerned. It was now one hour and thirty minutes. The box was opened. To everyone’s stunned disbelief, Houdini emerged out of it with flying colours. The reporter congratulated him. One key to his achievement lay in his ability to be calm in the face of any situation. The topic comes to close with the reporter’s fabulous remark, ‘You’re a fantastic man, Mr. Harry Houdini!’

ସାରାଂଶ :

Harry Houdini ଥିଲେ ଜଣେ ପ୍ରସିଦ୍ଧ ଯାଦୁକର । ତାଙ୍କ ସ୍ତ୍ରୀ Bessie ମଧ୍ୟ ଜଣେ ଯାଦୁକର ଥିଲେ । ସେମାନେ ଉଭୟ ଅନେକ ମ୍ୟାଜିକ୍ ସୋ ଦେଖାଇଛନ୍ତି । ତାଙ୍କର ସବୁଠୁ ଭଲ ସୋ ଥିଲା, Houdini ଙ୍କୁ ହ୍ୟାଣ୍ଟକପ୍ ପକାଇଦେଲେ ସେ ଆପଣାଛାଏଁ ହ୍ୟାଣ୍ଡକପ୍ ଖୋଲିଦେଇ ଚାଲିଆସୁଥିଲେ । ତାଙ୍କୁ ପାଣିଭିତରେ ବୁଡ଼ାଇ ରଖୁଲେ ସେ ଆପେ ଆପେ କୂଳରେ ଲାଗିଯାଉଥିଲେ । ତାଙ୍କର ଯାଦୁଖେଳର ୟୁରୋପରେ ଖୁବ୍ ନାଁ ଥିଲା । ଦିନେ Houdini ଘୋଷଣା କଲେ ଯେ ସେ ଗୋଟେ ନିବୁଜ ବାକ୍ସ ଭିତରେ ରହିବେ ଓ ବାକ୍ସଟି ପାଣିଭିତରେ ବୁଡ଼ାଯିବ । ବାକ୍ସ ଭିତରେ ଟେଲିଫୋନ୍, କଲିଂବେଲ ରହିବ । କିଛି ଲୋକ ବାକ୍ସଟିକୁ ପାଣିଭିତରେ ସ୍ଥିର ଭାବରେ ଧରି ରଖୁବେ । ବିଜ୍ଞାନ କୁହେ, ମଣିଷ ବିନା ବାୟୁରେ ମାତ୍ର ତିନିମିନିଟ୍ ରହିପାରିବ। ମାତ୍ର ସେ ପ୍ରମାଣ କରିଦେବେ ଯେ ସେ ବିନା ବାୟୁରେ ଏକ ଘଣ୍ଟାକାଳ ରହିପାରିବେ ।

Houdini ଙ୍କ ଅଲୌକିକ ଘଟଣା ଦେଖିବାକୁ ସେଦିନ ପୋଖରୀ କୂଳରେ ହଜାର ହଜାର ଲୋକ ଭିଡ଼ ଜମାଇଥାନ୍ତି । ତାଙ୍କ ସହକାରୀ Collins ସାମ୍ବାଦିକମାନଙ୍କୁ ତାଙ୍କ ଗୁରୁଙ୍କ ବିଷୟରେ ବୁଝାଉଥାଏ । Dr. Mcconnell ପ୍ରସ୍ତୁତ ଥାଆନ୍ତି, କାଳେ କିଛି ଅଘଟଣ ଘଟିବ । Rinn ପ୍ରସ୍ତୁତ ଥାଆନ୍ତି ସମୟ ଗଣିବେ । ପ୍ରତି ପାଞ୍ଚ ମିନିଟ୍‌ରେ ଥରେ ଲେଖାଏଁ ଘୋଷଣା କରିବେ । ଘଣ୍ଟାଏ ଡେଇଁଗଲାପରେ ଯଦି Houdini ନ ଆସନ୍ତି ଓ ଭଲ ଥାଆନ୍ତି, ତେବେ ସେ ତେଣିକି ପ୍ରତ୍ୟେକ ଏକ ମିନିଟ୍ ଘୋଷଣା କରିବେ । Houdini ଧୀର ଓ ହସହସ ମୁହଁରେ ଆସିଲେ । ସେ ସମବେତ ଜନତାଙ୍କୁ ଅଭିବାଦନ ଜଣାଇଲେ ଓ କହିଲେ, ମୁଁ ଆଜି ବିଜ୍ଞାନର ନିୟମକୁ ଭାଙ୍ଗିବାକୁ ଯାଉଛି । ଯଦି ସଫଳ ହେଲି ଭଲକଥା । ଯଦି ମରିଗଲି, ତା ହେଲେ ଜାଣିବ, ଈଶ୍ବର ବୋଧହୁଏ ମୋର ମୁର୍ଖମିକୁ ସହିଲେ ନାହିଁ । Houdini ବାକ୍ସରେ ପଶିଲେ । ତା’ପୂର୍ବରୁ ଡାକ୍ତର ତାଙ୍କର ହୃତ୍‌ସ୍ପନ୍ଦନ ନାଡ଼ି ପରୀକ୍ଷା କରି ଦେଖୁଥିଲେ ଯେ ସେ ସଂପୂର୍ଣ୍ଣ ସୁସ୍ଥ ଅଛନ୍ତି । ଏଥର ବାକ୍ସକୁ ଭଲଭାବରେ ନିବୁଜ କରାଗଲା । ତାକୁ ପାଣିଭିତରେ ବୁଡ଼ାଇ ରଖାଗଲା ଓ କିଛିଲୋକ ବାସଟିକୁ ପାଣିଭିତରେ ସ୍ଥିରକରି ରଖୁଲେ । ଏଣେ Rinn ଗଣନା ଆରମ୍ଭ କଲେ ।

ମିନିଟ୍ ପରେ ମିନିଟ୍ ଗଡ଼ିଚାଲିଲା । ସାମ୍ବାଦିକ ଜଣକ Collins ଓ ଅନ୍ୟମାନଙ୍କଠାରୁ Houdini ଙ୍କ ବିଷୟରେ ବିଭିନ୍ନ ଖବର ସଂଗ୍ରହ କରୁଥାନ୍ତି । ସେ ଜଣେ ଦକ୍ଷ ଯାଦୁକର । ସେ ନିଜ ଦେହରୁ ହାତକାଢ଼ି ନେଉଥିଲେ । ସେ ୧୮୭୪ ମସିହାରେ Appleton ରେ ଜନ୍ମ ନେଇଥିଲେ । ତାଙ୍କର ପ୍ରକୃତ ନାଁ Enrich Weiss I ତାଙ୍କ ପରିବାର ଖୁବ୍‌ ଗରିବ ଥିଲା । ଏଣୁ ସେ ପିଲାଟି ଦିନରୁ ପରିବାର ପୋଷିବାପାଇଁ କାମ କରୁଥିଲେ । ସେତେବେଳେ ସେ ହ୍ୟାଣ୍ଡକପ୍ ଖୋଲିବାର କଳାକୌଶଳ ଶିଖୁଥିଲେ । ସେ ମଧ୍ୟ ବେଳେବେଳେ ଫରାସି ଯାଦୁକର Robert Houdini ଙ୍କ ରଚିତ କିଛି ବହି ପଢ଼ି ଖୁବ୍ ପ୍ରଭାବିତ ହେଲେ ଓ ନିଜ ନାଁ ବଦଳାଇ ନିଜକୁ Harry Houdini ନାମରେ ପରିଣତ କରାଇଲେ । ସମୟ ଗଡ଼ି ଚାଲିଥାଏ । Rinn ପାଞ୍ଚ, ଦଶ, ପନ୍ଦର, କୋଡ଼ିଏ ଏମିତି ଗଣିଚାଲିଥାଆନ୍ତି । ଯେଉଁ ଲୋକମାନେ ବାକ୍ସକୁ ଧରିଥିଲେ, ସେମାନେ ଖସି ପଡ଼ିଲେ । ଲୋକମାନେ ଉତ୍କଣ୍ଠିତ ହୋଇପଡ଼ିଲେ । ଭାବିଲେ, Houdini ବୋଧହୁଏ ବାହାରକୁ ଆସିବାକୁ ଚେଷ୍ଟା କରୁଛନ୍ତି । Collins ଟେଲିଫୋନ୍ ଲଗାଇ ପଚାରିଲେ Houdini! | 66 ଅଛନ୍ତି ? ସେ ଉତ୍ତର ଦେଲେ, ଭଲ ଅଛି’’ । ଡାକ୍ତରମାନେ ମତଦେଲେ, ବାକ୍ସଭିତରେ ଯେତିକି ବାୟୁ ଅଛି ସେଥ‌ିରେ ଜଣେ ଲୋକ ପଚାଶ ଥର ନିଃଶ୍ବାସ ନେଇପାରିବ । ତା’ପରେ ସେ ନିଶ୍ଚୟ ଅଶ୍ୱସ୍ତି ଅନୁଭବ କରିବ । ସେ ନିଶ୍ଚୟ ବାକ୍ସ ଖୋଲିବାକୁ କହିବେ । Collins କହିଲେ – ମୁଁ ତାଙ୍କୁ ଯେତିକି ଜାଣିଛି, ସେ ସମୟ କେବେ ଆସିବ ନାହିଁ । କାରଣ Houdini କୌଣସି କଥାକୁ ପାଞ୍ଚ ଦଶଥର ପରୀକ୍ଷା ନ କରି ଜଣଙ୍କ ଆଗରେ ପେଶ୍ କରନ୍ତି ନାହିଁ ।

ବେଳକୁ ବେଳ ସାମ୍ବାଦିକଙ୍କ ଭିଡ଼ ଜମିଲା । ଡାକ୍ତରମାନେ ମଧ୍ୟ ଆସି ପହଞ୍ଚିଲେ । ସେମାନେ କହିଲେ – ଏହା ଏକ ଆତ୍ମଘାତୀ କାର୍ଯ୍ୟ । Houdini ଙ୍କୁ ଏଥୁରୁ ନିବୃତ୍ତ କରାଯାଉ । ଏହା ଭିତରେ ଘଣ୍ଟାଏ ବି ହୋଇଗଲା । ଲୋକମାନେ କିଛି ସମୟପାଇଁ ଖୁସି ଦେଖାଗଲେ । ଭାବିଲେ, ଏଥର Houdini ବାହାରକୁ ଆସିବାକୁ କହିବେ । ମାତ୍ର ସେମିତି କିଛି ହେଲା ନାହିଁ । ଡାକ୍ତରମାନେ ବାଧ୍ୟ କରିବାରୁ Collins ପୁଣିଥରେ ଟେଲିଫୋନ୍ ଲଗାଇ ପଚାରିଲେ – Houdini ! ଡାକ୍ତରମାନେ ବାଧ୍ୟକରୁଛନ୍ତି ବାହାରକୁ ଆସିବାପାଇଁ । Houdini ଉତ୍ତର ଦେଲେ – ମୁଁ କହିଲେ ମତେ ଉପରକୁ ନେବ । ଏଥର ଲୋକମାନେ କୁହାକୁହି ହେଲେ – Houdini ନିଃଶ୍ଵାସ ପ୍ରଶ୍ଵାସ ରୋଧ କରିବା ତାଲିମ ପାଇଛନ୍ତି । ସେ ବିନା ବାୟୁରେ ରହିବାର କୌଶଳ ଶିଖ୍ଯାଇଛନ୍ତି । ଏହା ଭିତରେ ଏକ ଘଣ୍ଟା ତିରିଶ ମିନିଟ୍ ହେଲା । ଏଥର Houdini ଙ୍କ ଠାରୁ ସଂକେତ ଆସିଲା ତାଙ୍କ ବାହାରକୁ ଆଣିବାପାଇଁ । ଲୋକମାନେ ଏକଘଣ୍ଟା ଏକତିରିଶ ମିନିଟ୍‌ରେ ବାକ୍ସ ଖୋଲିଲେ । ଡାକ୍ତରମାନେ Houdiniଙ୍କ ସ୍ବାସ୍ଥ୍ୟ ପରୀକ୍ଷା କଲେ । କିଛି ବ୍ୟତିକ୍ରମ ହୋଇନଥିଲା । ସାମ୍ବାଦିକମାନେ Houdiniଙ୍କୁ ଅଭିନନ୍ଦନ ଜଣାଇଲେ ।

ଏଥର ସାମ୍ବାଦିକମାନେ Houdini ଙ୍କୁ ପଚାରିଲେ ଏହା କେମିତି ସଂଭବ ହେଲା ? Houdini କହିଲେ ଏଥ‌ିରେ ଯାଦୁବିଦ୍ୟା ନାହିଁ । ଏହା ଏକ ସାଧନା ମାତ୍ର । ମୁଁ ନିଃଶ୍ଵାସ ରୋଧ କରିବା ଶିଖୁ ଯାଇଛି । ଏହା ମତେ ଏତେ ସମୟ ବାକ୍ସ ଭିତରେ ବଞ୍ଚାଇ ରଖୁ । ଶ୍ଵାସକ୍ରିୟା ଖୁବ୍ ମନ୍ଥର ଥିଲା । ସାମ୍ବାଦିକମାନେ ପଚାରିଲେ – ଆପଣ ଯୁବଗୋଷ୍ଠୀଙ୍କୁ କି ବାର୍ତ୍ତା ଦେବେ ? Houdini କହିଲେ – ସେମାନେ ଭୟ କରିବା ଛାଡ଼ି ଦିଅନ୍ତୁ । ସେମାନେ ଯେ କୌଣସି ପରିସ୍ଥିତିକୁ ସହଜ, ସରଳ ଓ ଶାନ୍ତ ଭାବରେ ସାମ୍ନା କରିବାକୁ ଶିଖନ୍ତୁ । ସମସ୍ତେ Houdini ଙ୍କର ଜୟଗାନ କଲେ ।

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(k) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(k)

Evaluate the following Integrals:
Question 1.
(i) \(\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan x}\)dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.1(1)

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\)dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.1(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k)

(iii) \(\int_0^1 \frac{\ln (1+x)}{2+x^2}\)dx (x = tan θ)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.1(3)

(iv) \(\int_0^\pi \frac{x d x}{1+\sin x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.1(4)

Question 2.
(i) \(\int_{-a}^a\)x4 dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.2(1)

(ii) \(\int_{-a}^a\)(x5 + 2x2 + x) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.2(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k)

(iii) \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\)cos2 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.2(3)

(iv) \(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin5 x dx
Solution:
Let f(x) = sin5 x
Then f(-x) = sin5 (-x)
= -sin5 x = -f(x)
So f(x) is an odd function.
Thus \(\int_{-a}^a\)f(x) dx = 0
\(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin5 x dx = 0

Question 3.
(i) \(\int_0^\pi\)cos3 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.3(1)

(ii) \(\int_0^\pi\)cos2 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.3(2)

(iii) \(\int_0^\pi\)sin3 x cos x dx
Solution:
\(\int_0^\pi\)sin3 x cos x dx
[Put sin x = t, then cos x dx = dt
When x = 0, t = 0, when x = π, t = 0
\(\int_0^\pi\)t3 dt = 0

(iv) \(\int_0^\pi\)sin x cos2 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.3(4)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k)

Question 4.
Show that
(i) \(\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}\) dx = \(\frac{\pi}{2} \ln \frac{1}{2}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.4(1)

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x}\) dx = 0
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.4(2)

(iii) \(\int_0^\pi\)x ln sin x dx = \(\frac{\pi^2}{2} \ln \frac{1}{2}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.4(3)

Question 5.
(i) \(\int_0^{\pi / 2}\)ln (tan x + cot x) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.5(1)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k)

(ii) \(\int_0^\pi \frac{x \tan x-\sin x}{1+\sin x \cos x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.5(2)

(iii) \(\int_1^3 \frac{\sqrt{x} d x}{\sqrt{4-x}+\sqrt{x}}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.5(3)

(iv) \(\int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.5(4)

(v) \(\int_0^1\)x (1 – x)100 dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.5(5)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k)

(vi) \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.5(6)

(vii) \(\int_0^{50}\)ex-[x] dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(k) Q.5(7)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(h) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(h)

Evaluate the following Integrals.
Question 1.
(i) ∫\(\frac{d x}{4+5 \cos x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.1(1)

(ii) ∫\(\frac{d x}{3+\cos x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.1(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h)

(iii) ∫\(\frac{d x}{3+\sin x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.1(3)

(iv) ∫\(\frac{d x}{1+2 \sin x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.1(4)

(v) ∫\(\frac{d x}{2 \sin x+3 \cos x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.1(5)

BSE Odisha

(vi) ∫\(\frac{d x}{1+\cos x+\sin x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.1(6)

Question 2.
(i) ∫\(\frac{3 \sin x+28 \cos x}{5 \sin x+6 \cos x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.2(1)

(ii) ∫\(\frac{12 \sin x-2 \cos x+3}{\sin x+\cos x}\) dx
Solution:
Let 12 sin x – 2 cos x = A (sin x + cos x) + B ( cos x – sin x)
[Note that cos x – sin x is the derivative of sin x + cos x]
Then A – B = 12, A + B = -2
⇒ 2A = 10
⇒ A = 5, B = -7
Thus 12 sin x – 2 cos x = 5 (sin x + cos x) – 7 (cos x – sin x)
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.2(2)

(iii) ∫\(\frac{5 \sin x}{3-2 \sin x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.2(3)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h)

(iv) ∫\(\frac{2 \cos x+7}{4-\sin x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.2(4)
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.2(4.1)

Question 3.
(i) ∫\(\frac{d x}{2 \cos ^2 x+3 \cos x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.3(1)

(ii) ∫\(\frac{d x}{4 \sin ^2 x-\sin x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.3(2)
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.3(2.1)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h)

(iii) ∫\(\frac{\sin x \cos x}{x \sin ^2 x-2 \sin x+3}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.3(3)

(iv) ∫\(\frac{d x}{\cos x-\cos 3 x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.3(4)

Question 4.
(i) ∫\(\frac{d \theta}{4+3 \sin ^2 \theta}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.4(1)

(ii) ∫\(\frac{d \theta}{2-3 \cos ^2 \theta}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.4(2)

(iii) ∫\(\frac{d \theta}{4 \cos ^2 \theta+9 \sin ^2 \theta}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.4(3)

(iv) ∫\(\frac{d \theta}{2+3 \cos ^2 \theta-4 \sin ^2 \theta}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.4(4)

Question 5.
(i) ∫\(\frac{\sin 3 x}{\cos 7 x \cos 4 x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.5(1)

(ii) ∫\(\frac{\cos 2 x}{\sin 7 x \cos 5 x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.5(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h)

Question 6.
(i) ∫\(\frac{d x}{\cos x(5+3 \cos x)}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.6(1)

(ii) ∫\(\frac{d x}{\cos x(1+2 \sin x)}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(h) Q.6(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(g) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(g)

Evaluate the following Integrals.
Question 1.
(i) ∫\(\frac{\sqrt{2 x+3}}{x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(1)

(ii) ∫\(\frac{\sqrt{x^2-7}}{x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g)

(iii) ∫\(\frac{\sqrt{x}}{x+2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(3)

(iv) ∫x\((3 x+2)^{\frac{1}{3}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(4)

(v) ∫\(\frac{x+2}{(2 x-1)^{\frac{1}{3}}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(5)

(vi) ∫(x + 2)\((x+1)^{\frac{1}{4}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(6)

(vii) ∫\(\frac{x-1}{(x+2)^{\frac{3}{4}}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(7)

(viii) ∫\(\frac{d x}{\sqrt{x}-\sqrt[3]{x}}\) (x = t6)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.1(8)

Question 2.
(i) ∫\(\frac{3 x+4}{\sqrt{2 x-3}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.2(1)

(ii) ∫(7x + 4)\(\sqrt{3 x+2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.2(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g)

(iii) ∫(3x + 1)\((x-2)^{\frac{9}{2}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.2(3)

(iv) ∫\(\frac{2 x+5}{(x+2)^{\frac{7}{2}}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.2(4)

(v) ∫\(\frac{x^2+2 x+1}{\sqrt{x+4}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.2(5)

(vi) ∫(x2 + 2x + 7)\(\sqrt{x+1}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.2(6)

Question 3.
(i) ∫\(\frac{d x}{\sqrt{4 x^2-4 x+5}}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.3(1)

(ii) ∫\(\sqrt{4 x^2-4 x+5}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.3(2)

(iii) ∫\(\frac{d x}{\sqrt{x^2-6 x+5}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.3(3)

(iv) ∫\(\sqrt{x^2-6 x+5}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.3(4)

(v) ∫\(\frac{d x}{\sqrt{1+2 x-x^2}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.3(5)

(vi) ∫\(\sqrt{1+2 x-x^2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.3(6)

Question 4.
(i) ∫\(\frac{d x}{(1+x) \sqrt{1-x^2}}\) (1 + x = \(\frac{1}{t}\))
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.4(1)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g)

(ii) ∫\(\frac{d x}{(2-x) \sqrt{5-4 x+x^2}}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.4(2)

Question 5.
(i) ∫\(\frac{d x}{(2 x+5) \sqrt{x+2}}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.5(1)

(ii) ∫\(\frac{1+x^2}{x \sqrt{x^4+1}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.5(2)

(iii) ∫\(\sqrt{\frac{x-1}{2 x+1}}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.5(3)

(iv) ∫\(\frac{x}{\left(a^2-x^2\right)\left(x^2-b^2\right)}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(g) Q.5(4)

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

Odisha State Board CHSE Odisha Class 11 Invitation to English 2 Solutions Non-Detailed Chapter 2 After Twenty Years Textbook Exercise Questions and Answers.

CHSE Odisha 11th Class English Solutions Non-Detailed Chapter 2 After Twenty Years

CHSE Odisha Class 11 English After Twenty Years Text Book Questions and Answers

Unit – I

Gist:
One day, in a locality of New York a smart policeman was on the route doing patrol duty. The time was only 10 o’ clock at night. People in the streets had gone home early because of a drizzle and chilly weather. The majority of the houses in the locality were business places that were also closed early. The entire place was dark but a few lights were coming from a cigar store, an all-night hotel and one or two tailoring shops completing the day’s work. The policeman with his stalwart figure and smart movements was inspecting the closed doors as he went in the street.

Every now and then he was casting his watchful eyes here and there and was searching a man seriously. His boss had said him in the morning while giving the description of the man wanted : “Pale face, square jaws, deep and dark eyes and a little white scar near the right eyebrow.” He was ashamed that the name of the notorious criminal who Was printing counterfeit notes was not known to the Police Department. The criminal had fled away from Chicago and was moving in New York. The policeman had much confidence on himself and he felt pride of being a dutiful officer for the last eighteen years.

He had a feeling of luck with him as a debt. He had kept the reconstructed photograph of the criminal and a five-dollar note in his pocket which he looked at under a lamp-post. The policeman was in the habit of talking to himself when nobody was nearby. He suddenly looked at his watch and it was a quarter past ten. There was another thing in his mind. He hurried up to finish his duty. After a minute he saw a man standing near the doorway of a hardware store. He walked speedily to the man. He could not see the face of the man clearly as it was dark.

He was just going to address the man but suddenly changed his mind and waited the man to begin. The man, on the other hand, who thought to be looked suspicious in the eyes of the police started talking from his side. He said to the policeman that he was waiting for a friend with whom an appointment was made twenty years ago. Though it appears a little funny, he said to the policeman, it was the truth. About twenty years ago there was a restaurant where this store stands. Its name was ‘Brady’s Restaurant’. The policeman agreed to this and said it was changed into a store only in five years ago. The policeman had a chance to see the face of the man when he struck a match and lit a cigar.

ସାରାଂଶ :
ନିଉୟର୍କ ସହରର ଏକ ଗଳିରେ ସବୁଦିନ ଭଳି ଦିନେ ଜଣେ ପୋଲିସ୍ ପହରା ଦେଉଥିଲେ । ସମୟ ରାତ୍ରି ୧୦ଟା ହୋଇଥାଏ । ଲୋକମାନେ ସେଦିନ ଝିପିଝିପି ବର୍ଷା ଓ ଥଣ୍ଡା ଫେରି ଯାଇଥିଲେ । ପାଗ ଯୋଗୁଁ ସନ୍ଧ୍ୟା ସମୟରେ ବେଳାବେଳି ଗୃହକୁ ଦୋକାନ ଗୃହଗୁଡ଼ିକ ବନ୍ଦ ଥିଲା । ସ୍ଥାନଟି ପ୍ରାୟ ସେହ ଗଳିରେ ସେଦିନ ରାସ୍ତାକୁ ଲାଗିଥିବା ଅନ୍ଧକାରମୟ ଦିଶୁଥିଲା; କିନ୍ତୁ ଗୋଟିଏ ସିଗାରେଟ୍ ଷ୍ଟୋର, ଅହୋରାତ୍ର ଖୋଲାଥିବା ଏକ ହୋଟେଲ ଏବଂ ଗୋଟିଏ କିମ୍ବା ଦୁଇଟି ଦରଜି ଦୋକାନରୁ ଆଲୋକ ଆସୁଥୁଲୀ । ପୋଲିସ୍ ବାବୁଜଣକ ତାଙ୍କ ବଳିଷ୍ଠ ଚେହେରା ଓ କ୍ଷିପ୍ର ଗତି ବଳରେ ସେହି ଗଳିର ବନ୍ଦ ଦ୍ଵାରଗୁଡ଼ିକୁ ଅନୁସନ୍ଧାନ କରୁଥିଲେ ।

ବହୁ ସମୟରେ ସେ ତାଙ୍କର ତୀକ୍ଷ୍ଣଣ ଦୃଷ୍ଟିକୁ ଏଠି ସେଠି ନିକ୍ଷେପ କରୁଥିଲେ । ସେ ଜଣେ ଲୋକକୁ ଜରୁରୀ ଭାବରେ ଖୋଜୁଥିଲେ । ତାଙ୍କ ଉଚ୍ଚପଦସ୍ଥ ଅଫିସର ଆଜି ସକାଳେ ସେହି ଲୋକର ବର୍ଣ୍ଣନାରେ କହିଥିଲେ ‘ଲୋକଟିର ଶେତା ମୁଖମଣ୍ଡଳ, ବର୍ଗାକାର ମାଢ଼ି, ଗଭୀର ଓ କଳା ଆଖ୍ ଏବଂ ଡାହାଣ ଆସ୍ପତା ପାଖରେ ଏକ ଛୋଟ ଧଳା ଦାଗ ।’’ ଲୋକଟି ଜଣେ ବଡ଼ଧରଣର ଅପରାଧୀ ଥିଲା ଏବଂ ସେ ଚିକାଗୋରେ ଜାଲନୋଟ୍ ଛାପୁଥୁଲା । କିନ୍ତୁ ତାଙ୍କୁ ଲଜ୍ଜା ଲାଗୁଥିଲା କାରଣ ପୋଲିସ୍ ବିଭାଗକୁ ଏଭଳି ଏକ ଅପରାଧୀର ନାମ ଜଣା ନ ଥିଲା । ଏହି ଅପରାଧୀ ଜଣକ ଚିକାଗୋରୁ ଆସି ନିଉୟର୍କରେ ଥବର ସୂଚନା ମିଳିଥିଲା । ପୋଲିସ୍ ବାବୁଜଣଙ୍କର ନିଜ ଉପରେ ବହୁତ ବିଶ୍ୱାସ ଥିଲା ଏବଂ ସେ ଅଠର ବର୍ଷ ହେଲା ଜଣେ କର୍ତ୍ତବ୍ୟନିଷ୍ଠ ଅଫିସରଭାବେ କାର୍ଯ୍ୟ କରି ଆସିଥ‌ିବାରୁ ବହତୁ ଗର୍ବିତ ଥିଲେ । ତାଙ୍କ ସାଙ୍ଗରେ ଭାଗ୍ୟ ଏକ ଋଣ ଆକାରରେ ଅଛି ବୋଲି ସେ ଅନୁଭବ କରୁଥିଲେ ।

ସେ ତାଙ୍କ ପାଖରେ ଏକ ପାଞ୍ଚ ଡଲାର ନୋଟ୍ ଏବଂ ସେ ଅପରାଧୀର ଅଙ୍କା ଫଟୋ ରଖୁଥିଲେ ଏବଂ ସମୟ ସମୟରେ ତାକୁ ଦେଖୁଥିଲେ । ପୋଲିସ୍ ବାବୁଜଣକ କେହି ନ ଥିଲାବେଳେ ନିଜ ସହିତ କଥା ହେଉଥିଲେ । ହଠାତ୍‌ ହାତଘଣ୍ଟା ଉପରେ ନଜର ହେଉଥିଲେ । ହଠାତ୍ ହାତଘଣ୍ଟା ଉପରେ ନଜର ଗୋଟିଏ କଥା ମଧ୍ୟ ତାଙ୍କ ମନରେ ଥିଲା । ସେ ଶେଷ କରିବାପାଇଁ ତତ୍ପର ହୋଇଉଠିଲେ । କିଛି ସମୟ ପରେ ସେ ଦେଖ‌ିଲେ ଜଣେ ବ୍ୟକ୍ତି hardware ପୋଲିସ୍ ବାବୁଜଣକ କେହି ନ ଥିଲାବେଳେ ନିଜ ସହିତ କଥା ପକାଇ ଦେଖିଲେ ସମୟ ଦଶଟା ବାଜି ୧୫ ମିନିଟ୍ ହେଲାଣି । ଅନ୍ୟ ନିଜର କାର୍ଯ୍ୟକୁ ଶେଷ କରିବାପାଇଁ ତତ୍ପର ହୋଇଉଠିଲେ ।

କିଛି ସମୟ ପରେ ସେ ଦେଖ‌ିଲେ ଜଣେ ବ୍ୟକ୍ତି hardware ଦୋକାନ ଆଗରେ ଠିଆ ହୋଇଛନ୍ତି । ସେ ତାଙ୍କ ପାଖକୁ ଶୀଘ୍ର ଗଲେ, କିନ୍ତୁ ଅନ୍ଧାର ହେତୁ ସେ ତାଙ୍କ ମୁହଁ ଦେଖିପାରୁ ନ ଥିଲେ । ସେ ଲୋକଜଣକୁ କିଛି କହିବାକୁ ଚାହୁଁଥିଲେ; କିନ୍ତୁ ସଙ୍ଗେ ସଙ୍ଗେ ମନ ବଦଳାଇ ରହିଗଲେ । ଲୋକଜଣକ ନିଜକୁ ପୋଲିସ୍ ସନ୍ଦେହରୁ ମୁକ୍ତ କରିବାପାଇଁ ନିଜଆଡ଼ୁ କଥା ଆରମ୍ଭ କଲେ । ସେ ପୋଲିସ୍‌ବାବୁଙ୍କୁ କହିଲେ ସେ ଏଠି ଜଣେ ସାଙ୍ଗକୁ ଅପେକ୍ଷା କରିଛନ୍ତି ଯାହା ସହିତ ଆଜି ଦିନରେ ଭେଟ ହେବାପାଇଁ ୨୦ ବର୍ଷ ତଳେ ସେମାନେ କଥା ହୋଇଥିଲେ । କଥାଟା ବଡ଼ ମଜାଳିଆ ହେଲେ ବି ସତ୍ୟ ଥିଲା । ୨୦ ବର୍ଷ ତଳେ ଏହି ସ୍ଥାନରେ ଏକ ଭୋଜନାଳୟ ଥିଲା ଯାହାର ନାମ ଥିଲା Brady’s Restaurant । ପୋଲିସ୍ ବାବୁଜଣକ ତାଙ୍କ କଥାରେ ସମ୍ମତ ହେଲେ ଓ କହିଲେ ଏହି ପାଞ୍ଚ ବର୍ଷ ତଳେ ଏହାକୁ ଏକ ଷ୍ଟୋରରେ ପରିଣତ କରାଯାଇଛି । ଲୋକଟି ଠିକ୍ ଏହି ସମୟରେ ଦିଆସିଲି କାଠି ମାରି ସିଗାରେଟ୍ ଲଗାଇବାବେଳେ ପୋଲିସ୍ ଜଣଙ୍କ ତାଙ୍କ ମୁହଁ ସାମାନ୍ୟ ଦେଖୁ ପାରିଥିଲେ ।

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

Glossary:
wit (n): the ability to say or write things that are both clever and amusing She is a women of wit and intelligence.
wordplay: use of words (ଶବ୍ଦର ବ୍ୟବହାର) He makes better wordplay in his writings.
characterization: characters in a book or play seem real (ଖେଳ ବାସ୍ତବ ମନେହୁଏ). The success of the play depends much on its characterization.
surprise: unexpected event or happenings, a feeling of astonishment (ଆଶ୍ଚର୍ଯ୍ୟ) Her position in the top ten was a surprise in the last H.S.C. examination.
value: worth (ମୂଲ୍ୟ) We should better understand the value of life.
relationship: relation among people, friendship (ଲୋକମାନେ, ବନ୍ଧୁତା) He should regard our relationship.
triumph: victory, win (ଜିତନ୍ତୁ) Indians are happy with our World-Cup triumph.
on the beat: on the route doing patrol duty (ପାଟ୍ରୋଲିଂ କରୁଥିବା ମାର୍ଗରେ | କର୍ତ୍ତବ୍ୟ)
quiet road: road free from noise (ଶବ୍ଦରୁ ମୁକ୍ତ ରାସ୍ତା) He was walking freely on a quiet road.
habitual: doing something continuously (କ୍ରମାଗତ ଭାବରେ କିଛି କରିବା)
spectators: unlookers, viewers (ଦର୍ଶକ) Spectators sit in the gallery to watch the match.
kept early hours : went back home early in the night (ଶୀଘ୍ର ଘରକୁ ଫେରିଗଲେ ରାତିରେ)
chilly winds: too cold winds (ଅତ୍ୟଧିକ ଥଣ୍ଡା ପବନ) We could not go out because of chilly winds blowing outside.
drizzle: dribble, raining lightly (ହାଲୁକା ବର୍ଷା) There was a drizzle in the morning.
depeopled: emptied of people (ଲୋକମାନେ) The streets of the city were depeopled by the evening as curfew was declared.
examining : inspecting, searching, checking (ଯାଞ୍ଚ, ଖୋଜିବା, ଯାଞ୍ଚ କରୁଛି )
playful : full of fun (ମଜା ରେ ପରିପୂର୍ଣ୍ଣ ) His activities were mostly playful.
movements: moving (ଗତିଶୀଳ) We cannot feel the movements of earth.
small stick : (here) lathi (ଏଠାରେ) ଲାଥୀThe policeman always moves with a small stick or lathi.
stalwart: sturdy, well-built (ବଳିଷ୍ଠ, ସୁଗଠିତ) He is a man of stalwart figure.
guardian of the law: protector of law (ଆଇନର ରକ୍ଷକ) The policeman is a guardian of the law.
tailoring shop : dressmaker’s shop (ଡ୍ରେସମେକରଙ୍କ ଦୋକାନ)
watchful eyes: vigilant eyes (ସଜାଗ ଆଖି) Nobody can escape from his watchful eye.
pale face : whitish face (ଧଳା ଚେହେରା)
square jaws : jaws having equal sides (ସମାନ ପାର୍ଶ୍ୱ ଥିବା ଜହ୍ନଗୁଡିକ)
white scar: a white mark left on the skin (ଧଳା ଦାଗ)
ashamed: felt shame (ଲଜ୍ଜା ଅନୁଭବ କଲା) He is ashamed of his bad manner.
notorious: famous for bad quality (କୁଖ୍ୟାତ) Terrorists are notorious criminals.
counterfeit: illegal (ନକଲି)
serious affair: grave matter (ଗୁରୁତର ବିଷୟ) Terrorism is a serious affair of our country.
lamp-post: a post giving light (ଆଲୋକ ପ୍ରଦାନ କରୁଥିବା ଏକ ପୋଷ୍ଟ) He was standing under a lamp-post and waiting for a friend.
five-dollar note : ପାଞ୍ଚ ଡଲାରର ନୋଟ୍
reconstructed photograph: picture of a person drawn by experts basing on the reports about his or her face (ଚିତ୍ରିତ ବ୍ୟକ୍ତିଙ୍କ ଚିତ୍ର | ବିଶେଷଜ୍ଞଙ୍କ ଦ୍ ରା)The reconstructed photograph of the leader of the terrorist was broadcasted in T.V
described: saw (from a distance) ଦେଖ (ଦୂରରୁ)
suspicious: doubtful (ସନ୍ଦେହଜନକ) Police can detain suspicious people for investivation.

Think it out:

Question 1.
How does the writer describe the atmosphere of the story?
Answer:
The story begins with a policeman doing the patroling duty in a street of New York. The time was 10 o’ clock at night. The strict was almost calm and quiet and people had gone home in the early hours of the evening because of bad weather. The business houses were mostly closed and darkness has covered the major part of the street. It was all due to chilly winds and light raining. Few lights were coming from a cigar store, an whole-night hotel and one or two tailoring shops completing the day’s work. The area coming under the lamp-posts were little lighted. Thus, the atmosphere was somewhat unsuitable and unfavourable.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

Question 2.
How did the policeman perform his duties?
Answer:
The polieman was on the route doing patrol duty. He was moving smartly. His smartness was real and not for the show. It was night and most of the business houses in the street were closed early due to bad weather. There was darkness everywhere except light coming from few shops and lamp-posts. In spite of this, the policeman was inspecting every closed doors. He was holding a lathi and making some funny movements. The policeman, with his well-built figure and smart movements was proving himself as the guardian of law. He was very watchful and cast his eyes on the entire peaceful road. He was seriously searching a notorious criminal from Chicago. This duty was assigned to him by his boss and he was very confident as a dutiful officer for the last eighteen years to do the job perfectly.

Question 3.
What picture of the wanted criminal do you get from the text?
Answer:
The picture of the wanted criminal given to the polcieman in the morning of the said day was as follows. The criminal had a pale face, square jaws, deep and dark eyes, and a little white scar near the right eyebrow. The criminal’s name was not known and he was involved in a serious affair of printing counterfeit notes. The criminal was from Chicago. The policeman had a five dollar note and a reconstructed photograph of the criminal in his pocket.

Question 4.
How did the stranger try to interact with the policeman?
Answer:
The stranger, who was standing in front of a darkened hardware store, saw a policeman coming towards him. He thought that the policeman would suspect him as he did not know his story. So when the policeman walked upto him, he said to him that he was just waiting for a friend. It was an appointment made twenty years ago. It was a truth, though it seemed funny. To remove the doubts of the policeman, he narrated that about twenty years ago there was a restaurant where this store stands. Its name was “Brady’s Restaurant”. Then the policeman said that it was changed into a store just five years ago. In this way the stranger interacted with the policeman.

Question 5.
How did the policeman see the stranger’s face?
Answer:
The stranger standing at the doorway of the hardware store, was interacting with the policeman. As the place was dark the face of the stranger could not be seen. In the meanwhile, he wanted to smoke and thus struck a match and lit his cigar. This provided a chance to the policeman to see the face of the man.

Unit – II

Gist :
The man narrating about his friend said that he had last dined with his friend Jimmy Wells at Brady’s Restaurant twenty years ago tonight. Jimmy was his best friend and the finest man in the world. He said that they were bom and brought up in New York just like two brothers. When they left each other he was eighteen and Jimmy was twenty. He went to Chicago in the West to make my fortune. But Jimmy, as liked to stay at home, was unwilling to go the West with him. For Jimmy New York was the best place to live in on the earth.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

Accordingly they took our last parting dinner at Brady’s Restaurant that night and agreed that they would meet here again exactly twenty years from that date and time in spite of their distant living or whatever conditions of life of theirs. They also felt that in twenty years each of them have made his career and met the destiny of life. The policeman listening to it said that it was very interesting. He asked the man about their contacts between this long gap. The man said that they were in touch for a year or two. But after that they lost contact of each other and remained busy in their own affairs. The man said that though Chicago was his headquarter, he kept on moving here and there.

That time he had come to meet his friend after a long interval and he had deep faith to meet his friend there. He said that his friend was very sincere and true to his words. He would never forget it. His coming to that place from a distance of thousand miles would be fruitful if he could meet his old friend there. His friend would come difinitely. Saying so, he looked at his diamond watch and it was ten twenty-five. He remembered the time that was exactly half past ten when they parted here at the restaumt door. Then the policeman wanted to know whether he had earned lots of money in the West.

He admitted of his good earnings and expressed his hope that his friend Jimmy also had earned the half of his income. He said that his friend, though a nice man, was very slow in his earnings. But he had to compete with the most cunning people of the world to earn his dollars. A man in the New York becomes very ordinary but the West makes a man to face much competition. The policeman while leaving asked the man whether he would wait his friend or not, if he would not come at the appointed time. He said that he would give his friend half an hour to come. If he was alive on earth he would be there by that time. After that, the policeman took leave of this man by wishing ‘Good-night’.

ସାରାଂଶ :
ଲୋକଟି ତାଙ୍କ ସାଙ୍ଗ ବିଷୟରେ ବର୍ଣ୍ଣନା କରି କହିଲେ, ସେ ଆଜକୁ କୋଡ଼ିଏ ବର୍ଷତଳେ ଏହି ବ୍ରାଡ଼ି ଭୋଜନାଳୟରେ ଆଜି ରାତିରେ ଶେଷଥର ପାଇଁ ମିଶି ଖାଇଥିଲେ । ଜିମି ତାଙ୍କର ସବୁଠାରୁ ଉତ୍ତମ ବନ୍ଧୁ ଥିଲା ଏବଂ ପୃଥ‌ିବୀରେ ଅତି ଭଲ ଲୋକମାନଙ୍କ ମଧ୍ୟରେ ସେ ଜଣେ । ସେ କହିଲେ, ଆମେ ଦୁଇଜଣ ନିଉୟର୍କରେ ଦୁଇ ଭାଇ ଭଳି ଜନ୍ମ ହୋଇ ବଢ଼ିଥିଲୁ । ଆମେ ଯେତେବେଳେ ପରସ୍ପରଠାରୁ ଅଲଗା ହେଲୁ ମୋତେ ଅଠର ବର୍ଷ ଏବଂ ଜିହ୍ନିକୁ କୋଡ଼ିଏ ବର୍ଷ ହୋଇଥିଲା ।

ମୁଁ ମୋର ଭବିଷ୍ୟତ ଗଢ଼ିବାକୁ ଚିକାଗୋ ଗଲି । ଜିଛି କିନ୍ତୁ ଘର ଛାଡ଼ି କୁଆଡ଼େ ଯିବ ନାହିଁ ବୋଲି କହିଲା । ନିଉୟର୍କ ତା’ପାଇଁ ପୃଥ‌ିବୀର ସବୁଠାରୁ ଭଲ ସ୍ଥାନ ଥିଲା । ସେହି ଅନୁଯାୟୀ ଆମେ ଶେଷଥର ପାଇଁ ବ୍ରାଡ଼ି ଭୋଜନାଳୟରେ ରାତ୍ରିଭୋଜନ ଖାଇଲୁ ଏବଂ ରାଜି ହେଲୁ ଆଜକୁ ୨୦ ବର୍ଷ ପରେ ଏହି ଦିନ ଏହି ତାରିଖ ଏହି ସମୟ ଏବଂ ଏହି ସ୍ଥାନରେ ଆମେ ଯେକୌଣସି ପରିସ୍ଥିତିରେ ଥିଲେ ବି ଆସି ଭେଟ ହେବା । ସେଦିନ ପରସ୍ପରଠାରୁ ଅଲଗା ହେଲାବେଳକୁ ସମୟ ଥିଲା ରାତ୍ର ୧୦ ଟା ୩୦ ମିନିଟ୍ ।

ଆମେ ମଧ୍ୟ ଭାବିଲୁ ୨୦ ବର୍ଷ ପରେ ଆମେ ନିଜର ଭବିଷ୍ୟତ ଗଢ଼ି ସାରି । ଲୋକଟିର କଥାକୁ ପୋଲିସ୍ ବାବୁଜଣକ ଶୁଣୁଥିଲେ ଏବଂ କହିଲେ ଏହା ତ ବଡ଼ କୌତୁହଳପ୍ରଦ କଥା ! ସେ ପଚାରିଲେ, ଆଚ୍ଛା ଏହି କୋଡ଼ିଏ ବର୍ଷ ମଧ୍ୟରେ ଆପଣଙ୍କ ଭିତରେ କୌଣସି ସମ୍ପର୍କ ନ ଥିଲା କି ? ଲୋକଜଣକ କହିଲେ, ହଁ ଥିଲା, ତାହା କେବଳ ପ୍ରଥମ ବର୍ଷେ କିମ୍ବା ୨ ବର୍ଷ ମଧ୍ୟରେ ଏବଂ ପରେ ପରେ ଆମେ ନିଜ ନିଜ କାମରେ ବ୍ୟସ୍ତ ରହିଗଲୁ ।

ଲୋକଟି କହିଲା, ଚିକାଗୋ ମୋର ମୁଖ୍ୟ କାର୍ଯ୍ୟାଳୟ ହେଲେ ବି ମୁଁ ଚାରିଆଡ଼େ ଘୂରି ବୁଲୁଥୁଲି । ଆଜି ସେ ଆସିଛି ତା’ର ସାଙ୍ଗକୁ ଭେଟିବାପାଇଁ ସୁଦୂର ଚିକାଗୋରୁ ହଜାର ମାଇଲ ଅତିକ୍ରମ କରି ଏବଂ ସେ ନିଶ୍ଚୟ ଭେଟିବ । ସେ କହିଲେ ମୋର ସାଙ୍ଗ ଅତ୍ୟନ୍ତ ସତ୍ୟବାନ୍ ଏବଂ ସଚ୍ଚା ଏବଂ ସେ ନିଶ୍ଚୟ ମୋତେ ଭେଟିବାକୁ ଆସିବ । ଏହା କହି ଲୋକଟି ତାଙ୍କ ହାତରେ ଥ‌ିବା ହୀରା ଘଣ୍ଟାଟିକୁ ଚାହିଁଲେ । ସେତେବେଳକୁ ୧୦ଟା ୨୫ ମିନିଟ୍ ହୋଇଥିଲା । ସେ ମନେ ପକାଇଲେ ସେ ସେଦିନ ରାତ୍ର ୧୦ଟା ୩୦ ମିନିଟ୍‌ରେ ସେମାନେ ପରସ୍ପରଠାରୁ ଅଲଗା ହୋଇଥିଲେ । ତା’ପରେ ପୋଲିସ୍ ବାବୁଜଣଙ୍କ ଚିକାଗୋରେ ତାଙ୍କ ରୋଜଗାର ବିଷୟରେ ଜାଣିବାକୁ ଚାହିଁଲେ । ସେ ସେଠାରେ ତାଙ୍କ ଭଲ ରୋଜଗାର କଥା ମାନିଲେ ଏବଂ ଆଶା କଲେ ତାଙ୍କ ସାଙ୍ଗ ମଧ୍ଯ ଅତି କମ୍‌ରେ ତାଙ୍କ ରୋଜଗାରର।

Think it out

Question 1.
What was the appointment made between two friends twenty years before?
Answer:
The two friends in the story are Bob and Jimmy Wells. Both were born and brought up in New York and bred like two brothers. When they come of age, Bob went to the West to make his fortune but Jimmy stayed in New York. On the day of parting, they had dined together in a restaurant and agreed that night that they would meet here again exactly twenty years from that date and time in spite of their distant living and conditions. This appointment was made between two friends twenty years ago.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

Question 2.
What information about the friends do you gather from their conversation?
Answer:
After listening to the man about their appointment that was made twenty years ago, the policeman expressed his curiosity to know more about them. He asked the man about their correspondence, if any, during this long gap. The man said that they were in touch for a year or two but after that they lost contact of each other. It was perhaps due to increase of work and personal involvement. He said that though Chicago was his main place of living, he moved extensively in and outside of it. Then shifting his mind to his friend the man said that his friend Jimmy was always very sincere and true to his words.

He would never forget their appointment. He would meet him definitely. He had come from a distance of thousand miles to meet him and it would be worthwhile if his old friend comes back. Saying so he looked at his diamond watch and the time was ten twenty-five. He said that it was exactly half past ten when they parted there at the restaurant door twenty years ago. This Brady’s Restaurant had been changed into a hardware store. The policeman asked about his good income in the West which he admitted. He also expressed that his friend Jimmy would be earning at least half of his income.

Though a good human being, Jimmy was very slow in brain. He had to compete with the cleverest people of the world in Chicago for his livelihood. He said that life in the West was very fast and one had to struggle hard for earning and living. But in New York one became very ordinary. The policeman listened to all these and said him that his friend would be coming around. Before leaving him, he asked him whether he would wait for his friend or not. The man said that he would wait for his friend for an hour only. He would definitely come by that time. In short, two friends naturally exchanged pleasant words and talked old times without recognizing each other.

Unit – III

Introduction:
In this part of the story “After Twenty Years” the readers get surprised by coming across an unexpected situation. It gives a kind of displeasure as well as pleasure to the readers when they read about the arrest of a friend by another friend who has done it for the sake of his duty at the cost of personal relationship. Readers express sympathy on the friend who had come to meet his friend from a long distance after twenty years and got arrested by his friend. They also admire the another friend who out of his devotion to duty got his friend arrested by keeping personal relationship away from duty. Let us read the story to know the interesting happenings.

Gist:
The cold wind was blowing severely with the rain continuing all over the street. The few people who were out had hurried home. The man, who had come from thousand miles to meet an appointment at the doorway of the hardware store, was smoking and waiting his friend Jimmy Wells. About twenty minutes of his waiting a tall man wearing a long overcoat with collar turned upto his ears appeared before the man. “Is that you, Bob ? ” he asked doubtfully. “Is that you, Jimmy Wells ?” asked the man.

They shook their hands. Bob said that he was sure to find him there if he was alive. Jimmy said that they would have dined again if that old restaurant had existed. Then, he asked his friend about his days in the West. He said that the West had given him every thing. The man also enquired about Jimmy. Jimmy said that he was well and working in a departmental store as assitant manager. He was getting good salary and other benefits. Then he told Bob to go round to a place and talk about their old days. The two friends walked on the street holding arm in arm. Bob was talking of his possessions and of his important friend’s in Chicago elsewhere. The other friend wearing a overcoat was listening with deep attention.

They could not see each other’s face in the darkness. They came near a medicine store located in a lighted area. There they could see each other’s face. Bob stopped suddenly and freed his arm from his friend’s arm. He said to tha man that he was not Jimmy. Twenty years is a long time but not long enough to change a man’s sharp nose to a flat one. The tall man said that it sometimes changes a good man into a bad one. Then he arrested Bob and told that the Commissioner of Police, Chicago wanted to talk with Bob in connection with some bundles of five dollar counterfeit notes. He took Bob to the police station.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

While going to the police station, the tall man handed over a letter to Bob by saying that it was from Assistant Sub-Inspector Jimmy Wells. The friend from Chicago opened the letter and it was a short note. “Bob: I was at the apointed place on time. When you struck the match to light your cigar I saw it was the face of the man wanted in Chicago. Somehow I could not do it myself, so I went around and got a plain-clothes policeman to do the job.” It was revealed from the story that though Jimmy knew his friend, the wanted criminal he made him arrested. He ignored his personal relationship when it came in the way of his duty. His devotion to duty was so much that it undermined his personal relationship. Only a few people could do it and Jimmy Wells was one of them.

ସାରାଂଶ :
ନିଉୟର୍କ ସହରର ସେହି ଗଳିରେ ସେଦିନ ଥଣ୍ଡା ବଢ଼ି ବଢ଼ି ଯାଉଥିଲା ଏବଂ ତା’ ସାଙ୍ଗକୁ ବର୍ଷା ମଧ୍ଯ । ଯେଉଁ କେତେକ ଅଳ୍ପ ଲୋକ ବାହାରେ ରହିଥିଲେ ସେମାନେ ମଧ୍ୟ ଘରକୁ ଶୀଘ୍ର ଫେରି ଗଲେଣି । କିନ୍ତୁ ହଜାର ମାଇଲ ଅତିକ୍ରମ କରି ସୁଦୂର ଚିକାଗୋରୁ ନିଜ ବନ୍ଧୁଙ୍କୁ ୨୦ ବର୍ଷ ପରେ ଦେଖା କରିବାକୁ ଆସିଥିବା ଲୋକଟି ସେହି hardware ଷ୍ଟୋର ଆଗରେ ଛିଡ଼ା ହୋଇଥିଲା । ସେ ସିଗାରେଟ୍ ଟାଣୁଥିଲେ ଏବଂ ସାଙ୍ଗ ଜିୱି ୱେଲସ୍ ଆସିବା ବାଟକୁ ଅନାଇ ବସିଥିଲେ । ତାଙ୍କ ଅପେକ୍ଷା କରିବାର ପ୍ରାୟ ୨୦ ମିନିଟ୍ ପରେ ଜଣେ ଡେଙ୍ଗା ଲୋକ ସାଧା ପୋଷାକରେ ଏକ ଲମ୍ବାକୋର୍ଟ ପିନ୍ଧି ତାଙ୍କ ସମ୍ମୁଖରେ ପହଞ୍ଚିଲେ।

‘ତୁମେ ବବ୍ କି ?”’ ସେ ସହେନ୍ଦରେ ପଚାରିଲେ । ‘ତୁମେ କ’ଣ ଜିହ୍ନି ୱେଲସ୍ ?”’ ଲୋକଟି ପଚାରିଲା । ପରିଚୟ ମିଳିଥିବାରୁ ସେମାନେ ହାତ ମିଳାଇଲେ । ବବ୍ କହିଲେ, ‘ମୁଁ ନିଶ୍ଚିନ୍ତ ଥୁଲି, ତୁମେ ଯଦି ବଞ୍ଚାଅ ଆସିବ ।’’ ଜିହ୍ନି କହିଲେ, ‘ଆଜି ଯଦି ସେ ପୁରୁଣା ଭୋଜନାଳୟଟି ଥାଆନ୍ତା ଆମେ ପୁଣିଥରେ ସେଠି ଭୋଜନ କରିଥା’ନ୍ତେ । ଆଚ୍ଛା ହେଉ, ତୁମେ ପଶ୍ଚିମରେ କେମିତି ଅଛ କୁହ।’’ ଉତ୍ତରରେ ବବ୍ କହିଲେ, ‘ପଶ୍ଚିମାଞ୍ଚଳ ମତେ ସବୁକିଛି ଦେଇଛି । ତୁମେ ନିଉୟର୍କରେ କେମିତି ଅଛ ?’’ ସେ କହିଲେ ‘ମୁଁ ଭଲରେ ଅଛି ଏବଂ ଏକ ଡିପାର୍ଟମେଣ୍ଟାଲ ଷ୍ଟୋରରେ ଉପମ୍ୟାନେଜର ଭାବରେ କାମ କରୁଛି । ମୁଁ ଭଲ ଦରମା ସହ ଅନ୍ୟାନ୍ୟ ସୁବିଧା ମଧ୍ଯ ପାଉଛି ।’’ ତା’ପରେ ସେ କହିଲେ ‘ବବ୍, ଆସ ଆଗକୁ ଟିକେ ବୁଲିଯିବା ଏବଂ ଆମର ସେହି ପୁରୁଣା ଦିନଗୁଡ଼ିକ ବିଷୟରେ କଥା ହେବା ।’’ ଦୁଇ ଜଣଯାକ ସାଙ୍ଗ ହାତ ଧରାଧରି ହୋଇ ଗଳି ରାସ୍ତାରେ ଆଗେଇ ଚାଲିଲେ ।

ବବ୍ ଚିକାଗୋରେ ଥିବା ତାଙ୍କର ଧନ ସମ୍ପତ୍ତି, କୋଠାବାଡ଼ି ଏବଂ ଧନୀ ଓ ପ୍ରତିପତ୍ତିଶାଳୀ ସାଙ୍ଗମାନଙ୍କ ବିଷୟରେ କହି ଚାଲିଲେ । ଅନ୍ୟ ସାଙ୍ଗଟି ସାଧା ପୋଷାକ ଉପରେ ଏକ ଓଭରକୋର୍ଟ ପିନ୍ଧି ବାଟ ଚାଲୁଥିଲାବେଳେ ତାଙ୍କ କଥା ମନଦେଇ ଶୁଣୁଥା’ନ୍ତି । ସେମାନେ ପରସ୍ପରର ମୁହଁକୁ ଅନ୍ଧାରରେ ଦେଖ୍ ପାରୁନଥିଲେ । ସେମାନେ ଆସ୍ତେ ଆସ୍ତେ ଆଲୋକିତ ସ୍ଥାନରେ ଥ‌ିବା ଏକ ଔଷଧ ଦୋକାନ ପାଖରେ ପହଞ୍ଚିଲେ । ସେଠାରେ ସେମାନେ ପରସ୍ପରର ମୁହଁକୁ ଦେଖିପାରିଲେ । ବବ୍ ହଠାତ୍ ରହିଗଲେ ଏବଂ ସାଙ୍ଗ ହାତରୁ ହାତ କାଢ଼ିନେଲେ ଏବଂ କହିଲେ ‘ଆପଣ ତ ଜିଛି ୱେଲସ୍ ନୁହଁନ୍ତି । କୋଡ଼ିଏ ବର୍ଷ ଏକ ଲମ୍ବା ସମୟ ହୋଇପାରେ, କିନ୍ତୁ ଏତେ ଲମ୍ବା ନୁହେଁ ଯେ ଏକ ବ୍ୟକ୍ତିର ଗୋଜିଆ ନାକକୁ ଚେପ୍‌ଟା ବବ୍ ହଠାତ୍ ରହିଗଲେ ଏବଂ ସାଙ୍ଗ ହାତରୁ ହାତ କାଢ଼ିନେଲେ ଏବଂ କହିଲେ ‘ଆପଣ ତ ଜିଛି ୱେଲସ୍ ନୁହଁନ୍ତି । କୋଡ଼ିଏ ବର୍ଷ ଏକ ଲମ୍ବା ସମୟ ହୋଇପାରେ, କିନ୍ତୁ ଏତେ ଲମ୍ବା ନୁହେଁ ଯେ ଏକ ବ୍ୟକ୍ତିର ଗୋଜିଆ ନାକକୁ ଚେପ୍‌ଟା ନାକରେ ବଦଳାଇ ଦେବ ।’’

ଡେଙ୍ଗା ଲୋକଟି କହିଲେ, ‘ଏହି ଲମ୍ବା ସମୟ ବେଳେବେଳେ ଜଣେ ଭଲ ଲୋକକୁ ଖରାପ ଲୋକ କରିଦିଏ । ବର୍ତ୍ତମାନ ତୁମକୁ ଆରେଷ୍ଟ କରାଗଲା। ଚିକାଗୋର ପୋଲିସ୍ କମିଶନର ତୁମ ସାଙ୍ଗରେ କଥା ହେବାପାଇଁ ଅପେକ୍ଷା କରିଛନ୍ତି । ସେ ତୁମ ସହିତ କିଛି ଜାଲନୋଟ୍ ବିଷୟରେ କଥା ହେବେ । ତୁମେ କୌଣସି ଚାଲବାଜି ନ କରି ମୋ ସାଙ୍ଗରେ ଆସ ।’’ ପୋଲିସ୍ ଷ୍ଟେସନ୍‌କୁ ଗଲାବାଟରେ ସାଧା ପୋଷାକଧାରୀ ପୋଲିସ୍‌ ତାଙ୍କ “‘ବବ୍, ମୁଁ ଠିକ୍ ସମୟରେ ନିର୍ଦ୍ଧାରିତ ସ୍ଥାନରେ ଥୁଲି । ତୁମେ ଯେତେବେଳେ ସିଗାରେଟ୍ ଲଗାଇବା ପାଇଁ ଦିଆସିଲି କାଠି ଲଗାଇଲ ମୁଁ ସେହି ଆଲୋକରେ ତୁମ ମୁହଁକୁ ଦେଖୁଲି ଓ ଜାଣିଲି ଏ ହେଉଛି ସେହି ମୁହଁ ଯାହାକୁ ଚିକାଗୋରେ ମୋଷ୍ଟ ୱାଣ୍ଟେଡ୍ ତାଲିକାରେ ରଖାଯାଇଛି ।ଯାହାହେଉ ମୁଁ ସେ କାମ କରିପାରିଲିନି । ତେଣୁ ମୁଁ ଚାଲିଗଲି ଏବଂ ଏକ ସାଧାପୋଷାକଧାରୀ ପୋଲିସ୍ ଜରିଆରେ ତୁମକୁ ବନ୍ଦୀ କଲି ।’’ ଏହି ଗଳ୍ପରୁ ପ୍ରତୀୟମାନ ହେଉଛି ଯେ ଜିୱି ଏକ ପୋଲିସ୍ ଭାବରେ ନିଜର କର୍ତ୍ତବ୍ୟ ସମ୍ପାଦନ କରିବାକୁ ଯାଇ ତାଙ୍କର ଜଣେ ଅପରାଧୀ ବନ୍ଧୁଙ୍କୁ ବନ୍ଦୀ କଲେ । କର୍ତ୍ତବ୍ୟ ଆଗରେ ବ୍ୟକ୍ତିଗତ ସମ୍ପର୍କ ଆସିଲେ ମଧ୍ୟ ସେ ପ୍ରଥମଟିକୁ ଗୁରୁତ୍ଵ ଦେଲେ । ଏହା କେବଳ ଖୁବ୍ କମ୍ ଲୋକଙ୍କ ପକ୍ଷରେ ସମ୍ଭବ ହୋଇଥାଏ ଏବଂ ଜିମି ୱେଲସ୍ ସେମାନଙ୍କ ମଧ୍ୟରୁ ଥିଲେ ଜଣେ ।

Glossary:
unexpected: not expected An unexpected (ଅପ୍ରତ୍ୟାଶିତ) danger puzzled him.
twist: put into confusion (ଦ୍ୱନ୍ଦ୍ୱରେ ପକାନ୍ତୁ) He is twist and is unable to take a decision about his future.
hardware store: deals with hardware (ହାର୍ଡୱେର୍) He is a manager of a hardware store.
puffed up: feeling proud (ଗର୍ବିତ ଅନୁଭବ କରୁଛି )He was puffed up by the his success in the civil examination.
sure as fate: very certain (ବହୁତ ନିଶ୍ଚିତ) His achievement was sure as fate.
rapt: deep (ଗଭୀର) He listened to the discussion with rapt attention.

Think it out:

Question 1.
How long did the man from the West wait? What was weather then?
Answer:
The man from the West waited about twenty minutes and then a tallman in a long overcoat came from the opposite side of the street to meet him. The weather then was very bad. The wind was too cold and the rain was continuing. People who were out hurried up home.

Question 2.
How did the man from the West and the man in a long overcoat greet each other?
Answer:
When the man from the West was waiting at the doorway of the hardware store, the man in a long overcoat appeared. Looking at the man at the hardware store he doubtfully asked, “Is that you, Bob ?” “Is that you, Jimmy Wells V asked Bob. Then they could know each other and shook hands. “Bless my heart !” exclaimed the new arrival. In this way they greeted each other.

Question 3.
When did Bob realise that the tall man he had met was not Jimmy Wells?
Answer:
When Bob and the tall man reached at the medicine store walking arm in arm, they could see each other’s face in the electric lights. Bob suddenly stopped by leaving his arm. He thought that how this tall man with flat nose would be Jimmy Wells who had a sharp nose. One could be 6 to 8 inch long in twenty years time but it is not long enough to convert a sharp nose to a flat nose. Thus Bob realised that the tall man he had met was not Jimmy Wells.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

Question 4.
Why did Jimmy get Bob arrested?
Answer:
Jimmy, who was a policeman, was a man of principle, honest and dutiful. Bob, on the other hand, was a professional criminal who printed counterfeit notes and earned easy money. Once upon a time in twenty years ago they were good friends. Before they got separated, they had promised to meet each other at the Brady’s Restaurant where they had taken their last dinner. But when they meet, Jimmy saw the
face of the man who was wanted in Chicago. The responsibility and devotion of his duty made him to ignore his personal relationship. Therefore he got him arrested.

Question 5.
Did Jimmy keep his appointment with Bob after twenty years? What consideration did he show for his old friend?
Answer:
Yes, Jimmy did so because he was exactly on time at the hardware store where they decided to meet after twenty years. He kept his appointment. At the same time he was assigned the duty to catch a notorious criminal from Chicago. He had the reconstructed photograph of the criminal with him which matched the face of his friend. So he did not disclose his identity. As a policeman he considered his duty more than his friendship. Thus instead of enjoying with his old friend Bob after twenty years, he made him arrested by another policeman’ who played the role of Jimmy Wells. Though he did not arrest his friend directly he did it by another policeman. This could be his only

CHSE Odisha Class 11 English After Twenty Years Important Questions and Answers

Question 1.
Read through the extract and answer the questions that follow.
Examining closed doors as he went, making various interesting and playful movements with his small stick, the officer, with his stalwart figure and smart movements made a fine picture of the guardian of the law. He could see a few lights coming from a cigar store, an all-night hotel, and one or two tailoring shops completing the day’s work. The majority of the houses were business places that had long since been closed. Now and then he would suddenly turn about, and cast his watchful eye along the peaceful road. He was thinking of what his boss had said in the morning: “Pale face, square jaws, deep and dark eyes, and a little white scar near the right eyebrow.” As a part of the police organization, he felt a little ashamed that the name of this notorious criminal was still unknown, and that he was still at large. Printing counterfeit notes was a serious affair, he knew, and the criminal must be caught. He took out his pocket-book under a lamp-post and looked at the five-dollar note and the reconstructed photograph. He became thoughtful: Chicago – a thousand miles away. What chance ?” Then he said to himself aloud, ‘Never mind, my boy, you have been a dutiful one these last eighteen years, and luck owes you a debt.’ He was in the habit of talking to himself, sometimes,- when nobody was by. And he liked to refer to himself as a boy, though he was forty. There was another thing in his mind. He looked at his watch, – a quarter past ten; fifteen minutes more. It was a long road; but there was enough time. After a minute he descried about five hundred metres ahead the outline of a man near the doorway of a darkened hardware store. He quickened his steps. The place was slightly dark, lying exactly midway between two lamp-posts. He was about to address the man, but changed his mind and allowed the man to begin. The man realised that the way he was standing there must look suspicious to one who didn’t know his story. So he said, as the policeman walked up to him: ‘It’s all right, officer; I’am just waiting for a friend. It is an appointment made twenty years ago. It sounds a little funny to you, doesn’t it? Well, I’ll explain so as to remove all suspicion from your mind: About twenty years ago there used to be a restaurant where this store stands, “Brady’s Restaurant.”

Questions :
(i) Describe the policeman’s feelings while doing patrol duty?
(ii) ‘There was another thing in his mind.’ What was that?

Answers :
(i) The policeman was assigned a duty to catch a die-hard criminal. He was in search of that notorious criminal. He was quite sure to be successful in performing his duty. He thought that he had been a sincere and dutiful policeman since last eighteen years and luck always remained with him. At the same time he felt a little ashamed that the name of such a notorious criminal was still unknown to the police department. Thus, the policeman had serious feelings while doing patrol duty.
(ii) Though police man was assigned a duty to catch a notorious criminal, still there was another thing in his mind. Perhaps he had to meet someone or to do something on a particular time. Therefore, he looked at his watch and said that there was more fifteen minutes in his hand.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 2 After Twenty Years

Question 2.
Read through the extract and answer the questions that follow.
The two men started up the streets, arm in arm. The man from the West, somewhat puffed up by success, was talking of his possessions and of his important friends in Chicago and elsewhere. The other, covered with his overcoat, was listening with rapt attention. They had not yet taken a good look at each other’s face. They neared a medicine store lit brilliantly, with electric lights. There each of them turned simultaneously to gaze at the other’s face. The man from the West stopped suddenly and let go the other’s arm.
‘You’re not Jimmy Wells’, he faltered. ‘Twenty year is a long time, but not long enough to change a man’s sharp nose to a flat one?’
‘It sometimes changes a good man into a bad one’, said the tall man.
‘You’ve been under arrest for ten minutes, Bob, or whatever your name is. The Commissioner of police, Chicago, longs to have a chat with you in connection with some bundles of five-dollar notes, which have come his way. Come quietly, please, and no tricks. Before we gQ to the station, here’s a note I was asked to give you. It’s from Assistant Sub-Inspector Jimmy Wells.’ The man from the West unfolded the little piece of paper handed to him. His hand was steady when he began to read, but it trembled a little by the time he had finished. The note was rather short.
‘Bob: I was at the appointed place on time. When you struck the match to light your cigar I saw it was theface of the man wanted in Chicago. Somehow I couldn ’t do it myself, so I went around and got a plain-clothes policeman to do the job.’

Questions :
(i) Give a picture of Bob and Jimmy when they were in the streets?
(ii) Comment on the ending of the story.

Answers :
(i) We find the two friends, Bob and Jimmy, in the streets walking arm in arm. Bob who had now come from the West was in high spirits. Unprecedented success had made him as if he were over the moon. He kept on talking of his wealth and of his great friends in Chicago. Jimmy, covered with his overcoat, was listening with deep attention. In spite of being with each other for some time, they had not looked distinctly at each other. The inevitable happened. They gazed at each other’s face at the same time.
(ii) The two friends, Bob and Jimmy, have kept their appointment after twenty years, but there is a turning point in the end. Bob is shocked in disbelief when the man tells him that he has been under arrest. The man talking to him is asked to hand over him a note which outlines Jimmy’s arrival at the appointed place on time and recognition of his face wanted in Chicago while lighting his cigar and his inability to arrest his friend and hence through a plain-clothes policeman. This is the consideration Jimmy Wells showed for his old friend Bob. Honest and dutiful as he is, Jimmy Wells has done his job perfectly. The ending is superb for its ‘O’ Henry Twist’ or suprising ending.

Introducing the Author:
William Sydney Porter was an American writer. He was known by his pen name O’ Henry. He lived for a successful period of forty-eight years. Before his death, he had written a good number of short stories. All his short stories are known for their wit, wordplay, warm characterization and surprise.

About the Story: 
Now-a-days people mostly value their personal relationship. They even undermine their duty when it clashes with their personal relationship. But there are still people who prefer duty to their relationship. Duty is god to them. O’ Henry describes this story to show – ‘Devotion to duty triumps over personal relationship’. In the story, there are two friends who gets separated at a point of time and meet again after a period of twenty years. But the duty of a friend compels him to undermine his relationship. A friend in the way of his duty has been forced to arrest his friend. The story is narrated in three units which depicts the irony and pathos of life in a subtle and dramatic manner.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(f) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(f)

Evaluate the following Integrals.
Question 1.
(i) ∫\(\frac{4 x-9}{x^2-5 x+6}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.1(1)

(ii) ∫\(\frac{3 x}{(x-4)(x+2)}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.1(2)

(iii) ∫\(\frac{5 x-12}{(2 x-3)(x-6)}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.1(3)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f)

(iv) ∫\(\frac{20 x+3}{6 x^2-x-2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.1(4)

(v) ∫\(\frac{2 x^2}{(x-1)(x-2)(x-3)}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.1(5)

(vi) ∫\(\frac{12 x^4-2 x^3-4 x^2+x-3}{6 x^2-x-2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.1(6)

Question 2.
(i) ∫\(\frac{2 x+9}{(x+3)^2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.2(1)

(ii) ∫\(\frac{5 x^2+4 x+4}{(x+2)(x+2)^2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.2(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f)

(iii) ∫\(\frac{x^2+7 x+4}{x^3+x^2-x-1}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.2(3)

(iv) ∫\(\frac{x^4+3 x^3+x^2-1}{x^3+x^2-x-1}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.2(4)

Question 3.
(i) ∫\(\frac{4 x^2-x+3}{\left(x^2+1\right)(x-1)}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.3(1)

(ii) ∫\(\frac{5 x}{\left(x^2-2 x+2\right)(x+1)}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.3(2)

(iii) ∫\(\frac{3}{x^3-1}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.3(3)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f)

(iv) ∫\(\frac{x^5+x^4+x^3+x^2+4 x+1}{x^3+1}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.3(4)
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.3(4.1)

Question 4.
(i) ∫\(\frac{d x}{x^2-5}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.4(1)

(ii) ∫\(\frac{d x}{2 x^2+8 x+7}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.4(2)

(iii) ∫\(\frac{x+3}{2 x^2+8 x+7}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.4(3)

(iv) ∫\(\frac{4 x^2+20 x+25}{2 x^2+8 x+7}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.4(4)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f)

(v) ∫\(\frac{e^x}{e^{2 x}+3 e^x+1}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.4(5)

(vi) ∫\(\frac{\tan ^2 \theta+1}{\tan ^2 \theta-1}\) dθ
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.4(6)

Question 5.
(i) ∫\(\frac{d x}{3-x^2}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.5(1)

(ii) ∫\(\frac{d x}{7-x^2+6 x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.5(2)

(iii) ∫\(\frac{x-5}{7-x^2+6 x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.5(3)

(iv) ∫\(\frac{\cos \theta}{3-\sin ^2 \theta}\) dθ
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.5(4)

(v) ∫\(\frac{x^2 d x}{\left(x^2+3\right)\left(x^2+2\right)}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.5(5)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f)

(vi) ∫\(\frac{x^3 d x}{x^4+3 x^2+2}\) (Put x2 = t)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.5(6)

(vii) ∫\(\frac{d x}{\sin x(3+2 \cos x)}\) (Put cos x = z)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(f) Q.5(7)

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଧାତୁ ଓ ଅଧାତୁ

Odisha State Board BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଧାତୁ ଓ ଅଧାତୁ Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Physical Science Solutions Chapter 3 ଧାତୁ ଓ ଅଧାତୁ

Question 1.
ନିମ୍ନଲିଖତ କେଉଁ ଯୋଡ଼ା ବିସ୍ଥାପନ ପ୍ରତିକ୍ରିୟା କରେ ।
(a) ସୋଡ଼ିୟମ୍‌ କ୍ଲୋରାଇଡ୍ ଦ୍ରବଣ ଏବଂ ତମ୍ବାଧାତୁ ।
(b) ମ୍ୟାଗ୍ନେସିୟମ୍ କ୍ଲୋରାଇଡ୍ ଦ୍ରବଣ ଏବଂ ଏଲୁମିନିୟମ୍ ଧାତୁ ।
(c) ଫେରସ୍ ସଲ୍‌ଫେଟ୍ ଦ୍ରବଣ ଏବଂ ସିଲ୍‌ଭର୍ ଧାତୁ । ତମ୍ବା ଧାତୁ ।
(d) ସିଲ୍‌ଭର୍ ନାଇଟ୍ରେଟ୍ ଦ୍ରବଣ ଏବଂ ଧାତୁ
Answer:
(d) ସିଲ୍‌ଭର ନାଇଟ୍ରେଟ୍ ଦ୍ରବଣ ଓ ତମ୍ବା

Question 2.
ନିମ୍ନଲିଖ କେଉଁ ପ୍ରଣାଳୀ ଲୁହା ତାୱାକୁ କଳଙ୍କି ନିରୋଧପାଇଁ ବିନିଯୋଗ କରିହେବ ?
(a) ଗ୍ରୀକ୍ ପ୍ରୟୋଗ ଦ୍ଵାରା
(b) ରଙ୍ଗ ପ୍ରଲେପ ଦ୍ଵାରା
(c) ଜିଙ୍କର ଆବରଣ ପ୍ରୟୋଗ ଦ୍ଵାରା
(d) ତପତୋକ୍ର ସମସ୍ତ ପ୍ରଣାଳ1 ଦ୍ବ|ରା
Answer:
(c) ଜିଙ୍କର ଆବରଣ ପ୍ରୟୋଗ ଦ୍ଵାରା ।

Question 3.
ଗୋଟିଏ ମୌଳିକ ଅମ୍ଳଜାନ ସହ ପ୍ରତିକ୍ରିୟା କରି ଗୋଟିଏ ଉଚ୍ଚ ଗଳନାଙ୍କ ବିଶିଷ୍ଟ ଯୌଗିକ ସୃଷ୍ଟି କରେ । ଏହା ଜଳରେ ଦ୍ରବଣୀୟ । ନିମ୍ନ ଉତ୍ତରରୁ ଉକ୍ତ ମୌଳିକଟି ବାଛ ।
(a) କ୍ୟାଲ୍‌ସିୟମ୍‌
(b) କାରନ୍
(c) ପିଲିକନ୍
(d) ଆଇନ୍
Answer:
(a) କ୍ୟାଲ୍‌ସିୟମ୍

Question 4.
ଖାଦ୍ୟ ଡବାଗୁଡ଼ିକରେ ଜିଙ୍କଦ୍ୱାରା ପ୍ରଲେପ ନ ହୋଇ ଟିଣରେ ହୋଇଥାଏ କାରଣ–
(a) ଜିଙ୍କ୍ ଟିଣଠାରୁ ଅଧ‌ିକ ମୂଲ୍ୟବାନ ଅଟେ ।
(b) ଜିଙ୍କ୍ ଟିଣଠାରୁ ଉଚ୍ଚ ଗଳନାଙ୍କ ବିଶିଷ୍ଟ ।
(c) ଜିଜ୍ ଟିଣଠାରୁ ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ।
(d) ଜିଙ୍କ୍ ଟିଣଠାରୁ କମ୍ ପ୍ରତିକ୍ରିୟାଶୀଳ ।
Answer:
(c) ଜିଙ୍କ୍ ଟିଣଠାରୁ ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ।

Question 5.
ତୁମକୁ ଗୋଟିଏ ହାତୁଡ଼ି, ଗୋଟିଏ ବ୍ୟାଟେରୀ, ଗୋଟିଏ ବଲ୍‌ , ତାର ଏବଂ ଗୋଟିଏ ସ୍ଵିଚ୍ ଦିଆଯାଇଛି ।
(a) ଧାତୁ ଏବଂ ଅଧାତୁର ନମୁନାକୁ ଚିହ୍ନଟ କରିବାପାଇଁ ଏଗୁଡ଼ିକୁ କିପରି ବ୍ୟବହାର କରିବ ?
(b) ଏହି ପରୀକ୍ଷାର ଉପଯୋଗିତାକୁ ନିର୍ଣ୍ଣୟ (Assess) କରି ଧାତବ ଏବଂ ଅଧାତବ ମଧ୍ୟରେ ପାର୍ଥକ୍ୟ ଦର୍ଶାଅ |
Answer:
(a)
(i) ପ୍ରଦତ୍ତ ନମୁନାକୁ ହାତୁଡ଼ି ସାହାଯ୍ୟରେ ପିଟିଲେ ଯଦି ତାହା ପ୍ରସାରିତ ହୋଇ ପତଳା ତେବେ ତାହା ଧାତୁ । ଯଦି ନମୁନାଟି ଭାଙ୍ଗି ଗୁଣ୍ଡ ହୋଇଯାଏ ତେବେ ତାହା ଅଧାତୁ ।
(ii) ଧାତୁକୁ ହାତୁଡ଼ିରେ ଆଘାତ କଲେ ସେଥୁରୁ ଧ୍ଵନି ବାହାରେ, କିନ୍ତୁ ଅଧାତୁରୁ କୌଣସି ଧ୍ବନି ବାହାରେ ନାହିଁ ।

(b) ବ୍ୟାଟେରୀ, ବଲ୍‌ବ, ତାର ଓ ସ୍ଵିକୁ ନେଇ ଚିତ୍ରରେ ପ୍ରଦର୍ଶିତ ହେବାଭଳି ପରିପଥ ପ୍ରସ୍ତୁତ କରାଯାଉ, ପରିପଥର A ଓ B ପ୍ରାନ୍ତରେ ନମୁନାକୁ ଲଗାଇ ସୁଇଚ୍ ON କଲେ ଯଦି ବଲ୍‌ବ ଜଳିବ ତେବେ ନମୁନାଟି ଧାତୁ, ନଚେତ୍ ଅଧାତୁ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-1

Question 6.
ଏମ୍ପୋଟେରିକ୍ ଅକ୍‌ସାଇଡ୍ କହିଲେ କ’ଣ ବୁଝ ? ଏହି ଅକ୍‌ସାଇଡ୍‌ର ଦୁଇଟି ଉଦାହରଣ ଦିଅ ।
Answer:
ଯେଉଁ ଧାତବ ଅକ୍‌ସାଇଡ୍ ଉଭୟ ଅମ୍ଳ ଓ କ୍ଷାରକ ସହ ରାସାୟନିକ ପ୍ରତିକ୍ରିୟା କରି ଲବଣ ଓ ଜଳ ସୃଷ୍ଟିକରେ ତାହାକୁ ଉଭୟ ଧର୍ମୀ ଅକ୍‌ସାଇଡ୍ ବା ଏମ୍ପୋଟେରିକ୍ ଅକ୍‌ସାଇଡ କହନ୍ତି ।

ଉଦାହରଣ :
ଜିଙ୍ଗ୍ ଅକ୍‌ସାଇଡ୍ (ZnO) ଓ ଆଲୁମିନିୟମ୍ ଅକ୍‌ସାଇଡ୍ (Al203) ଦୁଇଟି ଏମୋଟେରିକ୍ ଅକ୍‌ସାଇଡ୍ ।
ZnO + 2HC → ZnCl2 + H2O
ZnO + 2NaOH → Na2 ZnO2 + H2O
Al2O3 + 6HCI → 2AlCl3 + 3H2O
Al2O3 + 2NaOH → 2NaAlO2 + H2O

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 7.
ଦୁଇଟି ଧାତୁର ନାମ ଦର୍ଶାଅ ଯାହାକି ହାଇଡ୍ରୋଜେନ୍‌କୁ ଲଘୁ ଅମ୍ଳରୁ ବିସ୍ଥାପନ କରେ ଏବଂ ଦୁଇଟି ଧାତୁର ନାମ ଦର୍ଶାଅ ଯାହା ଏପରି କରେ ନାହିଁ ।
Answer:
(i) ସକ୍ରିୟତା ଅନୁ କ୍ରମରେ ହାଇଡ୍ରୋଜେନ୍ ଉପରେ ଥ‌ିବା ଯେକୌଣସି ଦୁଇଟି ମୌଳିକ ନେଲେ ତାହା ହାଇଡ୍ରୋଜେନକୁ ବିସ୍ଥାପନ କରିବ । ଲଘୁ ଅମ୍ଳରୁ ହାଇଡ୍ରୋଜେନ୍, ମ୍ୟାଗ୍ନେସିୟମ୍ ଓ ଆଲୁମିନିୟମ୍‌କୁ ବିସ୍ଥାପନ କରିବ ।
(ii) କପର ଓ ସିଲ୍‌ଭର ଲଘୁ ଅମ୍ଳରୁ ହାଇଡ୍ରୋଜେନ୍‌କୁ ବିସ୍ଥାପନ କରେ ନାହିଁ ।

Question 8.
ବୈଦ୍ୟୁତିକ ଶୋଧନ ଗୋଟିଏ ଧାତୁ M ପାଇଁ, ଏନୋଡ୍, କ୍ୟାଥୋଡ୍ ଏବଂ ବୈଦ୍ୟୁତିକ ବିଶ୍ଳେଷ୍ୟ ରୂପେ କାହାକୁ ନିଆଯିବ ?
Answer:
ଅଶୋଧ ଧାତୁ Mକୁ ଏନୋଡ୍ ରୂପେ,ବିଶୁଦ୍ଧ ଧାତୁ Mକୁ କ୍ୟାଥୋଡ୍ ରୂପେ ଏବଂ ଧାତୁ Mର ଲବଣକୁ ବୈଦ୍ୟୁତିକ ବିଶ୍ଳେଷ୍ୟରୂପେ ନିଆଯିବ ।

Question 9.
ଜଣେ ଚେପ୍‌ଟା ଚାମଚରେ ସଲ୍ଫର ପାଉଡ଼ର ନେଇ ଉତ୍ତପ୍ତ କଲା ଏବଂ ଗୋଟିଏ ପରୀକ୍ଷାନଳୀକୁ ଓଲଟାଇ ବାହାରୁଥ‌ିବା ଗ୍ୟାସ୍‌କୁ ସଂଗ୍ରହ କଲା । (ଚିତ୍ରରେ ପ୍ରଦର୍ଶିତ ହେଲାପରି)
(a) ଗ୍ୟାସ୍‌ର କ୍ରିୟାଶୀଳତା
(i) ଶୁଷ୍କ ଲିଟମସ୍ କାଗଜ ଉପରେ କ’ଣ ହେବ ?
ଉ : ଶୁଷ୍କ ଲିଟମସ୍ କାଗଜ ଉପରେ କୌଣସି ପ୍ରତିକ୍ରିୟା ହେବ ନାହିଁ । ଅର୍ଥାତ୍ ଗ୍ୟାସ୍‌ର କ୍ରିୟାଶୀଳତା ଅପରିବତ୍ତତ ରହିବ ।

(ii) ଆର୍ଦ୍ର ଲିଟମସ୍ କାଗଜ ଉପରେ କ’ଣ ହେବେ ?
ଉ : ଆର୍ଦ୍ର ନୀଳ ଲିଟମସ୍ କାଗଜ ଲାଲ୍ ହେବ ।

(b) ସୃଷ୍ଟି ହୋଇଥ୍ ପ୍ରତିକ୍ରିୟାର ଏକ ସମତୁଲ
ଉ :
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-2
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-3

Question 10.
ଲୁହାର କଳଙ୍କି ନିରୋଧ ପାଇଁ ଦୁଇଟି ଉପାୟ ଲେଖ ।
Answer:
(a) ଗାନାଇଜିଙ୍ଗ୍ (ଜିଙ୍କ୍ ଲେପନ) : ଏହି ପ୍ରଣାଳୀରେ ଜିଙ୍କର ଏକ ପତଳା ସ୍ତରର ଆଚ୍ଛାଦନ ଦ୍ଵାରା ଲୁହାକୁ କଳଙ୍କି ନିରୋଧ କରାଯାଇ ପାରିବ ।
(b) ରଙ୍ଗ ଲେପନ: ଲୌହ ପଦାର୍ଥ ଉପରେ ରଙ୍ଗର ଲେପ ଦେଇ କଳଙ୍କିରୋଧ କରାଯାଇପାରିବ ।

Question 11.
ଅଧାତୁ ସହ ଅକ୍ସିଜେନ୍‌ର ସଂଯୋଗ ହେଲେ କେଉଁ ପ୍ରକାରର ଅକ୍‌ସାଇଡ୍ ଉତ୍ପନ୍ନ ହୁଏ ?
Answer:
ଅଧାତୁ ସହ ଅକ୍ସିଜେନ୍ ସଂଯୋଗ ହେଲେ ଅମ୍ଳୀୟ ବା ନିରପେକ୍ଷ ଅକ୍‌ସାଇଡ୍ ଉତ୍ପନ୍ନ ହୁଏ ।

ଉଦାହରଣ:
ଅମ୍ଳୀୟ ଅକ୍‌ସାଇଡ୍ – CO2, SO2
ନିରପେକ୍ଷ ଅକ୍‌ସାଇଡ୍ – H2O, CO, N2O

Question 12.
କାରଣ ଦର୍ଶାଅ ।
Answer:
(a) ପ୍ଲାଟିନମ, ସୁନା ଏବଂ ରୁପା ଗହଣା ତିଆରିରେ ବ୍ୟବହୃତ ହୁଏ ।
ଉ :
(i) ପ୍ଲାଟିନମ୍, ସୁନା ଓ ରୁପା ସବୁଠାରୁ କମ୍ ପ୍ରତିକ୍ରିୟାଶୀଳ ଧାତୁ । ଏଗୁଡ଼ିକ ସହଜରେ ବାୟୁଦ୍ବାରା କ୍ଷୟପ୍ରାପ୍ତ ହୁଅନ୍ତି ନାହିଁ ।
(ii) ଏଗୁଡ଼ିକ ଅତି ନମନୀୟ, ତନ୍ୟ ଓ ଉଜ୍ଜଳ । ତେଣୁ ଏଗୁଡ଼ିକ ଗହଣା ତିଆରିରେ ବ୍ୟବହୃତ ହୁଏ

(b) ତେଲ ଭିତରେ ସୋଡ଼ିୟମ୍, ପୋଟାସିୟମ୍କୁ ଓ ଲିଥ୍ୟମ୍‌କୁ ରଖାଯାଏ ।
ଉ :
(i) ପୋଟାସିୟମ୍, ସୋଡ଼ିୟମ୍ ଓ ଲିଥ୍ମ୍‌ ସାଧାରଣ ତାପମାତ୍ରାରେ ଏତେ ଜୋରରେ ପ୍ରତିକ୍ରିୟା କରେ ଯେ ସେଗୁଡ଼ିକରେ ନିଆଁ ଲାଗିଯାଏ ।
(ii) ତେଣୁ ଦୁର୍ଘଟଣାଜନିତ ନିଆଁରୁ ରକ୍ଷା ପାଇବା ପାଇଁ ସେଗୁଡ଼ିକୁ ସର୍ବଦା କିରୋସିନ୍‌ରେ ବୁଡ଼ାଇ ରଖାଯାଏ ।

(c) ଯଦିଓ ଏଲୁମିନିୟମ୍ ଏକ ଉଚ୍ଚ କ୍ରିୟାଶୀଳ ମୌଳିକ, ତଥାପି ଏହାକୁ ରନ୍ଧନ ବାସନକୁସନ ତିଆରିରେ ବ୍ୟବହାର କରାଯାଏ ।
(i) ଏଲୁମିନିୟମ୍ କ୍ଷୟରୋଧୀ ଓ ତାପର ସୁପରିବାହୀ । ଏହାର ପୃଷ୍ଠଦେଶ ଅକ୍‌ସାଇଡ୍‌ର ଏକ ପ୍ରତିକ୍ରିୟାହୀନ ସ୍ତର ସୃଷ୍ଟିକରେ ।
(ii) ଏହି ଅକ୍‌ସାଇଡ୍‌ର ଆସ୍ତରଣ ଧାତୁକୁ ସଂକ୍ଷାରଣରୁ ରକ୍ଷାକରେ । ତେଣୁ ଏହାକୁ ରନ୍ଧନ ବାସନକୁସନ ତିଆରିରେ କରାଯାଏ ।

(d) ଧାତୁ ନିଷ୍କାସନ ପ୍ରଣାଳୀରେ କାର୍ବୋନେଟ୍ ଓ ସଲଫାଇଡ୍ ଧାତୁପିଣ୍ଡକୁ ଅକ୍‌ସାଇଡ୍‌ରେ ପରିଣତ କରାଯାଏ ।
ଉ : ଧାତବ ଅକ୍‌ସାଇଡ୍‌କୁ ସହଜରେ ବିଜାରଣ କରି ସେଥୁରୁ ଧାତୁ ନିଷ୍କାସନ କରାଯାଇ ପାରିବ । କିନ୍ତୁ ସଲଫାଇଡ୍ ଓ କାର୍ବୋନେଟ୍ ଧାତୁପିଣ୍ଡରୁ ଧାତୁ ନିଷ୍କାସନ ସହଜରେ କରାଯାଇ ପାରିବ ନାହିଁ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 13.
ତୁମ୍ଭେମାନେ ଦେଖୁଥ‌ିବା ମଳିନ ପଡ଼ିଥିବା ତମ୍ବା ପାତ୍ରଗୁଡ଼ିକ ଲେମ୍ବୁ ଏବଂ ତେନ୍ତୁଳି ରସଦ୍ଵାରା ସଫା କରାଯାଏ । କାହିଁକି ଏହି ଖଟାଜାତୀୟ ପଦାର୍ଥ ଦ୍ଵାରା ପାତ୍ରଗୁଡ଼ିକ ସଫା କରିବା ସମ୍ଭବ ହୁଏ ? ବୁଝାଅ । ମଳିନ ପଡ଼ିଥିବା ତମ୍ବା ପାତ୍ରଗୁଡ଼ିକ ଉପରେ କପର କାର୍ବୋନେଟ୍‌ର ଏକ ସ୍ତର ଥାଏ ।
Answer:
(i) ମଳିନ ପଡ଼ିଥିବା ତମ୍ବା ପାତ୍ରଗୁଡ଼ିକ ଉପରେ କପର କାର୍ବୋନେଟ୍‌ର ଏକ ସ୍ତର ଥାଏ ।
(ii) କ୍ଷାରୀୟ କପର କାର୍ବୋନେଟ୍ ଜଳରେ ଅଦ୍ରବଣୀୟ । କିନ୍ତୁ ଲେମ୍ବୁରସ ଏବଂ ତେନ୍ତୁଳି ରସ ଭଳି ଅମ୍ଳରେ ଦ୍ରୁତ ଦ୍ରବଣୀୟ । ତେଣୁ ସହଜରେ ତମ୍ବା ପାତ୍ରଗୁଡ଼ିକ ସଫା ହୋଇଯାଏ ।

Question 14.
ଧାତୁ ଏବଂ ଅଧାତୁ ମଧ୍ୟରେ ଥ‌ିବା ରାସାୟନିକ ଧର୍ମଗୁଡ଼ିକର ପାର୍ଥକ୍ୟ ଦର୍ଶାଅ ।
Answer:
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-4

Question 15.
ଗୋଟିଏ ଲୋକ ଦୁଆର ଦୁଆର ବୁଲି ନିଜକୁ ବଣିଆ ବୋଲି କହିଲା । ସେ ପୁରୁଣା ଏବଂ ମାନ୍ଦା ସୁନା ଅଳଙ୍କାରକୁ ନୂତନ କରିଦେବ ବୋଲି ପ୍ରତିଶ୍ରୁତି ଦେଲା । ନିଃସନ୍ଦେହରେ ଜଣେ ଭଦ୍ର ମହିଳା ଏକ ଯୋଡ଼ା ଚୁଡ଼ି ତାକୁ ଦେବାରୁ ସେ ତାକୁ ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଦ୍ରବଣରେ ବୁଡ଼ାଇଦେଲା । ଏହାପରେ ସେହି ଚୁଡ଼ିଗୁଡ଼ିକ ନୂଆପରି ଚକ୍ ଚକ୍ ହୋଇଗଲା । ମାତ୍ର ତା’ର ଓଜନ ବହୁତ କମିଗଲା । ସେହି ସ୍ତ୍ରୀଲୋକଟି ଅଶାନ୍ତ ହୋଇ କିଛି ସମୟ ଯୁକ୍ତିତର୍କ କଲାପରେ
Answer:
(i) ଲୋକଟି ବ୍ୟବହାର କରିଥିବା ଉକ୍ତ ଦ୍ରବଣର ନାମ ଆକ୍କାରେଜିଆ ବା ଅମ୍ଳରାଜ। ସଦ୍ୟ ପ୍ରସ୍ତୁତ ଗାଢ଼ ହାଇଡ୍ରୋକ୍ଲୋରିକ୍ ଏସିଡ୍ ଏବଂ ଗାଢ଼ ନାଇଟ୍ରିକ୍ ଏସିଡ୍‌ର 3 : 1 ମିଶ୍ରଣ । ଏ ଦୁଇ ଅମ୍ଳ ମଧ୍ୟରୁ କୌଣସିଟି ସୁନାକୁ ଦ୍ରବୀଭୂତ କରିପାରେ ନାହିଁ । କିନ୍ତୁ ଏହାର ମିଶ୍ରଣ ସୁନାକୁ ଦ୍ରବୀଭୂତ କରିପାରେ ।
(ii) ଅମ୍ଳରାଜରେ ଏକ ଯୋଡ଼ା ଚୁଡ଼ିକୁ ପକାଇବା ଦ୍ଵାରା କିଛି ସୁନା ଦ୍ରବୀଭୂତ ହୋଇଗଲା, ଫଳରେ ଓଜନରେ

Question 16.
ଗରମ ପାଣି ଟାଙ୍କି ପାଇଁ ଷ୍ଟିଲ୍‌ ପରିବର୍ତ୍ତେ ତମ୍ବା କାହିଁକି ବ୍ୟବହାର କରାଯାଏ ?
Answer:
(i) ଷ୍ଟିଲ୍ ଲୁହାରେ ତିଆରି ଯାହାକି ଗରମପାଣିର ବାମ୍ଫ ସହ ପ୍ରତିକ୍ରିୟା କରେ; କିନ୍ତୁ କପର ପ୍ରତିକ୍ରିୟା କରେ ନାହିଁ ।
3Fe(s) + 4H2O → Fe3O4(s) + 4H2(g)
( ଗରମ)
(ii) ଷ୍ଟିଲ୍ ତୁଳନାରେ ତମ୍ବାର ଗଳନାଙ୍କ ଅତି ଉଚ୍ଚ ଏବଂ ତମ୍ବା ମଧ୍ୟ ଅଧିକ ତାପ ସୁପରିବାହୀ । ତେଣୁ ଗରମ ପାଣି ଟାଙ୍କି ପାଇଁ ଷ୍ଟିଲ୍‌ ପରିବର୍ତ୍ତେ ତମ୍ବା ବ୍ୟବହାର କରାଯାଏ ।

ପ୍ରଶ୍ନବଳୀ ଓ ଉତ୍ତର:

Question 1.
ଗୋଟିଏ ଧାତୁର ଉଦାହରଣ ଦିଅ ଯାହାକି
(i) ସାଧାରଣ ତାପମାତ୍ରାରେ ତରଳ ଅଟେ
(ii) ସହଜରେ ଛୁରୀରେ କଟାଯାଇ ପାରିବ
(iii) ତାପର ସୁପରିବାହୀ
(iv) ତାପର କୁପରିବାହୀ
Answer:
(i) ପାରଦ, (ii) ସୋଡ଼ିୟମ୍, (iii) ରୁପା, ଲୁହା, (iv) ଲେଡ଼

Question 2.
ନମନୀୟ ଓ ତନ୍ୟର ଅର୍ଥ ବୁଝାଅ ।
Answer:
ନମନୀୟ – ଧାତୁକୁ ଆଘାତ କରି ଚଦରରେ ପରିଣତ କରିବା ଗୁଣକୁ ନମନୀୟ ଗୁଣ କହନ୍ତି ।
ତନ୍ୟ – ଧାତୁଗୁଡ଼ିକ ତାରରେ ରୂପାନ୍ତରଣ ହେବା ଗୁଣକୁ ତନ୍ୟ ଗୁଣ କହନ୍ତି ।

Question 3.
ସୋଡ଼ିୟମ୍‌କୁ କାହିଁକି କିରୋସିନ୍ ତେଲରେ ବୁଡ଼ାଇ ରଖାଯାଏ ?
Answer:
ସୋଡ଼ିୟମ ଅତ୍ୟନ୍ତ ପ୍ରତିକ୍ରିୟାଶୀଳ ଧାତୁ ହୋଇଥିବାରୁ ଏହା ବାୟୁ ସଂସ୍ପର୍ଶରେ ଆସିଲେ ଜଳିଯାଏ । ସୋଡ଼ିୟମ୍ ସାଧାରଣ ତାପମାତ୍ରାରେ ଅମ୍ଳଜାନ ଓ ଜଳୀୟବାଷ୍ପ (ଆର୍ଦ୍ର ବାୟୁ) ସହ ପ୍ରତିକ୍ରିୟା କରେ । କିନ୍ତୁ କିରୋସିନ୍‌ରେ ବୁଡ଼ାଇ ରଖିଲେ ସୋଡ଼ିୟମ୍ ପ୍ରତିକ୍ରିୟା କରେ ନାହିଁ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 4.
ନିମ୍ନଲିଖୁତ ପ୍ରତିକ୍ରିୟା ପାଇଁ ରାସାୟନିକ ସମୀକରଣ ଲେଖ ।
(i) ଆଇରନ୍ ସହିତ ବାମ୍ଫ
(ii) କ୍ୟାଲ୍‌ସିୟମ୍ ଏବଂ ପୋଟାସିୟମ୍ ସହିତ ଜଳ ।
Answer:
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-5

Question 5.
ଚାରୋଟି ଧାତୁର ନମୁନା A, B, C ଏବଂ D ନିଆଯାଇଛି ଏବଂ ଗୋଟିଏ ଧାତୁକୁ ପକାଯାଇଛି । ଫଳାଫଳକୁ ନିମ୍ନସାରଣୀରେ ଲେଖା ଧାତୁ |
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-6
ଉପରୋକ୍ତ ସାରଣୀ ବ୍ୟବହାର କରି A, B, C ଏବଂ D ଧାତୁ ବିଷୟରେ ନିମ୍ନଲିଖ୍ତ ପ୍ରଶ୍ନର ଉତ୍ତର ଦିଅ |
(i) କେଉଁ ଧାତୁଟି ସବୁଠାରୁ ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ଅଟେ ?
ଉ : B ଧାତୁଟି ସବୁଠାରୁ ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ଅଟେ

(ii) B ଧାତୁକୁ କପର (II) ସଲଫେଟ୍ ଦ୍ରବଣ ସହ ମିଶାଇଲେ କ’ଣ ଦେଖୁବ ?
ଉ : B ଧାତୁ ଆଇରନ୍‌ଠାରୁ ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ଏକ ଆଇରନ୍ କପରଠାରୁ ଅଧ୍ଵ ପ୍ରତିକ୍ରିୟାଶୀଳ । ତେଣୁ B ଧାତୁ କର୍ପରଠାରୁ ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ । ତେଣୁ କପର ସଲଫେଟ୍ ଦ୍ରବଣରୁ କପ୍‌କୁ ବିସ୍ଥାପିତ କରିବ । :
B + CuSO4 → Cu + BSO4

(iii) ପ୍ରତିକ୍ରିୟାଶୀଳତାର ଅଧଃକ୍ରମରେ A, B, C ଏବଂ D ଧାତୁକୁ ସଜାଇ ଲେଖ ।
ଉ : ଧାତୁ B > ଧାତୁ A > ଧାତୁ C > ଧାତୁ D

Question 6.
ଗୋଟିଏ ପ୍ରତିକ୍ରିୟାଶୀଳ ଧାତୁ ସହ ଲଘୁ ହାଇଡ୍ରୋକ୍ଲୋରିକ୍ ଏସିଡ୍ ମିଶାଇଲେ କେଉଁ ଗ୍ୟାସ୍ ପାଇବ ? ଲୁହା ସହିତ ଲଘୁ H2SO4 ର ରାସାୟନିକ ପ୍ରତିକ୍ରିୟାକୁ ସମୀକରଣ ସହ ଲେଖ ।
Answer:
ଗୋଟିଏ ପ୍ରତିକ୍ରିୟାଶୀଳ ଧାତୁ ସହ ଲଘୁ ହାଇଡ୍ରୋକ୍ଲୋରିକ୍ ଏସିଡ୍ ମିଶାଇଲେ ହାଇଡ୍ରୋଜେନ୍ ଗ୍ୟାସ୍ ସୃଷ୍ଟି ହେବ
Fe(s) + H2SO4 (aq) → FeSO4(aq) + H2(g)

Question 7.
ଜିକ୍ ସହିତ ଆଇରନ୍ (II) ସଲ୍‌ଫେଟ୍‌କୁ ମିଶାଇଲେ କ’ଣ ପର୍ଯ୍ୟବେକ୍ଷଣ କରିବ ? ଏହି ରାସାୟନିକ ପ୍ରତିକ୍ରିୟାଟିକୁ ସମୀକରଣ ସହ ଲେଖ ।
Answer:
ଜିଙ୍କ୍ ଆଇରନ୍‌ଠାରୁ ଅଧିକ ପ୍ରତିକ୍ରିୟାଶୀଳ । ତେଣୁ ଜିଙ୍କ୍କୁ ଆଇରନ୍‌ ସଲଫେଟ୍ ସହ ମିଶାଇଲେ ଜିଙ୍କ୍ ଆଇରନକୁ
Zn(s) + FeSO4(aq) → ZnSO4(aq) + Fe

Question 8.
(i) ସୋଡ଼ିୟମ୍‌, ଅକ୍‌ସିଜେନ୍ ଏବଂ ମ୍ୟାଗ୍ନେସିୟମ୍‌ର ଇଲେକ୍‌ଟ୍ରନ୍‌ ସଂରଚନା ଲେଖ । (ଏହାକୁ ବିନ୍ଦୁଦ୍ଵାରା ଚିହ୍ନିତ କର)
ଉ :
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-7

(ii) ଇଲେକ୍‌ଟ୍ରନ ସ୍ଥାନାନ୍ତରଦ୍ୱାରା (Na2O) ଏବଂ (MgO)ର ଗଠନ ଦର୍ଶାଅ ।
ଉ :
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-8

(iii) ଏହି ଯୌଗିକଗୁଡ଼ିକରେ କେଉଁ ଆୟନସବୁ ରହିଅଛି ?
ଉ : Na2O ରେ Na+ ଓ O2- ଏବଂ MgO ରେ Mg2+ ଓ O2- ଏନାୟନ୍ ସବୁ ରହିଅଛି ।

Question 9.
ଆୟନିକ ଯୌଗିକର କାହିଁକି ଉଚ୍ଚ ଗଳନାଙ୍କ ହୁଏ ?
Answer:
ଆୟନିକ ଯୌଗିକର ପରମାଣୁଗୁଡ଼ିକ ପରସ୍ପର ସହିତ ଦ୍ରୁତ ରାସାୟନିକ ବନ୍ଧଦ୍ୱାରା ବାନ୍ଧିହୋଇ ରହିଥାନ୍ତି । ଆୟନିକ ଯୌଗିକର ଶକ୍ତ ଅନ୍ତଃ ଆୟନୀୟ ଆକର୍ଷଣ ଭାଙ୍ଗିବା ପାଇଁ ପ୍ରଚୁର ଶକ୍ତି ଆବଶ୍ୟକ । ତେଣୁ ଆୟନିକ ଯୌଗିକର ଗଳନାଙ୍କ ଉଚ୍ଚ ହୋଇଥାଏ ।

Question 10.
ନିମ୍ନଲିଖୂତ ପଦଗୁଡ଼ିକୁ ବୁଝାଇ ଲେଖ ।
(i) ଖଣିଜ
(ii) ଧାତୁପିଣ୍ଡ
(iii) ଗାର୍ଲ୍
Answer:
(i) ଖଣିଜ – ପ୍ରକୃତିରେ ଭୂତ୍ବକ୍‌ରୁ ମିଳୁଥିବା ମୌଳିକ ବା ଯୌଗିକକୁ ଖଣିଜ କୁହାଯାଏ ।
(ii) ଧାତୁପିଣ୍ଡ – ଯେଉଁ ଖଣିଜରୁ ଧାତୁକୁ ନିଷ୍କାସନ କରିବା ଲାଭଜନକ ହୋଇଥାଏ, ସେହି ଖଣିଜକୁ ଓର୍ ବା ଧାତୁପିଣ୍ଡ କୁହାଯାଏ ।
(iii) ଗାଙ୍ଗ୍ – ଧାତୁପିଣ୍ଡରେ ମିଶିରହିଥ‌ିବା ମାଟି, ବାଲି ଆଦି ଅପଦ୍ରବକୁ ଗାଙ୍ଗୁ କୁହାଯାଏ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

Question 11.
ମୁକ୍ତ ଭାବରେ ପ୍ରକୃତିରୁ ମିଳୁଥ‌ିବା ଦୁଇଟି ଧାତୁର ନାମ ଲେଖ ।
Answer:
ସୁନା, ପ୍ଲାଟିନମ୍

Question 12.
କେଉଁ ରାସାୟନିକ ପ୍ରଣାଳୀ ବ୍ୟବହାର କରି ଧାତବ ଅକ୍‌ସାଇଡ୍‌ରୁ ଧାତୁ ନିଷ୍କାସନ କରାଯାଏ ?
Answer:
ବିଜାରଣ ପଦ୍ଧତି ବ୍ୟବହାର କରି ଧାତବ ଅକ୍‌ସାଇଡ୍‌ରୁ ଧାତୁ ନିଷ୍କାସନ କରାଯାଏ ।

Question 13.
ଜିଙ୍କ୍, ମ୍ୟାଗ୍ନେସିୟମ୍ ଏବଂ ତମ୍ବାର ଧାତବ ଅକ୍‌ସାଇଡ୍‌ଗୁଡ଼ିକୁ ନିମ୍ନଧାତୁ ସହ ଉତ୍ତପ୍ତ କରାଗଲା । ମ୍ୟାଗ୍ନେସିୟମ୍
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-9
କେଉଁ କ୍ଷେତ୍ରରେ ଅପସାରଣ ପ୍ରତିକ୍ରିୟା ହେବାର ଦେଖିପାରିବ ?
Answer:
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-10

Question 14.
କେଉଁ ଧାତୁ ସହଜରେ ସଂକ୍ଷାରଣ ହୋଇପାରେ ନାହିଁ ?
Answer:
ସୁନା, ରୁପା ଓ ପ୍ଲାଟିନମ୍

Question 15.
ମିଶ୍ରଧାତୁ କହିଲେ କ’ଣ ବୁଝ ?
Answer:
ଧାତୁ ଓ ଅଧାତୁ କିମ୍ବା ଧାତୁ ଓ ଧାତୁ କିମ୍ବା ଧାତୁ ଓ ଉପଧାତୁର ମିଶ୍ରଣକୁ ମିଶ୍ରଧାତୁ କୁହାଯାଏ ।

କାର୍ଯ୍ୟାବଳୀ (Activity):

କାର୍ଯ୍ୟାବଳୀ -1 (Activity-1)
(a) ଧାତବ ଦୀପ୍ତି ବା ଧାତବ ଉଜ୍ଜଳତା :
ବିଶୁଦ୍ଧ ଅବସ୍ଥାରେ ଧାତୁଗୁଡ଼ିକ ଚକ୍ ଚକ୍ କରୁଥବାରୁ, ଏହି ଗୁଣକୁ ଧାତବ ଦୀପ୍ତି କୁହାଯାଏ

ପରୀକ୍ଷା ପଦ୍ଧତି:
ବାଲିକାଗଜ ଦ୍ବାରା ପ୍ରତ୍ୟେକ ନମୁନା ଧାତୁକୁ ଭଲଭାବରେ ଘଷ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଘଷାଯାଇଥିବା ପାର୍ଶ୍ଵ ଓ ଘଷା ଯାଇନଥ‌ିବା ଅଂଶକୁ ଦେଖୁଲେ ଜଣାଯାଏ ଯେ ଘଷାଯାଇଥିବା ପାର୍ଶ୍ୱ

ସିଦ୍ଧାନ୍ତ:
ବିଶୁଦ୍ଧ ଅବସ୍ଥାରେ ଧାତୁର ପୃଷ୍ଠ ଚକ୍ ଚକ୍ କରେ । ଧାତୁର ଏହି ଧର୍ମକୁ ଧାତବ ଦୀପ୍ତି ବା ଧାତବ ଉଜ୍ଜ୍ବଳତା କୁହାଯାଏ ।

କାର୍ଯ୍ୟାବଳୀ -2 (Activity-2)
(b) ଧାତୁଗୁଡ଼ିକ ଦୃଢ଼ ବା ଶକ୍ତ:

ଆବଶ୍ୟକ ଉପକରଣ:
ଲୁହା, ତମ୍ବା, ଏଲୁମିନିୟମ୍ ଏବଂ ମ୍ୟାଗ୍ନେସିୟମର ଛୋଟ ଖଣ୍ଡ ନିଅ । ସୋଡ଼ିୟମ ଖଣ୍ଡ ଚାପି ଶୁଖାଅ ଓ ୱାସ ଉପରେ ରଖ ।

ପରୀକ୍ଷା ପଦ୍ଧତି:
ପ୍ରତ୍ୟେକ ଖଣ୍ଡକୁ ଧାରୁଆ ଛୁରିରେ କାଟିବାକୁ ଚେଷ୍ଟାକର ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଲୁହା, ତମ୍ବା, ଏଲୁମିନିୟମ୍ ସହଜରେ ଛୁରୀଦ୍ଵାରା କାଟି ହେଉନାହିଁ; କିନ୍ତୁ ସୋଡ଼ିୟମ ଧାତୁ ସହଜରେ

ସିଦ୍ଧାନ୍ତ:
ସାଧାରଣତଃ ଧାତୁ ଗୁଡ଼ିକ ଦୃଢ଼ ବା ଶକ୍ତ । କିନ୍ତୁ ସୋଡ଼ିୟମ ପରି ଧାତୁର ଦୃଢ଼ତା ଅନ୍ୟଧାତୁଠାରୁ କମ୍ ( ନରମ ) ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -3 (Activity-3)
(c) ଧାତୁଗୁଡ଼ିକ ଦୃଢ଼ ବା ଶକ୍ତ:
ଆବଶ୍ୟକୀୟ ଉପକରଣ:
ଲୁହା, ଜିଙ୍କ୍, ଲେଡ୍ ଏବଂ ତମ୍ବାର ଧାତୁ ଖଣ୍ଡ ମାନ ନିଅ ।

ପରୀକ୍ଷା ପଦ୍ଧତି:
ଏକ ଲୁହା ଖଣ୍ଡ ଉପରେ ଯେ କୌଣସି ଧାତୁକୁ ରଖ୍ ହାତୁଡ଼ିରେ ଚାରିପାଞ୍ଚ ଥର ଆଘାତ କର । ଏହି ପରି ଅନ୍ୟ ଧାତୁ ଖଣ୍ଡକୁ ରଖ୍ ଆଘାତ କରି ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଧାତୁଖଣ୍ଡକୁ ବାରମ୍ବାର ଆଘାତକଲେ ତାହା ପତଳା ଚଦର ପରି ହୋଇଯାଉଛି ।

ସିଦ୍ଧାନ୍ତ:
ଧାତୁ ପ୍ରସାରଣଶୀଳ ବା ନମନୀୟ । ଧାତୁର ଏହି ଗୁଣକୁ ନମନୀୟତା କୁହାଯାଏ । ସୁନା ଓ ରୁପାର ନମନୀୟଗୁଣ ସର୍ବାଧ‌ିକ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-11

କାର୍ଯ୍ୟାବଳୀ -4 (Activity-4)
(d) କାର୍ଯ୍ୟ|ଚ୍ଳ| (Ductility):

ଆବଶ୍ୟକ ଉପକରଣ:
ଲୁହା, ତମ୍ବା, ଏଲୁମିନିୟମ, ଲେଡ୍ ଇତ୍ୟାଦି ଧାତୁ ନିଆଯାଉ । ପରୀକ୍ଷା ପଦ୍ଧତି: ଧାତୁଗୁଡ଼ିକୁ ଗରମ କରି ଚିମୁଟାରେ ଟଣାଯାଉ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଧାତୁଗୁଡ଼ିକୁ ଚିମୁଟାରେ ଟାଣିଲେ ଧାତୁରୁ ତାର ମିଳୁଛି ।

ସିଦ୍ଧାନ୍ତ:
ଧାତୁ ତାରରେ ପରିଣତ ହେବା ଗୁଣକୁ ତନ୍ୟତା କୁହାଯାଏ ।

କାର୍ଯ୍ୟାବଳୀ -5 (Activity-5)
(e) ଧାତୁ ଗୁଡ଼ିକ ତାପ ସୁପରି ବାହୀ ଓ ଭଚ୍ଚଗଳନାକ ଚିଣିଷ୍ଟ:

ଆବଶ୍ୟକ ଉପକରଣ:
ଟଣ୍ଡେ ତମ୍ବା ଜିମ୍ବା ଏଲୁମିନିୟମ୍ ତାର, କ୍ଲାମ୍ପ ଷ୍ଟାଣ୍ଡ, ସ୍ପିରିଟ୍ ଲ୍ୟାମ୍ପ କିମ୍ବା ମହମବତୀ, ମହମ, ପିନ୍ କଣ୍ଟା ।

ପରୀକ୍ଷା ପଦ୍ଧତି:
ତମ୍ବା ତାରଖଣ୍ଡ ନେଇ ଗୋଟିଏ କ୍ଲାମ୍ପ ସାହାଯ୍ୟରେ ଷ୍ଟାଣ୍ଡରେ ଧରି ରଖାଯାଉ । ମହମ ସାହାଯ୍ୟରେ ଗୋଟିଏ ପିନ୍ କଣ୍ଟାକୁ ତାରର ମୁକ୍ତ ପ୍ରାନ୍ତରେ ଯୋଡ୍ ରଖାଯାଉ । ସ୍ପିରିଟ୍ ଲ୍ୟାମ୍ପ/ ମହମବତୀ ସାହାଯ୍ୟରେ ତାରର ମଝି ସ୍ଥାନରେ ଗରମ କରାଯାଉ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
କିଛି ସମୟପରେ ମହମ ତରଳିବାକୁ ଆରମ୍ଭକରିବ ଏବଂ ପିକଣ୍ଟାଟି ତଳକୁ ଖସି ପଡ଼ିବ । କିନ୍ତୁ ଧାତବ ତାର, ତରଳୁ ନାହିଁ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-12
ସିଦ୍ଧାନ୍ତ:
ଧାତୁ ଗୁଡ଼ିକ ତାପ ସୁପରିବାହୀ ଏବଂ ଉଚ୍ଚ ଗଳନାଙ୍କ ବିଶିଷ୍ଟ ।

ରୁପା ଓ ତମ୍ବା ହେଉଛି ସବୁଠାରୁ ଉତ୍ତମ ତାପ ପରିବାହୀ । ଲେଡ୍ ଓ ପାରଦ ଅପେକ୍ଷାକୃତ କମ୍ ତାପ ପରିବାହୀ

କାର୍ଯ୍ୟାବଳୀ -6 (Activity-6)
(f) ଧାତୁ ବିଦ୍ୟୁତ୍ ପରିବହନ କରେ:
ଆବଶ୍ୟକ ଉପକରଣ: ବ୍ୟାଟେରୀ, ସୁଇଚ୍, ନମୁନା ଧାତୁ, କ୍ଳିପ୍ , ରବର କିମ୍ବା PVC ଭଳି ପଦାର୍ଥ ଦ୍ଵାରା ଆବୃତ ପରିବାହୀ ତାର, ବିଦ୍ୟୁତ୍ ହୋଲଡର ସହ ବଲ୍‌ବ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-13
ପରୀକ୍ଷା ପଦ୍ଧତି :
ଚିତ୍ରରେ ଦର୍ଶାଯାଇଥ‌ିବା ଭଳି ଉପକରଣଗୁଡ଼ିକୁ ବିଦ୍ୟୁତ୍ ପରିପଥରେ ସଂଯୋଗ କରାଯାଉ । A ଓ B ଅଗ୍ରମଧ୍ଯରେ ପରୀକ୍ଷା ପାଇଁ ଥିବା ନମୁନା ଧାତୁଟିକୁ ସଂଯୋଗ କରାଯାଉ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ :
ବିଦ୍ୟୁତ୍ ବଲବଟି ଜଳୁଛି ।

ସିଦ୍ଧାନ୍ତ :
ଧାତୁ ବିଦ୍ୟୁତ୍ ପରିବହନ କରେ ।

(g) ଧାତୁ ଧ୍ୱନି ସୃଷ୍ଟିକରେ:
ଆବଶ୍ୟକ ଉପକରଣ: ନମୁନା ଧାତୁ କିମ୍ବା ବିଦ୍ୟାଳୟ ଧାତବ ପିଟା ଘଣ୍ଟା କିମ୍ବା କଠିନ ପଦାର୍ଥ । ମନ୍ଦିରଗୁଡ଼ିକର ଘଣ୍ଟ,

ପରୀକ୍ଷା:
ନମୁନା ଧାତୁଖଣ୍ଡ ନେଇ ଯେ କୌଣସି କଠିନ ପଦାର୍ଥ ଦ୍ଵାରା ଆଘାତ କରାଯାଉ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଧାତୁକୁ କଠିନ ବସ୍ତୁରେ ଆଘାତ କଲେ ଧ୍ଵନି ସୃଷ୍ଟି ହୁଏ ।

ଅଧାତୁ (Non-metals):
ଅଧାତୁ ଗୁଡ଼ିକ ତିନୋଟି (କଠିନ ତରଳ ଓ ଗ୍ୟାସୀୟ) ଅବସ୍ଥାରେ ଦେଖାଯାନ୍ତି । କଠିନ ଅବସ୍ଥାରେ ମିଳୁଥ‌ିବା ଅଧାତୁ – କାର୍ବନ, ସଲଫର୍‌, ଆୟୋଡ଼ିନ୍ ଇତ୍ୟାଦି । ତରଳ ଅବସ୍ଥାରେ ମିଳୁଥିବା ଅଧାତୁ – ବ୍ରୋମିନ୍ ଗ୍ୟାସୀୟ ଅବସ୍ଥାରେ ମିଳୁଥ‌ିବା ଅଧାତୁ – ହାଇଡ୍ରୋଜେନ୍, ଅକ୍‌ସିଜେନ୍, କ୍ଲୋରିନ୍, ନାଇଟ୍ରୋଜେନ୍, ଫ୍ଲୋରିନ୍,  ହିଲିୟମ, ନିୟନ ଓ ଆର୍ଗନ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -7 (Activity-7)
ଆବଶ୍ୟକ ଉପକରଣ:
କାର୍ବନ, ସଲ୍‌ଫର ଓ ଆୟୋଡ଼ିନ୍‌ର କିଛି ନମୁନା ସଂଗ୍ରହ କରାଯାଉ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଏହି ଅଧାତୁଗୁଡ଼ିକୁ ନେଇ କାର୍ଯ୍ୟାବଳୀ-1ରୁ କାର୍ଯ୍ୟାବଳୀ-7 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ପରୀକ୍ଷା କରି ନିମ୍ନ ସାରଣୀରେ ଲିପିବଦ୍ଧ କରାଯାଉ ।

ଅଧାତୁ ସମ୍ବନ୍ଧୀୟ ପର୍ଯ୍ୟବେକ୍ଷଣ ସାରଣୀ
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-14

ସିଦ୍ଧାନ୍ତ :
ଧାତୁ ଓ ଅଧାତୁକୁ କେବଳ ଭୌତିକ ଧର୍ମ ଅନୁଯାୟୀ ବର୍ଗୀକରଣ କରାଯାଇ ପାରିବ ନାହିଁ କାରଣ ଏଥରେ ଅନେକ

  • ସାଧାରଣ ତାପମାତ୍ରାରେ ସମସ୍ତ ଧାତୁ କଠିନ; କିନ୍ତୁ ପାରଦ ତରଳ ।
  • ଧାତୁଗୁଡ଼ିକ ଉଚ୍ଚ ଗଳନାଙ୍କ ବିଶିଷ୍ଟ କିନ୍ତୁ ଗାଲିୟମ୍ ଏବଂ ସୀସିୟମ୍ ଅତି ନିମ୍ନ ଗଳନାଙ୍କ ବିଶିଷ୍ଟ ।
  • ଆୟୋଡ଼ିନ୍ ଏକ ଅଧାତୁ ହେଲେ ମଧ୍ୟ ଏହାର ଧାତବ ଔଜଲ୍ୟ ଅଛି ।
  • ମୁଖ୍ୟତଃ କଠିନ ଅଧାତୁଗୁଡ଼ିକ ନରମ; କିନ୍ତୁ କାର୍ବନର ଏକ ରୂପ ହୀରା ଯାହାକି ଗୋଟିଏ କଠିନତମ ପ୍ରାକୃତିକ ଅପରରୂପ (allotrope) ।
  • ଧାତୁଗୁଡ଼ିକ କଠିନ; କିନ୍ତୁ କ୍ଷାରୀୟ ଧାତୁ (Alkali metals) ( ସୋଡ଼ିୟମ୍‌, ଲିଥ୍ୟମ, ପୋଟାସିୟମ୍ ଇତ୍ୟାଦି) ନରମ ଯାହାକୁ ଛୁରୀରେ କଟାଯାଇପାରିବ।
  • ଅଧାତୁ ବିଦ୍ୟୁତ୍ କୁପରିବାହୀ; କିନ୍ତୁ କାର୍ବନର ଏକ ଅପର ରୂପ ଗ୍ରାଫାଇଟ୍ ବିଦ୍ୟୁତ୍ ପରିବହନ କରେ ।

କାର୍ଯ୍ୟାବଳୀ -8 (Activity-8)
ଆବଶ୍ୟକ ଉପକରଣ:
ସଲଫର ପାଉଡ଼ର, ଜଳ, ଲିଟ୍‌ସ୍ କାଗଜ, ପରୀକ୍ଷାନଳୀ

ପରୀକ୍ଷଣ:
ସଲଫର ପାଉଡ଼ରକୁ ଜାଳି ଉତ୍ପନ୍ନ ହେଉଥ‌ିବା ଧୂଆଁକୁ ଏକ ପରୀକ୍ଷାନଳୀରେ ସଂଗ୍ରହ କରାଯାଉ । ଏହି ପରୀକ୍ଷାନଳୀରେ କିଛି ଜଳ ମିଶାଇ ଜୋରରେ ହଲାଇଦିଅ । ଏହି ଦ୍ରବଣରେ ନୀଳ ଓ ନାଲି ଲିଟମସ୍ କାଗଜ ବୁଡ଼ାଇ ପରୀକ୍ଷା କରାଯାଉ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
ଦ୍ରବଣ ନୀଳ ଲିଟମସ୍ କାଗଜକୁ ନାଲି କରିବ । କିନ୍ତୁ ନାଲି ଲିଟ୍‌ସ୍ କାଗଜର ରଙ୍ଗରେ କୌଣସି ପରିବର୍ତ୍ତନ ହେବ ନାହିଁ ।

ସିଦ୍ଧାନ୍ତ:
ଅମ୍ଳ ନୀଳ ଲିଟମସ୍ କାଗଜକୁ ଲାଲ୍ କରିଥାଏ । ଏଠାରେ SO2 ର ଜଳୀୟ ଦ୍ରବଣ ଅମ୍ଳୀୟ ଅଟେ । ଅଧାତବ ଅକ୍‌ସାଇଡ୍‌ର ଜଳୀୟ ଦ୍ରବଣ ଅମ୍ଳୀୟ ।
ସମୀକରଣ S + 02 → SO2
SO2 + H2O → H2SO3 (AM&NA UF)

କାର୍ଯ୍ୟାବଳୀ -9 (Activity-9)
ଆବଶ୍ୟକ ଉପକରଣ:
ଏଲୁମିନିୟମ୍, ତମ୍ବା, ଲୁହା, ଲେଡ୍, ମ୍ୟାଗ୍ନେସିୟମ, ଜିଙ୍କ୍, ସୋଡ଼ିୟମ୍, ସୁନା ଓ ରୁପା ଧାତୁର ନମୁନା ସଂଗ୍ରହ କର ।

ପରୀକ୍ଷଣ:
ଉପର ଲିଖ୍ ଧାତୁର ନମୁନା ଗୁଡ଼ିକୁ ଚିମୁଟାରେ ଧରି ନିଆଁରେ ଜଳାଯାଉ । ସୃଷ୍ଟି ହୋଇଥ‌ିବା ଉତ୍ପାଦଗୁଡ଼ିକୁ ସଂଗ୍ରହ କରାଯାଉ । ଧାତୁର ପୃଷ୍ଠ ଏବଂ ଉତ୍ପାଦକ ଥଣ୍ଡା ହେବାକୁ ଦିଅ । ଉତ୍ପାଦଗୁଡ଼ିକୁ ଜଳରେ ଦ୍ରବୀଭୂତ କରିବାକୁ ରେଷ୍ଠ| ଇର|ତ୍ପ|ଉ |

ପର୍ଯ୍ୟବେକ୍ଷଣ:

  • ମ୍ୟାଗ୍ନେସିୟମ ଧାତୁ ସହଜରେ ଜଳିଲା ଏବଂ ଉଜ୍ଜ୍ଵଳ ଆଲୋକ ଶିଖା ଦେଖାଗଲା ।
  • ଧାତୁର ପୃଷ୍ଠଟି ରୂପେଲୀ ଧଳା ରଙ୍ଗର ଦେଖାଗଲା ।
  • ସୋଡ଼ିୟମ୍ ସହଜରେ ଜଳିଲା ଏବଂ ସୁନେଲୀ ହଳଦିଆ ରଙ୍ଗର ଶିଖା ଦେଖାଗଲା ।

ସିଦ୍ଧାନ୍ତ:

  • ଅକ୍‌ସିଜେନ ସହ ପ୍ରତିକ୍ରିୟାଶୀଳତାର ହ୍ରାସ ଅନୁସାରେ
    Na > Mg > Al > Zn > Fe > Pb > Cu
  • ସୋଡ଼ିୟମ ଅକ୍‌ସାଇଡ ଜଳରେ ଦ୍ରବଣୀୟ ।
  • ଆଲୁମିନିୟମ, କପର, ଲୌହ, ଲେଡ୍, ମ୍ୟାଗ୍ନେସିୟମ୍, ଜିଙ୍କ୍ ଆଦି ଧାତୁର ଅକ୍‌ସାଇଡ୍ ଜଳରେ ଅଦ୍ରବଣୀୟ ।
  • ଅଧିକାଂଶ ଧାତୁ ଅକ୍‌ସିଜେନ୍ ସହିତ ପ୍ରତିକ୍ରୟାକରି ଧାତବ ଅକ୍‌ସାଇଡ ସୃଷ୍ଟିକରେ ।
    ଧାତୁ + ଅକ୍‌ସିଜେନ → ଧାତବ ଅକ୍‌ସାଇଡ୍

ଉଦାହରଣ:
(i) କପର ବାୟୁର ଉପସ୍ଥିତିରେ ଉତ୍ତପ୍ତ ହେଲେ ଅକ୍‌ସିଜେନ ସହ ମିଶି କଳାରଙ୍ଗର କପର
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-15

(ii) ଏଲୁମିନିୟମ୍ ଅକ୍‌ସିଜେନ ସହ ମିଶି ଏଲୁମିନିୟମ୍ ଅକ୍‌ସାଇଡ ସୃଷ୍ଟି କରେ ।
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-16

ଉଭୟଧର୍ମୀ ଅକ୍‌ସାଇଡ୍ ବା ଏମ୍ପୋଟେରିକ୍ ଅକ୍‌ସାଇଡ୍ :
ଯେଉଁ ଧାତବ ଅକ୍‌ସାଇଡ୍ ଉଭୟ ଅମ୍ଳସହ ଓ କ୍ଷାରସହ ରାସାୟନିକ ପ୍ରତିକ୍ରିୟା କରି ଲବଣ ଓ ଜଳ ସୃଷ୍ଟିକରେ ତାହାକୁ ଉଭୟଧର୍ମୀ ଅକସାଇଡ୍ ବା ଏମ୍ପୋଟେରିକ୍ ଅକ୍‌ସାଇଡ୍ (amphoteric oxide) କହନ୍ତି ।
Al2O3 + 6HCI → 2ACl3 + 3H2O
Al2O3 + 2NaOH → 2NaAlO3 + H2O
(ସୋଡ଼ିୟମ୍‌ ଏଲୁମିନେଟ୍)
ଅଧିକାଂଶ ଧାତବ ଅକ୍‌ସାଇଡ୍ ଜଳରେ ଅଦ୍ରବଣୀୟ ; କିନ୍ତୁ କିଛି ଧାତବ ଅକ୍‌ସାଇଡ ଜଳରେ ଦ୍ରବୀଭୂତ ହୋଇ କ୍ଷାର (Alkali) ସୃଷ୍ଟି କରନ୍ତି । ସୋଡ଼ିୟମ ଅକ୍‌ସାଇଡ୍ ଏବଂ ପୋଟାସିୟମ୍ ଅକ୍‌ସାଇଡ୍ ଜଳରେ ଦ୍ରବୀଭୂତ ହୋଇ କ୍ଷାର ସୃଷ୍ଟି କରନ୍ତି ।
Na2O(s) + H2O(l) → 2NaOH(aq)
K2O(s) + H2O(l) → 2KOH(aq)

(iii) ପୋଟାସିୟମ୍ ଓ ସୋଡ଼ିୟମ୍ ପରି ଧାତୁକୁ ବାହାରେ ରଖୁଦେଲେ ଜୋର୍‌ରେ ପ୍ରତିକ୍ରିୟା ହୁଏ ଏବଂ ସେଗୁଡ଼ିକରେ ନିଆଁ ଲାଗିଯାଏ ।

(iv) ଲୁହାକୁ ଉତ୍ତପ୍ତକଲେ ଜଳେ ନାହିଁ କିନ୍ତୁ ଲୁହାର ଗୁଣ୍ଡକୁ ଅଗ୍ନିଶିଖାରେ ଛିଞ୍ଚିଦେଲେ ଖୁବ୍‌ଶୀଘ୍ର ଜଳିଯାଏ ।

ରୁପା ଓ ସୁନା ଅକ୍‌ସିଜେନ ସହ ଏପରିକି ଉଚ୍ଚ ତାପମାତ୍ରାରେ ମଧ୍ୟ ପ୍ରତିକ୍ରିୟା କରନ୍ତି ନାହିଁ ।

ସିଦ୍ଧାନ୍ତ:

  • ଧାତୁଗୁଡ଼ିକ ଅକ୍‌ସିଜେନ ସହ ସମାନ ବେଗରେ ପ୍ରତିକ୍ରିୟା କରନ୍ତି ନାହିଁ ।
  • ବିଭିନ୍ନ ଧାତୁ ଅକ୍‌ସିଜେନ୍ ସହ ଭିନ୍ନ ଭିନ୍ନ ପ୍ରତିକ୍ରିୟାଶୀଳତା ପ୍ରଦର୍ଶନ କରନ୍ତି ।
  • ପୋଟାସିୟମ୍ ଓ ସୋଡ଼ିୟମ୍ ସବୁଠାରୁ ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ଧାତୁ ।
  • ମ୍ୟାଗ୍ନେସିୟମ୍‌ର ପ୍ରତିକ୍ରିୟା କମ୍ ।
  • ଜିଙ୍କ, ଲୁହା, ତମ୍ବା ଏବଂ ଲେଡ୍‌କୁ ଅକ୍‌ସିଜେନ ଉପସ୍ଥିତିରେ ଜଳାଇଲେ ଏଗୁଡ଼ିକର ପ୍ରତିକ୍ରିୟାଶୀଳତା କ୍ରମ ବିଷୟରେ କୌଣସି ସୂଚନା ମିଳେ ନାହିଁ ।
  • ସାଧାରଣ ତାପମାତ୍ରାରେ ମ୍ୟାଗ୍ନେସିୟମ୍, ଏଲୁମିନିୟମ୍, ଜିଙ୍କ୍, ଲେଡ୍ ଇତ୍ୟାଦି ଧାତୁଗୁଡ଼ିକର ପୃଷ୍ଠତଳରେ ଏକ ପତଳା ଅକ୍‌ସାଇଡ୍ ଆବରଣ ରହିଥାଏ । ସଂରକ୍ଷୀ ଅକ୍‌ସାଇଡ୍‌ର ସ୍ତର ଧାତୁକୁ ଅଧିକ ଜାରଣରୁ ରକ୍ଷାକରେ ।
  • ତମ୍ବା ନିଆଁରେ ଜଳେ ନାହିଁ । ମାତ୍ର ଉତ୍ତପ୍ତ ଧାତୁ କପର (II) ଅକ୍‌ସାଇଡ୍‌ର ଏକ କଳା ଆବରଣ ଦ୍ଵାରା ଆଚ୍ଛାଦିତ ହୋଇଥାଏ ।

ମନେତଟ:
ଏନୋଡ଼ାଇଜିଂ ଏଲୁମିନିୟମ୍‌ରେ ଏକ ମୋଟା ଅକ୍‌ସାଇଡ୍ ପ୍ରଲେପ ଦେବାର ଏକ ପ୍ରଣାଳୀ । ଏଲୁମିନିୟମ୍ ବାୟୁରେ ରହିଲେ ଏକ ପତଳା ଅକ୍‌ସାଇଡ଼ର ଆବରଣ ସୃଷ୍ଟି କରେ । ଏହି ଏଲୁମିନିୟମ୍ ଅକ୍‌ସାଇଡ୍‌ର ପ୍ରଲେପ ଅଧ‌ିକ ସଂକ୍ଷାରଣ ହେବାକୁ ପ୍ରତିରୋଧ କରେ ।
ଏନୋଡ଼ାଇଜିଂ ପଦ୍ଧତିରେ ପରିଷ୍କାର ଏଲୁମିନିୟମ୍‌କୁ ଏନୋଡ଼ରେ ସଂଯୁକ୍ତ କରି ଲଘୁ ସଫ୍ୟୁରିକ୍ ଅମ୍ଳରେ ବୈଦ୍ୟୁତିକ ବିଶ୍ଳେଷଣ କରାଯାଏ । ଏନୋଡ଼ରେ ସୃଷ୍ଟି ହୋଇ ଅକ୍‌ସିଜେନ୍ ଏଲୁମିନିୟମ୍ ସହ ପ୍ରତିକ୍ରିୟା କରି ଏହା ଉପରେ ଏକ ମୋଟା ପ୍ରତିରୋଧ ଜାରଣ ଅକ୍‌ସାଇଡ୍ ଆବରଣ ସୃଷ୍ଟି କରେ ।

(b) ଧାତୁର ଜଳ ସହିତ ପ୍ରତିକ୍ରିୟା | (Reaction of Metals with Water):

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -10 (Activity-10)
ଆବଶ୍ୟକ ଉପକରଣ:
ଏଲୁମିନିୟମ୍, ତମ୍ବା, ଲୁହା, ଲେଡ୍ ମ୍ୟାଗ୍ନେସିୟମ୍, ଜିଙ୍କ୍, ସୋଡ଼ିୟମ, ସୁନା, ରୁପା ଧାତୁର ନମୁନା, ବିକର ଓ ଥଣ୍ଡା ଜଳ ।

ପରୀକ୍ଷଣ:
ସଂଗୃହୀତ ନମୁନାର ଛୋଟ ଖଣ୍ଡକୁ ଅଲଗା ଅଲଗା ବିକର୍‌ରେ ଅଧା ପର୍ଯ୍ୟନ୍ତ ଥଣ୍ଡାଜଳ ପୂରାଇ ରଖ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:

  • ସୋଡ଼ିୟମ, ପୋଟାସିୟମ ଏବଂ କ୍ୟାଲସିୟମ ଜଳ ସହିତ ପ୍ରତିକ୍ରିୟା କଲେ ବର୍ଦ୍ଧିତ ପ୍ରତିକ୍ରିୟାଶୀଳତା ଅନୁଯାୟୀ
    କ୍ୟାଲସିୟମ୍ < ପୋଟାସିୟମ୍ < ସୋଡ଼ିୟମ୍
  • ହଁ ସୋଡ଼ିୟମ୍ ଓ ପୋଟାସିୟମ ଧାତୁ ଜଳ ସହ ପ୍ରତିକ୍ରିୟା କରି ନିଆଁ ସୃଷ୍ଟି କଲା ।
  • କିଛି ସମୟ ପରେ କ୍ୟାଲସିୟମ୍ ଭାସିବାକୁ ଆରମ୍ଭ କଲା ।

ପରୀକ୍ଷଣ:
ଯେଉଁ ଧାତୁଗୁଡ଼ିକ ଥଣ୍ଡା ଯେଉଁ ଧାତୁଗୁଡ଼ିକ ଥଣ୍ଡା କିମ୍ବା ଜଳରେ ପ୍ରତିକ୍ରିୟା କଲାନାହିଁ, ସେଗୁଡ଼ିକୁ ଅଧା ବିକର ଗରମ ପାଣିରେ ପକାଅ । ଗରମ ପାଣିରେ ପ୍ରତିକ୍ରିୟା କଲାନାହିଁ, ସେଗୁଡ଼ିକ ବାମ୍ଫ ସହ ପ୍ରତିକ୍ରିୟା କରେ । ମ୍ୟାଗ୍ନେସିୟମ୍ ଓ କ୍ୟାଲସିୟମ୍ ଥଣ୍ଡାଜଳରେ କମ୍ ମାତ୍ରାରେ ପ୍ରତିକ୍ରିୟା କରନ୍ତି କିନ୍ତୁ ଫୁଟନ୍ତା ପାଣିରେ ତୀବ୍ର ବେଗରେ ପ୍ରତିକ୍ରିୟା କରନ୍ତି ।
Mg + H2O → MgO + H2
Zn + H2O → ZnO + H2
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-17
ଏଲୁମିନିୟମ୍ ଲୌହ, ଜିଙ୍କ୍ ବାମ୍ଫସହ ରାସାୟନିକ ପ୍ରତିକ୍ରିୟା କରେ ।
ଲେଡ୍, କପର, ସୁନା ଓ ରୁପା ଆଦି ଧାତୁ ଜଳସହ ରାସାୟନିକ ପ୍ରତିକ୍ରିୟା କରନ୍ତି ନାହିଁ ।
ସୋଡ଼ିୟମ୍ > ପୋଟାସିୟମ୍ > କ୍ୟାଲସିୟମ୍ > ମ୍ୟାଗ୍ନେସିୟମ୍ > ଏଲୁମିନିୟମ୍ > ଲୌହ > ସୀସା > ତମ୍ବା > ରୁପା > ପାରଦ > ସୁନା ।

ସିଦ୍ଧାନ୍ତ :
(i) ଧାତୁ + ଜଳ → ଧାତବ ଅକ୍‌ସାଇଡ୍ + ହାଇଡ୍ରୋଜେନ

(ii) ଧାତବ ଅକ୍‌ସାଇଡ୍ + ଜଳ → ଧାତବ ହାଇଡ୍ରକ୍‌ସାଇଡ୍

(iii) ସୋଡ଼ିୟମ୍ ପୋଟାସିୟମ୍ ଭଳି ଧାତୁ ଥଣ୍ଡାଜଳ ସହ ତୀବ୍ର ପ୍ରତିକ୍ରିୟା କରେ ।
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) + ତାପଶକ୍ତି
(iv) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) + ତାପଶକ୍ତି
କ୍ୟାଲସିୟମ୍ ଜଳସହ କମ୍ ତୀବ୍ରତାରେ ପ୍ରତିକ୍ରିୟା କରେ ।
Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)

(v) କ୍ୟାଲ୍‌ସିୟମ୍ ଜଳରେ ଭାସେ କାରଣ ଉତ୍ପନ୍ନ ହେଉଥିବା ହାଇଡ୍ରୋଜେନ୍ ଗ୍ୟାସ୍‌ର ଫୋଟକାଗୁଡ଼ିକ ଧାତୁର ଉପର ଭାଗରେ ଲାଖ୍ ରହିଯାଏ ।

(vi) ମ୍ୟାଗ୍ନେସିୟମ୍ ଗରମ ପାଣି ସହିତ ପ୍ରତିକ୍ରିୟା କରି ମ୍ୟାଗ୍ନେସିୟମ୍‌ ହାଇଡ୍ରକ୍‌ସାଇଡ୍ ଓ ହାଇଡ୍ରୋଜେନ ଗ୍ୟାସ୍ ଉତ୍ପନ୍ନ କରେ । ଏହାର ପୃଷ୍ଠଭାଗରେ ଲାଗିଥିବା ହାଇଡ୍ରୋଜେନ୍ ଗ୍ୟାସ୍‌ର ଫୋଟକା ଯୋଗୁଁ ଏହା ଭାସିବାକୁ ଆରମ୍ଭ କରେ ।

(vii) ଏଲୁମିନିୟମ୍, ଲୌହ, ଜିଙ୍କ୍ ପରି ଧାତୁ ବାମ୍ଫସହ ପ୍ରତିକ୍ରିୟା କରି ଧାତବ ଅକ୍‌ସାଇଡ୍ ଏବଂ ହାଇଡ୍ରୋଜେନ୍ ଗ୍ୟାସ୍ ସୃଷ୍ଟିକରେ ।
2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g)
3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)
ଲେଡ୍, ତମ୍ବା, ରୁପା, ସୁନା ଆଦୌ ଜଳସହ ପ୍ରତିକ୍ରିୟା କରନ୍ତି ନାହିଁ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -11 (Activity-11)
ଆବଶ୍ୟକ ଉପକରଣ:
ସୋଡ଼ିୟମ୍ ଓ ପୋଟାସିୟମ ବ୍ୟତୀତ ଅନ୍ୟ ଧାତୁର ନମୁନା ସଂଗ୍ରହକର । ନମୁନା ମଳିନ ପଡ଼ିଥିଲେ ବାଲି କାଗଜରେ ଘଷି ସଫାକର ।

ପରୀକ୍ଷା:

  • ଲଘୁ ହାଇଡ୍ରୋକ୍ଲୋରିକ୍ ଏସିଡ୍ ଥ‌ିବା ପରୀକ୍ଷାନଳୀରେ ନମୁନାଗୁଡ଼ିକୁ ଅଲଗା ଅଲଗା ଭର୍ତ୍ତି କର ।
  • ଥର୍ମୋମିଟରକୁ ପରୀକ୍ଷାନଳୀ ମଧ୍ଯରେ ରଖି । ଏହାର ବଲ୍‌ବ ଏସିଡ୍ ଭିତରେ ବୁଡ଼ିରହୁ ।

ପର୍ଯ୍ୟବେକ୍ଷଣ:
କେତେକ ପରୀକ୍ଷାନଳୀରୁ ଫୋଟକା ବିଭିନ୍ନ ବେଗରେ ବାହାରୁଛି ଲକ୍ଷ୍ୟକର ।

ସିଦ୍ଧାନ୍ତ :
(i) ମ୍ୟାଗ୍ନେସିୟମ୍ ତୀବ୍ରଭାବରେ ପ୍ରତିକ୍ରିୟା କରୁଛି ଏବଂ ସବୁଠାରୁ ଦ୍ରୁତବେଗରେ ଫୋଟକା ସୃଷ୍ଟି ହେଉଛି ।

(ii) ମ୍ୟାଗ୍ନେସିୟମ୍ ଥ‌ିବା ପରୀକ୍ଷାନଳୀର ତାପମାତ୍ରା ଅଧ‌ିକ ।

(iii) Mg > Al > Zn > Fe (କ୍ରମରେ ପ୍ରତିକ୍ରିୟାଶୀଳତା କମିଥାଏ)

(iv) ତମ୍ବା, ରୁପା, ସୁନା, ଲଘୁ ହାଇଡ୍ରୋକ୍ଲୋରିକ୍ ଅମ୍ଳ ସହ ପ୍ରତିକ୍ରିୟା କରନ୍ତି ନାହିଁ ।

(v)
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-18

କୌଣସି ଧାତୁ HNO3 ସହ ପ୍ରତିକ୍ରିୟା କଲେ ହାଇଡ୍ରୋଜେନ୍ ଗ୍ୟାସ୍ ନିର୍ଗତ ହୁଏ ନାହିଁ କାରଣ HNO3 ସବଳ ଜାରକ ।
ଉତ୍ପନ୍ନ ହେଉଥ‌ିବା ହାଇଡ୍ରୋଜେନକୁ ଜାରଣ କରି ଜଳ ସୃଷ୍ଟିକରେ ଏବଂ ନିଜେ ବିଜାରିତ ହୋଇ ଯେକୌଣସି ନାଇଟ୍ରୋଜେନ୍ ଅକ୍‌ସାଇଡ୍‌ (N2O, NO, NO2,) ହୁଏ ।
କିନ୍ତୁ ମ୍ୟାଗ୍ନେସିୟମ୍ (Mg) ଓ ମାଙ୍ଗାନିଜ (Mn) ଲଘୁ ନାଇଟ୍ରିକ ଏସିଡ଼୍ ସହ ପ୍ରତିକ୍ରିୟା କରି H2 ଗ୍ୟାସ୍ ନିର୍ଗତ

ଅମ୍ଳରାଜ :
ଅମ୍ଳରାଜ(Aquaregia) ହେଉଛି ସଦ୍ୟ ପ୍ରସ୍ତୁତ ଗାଢ଼ ହାଇଡ୍ରୋକ୍ଲୋରିକ୍ ଏସିଡ୍ ଏବଂ ଗାଢ଼ ନାଇଟ୍ରିକ ଏସିଡ୍ 3 : 1 ଅନୁପାତର ଏକ ମିଶ୍ରଣ । ଏହି ଦୁଇ ଅମ୍ଳମଧ୍ୟରୁ କୌଣସିଟି ସୁନାକୁ ଦ୍ରବୀଭୂତ କରିପାରେ ନାହିଁ । କିନ୍ତୁ ଏହାର ମିଶ୍ରଣ ସୁନାକୁ ଦ୍ରବୀଭୂତ କରିପାରେ । ଆକ୍ସାରେଜିଆ (ଅମ୍ଳରାଜ) ଏକ ଭଲ ସଂକ୍ଷାରଣ, ଦ୍ରବୀଭୂତ କରିବାର କ୍ଷମତା ରହିଛି ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -12 (Activity-12)
ଆବଶ୍ୟକ ଉପକରଣ:
ଖଣ୍ଡିଏ ସଫା ତମ୍ବାତାର, ଲୁହା କଣ୍ଟା, ଆଇରନ୍ ସଲଫେଟ୍ ଦ୍ରବଣ, କପର ସଲ୍‌ଫେଟ ଦ୍ରବଣ । ଆଇରନ୍ ସଲଫେଟ୍ ଦ୍ରବଣ ନେଇ ସେଥୁରେ ତମ୍ବାତାରଟିକୁ

ପରୀକ୍ଷଣ:
ଗୋଟିଏ ପରୀକ୍ଷାନଳୀରେ ଆଇରନ୍ ସଲଫେଟ୍ ଦ୍ରବଣ ନେଇ ସେଥ‌ିରେ ତମ୍ବାତାରଟିକୁ ଏବଂ ଅନ୍ୟ ଏକ ପରୀକ୍ଷାନଳୀରେ କପଟ ସଲଫେଟ୍ ଦ୍ରବଣ ନେଇ ସେଥ‌ିରେ ଲୁହାକଣ୍ଟା ଭର୍ତ୍ତି କର ।

BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-19

ପର୍ଯ୍ୟବେକ୍ଷଣ:
20 ମିନିଟ୍ ପର୍ଯ୍ୟନ୍ତ ପରୀକ୍ଷଣଟିକୁ ଲକ୍ଷ୍ୟକର ।

ସିଦ୍ଧାନ୍ତ:
(i) ଯେଉଁ ପରୀକ୍ଷାନଳୀରେ କପର ସଲଫେଟ୍ ଦ୍ରବଣରେ ଲୁହାକଣ୍ଟାଟିକୁ ଭର୍ତି କରାଯାଇଥିଲା, ସେହି ପରୀକ୍ଷାନଳୀରେ ନିମ୍ନ ପ୍ରତିକ୍ରିୟା ଲକ୍ଷ୍ୟ କରାଗଲା
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-20
(ii) ନୀଳରଙ୍ଗର CuSO4 ଦ୍ରବଣ କ୍ରମଶଃ ସବୁଜ ଦ୍ରବଣରେ ପରିଣତ ହୋଇ ମାଟିଆଲାଲ୍ କପର ବାହାରିବ । ଏ ପ୍ରକାର ପ୍ରତିକ୍ରିୟାର ନାମ ହେଉଛି ପ୍ରତିସ୍ଥାପନ ପ୍ରତିକ୍ରିୟା ।

(iii) ଏ ପ୍ରକାର ପ୍ରତିକ୍ରିୟାର ନାମ ହେଉଛି ପ୍ରତିସ୍ଥାପନ ପ୍ରତିକ୍ରିୟା ଅଧ୍ଵ ପ୍ରତିକ୍ରିୟାଶୀଳ ଧାତୁ କମ୍ ପ୍ରତିକ୍ରିୟାଶୀଳ ଧାତୁକୁ ସେମାନଙ୍କର ଦ୍ରବଣୀୟ ଯୌଗିକରୁ କିମ୍ବା ତରଳ ଅବସ୍ଥାରୁ ବିସ୍ଥାପନ କରେ । ସାଧାରଣ ଭାବେ କୁହାଯାଇପାରିବ ଯେ ଯଦି ଧାତୁ A, ଧାତୁ Bକୁ ଏହାର ଦ୍ରବଣରୁ ବିସ୍ଥାପନ କରେ ତେବେ Bଠାରୁ A ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ହେବ ।
ଧାତୁ A + Bର ଲବଣ ଦ୍ରବଣ → Aର ଲବଣ ଦ୍ରବଣ + ଧାତୁ B B ଠାରୁ A ଅଧ‌ିକ ପ୍ରତିକ୍ରିୟାଶୀଳ ।

ପ୍ରତିକ୍ରିୟାଶୀଳତାର ଅନୁକ୍ରମ (The Reactivity Series):
ପ୍ରତିକ୍ରିୟାଶୀଳତାର ଅନୁକ୍ରମ ହେଉଛି କେତେକ ଧାତୁର ସକ୍ରିୟତାର ଅଧଃକ୍ରମ ସଜ୍ଜାର ସାରଣୀ ।
ସକ୍ରିୟତାର ଅନୁକ୍ରମ : ଧାତୁର ଆପେକ୍ଷିକ ଓ ପ୍ରତିକ୍ରିୟାଶୀଳତା
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-21

କାର୍ଯ୍ୟାବଳୀ -13 (Activity-13)
ଆବଶ୍ୟକ ଉପକରଣ :
ବିଜ୍ଞାନାଗାରରେ ଯେକୌଣସି ଲବଣର ନମୁନା ; ଯଥା – ସୋଡ଼ିୟମ କ୍ଲୋରାଇଡ୍, ପୋଟାସିୟମ ଆୟୋଡାଇଡ୍, ବ୍ରେରିୟମ୍ କ୍ଲୋରାଇଡ୍ ସଂଗ୍ରହ କର ।
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-22

ପରୀକ୍ଷଣ:

  • ଏହିସବୁ ଲବଣର ଭୌତିକ ସ୍ଥିତି ଲକ୍ଷ୍ୟକର ।
  • ଧାତବ ଚେପ୍‌ଟା ଚାମଚରେ ଗୋଟିଏ ଧାତୁର କିଛି ନମୁନା ନେଇ ସିଧାସଳଖ ଗରମ କର । ସେହିପରି ଅନ୍ୟ ନମୁନାଗୁଡ଼ିକୁ ମଧ୍ୟ ନେଇ ଅନୁରୂପ ଭାବେ ଗରମ କର ।
  • ନମୁନାଗୁଡ଼ିକୁ ଜଳ, ପେଟ୍ରୋଲ ଏବଂ କିରୋସିନ୍‌ରେ ଦ୍ରବୀଭୂତ କରିବା ପାଇଁ ଚେଷ୍ଟାକର ।
  • ଚିତ୍ରରେ ପ୍ରଦର୍ଶିତ ହେଲାପରି ବିଦ୍ୟୁତ୍ ପରିପଥ ତିଆରି କରି ଲବଣର ଦ୍ରବଣରେ ଇଲେକ୍‌ଟ୍ରୋଡ୍ ଦଣ୍ଡକୁ ବୁଡ଼ାଅ ।

BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-23

ପର୍ଯ୍ୟବେକ୍ଷଣ:

  • ନମୁନାଗୁଡ଼ିକରେ ଭୌତିକ ସ୍ଥିତି ହେଉଛି କଠିନ । ଏହି ନମୁନାଗୁଡ଼ିକ ଶିଖାର ବର୍ଣ୍ଣ ବଦଳାଇଲା । ଯୌଗିକଗୁଡ଼ିକ ତରଳିଲା ନାହିଁ ।
  • ଏଗୁଡ଼ିକ ଜଳରେ ଦ୍ରବଣୀୟ କିନ୍ତୁ ପେଟ୍ରୋଲ ଏବଂ କିରାସିନ୍‌ରେ ଦ୍ରବଣୀୟ ନୁହଁନ୍ତି ।
  • ଆୟନିକ, ଯୌଗିକ କଠିନ ଅବସ୍ଥାରେ ବିଦ୍ୟୁତ୍ ପରିବହନ କରନ୍ତି ନାହିଁ; କିନ୍ତୁ ଲବଣ ଦ୍ରବଣରେ ବିଦ୍ୟୁତ୍ ପରିବହନ କରେ ।

ସିଦ୍ଧାନ୍ତ:
BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-24

ଆୟନିକ୍ ଯୌଗିକର ଧର୍ମ;

  • ୟନିକ୍ ଯୌଗିକଗୁଡ଼ିକ କଠିନ : ଆୟନିକ ଯୌଗିକଗୁଡ଼ିକ ଯୁକ୍ତାତ୍ମକ ଏବଂ ବିଯୁକ୍ତାତ୍ମକ ଆୟନ ମଧ୍ୟରେ ଦୃଢ଼ ଆକର୍ଷଣ ବଳ ଯୋଗୁଁ ଏଗୁଡ଼ିକ କଠିନ ଏବଂ କିଛି ପରିମାଣରେ ଶକ୍ତ । ଏହି ଯୌଗିକଗୁଡ଼ିକ ସାଧାରଣତଃ ଭଙ୍ଗୁର ଏବଂ ଚାପ ପ୍ରୟୋଗ ଫଳରେ ଖଣ୍ଡ ଖଣ୍ଡ ହୋଇ ଭାଙ୍ଗିଯାଏ ।
  • ଗଳନାଙ୍କ ଓ ସ୍ଫୁଟନାଙ୍କ : ଆୟନିକ ଯୌଗିକଗୁଡ଼ିକର ଉଚ୍ଚ ଗଳନାଙ୍କ ଏବଂ ଉଚ୍ଚ ସ୍ଫୁଟନାଙ୍କ ଥାଏ । ଆୟନଗୁଡ଼ିକ ମଧ୍ୟରେ ଥ‌ିବା ଶକ୍ତ ଆନ୍ତଃ ଆୟନୀୟ ଆକର୍ଷଣ ଭାଙ୍ଗିବା ପାଇଁ ବିପୁଳ ପରିମାଣର ଶକ୍ତି ଆବଶ୍ୟକ ହୋଇଥାଏ ।
  • ଦ୍ରବଣୀୟତା : ଏହି ଯୌଗିକଗୁଡ଼ିକ ସାଧାରଣତଃ ଜଳରେ ଦ୍ରବଣୀୟ କିନ୍ତୁ କିରୋସିନ୍, ପେଟ୍ରୋଲ ଆଦି ଦ୍ରାବକରେ ଅଦ୍ରବଣୀୟ ।
  • ବିଦ୍ୟୁତ୍ ପରିବହନ କରନ୍ତି : କଠିନ ଅବସ୍ଥାରେ ଆୟନିକ ଯୌଗିକଗୁଡ଼ିକ ବିଦ୍ୟୁତ୍ ପରିବହନ କରେ ନାହିଁ । କିନ୍ତୁ ତରଳ ଅବସ୍ଥାରେ ଆୟନ ମୁକ୍ତ ହେବାରୁ ଏଗୁଡ଼ିକ ବିଦ୍ୟୁତ୍ ପରିବହନ କରିପାରେ ।

BSE Odisha 10th Class Physical Science Solutions Chapter 3 ଅମ୍ଳ, କ୍ଷାରକ ଓ ଲବଣ

କାର୍ଯ୍ୟାବଳୀ -14 (Activity-14)
ଆବଶ୍ୟକ ଉପକରଣ : ତିନୋଟି ପରୀକ୍ଷାନଳୀ, ପ୍ରତ୍ୟେକଟିରେ ପରିଷ୍କୃତ ଲୁହା କଣ୍ଟା, କର୍କ, ପାତିତ ଜଳ, ନିର୍ଜଳୀୟ କ୍ୟାଲସିୟମ୍ କ୍ଲୋରାଇଡ୍ , ତେଲ

ପରୀକ୍ଷଣ:

  • ପରୀକ୍ଷାନଳୀ ତିନୋଟିର ନାମକରଣ A, B ଓ C ଦିଅ । ପ୍ରତ୍ୟେକ ପରୀକ୍ଷାନଳୀରେ ଲୁହା କଣ୍ଟା ରଖ । À ନଳୀରେ କିଛି ଜଳ ଢାଳି କର୍କଦ୍ଵାରା ବନ୍ଦ କର।
  • B ନଳୀରେ ଫୁଟାହୋଇଥିବା ପାତିତ ଜଳ ଢାଳ । ପ୍ରାୟ 1 ମିଲିଲି ତେଲ ମିଶାଇ କର୍କ ଦ୍ଵାରା ବନ୍ଦ କର । ତେଲ ଜଳରେ ଭାସିବ ଏବଂ ବାୟୁ ଜଳରେ ଦ୍ରବୀଭୂତ କରାଇ ଦେବ ନାହିଁ ।
  • C ପରୀକ୍ଷାନଳୀରେ କିଛି ନିର୍ଜଳୀୟ କ୍ୟାଲସିୟମ୍ କ୍ଲୋରାଇଡ ନେଇ ନଳୀରେ | ରଖ ଓ କର୍କଦ୍ଵାରା ବନ୍ଦକର । ପରୀକ୍ଷା ନଳୀକୁ କିଛିଦିନ ରଖିଦିଅ । ।

BSE Odisha Class 10 Physical Science Solutions Chapter 3 img-25

ପର୍ଯ୍ୟବେକ୍ଷଣ:

  • A ପରୀକ୍ଷାନଳୀରେ ଥ‌ିବା ଲୁହାକଣ୍ଟାରେ କଳଙ୍କି ଲାଗିଛି ।
  • B ଓ C ପରୀକ୍ଷାନଳୀରେ ଥିବା ଲୁହାକଣ୍ଟାଗୁଡ଼ିକରେ କଳଙ୍କି ଲାଗିନାହିଁ ।
  • À ପରୀକ୍ଷାନଳୀରେ କଣ୍ଟାଗୁଡ଼ିକ ଉଭୟ ଜଳ ଓ ବାୟୁ ସଂସ୍ପର୍ଶ ଆସିବା ଫଳରେ କଳଙ୍କି ଲାଗିଲା ।
  • B ପରୀକ୍ଷାନଳୀରେ କଣ୍ଟାଗୁଡ଼ିକ କେବଳ ଜଳ ସଂସ୍ପର୍ଶରେ ଆସିଲା । C ପରୀକ୍ଷାନଳୀର କଣ୍ଟାଗୁଡ଼ିକ କେବଳ ଶୁଷ୍କବାୟୁ ସଂସ୍ପର୍ଶରେ ଆସିଲା ।

ସିଦ୍ଧାନ୍ତ:
ଉଭୟ ବାୟୁ ଓ ଜଳ ସଂସ୍ପର୍ଶରେ ଆସିଲେ ଲୌହ ବସ୍ତୁରେ କଳଙ୍କି ଲାଗେ ।

ସଂକ୍ଷାରଣର ପ୍ରତିରୋଧ (Prevention of Corrosion):
ସଂକ୍ଷାରଣର ପ୍ରତିରୋଧ ପାଇଁ ନିମ୍ନଲିଖତ ଉପାୟଗୁଡ଼ିକ ଅବଲମ୍ବନ କରାଯାଇଥାଏ ।

  • ତୈଳଦ୍ଵାରା: ଧାତୁ ନିର୍ମିତ ପଦାର୍ଥ ଉପରେ ତୈଳ ଜାତୀୟ ପଦାର୍ଥ ଲେପନ କଲେ ଧାତୁ ଆର୍ଦ୍ର ବାୟୁର ସଂସ୍ପର୍ଶରେ ଆସେନାହିଁ । ଫଳରେ ସଂକ୍ଷାରଣ ହୁଏ ନାହିଁ।
  • ଗ୍ରୀଜଦ୍ଵାରା: ଶ୍ରୀଜଲେପ ଦେଲେ କଳଙ୍କି ଲାଗେ ନାହିଁ ।
  • ରଙ୍ଗ ଲେପନଦ୍ଵାରା କ୍ଷୟରୋଧ କରାଯାଏ ।
  • ଜିଙ୍କ୍ ଲେପନ (ଗାନାଇଜିଙ୍ଗ୍) – ଜିଙ୍କର ଏକ ପତଳା ସ୍ତରର ଆଚ୍ଛାଦନ ଦ୍ବାରା ଷ୍ଟିଲ ଓ ଲୁହାକୁ କଳଙ୍କି ଲାଗିବାରେ ପ୍ରତିରୋଧ କରାଯାଇ ପାରିବ । ଜିଙ୍କ ଆବରଣ ନଷ୍ଟ ହେଲେ ମଧ୍ୟ ଜିଙ୍କ ଲେପିତ ଜିନିଷ ଗୁଡ଼ିକ କଳଙ୍କି ଲାଗିବାରୁ ସୁରକ୍ଷିତ ରହିଥାଏ ।
  • ଇଲେକ୍‌ଟ୍ରୋସ୍ଟେଟିଂ ଦ୍ଵାରା : ଲୁହାରେ ନିର୍ମିତ ଦ୍ରବ୍ୟାଦିଗୁଡ଼ିକ ବିଦ୍ୟୁତ୍ ଶକ୍ତି ସାହାଯ୍ୟରେ ଅନ୍ୟଧାତୁର ପ୍ରଲେପଦ୍ବାରା କ୍ଷୟରୋଧ କରାଯାଏ ।
  • ଟିନିଂ ଦ୍ଵାରା: ବଜାରରେ ମିଳୁଥିବା ଟିଣ ଡବା ଲୁହାରେ ତିଆରି । ଏହି ଲୁହାଡ଼ବା ଉପରେ ଟିଣର ଲେପ ଦିଆଯାଇଛି । ତେଣୁ ସେହିଡବାରେ କଳଙ୍କି ଲାଗିପାରେ ନାହିଁ ।
  • କେତକ ମିଶ୍ରଧାତୁରେ ଆଦୌ କଳଙ୍କି ଲାଗେନାହିଁ । ବିଶୁଦ୍ଧ ଲୁହା ନରମ ଅଟେ । ମାତ୍ର ଅଳ୍ପ ପରିମାଣର କାର୍ବନ (0.05%) ସହ ମିଶାଇଲେ ତାହା ଶକ୍ତ ଓ କଠିନ ହୋଇଯାଏ ।

ଲୁହାରେ ନିକେଲ୍‌ ଓ କ୍ରୋମିୟମ୍ ମିଶିଲେ ଷ୍ଟେନ୍‌ଲେସ୍‌ ଷ୍ଟିଲ୍‌ ମିଳେ ଯାହାକି ଶକ୍ତ ହୁଏ ଓ କଳଙ୍କି ଲାଗେନାହିଁ ।

ମିଶ୍ରଧାତୁ (Alloy) – ଦୁଇ ବା ଅଧ୍ଵ ଧାତୁର କିମ୍ବା ଧାତୁ ଓ ଉପଧାତୁ କିମ୍ବା ଧାତୁ ଓ ଅଧାତୁର ମିଶ୍ରଣକୁ ମିଶ୍ରଧାତୁ କହନ୍ତି । ଏହା ସମଜାତୀୟ ଅଟେ ।

ପାରଦର ମିଶ୍ରଧାତୁକୁ ଅମାଲ୍‌ଗାମ୍ କୁହାଯାଏ । ପିତ୍ତଳ ହେଉଛି ତମ୍ବା ଏବଂ ଜିଙ୍କ୍ (Cu ଓ Zn) ର ଏକ ମିଶ୍ର ଧାତୁ ।
ବ୍ରୋଞ୍ଜ୍ ହେଉଛି ତମ୍ବା ଏବଂ ଟିଣ (Cu ଓ Sn) ର ଏକ ମିଶ୍ରଧାତୁ ।
ସୋଲ୍ଡର ହେଉଛି ସୀସା ଓ ଟିଣ (Pb ଓ Sn) ର ଏକ ମିଶ୍ରଧାତୁ ।

ବିଶୁଦ୍ଧ ସୁନା 24 କ୍ୟାରେଟ୍ ନାମରେ ଜଣାଶୁଣା । ତାହା ଅତ୍ୟନ୍ତ ନରମ ହୋଇଥିବାରୁ ସେଥିରେ ଅଳଙ୍କାର ତିଆରି ହୋଇପାରେ ନାହିଁ । ଏହାକୁ ଶକ୍ତ କରିବା ପାଇଁ ଏଥିରେ ରୁପା କିମ୍ବା ତମ୍ବା ମିଶାଯାଏ । ସାଧାରଣତଃ 22 କ୍ୟାରେଟ୍‌ ସୁନା ଅଳଙ୍କାର ତିଆରି ପାଇଁ ବ୍ୟବହୃତ ହୁଏ । ଅର୍ଥାତ୍‌ 22 ଭାଗ ବିଶୁଦ୍ଧ ସୁନା ସହ 2 ଭାଗ ତମ୍ବା କିମ୍ବା ରୁପା ମିଶାଯାଇ ମିଶ୍ରଧାତୁ ପ୍ରସ୍ତୁତ କରାଯାଏ । 1600 ବର୍ଷ ପୂର୍ବେ ଦିଲ୍ଲୀର କୁତୁବ୍‌ମିନାର ଠାରେ ନିର୍ମିତ ହୋଇଥିବା ଲୌହସ୍ତମ୍ଭର ଉଚ୍ଚତା 8 ମିଟର ଏବଂ ଓଜନ 6 ଟନ୍ । ଏହି ଲୌହସ୍ତମ୍ଭ ଏପରି ପ୍ରଣାଳୀରେ ନିର୍ମିତ ହୋଇଥିଲା ଯେ ଏହା ଏବେବି କଳଙ୍କହୀନ ଅବସ୍ଥାରେ ରହିଛି ।

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(e) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(e)

Evaluate the following:
Question 1.
(i) ∫(1 + x) ex dx
Solution:
∫(1 + x) ex dx
[Choolse 1 + x as first and ex as second function
= (1 + x) ex – ∫1 . ex dx
= ( 1 + x) ex – ex + C = xex + C

(ii) ∫x3 ex dx
Solution:
∫x3 ex dx = x3 ex – ∫3x2 ex dx
= x3 ex – 3{x2 ex – ∫2x ex dx}
= x3 ex – 3x2 ex + 6 ∫x ex dx
= x3 ex – 3x2 ex +6 {x . ex – ∫1 . ex dx}
= x3 ex – 3x2 ex + 6x ex – 6ex + C

(iii) ∫x2 eax dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.1(3)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

(iv) ∫(3x + 2)2 e2x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.1(4)

Question 2.
(i) ∫x sin x dx
Solution:
∫x sin x dx
[x = first function
sin x = 2nd function]
= x (-cosx) – ∫\(\frac{d}{d x}\)(x) . (-cos x) dx
= -x cos x + ∫cos x dx
= -x cos x + sin x + C

(ii) ∫x2 cos x dx
Solution:
∫x2 cos x dx
[x2 = 1st
cos x = 2nd]
= x2 . sin x – ∫\(\frac{d}{d x}\)(x2) sin x dx
= x2 sin x – ∫2x . sin x dx
[x = 1st
sin x = 2nd]
= x2 sin x – 2 {x . (-cos x) – ∫1 . (-cos x) dx}
= x2 sin x + 2x cos x – 2∫cos x dx
= x2 sin x + 2x cos x – 2 sin x + C

(iii) ∫x2 sin ax dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.2(3)

(iv) ∫x cos2 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.2(4)

(v) ∫x sin3 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.2(5)

(vi) ∫2x sin 2x cos x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.2(6)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

(vii) ∫2x cos 3x cos 2x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.2(7)

(viii)∫2x3 cos x2 dx
Solution:
∫2x3 cos x2 dx
[Put x2 =t
Then 2x dx = dt]
= ∫x2 . cos x2 . 2x dx
= ∫t . cos t dt
= t . sin t – ∫1 . sin t dt
= t sin t + cos t + C
= x2 sin x2 + cos x2 + C

(ix) ∫x cosec2 x dx
Solution:
∫x cosec2 x dx
[x = 1st
cosec2 x = 2nd]
= x ∫cosec2 x dx – ∫[\(\frac{d}{d x}\)(x) × ∫cosec2 x dx] dx
= -x cot x + ∫cot x dx
= -x cot x + ln |sin x| + C

(x) ∫x tan2 x dx
Solution:
∫x tan2 x dx = ∫x (sec2 x – 1) dx
= ∫x sec2 x dx – ∫x dx
= x tan x – ∫1 . tan x dx – \(\frac{1}{2}\)x2
[x = 1st
sec2 x = 2nd]
= x tan x + ln |cos x| – \(\frac{x^2}{2}\) + C

Question 3.
(i) ∫x ln (1 + x) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(1)

(ii) ∫x7 ln x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(2)

(iii) ∫(ln x)3 dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(3)

(iv) ∫ln(x2 + 1) dx
Solution:
∫ln (x2 + 1) dx
= ∫ln (x2 + 1) . 1 dx
[Put ln (x2 + 1 ) as first function and 1 as the second function.]
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(4)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

(v) ∫\(\frac{\ln x}{x^5}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(5)

(vi) ∫ln (x2 + x + 2) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(6)

(vii) ∫ln (x + \(\sqrt{x^2+a^2}\)) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(7)

(viii) ∫ln (x + \(\sqrt{x^2-a^2}\)) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.3(8)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

Question 4.
(i) ∫sin-1 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.4(1)

(ii) ∫x sin-1 dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.4(2)

(iii) ∫cos-1 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.4(3)

(iv) ∫x tan-1 dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.4(4)

(v) ∫x2 tan-1 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.4(5)

(vi) ∫sec-1 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.4(6)

(vii) ∫x cosec-1 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.4(7)

Question 5.
(i) ∫e3x cos 2x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.5(1)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

(ii) ∫ex sin x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.5(2)

(iii) ∫ex cos2 x dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.5(3)

(iv) ∫x \(e^{x^2}\) sin x2 dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.5(4)

(v) ∫eax sin (bx + c) dx
Solution:
Let I = ∫eax sin (bx + c) dx
Integrating by parts we get
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.5(5)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

(vi) ∫(2x2 + 1)\(e^{x^2}\) dx
Solution:
I = ∫(2x2 + 1)\(e^{x^2}\) dx
= ∫2x2 \(e^{x^2}\) dx + ∫\(e^{x^2}\) . 1 dx
= ∫2x2 \(e^{x^2}\) dx + x2 \(e^{x^2}\) ∫2x\(e^{x^2}\) .x dx
= x\(e^{x^2}\) + C

Question 6.
(i) ∫\(\sqrt{9-x^2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.6(1)

(ii) ∫\(\sqrt{5-4 x^2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.6(2)

(iii) ∫\(\sqrt{1-x^2-2 x}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.6(3)

(iv) ∫ez \(\sqrt{4-e^{2 z}}\) dz
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.6(4)

(v) ∫cos θ \(\sqrt{5-\sin ^2 \theta}\) dθ
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.6(5)

Question 7.
(i) ∫\(\sqrt{x^2+4}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.7(1)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

(ii) ∫\(\sqrt{7 x^2+2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.7(2)

(iii) ∫\(\sqrt{4 x^2+12 x+13}\) dx (2x + 3 = z)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.7(3)

(iv) ∫e2z \(\sqrt{e^{4 z}+6}\) dz
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.7(4)

(v) ∫sec2 θ \(\sqrt{\sec ^2 \theta+3}\) dθ
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.7(5)

(vi) ∫(2x2 +1) \(e^{x^2}\) dx
Solution:
Same as No. 5 (vi).

Question 8.
(i) ∫\(\sqrt{x^2-8}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.8(1)

(ii) ∫\(\sqrt{3 x^2-2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.8(2)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

(iii) ∫\(\sqrt{x^2-4 x+2}\) dx (x – 2 = z)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.8(3)

(iv) ∫az \(\sqrt{a^{2 z}-4}\) dz
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.8(4)

(v) ∫sec θ tan θ \(\sqrt{\tan ^2 \theta-3}\) dθ
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.8(5)

Question 9.
(i)∫ex (tan x + ln sec x) dx
Solution:
∫ex (tan x + ln sec x) dx
= ∫ex tan x dx + ∫ex ln sec x dx
(Integrating by parts)
= ∫ex tan x dx + ex ln sec x – ∫ex tan x dx
= ex ln(sec x) + C

(ii) ∫ex (cot x + ln sin x) dx
Solution:
∫ex (cot x + ln sin x) dx
[Integrating by parts taking ex as first function and cot x as second function.]
= ∫ex ln sin x – ∫ex ln sin x dx + ∫ex ln sin x dx + C
= ex ln sin x + C

(iii) ∫\(\frac{e^x}{x}\) (1 + x ln x) dx
Solution:
∫\(\frac{e^x}{x}\) (1 + x ln x) dx
= ∫\(\frac{e^x}{x}\) dx + ∫\(\frac{e^x}{x}\) ex ln x dx + C
= ex ln x + C

(iv) ∫\(\frac{x e^x}{(1+x)^2}\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.9(4)

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e)

Question 10.
(i) ∫\(\left[\frac{1}{\ln x}-\frac{1}{(\ln x)^2}\right]\) dx
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.10(1)

(ii) ∫sin (ln x) dx
Solution:
Let I = ∫sin (ln x) dx
[Integrating by parts taking sin (ln x) as first and 1 as second function.]
CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(e) Q.10(2)

(iii) ∫sin x ln (cosec x – cot x) dx
Solution:
∫sin x ln (cosec x – cot x) dx
[Integrating by parts taking In (cosec x cot x) as first function and sin x as second function.]
= ln (cosec x – cot x) . – cos x – ∫\(\frac{1}{{cosec} x-\cot x}\)× – cosec x . cot x + cosec2 x × – cos x dx
= -cos x . ln (cosec x – cot x) + ∫\(\frac{{cosec} x({cosec} x-\cot x)}{{cosec} x-\cot x}\) . cos x dx
= -cos x . ln (cosec x – cot x) + ∫cot x dx
= -cos x . ln (cosec x – cot x) + ln sin x + C

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Odisha State Board CHSE Odisha Class 11 Invitation to English 2 Solutions Non-Detailed Chapter 1 Three Questions Textbook Exercise Questions and Answers.

CHSE Odisha 11th Class English Solutions Non-Detailed Chapter 1 Three Questions

CHSE Odisha Class 11 English Three Questions Text Book Questions and Answers

Unit – 1

Gist :
A king was worried, because he had three questions constantly troubling him. He wanted to know what was the right time for the right work with the right person, who were the most important men whom he should consult and what was the most important work. Many learned persons came but failed to satisfy the king with their answers. Some said that the right time could be ascertained if the king consulted a table of days, months and years strictly. Others said the king should seek the help of a magician. In reply to the second question, some said the people the king most needed were the council of ministers, others said they were priests or doctors. To the third question, the answer was equally confusing. The king could not agree with anyone of them. So he decided to approach a learned hermit living in a wood. In other words, the hermit was living far from the madding crowd. He was famous for wisdom.

Glossary:
above all: more than anything else (ଅନ୍ୟ ସମସ୍ତ ବିଷୟଠାରୁ ଅଧିକ )
occurred: came to mind (ମୁଣ୍ଡକୁ )
proclaimed:  made known publicly or officially (ଆନୁଷ୍ଠାନିକ ଭାବରେ)
pastimes: things done to pass time pleasantly (ସମୟ)
Reward: prize (ପୁରସ୍କାର)
beforehand: in advance (ଆଗୁଆ)
letting: allowing (ଅନୁମତି)
absorbed: giving the whole mind to (ମଜ୍ଜିଯିବା)
skill: expertness (ବିଶେଷଜ୍ଞତା)
warfare: the state of being at war (ଯୁଦ୍ଧ)

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Think it out

Question 1.
What were the three questions that occurred to the king? What did he do get the answers to his questions?
Answer:
The three questions that occurred to the king were if he was always aware of the right time to start everything, whether he had the knowledge of who the perfect people to listen to, and whom to get rid of, and at last if he was alive to do what the most genuine work to perform. To get the answers to his questions, he announced a great reward for anybody who would answer these three questions satisfactorily.

Question 2.
What answers did the wise men give to his first question?
Answer:
The happy prospect of being rewarded by the king fabulously attracted many wise men. They made a beeline to the presence of the king hoping to answer his questions. In reply to the first question, some suggested that the king ought to consult a table of days, months and years in order to know the right time to do the right work. Others advised him to consult a council. Yet there were some who even suggested to the king to seek the help of a magician in the matter.

Question 3.
How did the wise men answer his second question?
Answer:
In reply to the second question, the wise men were not unanimous in their answers. Some said that the king should consult his councillors. Others said he should consult priests and doctors; while some said the warriors were the most necessary.

Question 4.
What answers did the king get for his third question?
Answer:
The king got various answers for his third question from the wise men. In their opinion they differ from each other. Some wise men replied that the important thing in the world was science. To others, it was skill in warfare. The answer did not end there. Some other wise men replied that it was religious worship.

Question 5.
Why did he decide to consult a hermit?
Answer:
Different answers to his three important questions failed to satisfy the king. In other words, he did not subscribe to any of the answers provided by the wise men. There was a hermit who was exceedingly famous for his wisdom. Therefore, the king decided to consult a hermit.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Unit – II

Gist :
The king approached the hermit in the guise of a common man. The holy man was then digging a hole outside. He blessed the visitor and went about his work. The king asked him the three questions seeking his answers. But the hermit didn’t bother to answer the questions. He went on digging the hole as before. He looked very tired. The king took pity on the old man and offered his help. The hermit took rest and the king went on digging the hole till it was evening. He was now impatient and begged the holy man to answer his questions. But the holy man was as silent as before. In the mean time, a man came running towards them.

He was bleeding profusely from a wound in his stomach. The king was moved and washed the man’s wound and bandaged it. The hermit served the man too. Both of them carried the wounded man inside and laid him on a bed. Being tired the king fell asleep on the threshold. When he woke up, he heard the wounded man aplogising to him. It was now revealed that the man was a former enemy of the king who had followed him to kill him, but he was found out by the king’s bodyguards and was mortally wounded by them. The man said that if the king had been late, he would have killed him. The king was happy to know that a die hard enemy of his had been won over so easily. The wounded man swore to devote the rest of his life serving the king.

Glossary:
quitted : left ବାମ
cell : a single room ଗୋଟିଏ କୋଠରୀ
hermit : a holy person living alone ଏକ ପବିତ୍ର ବ୍ୟକ୍ତି
dismounted : climbed down ବିସର୍ଜନ
approached : went towards ଆଡକୁ ଗଲା |
unfastened : opened ଖୋଲିଲା
soaked : wet ଓଦା
ceased : stopped ଅଟକି ଗଲା
revived : got well again ପୁଣି ଭଲ ହୋଇଗଲା
crouched down : ଆଣ୍ଠେଇ
threshold : doorstep ସୀମା
frail : weak and thin ଦୁର୍ବଳ ଏବଂ ପତଳା
spade: an instrument for digging (କୋଡ଼ି)
fainting : collapsing (ଅଚେତ, ବେହୋସ )
moaning : making a low mournful sound ଏକ କମ୍ ଶୋକର ଶବ୍ଦ ଶୁଣିବା
feebly: slowly (ଧୀରେ ଧୀରେ)
intently : eagerly ଉତ୍ସାହର ସହିତ
executed : punished by death ମୃତ୍ୟୁ ଦ୍ୱାରା ଦଣ୍ଡିତ
ambush : hiding ଲୁଚି
restore : to give back ଫେରାଇବାକୁ

Think it out :

Question 1.
Where did the king meet the hermit? How did the hermit receive the king?
Answer:
The king clad in simple clothes met the hermit in the wood which was his permanent abode. The hermit’s meeting was only confined to the common folk. However, he received the king as usual. There was not a touch of extra-ordinariness about it. After greeting the king, the hermit kept on digging the ground in front of his hut.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Question 2.
How did the king help the hermit?
Answer:
The king saw the hermit digging the ground single-handedly. The old man was weak and frail and skinny. He seemed exhausted. The king took pity on him and kind as he was, he wished to lend a helping hand to the holy man. He took the spade from him and dug the ground himself. Thus, the king helped the hermit.

Question 3.
How did the king nurse the wounded stranger?
Answer:
As soon as the king saw the wounded stranger, he along with the hermit opened his clothing. He carefully washed and bandaged the large wound in his stomach with his handkerchief, yet the blood kept on flowing. Therefore, the king again and again removed the bandage soaked with blood, and washed and bandaged it once again. At last blood stopped flowing. The injured man came to his senses and asked for water. The king rose to the occasion. At last, he along with the hermit carried the wounded stranger to the hut.

Question 4.
Why did the wounded person desire to serve the king as his most faithful slave?
Answer:
The wounded man was a die-hard enemy of the king. The king had executed his brother and confiscated his property. So the man wanted to take revenge by killing the king while he was paying visit to the hermit alone. But the way with which the king nursed him and saved his life moved the man. It was unbelievable. The hostility he had nurtured for the king vanished. His heart was filled with repentance. Therefore, the wounded person desired to serve the king as his most faithful servant

Unit: III

Gist :
The king wanted to return to the palace. He again approached the hermit seeking answers to his questions. But the holy man replied smilingly that he had already had his answers. The right time is the time which is ‘now’, the right work is the work before you and the right person is the one with whom you are. The most important affair is to do good to the man who needs your help at the moment. The king understood everything. He took up the work that the holy man had been doing and that work and that time were the most important ones. The hermit who was before him was the most important man needing his care and attention.

Glossary :
taken leave of : ଛୁଟି
wounded : ଆହତ
injured : ଆହତ
porch : a covered entrance to a house ଗୋଟିଏ ଘରର ଆଚ୍ଛାଦିତ ପ୍ରବେଶ
sowing : scattering seeds over ବୁଣିବା
pitied : showed pity ଦୟା
attended to : treated ଚିକିତ୍ସିତ
made peace : ଶାନ୍ତି ସ୍ଥାପନ କଲା
dealings : treatment କାରବାର
that…. life ତାହା …. ଜୀବନ: God had sent man into this life for the welfare of all ସମସ୍ତଙ୍କ କଲ୍ୟାଣ ପାଇଁ man ଶ୍ବର ମନୁଷ୍ୟକୁ ଏହି ଜୀବନରେ ପଠାଇଥିଲେ

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Think it out:

Question 1.
How did the hermit answer the king’s questions?
Answer:
Before taking leave of him, the king again asked the hermit to answer his three questions, but he was told that he had already been answered. The most important time for him the day before was when he was digging the bed, the most important man was the hermit. Afterwards when he attended to the wounded stranger that was the right time, the stranger the right man and the help given was the most important work otherwise the injured man would have died. Therefore the right time is always the present time, the most important man is one whom one deals with at that time and the right occupation is to help others.

Question 2.
Was the king satisfied with the hermit’s answers? Give a reasoned answer.
Answer:
The king was greatly satisfied with the hermit’s answers. Each answer the hermit gave to his questions was relevant. For instance, the enemy would have killed the king, had he not digged the ground and consequently, staying with him. The hermit’s focus on ‘now’ impressed him most. The king was satisfied, when the hermit wanted him to do good to the man of that moment is the most important affair, because
God has created man only for this purpose.

CHSE Odisha Class 11 English Three Questions Important Questions and Answers

Question 1.
Read through the extract and answer the questions that follow.
The hermit lived in a wood which he never quitted, and he received none but common folk. So the King put on simple clothes, and before reaching the hermit’s cell dismounted from his horse, and, leaving his body-guard behind, went on alone. When the King approached, the hermit was digging the gound in front of his hut. Seeing the King, he greeted him and went on digging. The hermit was frail and weak, and each time he stuck his spade into the ground and turned a little earth, he breathed heavily. The King went up to him and said : “I have come to you, wise hermit, to ask you to answer three questions: How can I learn to do the right thing at the right time ? Who are the people I most need, and to whom should I, therefore, pay more attention than to the rest ? And, what affairs are the most important, and need my first attention ?” The hermit listened to the King, but answered nothing. He just spat on his hand and recommenced digging.
“You are tired,” said the King, “let me take the spade and work awhile for you.”
“Thanks !” said the hermit, and, giving the spade to the King, he sat down on the ground.
When he had dug two beds, the King stopped and repeated his questions.
The hermit again gave no answer, but rose, stretched out his hand for the spade, and said: “Now rest awhile and let me work a bit.”
But the King did not give him the spade, and continued to dig. One hour passed, and another. The sun began to sink behind the trees, and the king at last stuck the spade into the ground, and said :
“I came to you, wise man, for an answer to my questions. If you can give me none, tell me so, and I will return home.”
“Here comes some one running,” said the hermit, “let us see who it is.”
The King turned round, and saw a bearded man come running out of the wood. The man held his hands pressed against his stomach, and blood was flowing from under them. When he reached the King, he fell fainting on the ground moaning feebly. The King and the hermit unfastened the man’s clothing. There was a large wound in his stomach.
Questions :
(i) Throw light on the hermit.
(ii) What picture of the king do you find in the extract?

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Answers :
(i) The hermit lived in a forest. He was very fond of this place. He identified himself with the common man and shared their feelings. Against this backdrop, he received none except the common folk. He was the epitome of great values. He cordially greeted the king clad in common clothes, the hermit was thin and weak. He believed in the philosophy of duty. In spite of being physically weak, he was digging the soil with a spade. In him, the king found a wise man. He was interested to ask his questions to the hermit.

(ii) The extract presents the king in a bright light. He is averse to exhibiting his royal glamour. He was simple to the core. He meets the hermit, putting on simple clothes. He understands the feelings of the hermit. He is a kindhearted person. He cannot stand the sight of the hermit digging the soil with a spade. The king takes the tool from him and performs his work. Despite everything, his curiosity to ask his questions to the hermit and find answers does not diminish. The king also shows kidness to the injured bearded man.

Question 2.
Read through the extract and answer the questions that follow.
The King turned round, and saw a bearded man come running out of the wood. The man held his hands pressed against his stomach, and blood was flowing from under them. When he reached the King, he fell fainting on the ground moaning feebly. The King and the hermit unfastened the man’s clothing. There was a large wound in his stomach. The King washed it as best he could, and bandaged it with his handkerchief and with a towel the hermit had. But the blood would not stop flowing, and the King again and again removed the bandage soaked with warm blood, and washed and rebandaged the wound. When at last the blood ceased flowing, the man revived and asked for something to drink. The King brought fresh water and gave it to him. Meanwhile the sun had set, and it had become cool. So the King, with the hermit’s help, carried the wounded man into the hut and laid him on the bed. Lying on the bed the man closed his eyes and was quiet, but the King was so tired with his walk and with the work he had done, that he crouched down on the threshold, and also fell asleep- so soundly that he slept all through the short summer night. When he awoke in the morning, it was long before he could remember where he was, or who was the strange bearded man lying on the bed and gazing intently at him with shining eyes.
“Forgive me !” said the bearded man in a weak voice, when he saw that the King was awake and was looking at him.
“I do not know you, and have nothing to forgive you for,” said the King.
“You do not know me, but I know you. I am that enemy of yours who swore to revenge himself on you, because you executed his brother and seized his property. I knew you had gone alone to see the hermit, and I resolved to kill you on your way back. But the day passed and you did not return. So I came out from my ambush to find you, and I came upon your bodyguard, and they recognized me, and wounded me. I escaped from them, but should have bled to death had you not dressed my wound. I wished to kill you, and you have saved my life. Now, if I live, and if you wish it, I will serve you as your most faithful slave, and will bid my sons do the same. Forgive me !” The King was very glad to have made peace with his enemy so easily, and to have gained him for a friend, and he not only forgave him, but said he would send his servants and his own physician to attend him, and promised to restore his property.

Questions :
(i) What made the king enjoy a sound sleep?
(ii) Describe the meeting between the injured person and the king.

Answers :
(i) The king came forward to relieve the wise hermit of his work, because the latter was thin and weak. He did not allow the hermit to go on digging with the spade. The king requested him to give him the spade. He kept on doing his work. The hermit want him to take rest, but in vain. In the meanwhile. a bearded person came running out of the wood, with blood gushing from the stomach. The king and the hermit rose to the occasion. At last, they helped him carry to the hut. Walk and work made the king dog-tired. At last, he fell a sound sleep.
(ii) The king learnt that the bearded man was his enemy, because of his act of executing his brother and seizing his property. Knowing that he had come alone to meet the hermit, the man had promised to slay him. On the way his soldiers recognised him and attacked him. But for the king’s compassion, he would have been bled to death. Again, the man expressed his wish to serve the king and be his most trusted servant, and begged the king’s forgiveness. The king was glad to establish peace with his enemy and promised to give his property back.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Question 3.
Read through the extract and answer the questions that follow.
Having taken leave of the wounded man, the King went out into the porch and looked around for the hermit. Before going away he wished once more to beg an answer to the questions he had put. The hermit was outside, on his knees sowing seeds in the beds that had been dug the day before.
The King approached him, and said:
“For the last time, I pray you to answer my questions, wise man.”
“You have already been answered !” said the hermit, still crouching on his thin legs, and looking up at the King, who stood before him.
“How answered? What do you mean ?” asked the King.
“Do you not see,” replied the hermit. “If you had not pitied my weakness
yesterday, and had not dug those beds for me, but had gone your way, that man would have attacked you, and you would have repented of not having stayed with me. So the most important time was when you were digging the beds, and I was the most important man, and to do me good was your most important business. Afterwards when that man ran to us, the most important time was when you were attending to him, for if you had not bound up his wounds he would have died without having made peace with you. So he was the most important man, and what you did for him was your most important business. Remember then : there is only one time that is important Now! It is the most important time because it is the only time when we have any power. The most necessary man is he with whom you are, for no man knows whether he will ever have dealings with any one else: and the most important affair is, to do him good, because for that purpose alone was man sent into this life !”

Questions :
(i) How was the king’s life saved?
(ii) What idea of the writer’s message do you get in the extract?

Answers :
(i) Kind hearted as the was, the king took pity on the hermit and wished to lend him a helping hand. He took the spade from the hermit and dug the ground till sunset. If he had not stayed with the holy person and gone away, he would have attacked by his enemy, who resolved to kill him. Because of his kindness and helpful nature his life was saved.
(ii) This extract forms the essence of the story ‘Three Questions”. Tolstoy here conveys a very salutary message to mankind. His emphasis on ‘now’ is a case in point. Man’s present moment always needs a careful attention. The writer brings home another fact that man’s life on this earth should not be meaningless. God has created him to serve mankind. Man should not lose sight of this fact.

Introducing the Author
Count Leo Tolstoy (1828-1910), a great moral teacher of recent times, was an eminent writer and thinker of Russia. In 1879, he underwent a spiritual transformation which he has described in his ‘Confession’. He was a staunch believer in God: his love for men was matchless. His amazing creativity finds expression in his books novels, plays and exhortations. His writings were censored, but nothing could stop him from preaching the virtues of self-purification, love and compassion for all forms of life. He had a great impact on Gandhiji, the sage-politician of India. His short stories are remarkable for great moral conviction and deep religious spirit. The style is simple and is packed with realistic details and colloquial diction purged of all rhetoric.

About the Story
Tolstoy’s ‘Three Questions’ gives us a piece of practical wisdom. It prescribes for man three moral ways of living. The scriptural persuasiveness of tone accounts for its impact and appeal. We are ever in need of guidance and wisdom to know the right way and time to do the right thing with the right person. But man is often at a loss to know this. So he gropes in moral and spiritual darkness and in his confusion and bewilderment does the wrong things detrimental to him and the society as well. Faced with such a situation, a king had to go to a hermit in the guise of a common man where he had to do hard work, spend the night and know things the hard way. As a reward, however, the answers he found were fully satisfactory and convincing.

Summary
Once, a king was in a pensive and reflective mood. He was racking his brain to find apt answers to three important questions – how to know the right time to do the right thing* how to know who the right people were to listen to, and what was the most important things to do. He thought that if he got answers to these tricky questions he could conduct his affairs smoothly and wisely.

Many learned people came to the palace to answer these questions. But the king found their answers quite unsatisfactory. The answers were various and even contradictory. In reply to the first question, some said that the king ought have to draw up a table of days, months and years and live strictly according to it. Others suggested that he should abandon idle pastimes and the habit of procrastination. In reply to the second question, the answers were equally confusing.

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

Some said that a council of wise men must be handy for ready consultation. Yet others suggested that the help of a magician ought to be sought in this regard. To the third question, as to what was the most important work or occupation, some replied that it was science; others said it was skill in warfare or religious worship. The king could agree with none of them. So he sought the advice of a learned sage living in a wood. The king went to the holy hermit in the guise of a common man. When he reached the hermitage, he saw the skinny hermit digging a hole. He was in sweats. The king asked him the three questions and begged him to answer them. But the hermit paid little heed to the king’s repeated queries. He went on digging till he was damn tired.

The king offered to dig the hole himself. The hermit handed over the spade to the king and took rest. The king dug and dug till it was night. The hermit was as silent as before. At this moment, a wounded man came running. The hermit and the king nursed him. The king bandaged his wound and forgot all about his questions. They carried the wounded man inside and laid him on the bed. The man fell silent. He seemed to have felt better. The king was so tired that he too fell asleep on the threshold. When he woke up, he saw that the wounded man was fixing his gaze on him. The identity of the wounded man was revealed.

He was a die-hard enemy of the king, who had followed him to kill him on his way to the hermitage. But as the king was detained long at the hermitage, he came out of his hide and was wounded mortally by the king’s bodyguards. The man apologised to the king and swore to be his faithful slave the rest of his life. The king was happy to know that he had won over his enemy with such ease. The king took leave of the wounded man and before leaving the hermitage, approached the hermit again.

He was now disgusted with the holy man’s puzzling silence. When he asked the questions for the last time, the hermit replied that he had already got the answers. The king was startled. But the holy man explained to him that the right time was the time at hand. It was ‘now’ which was most important. The king took pity on the hermit and helped him in the work. For him, that time was the most opportune one and that work was the most important one. The hermit was the most necessary man at that time. The most necessary man is he with whom one is at the moment and the most important affair is, to do him good. Man is sent into this life to serve his fellow beings. The king got his answers and returned to the palace wiser and more enlightened.

ସାରାଂଶ :

ଲିଓ ଟଲ୍‌ଷ୍ଟୟଙ୍କଦ୍ବାରା ରଚିତ ଉକ୍ତ କ୍ଷୁଦ୍ରଗଳ୍ପଟିରେ ମଣିଷମାନଙ୍କ ପାଇଁ ଏକ ଅମୂଲ୍ୟ ଉପଦେଶ ପ୍ରଦାନ କରାଯାଇଅଛି । ପ୍ରତ୍ୟେକ ମଣିଷ ଆପଣାର ଜୀବନ କ୍ଷେତ୍ରରେ ବହୁ ଘାତ ପ୍ରତିଘାତର ସମ୍ମୁଖୀନ ହୋଇ ଭୂଲୁଣ୍ଠିତ ହେବାବେଳେ ଭାବିବସେ ଯେ ଯଦି ସେ ସଫଳତା ହାସଲ କରିବାର ଉପାୟ ଓ କଳାଟିକୁ ଜାଣିପାରନ୍ତା, ସେ ହୁଏତ ଆପଣାର ଜୀବନଟିକୁ ସରସ, ସୁନ୍ଦର ଓ ଋଦ୍ଧିମନ୍ତ କରିପାରନ୍ତା । ସେ ଜାଣେନାହିଁ ଯେ ସାଫଲ୍ୟର ଚାବିକାଠି ତା’ର ହାତ ପାହାନ୍ତାରେ, ତା’ର ଚେତନାର ଉପର ସ୍ତରରେ – ଖାଲି ଯାହା ତାହାର ଦୃହକ୍ଳିଷ୍ଟ, ବିକ୍ଷୁବ୍‌ଧ ମାନସିକତା ଯୋଗୁଁ ସେ ଏବେ ବି ଅଜ୍ଞାନ ଅନ୍ଧକାରରେ ଆଚ୍ଛନ୍ନ, ମରୀଚିକାର ପଛରେ ନିୟତ ଧାବମାନ ।

ଏକଦା ଜଣେ ରାଜାଙ୍କର ମନେହେଲା ଯେ ଯଦି ସେ ତିନୋଟି ପ୍ରଶ୍ନର ଉତ୍ତର ପାଇପାରନ୍ତେ, ତେବେ ସେ କେବେହେଲେ ବିଫଳତାର ଗହ୍ଵରରେ ପତିତ ହୁଅନ୍ତେ ନାହିଁ । ପ୍ରଥମ ପ୍ରଶ୍ନ – କାର୍ଯ୍ୟ ଆରମ୍ଭ କରିବାର ପ୍ରକୃଷ୍ଟ ସମୟ କ’ଣ ? ଦ୍ଵିତୀୟ ପ୍ରଶ୍ନ – ଉତ୍ତମ ପରାମର୍ଶଦାତା ଓ ଉତ୍ତମ ବ୍ୟକ୍ତି କିଏ ଯାହାଙ୍କ ଉପରେ ସେ ନିର୍ଭର କରିପାରିବେ ଏବଂ ଯାହାଙ୍କ ପାଇଁ ସେ ଆପଣାକୁ ଉତ୍ସର୍ଗ କରିପାରିବେ ? ତୃତୀୟ ପ୍ରଶ୍ନ – ସର୍ବୋତ୍କୃଷ୍ଟ କାର୍ଯ୍ୟଟି କ’ଣ ? ଯଥା ସମୟରେ ରାଜ୍ୟସାରା ଡେଙ୍ଗୁରା ଦିଆଗଲା । ଯେଉଁ ବିଜ୍ଞବ୍ୟକ୍ତି ରାଜାଙ୍କର ଏହି ତିନୋଟି ପ୍ରଶ୍ନର ଉତ୍ତର ଦେଇପାରିବେ ତାଙ୍କୁ ବିପୁଳ ଭାବରେ ପୁରସ୍କୃତ କରାଯିବାର ଘୋଷଣା କରାଗଲା । କିନ୍ତୁ ହାୟ, ଅନେକ ବିଜ୍ଞବ୍ୟକ୍ତି ପୁରସ୍କାର ଲୋଭରେ ପ୍ରଶ୍ନର ଉତ୍ତର ଦେବାକୁ ଚେଷ୍ଟାକଲେ ମଧ୍ୟ ରାଜା କୌଣସି ଉତ୍ତରରେ ସନ୍ତୁଷ୍ଟ ହୋଇପାରିଲେ ନାହିଁ । ଭିନ୍ନ ଭିନ୍ନ ବିଜ୍ଞବ୍ୟକ୍ତି ଭିନ୍ନ ଭିନ୍ନ ଉତ୍ତର ଦେଲେ ।

ପ୍ରଥମ ପ୍ରଶ୍ନର ଉତ୍ତରରେ କିଏ ଉପଦେଶ ଦେଲେ ଗୋଟିଏ ଉତ୍ତମ ପଞ୍ଜିକାରୁ କାର୍ଯ୍ୟାରମ୍ଭର ସଠିକ୍ ତିଥି, ବାର, ନକ୍ଷତ୍ର ଜାଣିବାକୁ ତ ଅନ୍ୟ କେତେକ ବେଶ୍ ଦାର୍ଶନିକ ଭାବରେ ଉପଦେଶ ଦେଲେ ଯେ ରାଜା ଆଳସ୍ୟର ବଶବର୍ତ୍ତୀ ନ ହୋଇ ଯେଉଁ କାର୍ଯ୍ୟ ଯେତେବେଳେ କରିବାର କଥା ତାହା ଯଥାଶୀଘ୍ର ସମ୍ପାଦନ କରିବା ଶ୍ରେୟସ୍କର । ସେହିଭଳି ଦ୍ୱିତୀୟ ପ୍ରଶ୍ନର ଉତ୍ତରରେ କିଏ ରାଜାଙ୍କୁ ମନ୍ତ୍ରିପରିଷଦ ଉତ୍ତମ ପରାମର୍ଶ ଦେଇପାରିବେ ବୋଲି କହିଲେ ତ ଆଉ କେତେକ ରାଜପୁରୋହିତ କିମ୍ବା ରାଜବୈଦ୍ୟ କିମ୍ବା ସୈନ୍ୟସାମନ୍ତ ଉପଯୁକ୍ତ ବ୍ୟକ୍ତି ବା ପରାମର୍ଶଦାତା ବୋଲି ଘୋଷଣା କଲେ । ତୃତୀୟ ପ୍ରଶ୍ନର ଉତ୍ତରରେ କିଏ ଜ୍ଞାନାର୍ଜନ ସର୍ବଶ୍ରେଷ୍ଠ ବୃତ୍ତି ବୋଲି କହିଲେ ତ ଆଉ କିଏ ଯୁଦ୍ଧବିଗ୍ରହ କିମ୍ବା ଧାର୍ମିକ ଅନୁଷ୍ଠାନମାନ ଉତ୍ତମ ବୃତ୍ତି ବୋଲି କହିଲେ । ସୁତରାଂ ରାଜା ତିନୋଟି ପ୍ରଶ୍ନର କୌଣସି ସଠିକ୍ ସନ୍ତୋଷଜନକ ଉତ୍ତର ପାଇପାରିଲେ ନାହିଁ ଏବଂ ତାଙ୍କର ଜିଜ୍ଞାସା ପ୍ରବଳରୁ ପ୍ରବଳତର ହେଲା । ପରିଶେଷରେ ରାଜା ରାଜ୍ୟର ଏକ ଅରଣ୍ୟର ନିକାଞ୍ଚନ ପରିବେଶରେ ଏକ କୁଟୀରରେ ରହୁଥ‌ିବା ଜଣେ ଜ୍ଞାନୀ ସାଧୁଙ୍କ ପାଖକୁ ତାଙ୍କ ପ୍ରଶ୍ନର ଉତ୍ତର ପାଇବାପାଇଁ ଛଦ୍ମବେଶରେ ଗଲେ ।

ସାଧୁଙ୍କ ବାସସ୍ଥାନର ଅନତି ଦୂରରେ ରାଜାଙ୍କର ଅଙ୍ଗରକ୍ଷକମାନେ ରହିଲେ । ରାଜା ଘୋଡ଼ାରୁ ଓହ୍ଲାଇ ଏକ ସାଧାରଣ ଜନତାର ଛଦ୍ମବେଶରେ ସାଧୁଙ୍କୁ ଭେଟିବାକୁ ଗଲେ । ଯେତେବେଳେ ରାଜା ସାଧୁଙ୍କୁ ଭେଟିଲେ, ସେତେବେଳେ ସେହି ଜ୍ଞାନୀ ବୃଦ୍ଧି ଆପଣାର କୁଟୀରର ବାହାର ପ୍ରଦେଶରେ ମାଟି ହାଣି ଏକ ଶସ୍ୟପଟାଳି ତିଆରି କରିବାରେ ବ୍ୟସ୍ତ ଥା’ନ୍ତି । ଦୁର୍ବଳ ଶରୀର ତାଙ୍କର କଠିନ କାର୍ଯ୍ୟ ପାଇଁ ଅନୁପଯୁକ୍ତ ଥିଲେ ବି ସାଧୁ ନିର୍ବିକାର ଭାବରେ ମାଟି ଖୋଳି ଚାଲିଥା’ନ୍ତି । ରାଜା ଖୁବ୍ ବିନମ୍ରତାର ସହ ସାଧୁଙ୍କୁ ନିଜର ଆସିବାର ଅଭିପ୍ରାୟ ଜଣାଇଲେ ଏବଂ ତିନୋଟି ପ୍ରଶ୍ନର ଉତ୍ତର ଦେବାପାଇଁ ଅନୁରୋଧ କଲେ; କିନ୍ତୁ ସାଧୁଜଣଙ୍କ ରାଜାଙ୍କ ପ୍ରଶ୍ନ ଶୁଣି ନ ଶୁଣିବାର ଅଭିନୟ କଲେ ଏବଂ ପୂର୍ବବତ୍ ମାଟିଖୋଳା କାମରେ ଲାଗିପଡ଼ିଲେ । ଦୟାଳୁ ରାଜା ସାଧୁଙ୍କୁ କାର୍ଯ୍ୟରେ ସାହାଯ୍ୟ କରିବାପାଇଁ ନିଜେ ମାଟି ଖୋଳିବାରେ ଲାଗିଲେ ଓ ସାଧୁଜଣଙ୍କ ବିଶ୍ରାମ ନେବାପାଇଁ ବସିପଡ଼ିଲେ । ଘଣ୍ଟା ଘଣ୍ଟା ଧରି ରାଜା ଶସ୍ୟପଟାଳି ହାଣି ଚାଲିଲେ ମଧ୍ୟ ସାଧୁ ତାଙ୍କ ପ୍ରଶ୍ନର ଉତ୍ତର ଦେଲେ ନାହିଁ ।

CHSE Odisha Class 11 English Solutions Non-Detailed Chapter 1 Three Questions

ସନ୍ଧ୍ୟା ଉପନୀତ । ଦୂରରୁ ଜଣେ ବ୍ୟକ୍ତି ବିକଳ ଚିତ୍କାର କରି ଦୌଡ଼ିଆସି ସାଧୁ ଓ ରାଜାଙ୍କ ପାଖରେ ଭୂପତିତ ହେଲେ । ତାଙ୍କର ପେଟରେ ଏକ ବିରାଟ କ୍ଷତ ଏବଂ ସେଥିରୁ ପ୍ରବଳ ରକ୍ତସ୍ରାବ ହେଉଥାଏ। ରାଜା ଓ ସାଧୁ ତା’ର ଉପଯୁକ୍ତ ସେବାଯତ୍ନ କଲାପରେ ଲୋକଟି ସାଷ୍ଟମ ହେଲା । ରାଜା କ୍ଲାନ୍ତ ହୋଇଥିବାରୁ କୁଡ଼ିଆର ଗୋଟିଏ କୋଣରେ ଶୋଇ ପଡ଼ିଲେ । ନିଦରୁ ଉଠିବା ପରେ ସେ ଜାଣିଲେ ଯେ ଯେଉଁ ବ୍ୟକ୍ତିର ସେ ସେବାଶ୍ରୁଶ୍ରୂଷା କରିଥିଲେ ସେ ତାଙ୍କର ଜଣେ ଘୋର ଶତ୍ରୁ ଯିଏକି ତାଙ୍କୁ ମାରିବାପାଇଁ ଅରଣ୍ୟକୁ ଆସିଥିଲା । ଯାହାହେଉ, ସେ ଲୋଟି ବର୍ତ୍ତମାନ ଅନୁତପ୍ତ ଏବଂ ସବୁଦିନ ପାଇଁ ରାଜାଙ୍କର ବିଶ୍ବସ୍ତ ଭୃତ୍ୟ ହୋଇ ରହିବାପାଇଁ ସେ ଶପଥବଦ୍ଧ ହେଲା । ଖୁବ୍ ସହଜରେ ଆପଣାର ଜଣେ ପ୍ରଚଣ୍ଡ ଶତ୍ରୁର ହୃଦୟ ଜୟ କରିପାରିଥିବାରୁ ରାଜା ଆନନ୍ଦିତ ହେଲେ । ତେବେ ଆଉ ବେଶି କାଳ ସାଧୁଙ୍କ କୁଟୀରରେ ରହିବା ଆବଶ୍ୟକତା ନ ଥ‌ିବାରୁ ସେ ଯିବାକୁ ବାହାରିଲେ; କିନ୍ତୁ ରାଜପ୍ରାସାଦକୁ ଫେରିଯିବା ପୂର୍ବରୁ ଶେଷଥର ପାଇଁ ସାଧୁଙ୍କଠାରୁ ତିନୋଟି ପ୍ରଶ୍ନର ଉତ୍ତର ଆଶାକରି ପୁନଶ୍ଚ ତାଙ୍କୁ ସେହି ପ୍ରଶ୍ନ ପଚାରି ବସିଲେ । ରହସ୍ୟମୟ ହସ ହସି ସାଧୁ କହିଲେ ଯେ ରାଜା ତାଙ୍କ ପ୍ରଶ୍ନର ଉତ୍ତର କେବେଠାରୁ ପାଇ ସାରିଛନ୍ତି । ଚକିତ ରାଜା ସାଧୁଙ୍କର ଏହି ଇଙ୍ଗିତ ବୁଝି ନ ପାରିବାରୁ ସାଧୁ ବୁଝାଇଦେଲେ ।

ଯେତେବେଳେ ରାଜା ସାଧୁଙ୍କୁ ମାଟି ହାଣିବାର ଦେଖ‌ିଲେ ଏବଂ ତାଙ୍କ ଉପରେ ଦୟାପ୍ରକାଶ କରି ତାଙ୍କୁ ସାହାଯ୍ୟ କରିବାପାଇଁ ନିଜେ ମାଟି ହାଣିଲେ, ସେ ତାଙ୍କର ଅଜ୍ଞାତସାରରେ ଆପଣାର ପ୍ରଥମ ପ୍ରଶ୍ନର ଉତ୍ତର ପାଇଗଲେ । କାର୍ଯ୍ୟ ଆରମ୍ଭ କରିବାର ପ୍ରକୃଷ୍ଟ ସମୟ ସର୍ବଦା ବର୍ତ୍ତମାନ । ତୁମ ସମ୍ମୁଖରେ ଯେଉଁ କାର୍ଯ୍ୟ ଉପସ୍ଥିତ, ସେହି ମୁହୂଉଁଟି ହିଁ ସେହି କାର୍ଯ୍ୟ ଆରମ୍ଭ କରିବାର ସର୍ବଶ୍ରେଷ୍ଠ ମୁହୂର୍ଭ । ତୁମ ସମ୍ମୁଖରେ ଯେଉଁ ବ୍ୟକ୍ତି ତୁମର ସାହାଯ୍ୟ ପାଇଁ ଉପସ୍ଥିତ, ସେହି ବ୍ୟକ୍ତି ହିଁ ତୁମ ପାଇଁ ସର୍ବଶ୍ରେଷ୍ଠ ବ୍ୟକ୍ତି ଏବଂ ସେହି ବ୍ୟକ୍ତିର ସେବା ହିଁ ତୁମର ସର୍ବଶ୍ରେଷ୍ଠ ବୃତ୍ତି । ଠିକ୍ ଯେମିତି ରାଜା ଆହତ ବ୍ୟକ୍ତିର ସେବା କରିଛନ୍ତି, ତାହାହିଁ ହେଉଛି ତାଙ୍କ ପାଇଁ ସର୍ବଶ୍ରେଷ୍ଠ କର୍ମ, ସର୍ବଶ୍ରେଷ୍ଠ ମୁହୂର୍ତ୍ତ ରାଜାଙ୍କର ଜ୍ଞାନୋଦୟ ହେଲା। ସେ ତାଙ୍କର ତିନୋଟି ପ୍ରଶ୍ନର ସନ୍ତୋଷଜନକ ଉତ୍ତର ପାଇ ପ୍ରାଜ୍ଞ ସାଧୁଙ୍କୁ ପ୍ରଣାମ କଲେ ଓ ସହର୍ଷ ମନରେ ରାଜପ୍ରାସାଦକୁ ଫେରିଆସିଲେ ।