A right circular cylinder is inscribed in a cone.

S = Curved Surface Area of Cylinder.

Based on the above information answer the following questions:

Case Based - Class 12 - A right circular cylinder is inscribed in cone - Case Based Questions (MCQ)

 

Question 1

(r )/r 1 = ?

(A) (h - h 1 )/h 1   

(B) (h 1 - h)/h 1  

(C) (h - h 1 )/h    

(D) (h + h 1 )/h 1

part 2 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

part 3 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

 

Question 2

Find the value of ‘S’?

(A) 2πr/h (h 1 - h)h  

(B) 2πr/h 1  (h 1 - h)h

(C) (2πr 1 )/h 1 (h 1 - h)h  

(D) (2πr 1 )/h 1 (h 1 + h)h

part 4 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 5 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

 

Question 3

Find the value of dS/dh ?

(A) (2πr 1 )/h (h 1 - 2h)    

(B) (2πr 1 )/h 1  (h - 2h 1 )

(C) 2πr/h (h 1 - 2h)  

(D) 2πr1/h 1 (h 1 - 2h)

part 6 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

 

Question 4

Find the value of (d^2 S)/(dh^2 )

(A) − (4πr 1 )/h 1  

(B) − 4πr/h

(C) − (4πr 1 )/h     

(D) (4πr 1 )/h   

part 7 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

 

Question 5

What is the relation between r 1 and r?

(A)  r 1 = r / 2    

(B) 2r 1 = 3r

(C) r 1 = 2r 

(D) r 1 /2 = r/3    

part 8 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 9 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 10 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Question A right circular cylinder is inscribed in a cone. S = Curved Surface Area of Cylinder. Based on the above information answer the following questions: Question 1 (š‘Ÿ )/š‘Ÿ1 = ? (A) (ā„Ž āˆ’ā„Ž1)/ā„Ž1 (B) (ā„Ž1āˆ’ā„Ž)/ā„Ž1 (C) (ā„Ž āˆ’ā„Ž1)/ā„Ž (D) (ā„Ž + ā„Ž1)/ā„Ž1 In Ī” OAD tan šœ¶ = š“š·/š‘‚š“ = š’“/(š’‰_šŸāˆ’š’‰) In Ī” OBC tan šœ¶ = šµš¶/š‘‚šµ = š’“_šŸ/š’‰_šŸ Comparing (1) and (2) š‘Ÿ/(ā„Ž_1 āˆ’ ā„Ž) = š‘Ÿ_1/ā„Ž_1 š’“/š’“_šŸ = (š’‰_šŸ āˆ’ š’‰)/š’‰_šŸ So, the correct answer is (B) Question 2 Find the value of ā€˜S’? (A) 2šœ‹š‘Ÿ/ā„Ž (ā„Ž_1āˆ’ā„Ž)ā„Ž (B) 2šœ‹š‘Ÿ/ā„Ž_1 (ā„Ž_1āˆ’ā„Ž)ā„Ž (C) (2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Ž_1āˆ’ā„Ž)ā„Ž (D) (2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Ž_1+ā„Ž)ā„Ž Now, S = Curved Surface Area of Cylinder = 2šœ‹š‘Ÿā„Ž We know that š’“/š’“_šŸ = (š’‰_šŸ āˆ’ š’‰)/š’‰_šŸ š’“ = ((š’‰_šŸ āˆ’ š’‰)/š’‰_šŸ ) Ɨ š’“_šŸ = 2šœ‹((š’‰_šŸ āˆ’ š’‰)/š’‰_šŸ ) Ɨ š’“_šŸ Ɨ ā„Ž =(2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Ž_1āˆ’ā„Ž)ā„Ž So, the correct answer is (C) Question 3 Find the value of š‘‘š‘†/š‘‘ā„Ž ? (A) (2šœ‹š‘Ÿ_1)/ā„Ž (ā„Ž_1āˆ’2ā„Ž) (B) (2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Žāˆ’2ā„Ž_1) (C) 2šœ‹š‘Ÿ/ā„Ž (ā„Ž_1āˆ’2ā„Ž) (D) 2šœ‹š‘Ÿ1/ā„Ž_1 (ā„Ž_1āˆ’2ā„Ž) Now, š‘†=(2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Ž_1āˆ’ā„Ž)ā„Ž š‘†=(2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Ž_1 Ɨ ā„Žāˆ’ā„Ž^2 ) Differentiating w.r.t h š‘‘š‘†/š‘‘ā„Ž=(šŸš…š’“_šŸ)/š’‰_šŸ (š’‰_šŸ āˆ’šŸš’‰) So, the correct answer is (D) Question 4 Find the value of (š‘‘^2 š‘†)/(š‘‘ā„Ž^2 ) (A) āˆ’ (4šœ‹š‘Ÿ_1)/ā„Ž_1 (B) āˆ’ 4šœ‹š‘Ÿ/ā„Ž (C) āˆ’ (4šœ‹š‘Ÿ_1)/ā„Ž (D) (4šœ‹š‘Ÿ_1)/ā„Ž Now, š‘‘š‘†/š‘‘ā„Ž=(2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Ž_1 āˆ’2ā„Ž)" " Differentiating w.r.t h (š‘‘^2 š‘†)/(š‘‘ā„Ž^2 )=(2šœ‹š‘Ÿ_1)/ā„Ž_1 (0āˆ’2) (š‘‘^2 š‘†)/(š‘‘ā„Ž^2 )=(āˆ’šŸ’š…š’“_šŸ)/š’‰_šŸ So, the correct answer is (A) Question 5 What is the relation between š‘Ÿ_1 and š‘Ÿ? (A) š‘Ÿ_1=š’“/šŸ (B) 2š‘Ÿ_1=3š‘Ÿ (C) š‘Ÿ_1=2š‘Ÿ (D) š‘Ÿ_1/2=š‘Ÿ/3 Putting š’…š‘ŗ/š’…š’‰=šŸŽ (2šœ‹š‘Ÿ_1)/ā„Ž_1 (ā„Ž_1 āˆ’2ā„Ž)"= 0 " ā„Ž_1=2ā„Ž And, (š‘‘^2 š‘†)/(š‘‘ā„Ž^2 )=(āˆ’šŸ’š…š’“_šŸ)/š’‰_šŸ ∓ (š‘‘^2 š‘†)/(š‘‘ā„Ž^2 ) < 0 for ā„Ž_1=2ā„Ž So, Surface area is maximum for ā„Ž_1=2ā„Ž Now, from Question 1 š’“/š’“_šŸ = (š’‰_šŸ āˆ’ š’‰)/š’‰_šŸ Putting ā„Ž_1=2ā„Ž š‘Ÿ/š‘Ÿ_1 = (2ā„Ž āˆ’ ā„Ž)/2ā„Ž š‘Ÿ/š‘Ÿ_1 = ā„Ž/2ā„Ž š‘Ÿ/š‘Ÿ_1 = 1/2 2š‘Ÿ=š‘Ÿ_1 š’“_šŸ=šŸš’“ So, the correct answer is (C)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo