A right circular cylinder is inscribed in a cone.

S = Curved Surface Area of Cylinder.

Based on the above information answer the following questions:

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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

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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

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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)

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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   

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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    

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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

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 (๐‘Ÿ )/๐‘Ÿ1 = ? (A) (โ„Ž โˆ’โ„Ž1)/โ„Ž1 (B) (โ„Ž1โˆ’โ„Ž)/โ„Ž1 (C) (โ„Ž โˆ’โ„Ž1)/โ„Ž (D) (โ„Ž + โ„Ž1)/โ„Ž1 In ฮ” OAD tan ๐œถ = ๐ด๐ท/๐‘‚๐ด = ๐’“/(๐’‰_๐Ÿโˆ’๐’‰) In ฮ” OBC tan ๐œถ = ๐ต๐ถ/๐‘‚๐ต = ๐’“_๐Ÿ/๐’‰_๐Ÿ Comparing (1) and (2) ๐‘Ÿ/(โ„Ž_1 โˆ’ โ„Ž) = ๐‘Ÿ_1/โ„Ž_1 ๐’“/๐’“_๐Ÿ = (๐’‰_๐Ÿ โˆ’ ๐’‰)/๐’‰_๐Ÿ So, the correct answer is (B) Question 2 Find the value of โ€˜Sโ€™? (A) 2๐œ‹๐‘Ÿ/โ„Ž (โ„Ž_1โˆ’โ„Ž)โ„Ž (B) 2๐œ‹๐‘Ÿ/โ„Ž_1 (โ„Ž_1โˆ’โ„Ž)โ„Ž (C) (2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Ž_1โˆ’โ„Ž)โ„Ž (D) (2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Ž_1+โ„Ž)โ„Ž Now, S = Curved Surface Area of Cylinder = 2๐œ‹๐‘Ÿโ„Ž We know that ๐’“/๐’“_๐Ÿ = (๐’‰_๐Ÿ โˆ’ ๐’‰)/๐’‰_๐Ÿ ๐’“ = ((๐’‰_๐Ÿ โˆ’ ๐’‰)/๐’‰_๐Ÿ ) ร— ๐’“_๐Ÿ = 2๐œ‹((๐’‰_๐Ÿ โˆ’ ๐’‰)/๐’‰_๐Ÿ ) ร— ๐’“_๐Ÿ ร— โ„Ž =(2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Ž_1โˆ’โ„Ž)โ„Ž So, the correct answer is (C) Question 3 Find the value of ๐‘‘๐‘†/๐‘‘โ„Ž ? (A) (2๐œ‹๐‘Ÿ_1)/โ„Ž (โ„Ž_1โˆ’2โ„Ž) (B) (2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Žโˆ’2โ„Ž_1) (C) 2๐œ‹๐‘Ÿ/โ„Ž (โ„Ž_1โˆ’2โ„Ž) (D) 2๐œ‹๐‘Ÿ1/โ„Ž_1 (โ„Ž_1โˆ’2โ„Ž) Now, ๐‘†=(2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Ž_1โˆ’โ„Ž)โ„Ž ๐‘†=(2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Ž_1 ร— โ„Žโˆ’โ„Ž^2 ) Differentiating w.r.t h ๐‘‘๐‘†/๐‘‘โ„Ž=(๐Ÿ๐…๐’“_๐Ÿ)/๐’‰_๐Ÿ (๐’‰_๐Ÿ โˆ’๐Ÿ๐’‰) So, the correct answer is (D) Question 4 Find the value of (๐‘‘^2 ๐‘†)/(๐‘‘โ„Ž^2 ) (A) โˆ’ (4๐œ‹๐‘Ÿ_1)/โ„Ž_1 (B) โˆ’ 4๐œ‹๐‘Ÿ/โ„Ž (C) โˆ’ (4๐œ‹๐‘Ÿ_1)/โ„Ž (D) (4๐œ‹๐‘Ÿ_1)/โ„Ž Now, ๐‘‘๐‘†/๐‘‘โ„Ž=(2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Ž_1 โˆ’2โ„Ž)" " Differentiating w.r.t h (๐‘‘^2 ๐‘†)/(๐‘‘โ„Ž^2 )=(2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (0โˆ’2) (๐‘‘^2 ๐‘†)/(๐‘‘โ„Ž^2 )=(โˆ’๐Ÿ’๐…๐’“_๐Ÿ)/๐’‰_๐Ÿ So, the correct answer is (A) Question 5 What is the relation between ๐‘Ÿ_1 and ๐‘Ÿ? (A) ๐‘Ÿ_1=๐’“/๐Ÿ (B) 2๐‘Ÿ_1=3๐‘Ÿ (C) ๐‘Ÿ_1=2๐‘Ÿ (D) ๐‘Ÿ_1/2=๐‘Ÿ/3 Putting ๐’…๐‘บ/๐’…๐’‰=๐ŸŽ (2๐œ‹๐‘Ÿ_1)/โ„Ž_1 (โ„Ž_1 โˆ’2โ„Ž)"= 0 " โ„Ž_1=2โ„Ž And, (๐‘‘^2 ๐‘†)/(๐‘‘โ„Ž^2 )=(โˆ’๐Ÿ’๐…๐’“_๐Ÿ)/๐’‰_๐Ÿ โˆด (๐‘‘^2 ๐‘†)/(๐‘‘โ„Ž^2 ) < 0 for โ„Ž_1=2โ„Ž So, Surface area is maximum for โ„Ž_1=2โ„Ž Now, from Question 1 ๐’“/๐’“_๐Ÿ = (๐’‰_๐Ÿ โˆ’ ๐’‰)/๐’‰_๐Ÿ Putting โ„Ž_1=2โ„Ž ๐‘Ÿ/๐‘Ÿ_1 = (2โ„Ž โˆ’ โ„Ž)/2โ„Ž ๐‘Ÿ/๐‘Ÿ_1 = โ„Ž/2โ„Ž ๐‘Ÿ/๐‘Ÿ_1 = 1/2 2๐‘Ÿ=๐‘Ÿ_1 ๐’“_๐Ÿ=๐Ÿ๐’“ So, the correct answer is (C)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.