Question 5 - Case Based Questions (MCQ) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
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
(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
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
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)
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
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
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 (𝑟 )/𝑟1 = ? (A) (ℎ −ℎ1)/ℎ1 (B) (ℎ1−ℎ)/ℎ1 (C) (ℎ −ℎ1)/ℎ (D) (ℎ + ℎ1)/ℎ1
In Δ OAD
tan 𝜶 = 𝐴𝐷/𝑂𝐴
= 𝒓/(𝒉_𝟏−𝒉)
In Δ OBC
tan 𝜶 = 𝐵𝐶/𝑂𝐵
= 𝒓_𝟏/𝒉_𝟏
Comparing (1) and (2)
𝑟/(ℎ_1 − ℎ) = 𝑟_1/ℎ_1
𝒓/𝒓_𝟏 = (𝒉_𝟏 − 𝒉)/𝒉_𝟏
So, the correct answer is (B)
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
Now,
S = Curved Surface Area of Cylinder
= 2𝜋𝑟ℎ
We know that
𝒓/𝒓_𝟏 = (𝒉_𝟏 − 𝒉)/𝒉_𝟏
𝒓 = ((𝒉_𝟏 − 𝒉)/𝒉_𝟏 ) × 𝒓_𝟏
= 2𝜋((𝒉_𝟏 − 𝒉)/𝒉_𝟏 ) × 𝒓_𝟏 × ℎ
=(2𝜋𝑟_1)/ℎ_1 (ℎ_1−ℎ)ℎ
So, the correct answer is (C)
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)
Now,
𝑆=(2𝜋𝑟_1)/ℎ_1 (ℎ_1−ℎ)ℎ
𝑆=(2𝜋𝑟_1)/ℎ_1 (ℎ_1 × ℎ−ℎ^2 )
Differentiating w.r.t h
𝑑𝑆/𝑑ℎ=(𝟐𝝅𝒓_𝟏)/𝒉_𝟏 (𝒉_𝟏 −𝟐𝒉)
So, the correct answer is (D)
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
Now,
𝑑𝑆/𝑑ℎ=(2𝜋𝑟_1)/ℎ_1 (ℎ_1 −2ℎ)" "
Differentiating w.r.t h
(𝑑^2 𝑆)/(𝑑ℎ^2 )=(2𝜋𝑟_1)/ℎ_1 (0−2)
(𝑑^2 𝑆)/(𝑑ℎ^2 )=(−𝟒𝝅𝒓_𝟏)/𝒉_𝟏
So, the correct answer is (A)
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
Putting 𝒅𝑺/𝒅𝒉=𝟎
(2𝜋𝑟_1)/ℎ_1 (ℎ_1 −2ℎ)"= 0 "
ℎ_1=2ℎ
And, (𝑑^2 𝑆)/(𝑑ℎ^2 )=(−𝟒𝝅𝒓_𝟏)/𝒉_𝟏
∴ (𝑑^2 𝑆)/(𝑑ℎ^2 ) < 0 for ℎ_1=2ℎ
So, Surface area is maximum for ℎ_1=2ℎ
Now, from Question 1
𝒓/𝒓_𝟏 = (𝒉_𝟏 − 𝒉)/𝒉_𝟏
Putting ℎ_1=2ℎ
𝑟/𝑟_1 = (2ℎ − ℎ)/2ℎ
𝑟/𝑟_1 = ℎ/2ℎ
𝑟/𝑟_1 = 1/2
2𝑟=𝑟_1
𝒓_𝟏=𝟐𝒓
So, the correct answer is (C)
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, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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