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Question

A sphere increases its volume at the rate of πcm3/s. The rate at which its surface area increases, when the radius is 1cm is


A

2πsqcm/s

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B

πsqcm/s

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C

3π2sqcm/s

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D

π2sqcm/s

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Solution

The correct option is A

2πsqcm/s


Explanation for the correct option:

Step 1. Given rate of volume increase of sphere is

dVdt=π …..(i)

As we know that,
Volume of sphere; V=43πr3

Step 2. Differentiate it with respect to r

dVdt=43π×3r2drdt

=4πr2drdt
π=4πr2drdt … [from Equation (i)]
drdt=14r2 …...(ii)

Also, we know that, surface of sphere; S=4πr2

Step 3. Differentiate it with respect to r

dSdt=8πrdrdt

=8πr14r2 …​ [from Equation (ii)]
dSdt=2πr

Step 4. Put the value r=1:

dSdt=2π

the rate of increase in surface area =2πsqcm/s

Hence, Option ‘A’ is Correct.


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