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A copper sphere is being heated. Its vol...

A copper sphere is being heated. Its volume as a result of increasing at the rate of `1mm^(3)` every second. Find the rate at which its surface area is increasing when its radius 10 cm.

A

`0.2mm^(2)//s`

B

`0.02mm^(2)//s`

C

`0.02cm^(2)//s`

D

none

Text Solution

Verified by Experts

The correct Answer is:
B
A copper ………..
`V = (4)/(3)piR^(3)` , `A = 4piR^(2)`
`implies (dV)/(dt)=(4pi)/(3)3R^(2)(dR)/(dt)=4piR^(2)(dR)/(dt)` …………(1)
`A = 4piR^(2)`
`(dA)/(dt) = 8piR(dR)/(dt)` ………..(2)
Using (1) and (2)
`implies (dV)/(dt) = (R )/(2)(dA)/(dt) implies`
`(dA)/(dt)=(2)/(R )(dV)/(dt)= (2)/(100mm)xx1mm^(3)//s= 0.02mm^(2)//s`
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