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Question

If there is an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is

A
k3%
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B
k%
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C
3k%
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D
2k%
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Solution

The correct option is C 3k%
Given that Δrr×100=kΔr=0.01kr
Volume of the sphere,
V=43πr3
Taking loagarithm on both sides
dVdr=34π(3r2)=4πr2
ΔV(dVdr)×Δr
ΔV4πr2×0.01kr
ΔV0.04kπr3
Percentage error in its period is :
ΔVV×100%0.04kπr343πr3×100%3k%
So, percentage error in its period is 3k%
Hence, Option (B) is correct.

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