Units & Dimensions

JEE Main level, single correct, medium

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Question 149 of 149

The percentage error in the calculated volume of a sphere, if there is 2%2\% error in its diameter measurement, is __________.

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Answer: (C) 66

The volume of a sphere in terms of its diameter DD is given by V=π6D3V = \dfrac{\pi}{6}D^3

Taking the natural logarithm on both sides, we get ln⁡V=ln⁡(π6)+3ln⁡D\ln V = \ln\left(\dfrac{\pi}{6}\right) + 3 \ln D

Differentiating for small errors, we obtain ΔVV=3ΔDD\dfrac{\Delta V}{V} = 3 \dfrac{\Delta D}{D}

Multiplying by 100100 to find the percentage error, ΔVV×100=3(ΔDD×100)\dfrac{\Delta V}{V} \times 100 = 3 \left(\dfrac{\Delta D}{D} \times 100\right)

Given that the percentage error in the diameter is 2%2\%, we have ΔDD×100=2\dfrac{\Delta D}{D} \times 100 = 2

Therefore, the percentage error in the volume is 3×2=6%3 \times 2 = 6\%

Answer: 66

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