Mathematics in Physics

JEE Main level, numerical answer, medium

90 of 121

Question 90 of 121

The radius of a sphere is measured to be (7.50±0.85)cm(7.50\pm 0.85)cm. Suppose the percentage error in its volume is xx. The value of xx, to the nearest xx, is ___ .

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Answer: 34

∵v=43πr3∵v=\dfrac{4}{3}\pi {r}^{3}
taking
\log & then differentiate
dVV=3drr\dfrac{dV}{V}=3\dfrac{dr}{r}=3×0.857.5×100=\dfrac{3\times 0.85}{7.5}\times 100%=34%

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