The radius of a sphere is measured to be ( 7 . 50 ± 0 . 85 )   cm . Suppose the percentage error…

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

Solution

∵v=43πr3

taking log & then differentiate

dVV=3drr=3×0.857.5×100%=34%

Asked in: JEE Main 2021 (18 Mar Shift 2)

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