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JEE Main 202118 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 18 March
Shift: Evening Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2021 · 18 March · Evening Shift · Physics · Mathematics in Physics

The radius of a sphere is measured to be ( 7 . 50 ± 0 . 85 ) cm ....

The radius of a sphere is measured to be

( 7 . 50 \pm 0 . 85 ) cm

. Suppose the percentage error in its volume is

x

. The value of

x

, to the nearest

x

, is ___ .

Correct answer

0

Step-by-step solution

$∵ v = \dfrac{4}{3}\pi{r}^{3}$

taking $log \text{ and }$ then differentiate

$\dfrac{dV}{V} = 3\dfrac{dr}{r}$ $= \dfrac{3 \times 0 . 85}{7 . 5} \times 100 % = 34 %$

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