JEE PYQ: Mathematics In Physics Question 3
Question 3 - 2021 (18 Mar Shift 2)
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 integer is _____.
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Answer: (34)
Solution
$\because V = \frac{4}{3}\pi r^3$ Taking log & then differentiate $\frac{dV}{V} = 3\frac{dr}{r}$ $= \frac{3 \times 0.85}{7.5} \times 100% = 34%$