JEE Main 2026 · 6 Apr · Shift 1 · Physics Q26

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PhysicsSingle correct+4 / -1Question 26 of 75

The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$ , length $l$ and diameter $d$ . The measured values of $m$ , $l$ and $d$ are $97.42\pm0.02$ g, $8.35\pm0.05$ mm and $20.20\pm0.02$ mm, respectively. Calculated percentage fractional error in $\rho$ is ________.

0.63%

0.82%

0.72%

0.25%

Answer and worked solution

Official answer: B

Density. $\rho=\dfrac{4m}{\pi d^2l}$ , so
$\frac{\Delta\rho}\rho=\frac{\Delta m}m+\frac{\Delta l}l+2\frac{\Delta d}d=\frac{0.02}{97.42}+\frac{0.05}{8.35}+\frac{0.04}{20.20}.$

Numbers. $0.00021+0.00599+0.00198=0.0082$ , or 0.82%.

(B)

Topics:Errors in measurement

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JEE Main 2026 · 6 Apr · Shift 1, Physics question 26