The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere

Q: The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left(\frac{x}{100}\right) \% .$ If the relative errors in measuring the mass and the diameter are $6.0 \%$ and $1.5 \%$ respectively, the value of $x$ is

d $\rho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3}}$ $\rho=\frac{6}{\pi} M D ^{-3}$ taking log $\ell n \rho=\ell n \left(\frac{6}{\pi}\right)+\ell n M -3 \ell m D$ Differentiates $\frac{ d p}{\rho}=0+\frac{ d M }{ M }-3 \frac{ d ( D )}{ D }$ for maximum error $100 \times \frac{ d \rho}{\rho}=\frac{ dM }{ M } \times 100+\frac{3 d D }{ D } \times 100$ $=6+3 \times 1.5$ $=10.5 \%$ $=\frac{1050}{100} \%$ $x=1050.00$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left(\frac{x}{100}\right) \% .$ If the relative errors in measuring the mass and the diameter are $6.0 \%$ and $1.5 \%$ respectively, the value of $x$ is

A$1000$
B$1075$
C$1060$
D$1050$

JEE MAIN 2020, Diffcult

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

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