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1.Units, Dimensions and Measurement
hard
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)
Solution
$\rho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3}}$
$\rho=\frac{6}{\pi} M D ^{-3}$
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
$100 \times \frac{ d \rho}{\rho}=\frac{ dM }{ M } \times 100+\frac{3 d D }{ D } \times 100$
$=6+3 \times 1.5$
$=10.5 \%$
$=\frac{1050}{100} \%$
$x=1050.00$
Standard 11
Physics
