Question:
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________
Solution:
$(1050)$
Density, $\rho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3}} \Rightarrow \rho=\frac{6}{\pi} M D^{-3}$
$\therefore \%\left(\frac{\Delta \rho}{\rho}\right)=\frac{\Delta m}{m}+3\left(\frac{\Delta D}{D}\right)=6+3 \times 1.5=10.5 \%$
$\%\left(\frac{\Delta \rho}{\rho}\right)=\frac{1050}{100} \%=\left(\frac{x}{100}\right) \%$
$\therefore x=1050.00$