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 $6.0\text{\%}$ and $1.5\text{\%}$ respectively, the value of $x$ is

Answer 1

Density is written as
$\rho =\frac{M}{V}=\frac{M}{\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3}\Rightarrow \rho =\frac{6}{\pi }M{D}^{-3}$
So the relative error can be written as
$\therefore \text{\%}\left(\frac{\Delta \rho }{\rho }\right)=\frac{\Delta m}{m}+3\left(\frac{\Delta D}{D}\right)=6+3\times 1.5=10.5\text{\%}$
or, $\text{\%}\left(\frac{\Delta \rho }{\rho }\right)=\frac{1050}{100}\text{\%}=\left(\frac{x}{100}\right)\text{\%}$
$\therefore x=1050.00$