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

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:

$\rho=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{M}}{\frac{4}{3} \pi\left(\frac{\mathrm{D}}{2}\right)^{3}}$

$\rho=\frac{6}{\pi} \mathrm{M} \mathrm{D}^{-3}$

taking log

$\ell \mathrm{n} \rho=\ell \mathrm{n}\left(\frac{6}{\pi}\right)+\ell \mathrm{nM}-3 \ell \mathrm{mD}$

Differentiates

$\frac{\mathrm{d} \rho}{\rho}=0+\frac{\mathrm{dM}}{\mathrm{M}}-3 \frac{\mathrm{d}(\mathrm{D})}{\mathrm{D}}$

for maximum error

$100 \times \frac{d \rho}{\rho}=\frac{d M}{M} \times 100+\frac{3 d D}{D} \times 100$

$=6+3 \times 1.5$

$=10.5 \%$

$=\frac{1050}{100} \%$ so $\mathrm{x}=1050.00$