Question

The density $\rho$ of a piece of metal of mass ${}^{\prime }{m}^{\prime }$ and volume $V$ is given by the relation, $\rho =\frac{m}{V}$. If $m=(275.35\pm 0.01)g$ and $V=(36.41\pm 0.01)c{m}^3$, then the percentage error in $\rho$ will be:

• A. $0.01\%$
• B. $0.02\%$
• C. $1\%$
• D. $0.03\%$

Answer 1

The correct option is D $0.03\%$

Density $\left(\rho \right)=\frac{\text{mass(m)}}{\text{volumer(v)}}$
$\therefore$ percentage error in density can be calculated as
$\Rightarrow \frac{\Delta \rho }{\rho }\times 100=(\frac{\Delta m}{m}+\frac{\Delta V}{V})\times 100$
$=(\frac{0.01}{275.32}+\frac{0.01}{36.41})\times 100=0.03\%$