Question

A certain body weighs $22.42 \,g$ and has a measured volume of $4.7 \,cc$. The possible error in the measurement of mass and volume are $0.01\, g$ and $0.1 \,cc$. Then maximum error in the densiy will be

• A. $22\%$
• B. $2\%$
• C. $0.2\%$
• D. $0.02\%$

Answer 1

Answer: (B)

Density $\rho=\frac{\text{mass m}}{\text{volume V}} .....(i)$
Take logarithm to take base $e$ on the both sides of eqn $(i)$, we get
$\ln \rho=\ln m-\ln V .....(ii)$
Differentiate eqn $(ii)$, on both sides, we get
$ \frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}-\frac{\Delta V}{V}$
Errors are always added, Error in the density $\rho$ will be
$ =\big[\frac{\Delta m}{m}+\frac{\Delta V}{V}\big]\times100 \%$
$=\big[\frac{0.01}{22.42}+\frac{0.1}{4.7}\big]\times100 \%$
$=2 \%$