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 $0.01\,gm$ and $0.1\,cc$. Then maximum error in the density will be

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

Answer 1

Answer: (B)

$2\%$

Answer 2

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