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# The measured mass and volume of a bosy are $2.42\; g$ and $4.8\; cm^3$ respectively. With possible errors 0.01 g and $0.1 cm^3$ the maximum error in density is about

$\begin{array}{1 1}(A)\;0.2 \% \\(B)\;2 \% \\(C)\;5 \% \\(D)\;10 \% \end{array}$

$\rho = \large\frac{M}{V}$
$\large\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} +\frac{\Delta V}{V}$
$\qquad= \large\frac{0.01}{2.42} +\frac{0.1}{4.8}$
Hence $\bigg(\large\frac{\Delta \rho}{\rho} \bigg)$$\times 100$ is about $2 \%$
Hence B is the correct answer.