Question

If radius of the sphere is (5.3 ± 0.1) cm. Then percentage error in its volume will be ______.

Options

• \[3+6.01\times\frac{100}{5.3}\]

• \[\frac{1}{3}\times0.01\times\frac{100}{5.3}\]

• \[\left(\frac{3\times0.1}{5.3}\right)\times100\]

• \[\frac{0.1}{5.3}\times100\]

Answer 1

If radius of the sphere is (5.3 ± 0.1) cm. Then percentage error in its volume will be \[\left(\frac{3\times0.1}{5.3}\right)\times100\].

Explanation: 

\[Volumeofsphere(V)=\frac{4}{3}\pi r^3\]
\[\begin{aligned} \%\text{error in volume} & =3\times\frac{\Delta\mathrm{r}}{\mathrm{r}}\times100 \\ & =\left(3\times\frac{0.1}{5.3}\right)\times100 \end{aligned}\]