Question

The radius of a sphere is measured to be $5.3\pm 0.1cm$. Calculate the percentage error in the measurement of its volume.

• A. $3.45$%
• B. $5.66$%
• C. $2.4$%
• D. $1.05$%

Answer 1

The correct option is B $5.66$%

$\text{Step 1: Differentiating the volume}$

Volume of the sphere: $V=\frac{4}{3}\pi r^3$

Differentiating with respect to $r$

$\frac{dV}{dr}=\frac{4}{3}\pi .3r^2$

$\Rightarrow \frac{dV}{dr}=\frac{4}{3}\pi r^3.\frac{3}{r}$

$\Rightarrow \frac{dV}{V}=\frac{3dr}{r}$

Or

$\Rightarrow \frac{\Delta V}{V}=\frac{3\Delta r}{r}$

$\text{Step 2: Percentage error in the volume}$

Given: $r=5.3\pm 0.1cm$

So, $\Delta r=0.1cm$ and $r=5.3cm$

Percentage error in volume= $\frac{\Delta V}{V}\times 100\%=\frac{3\times 0.1cm}{5.3cm}\times 100\%=5.66\%$

Hence, Option B is correct.