Question

The radius of a sphere is $(2.6\pm 0.1)$ cm. The percentage error in its volume is

• A. $\frac{0.1}{2.6}\times 100$
• B. $3\times \frac{0.1}{2.6}\times 100$
• C. $\frac{8.49}{73.6}\times 100$
• D. $\frac{0.1}{2.6}$

Answer 1

The correct option is C $\frac{8.49}{73.6}\times 100$

Given,

Radius$=\left(2.6\pm 0.1\right)cm$

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\times 3.14\times {\left(2.6\right)}^3=73.6c{m}^3$

Now, in error

$\begin{array}{c}\frac{\Delta V}{V}=3\left(\frac{\Delta r}{r}\right)\\ \Delta V=v\propto 3\left(\frac{\Delta r}{r}\right)\\ \Delta V=73.6\propto 3\left(\frac{0.1}{2.6}\right)\\ \Delta V=8.49c{m}^3\\ \frac{\Delta V}{V}\times 100=\frac{8.49}{73.6}\times 100\end{array}$

$=11.53$%

$\therefore$ Option $C$ is correct.