Question

If an error of $\left(\frac{1}{10}\right)\%$ is made in measuring the radius of a sphere then percentage error in its volume is

• A. $0.3\%$
• B. $0.03\%$
• C. $0.003\%$
• D. $0.0003\%$

Answer 1

The correct option is A $0.3\%$

Solution:- (A) $0.3$

As we know that, volume of sphere $\left(V\right)$ is-

$V=\frac{4}{3}\pi r^3$

Taking $\log$ both sides, we have

$\log V=\log \left(\frac{4}{3}\pi \log r^3\right)$

$\log V=\log \left(\frac{4}{3}\pi \right)+\log r^3$

$\lg V=\log \left(\frac{4}{3}\pi \right)+3\log r$

Differentiating the above equation, we have

$\frac{1}{V}dV=0+\left(\frac{1}{r}dr\right)$

$\Rightarrow \frac{dV}{V}=3\frac{dr}{r}.....\left(1\right)$

Given that the error in measuring the radius of the sphere is $\frac{1}{10}\%$.

$\therefore \frac{dr}{r}=\frac{1}{10}\%=0.001$

Substituting the value of $\frac{dr}{r}$ in ${eq}^n\left(1\right)$, we have

$\frac{dV}{V}=3\times 0.001=0.003=0.3\%$

Hence the error in measurement of volume of sphere will be $0.3\%$.