Question

The radius of a sphere is $(6.3\pm 0.1)\text{cm}$. The percentage error in its volume is

• A. $\frac{0.1}{6.3}\times 100$
• B. $3\times \frac{0.1}{6.3}\times 100$
• C. $\frac{0.1\times 100}{3.53}$
• D. $3+(\frac{0.1}{5.3}\times 100)$

Answer 1

The correct option is B $3\times \frac{0.1}{6.3}\times 100$

Fractional error in radius is given by $\frac{dr}{r}$
Given,
Measurement of Radius of the sphere $r\pm dr=6.3\pm 0.1\text{cm}$
From this,
$dr=0.1\text{cm},r=6.3\text{cm}$
We know that, Volume of sphere $V=\frac{4}{3}\pi r^3$
Applying logarithm on both sides we get,
$\ln V=\ln \left(\frac{4\pi }{3}\right)+\ln r^3$
Differentiating on both sides we get,
$\frac{dV}{V}=3\times \frac{dr}{r}$
$\%$ error in volume will be,
$\frac{dV}{V}\times 100=3\times \frac{dr}{r}\times 100$
$\Rightarrow \%$ error in volume $=3\times \frac{0.1}{6.3}\times 100$
Thus, option (b) is the correct answer.