Error in the measurement of radius of a sphere is 1%. The error in the calculated value of its volume is

The correct option is 3%.

Step-by-Step Explanation:

Let us denote the radius of the sphere as r and its volume as V.

The formula for the volume of a sphere is given by:

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

To find the relation between the relative error in volume and the relative error in radius, we can take the natural logarithm (ln) on both sides of the equation:

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

Using the properties of logarithms, this simplifies to:

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

Now, differentiating both sides with respect to their respective variables, we get:

$\frac{dV}{V}=0+3\frac{dr}{r}$

(Since $\ln \left(\frac{4}{3}\pi \right)$ is a constant value, its derivative is zero.)

For small fractional changes or errors, we can write the relation in terms of errors as:

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

To express this as a percentage error, we multiply both sides by 100:

$\left(\frac{\Delta V}{V}\times 100\right)=3\times \left(\frac{\Delta r}{r}\times 100\right)$

Given that the percentage error in the measurement of the radius is 1%:

$\frac{\Delta r}{r}\times 100=1\%$

Substituting this value back into our equation for the percentage error of the volume:

$\text{Percentage error in volume}=3\times 1\%=3\%$

Therefore, the calculated value of its volume has a percentage error of 3%.