Question

The radius of a sphere is $(2.6 \pm 0.1)$ cm. What is the percentage error in its volume?
(A) $\dfrac{{0.1}}{{2.6}} \times 100\%$
(B) $3 \times \dfrac{{0.1}}{{2.6}} \times 100\%$
(C) $\dfrac{{8.49}}{{73.6}} \times 100\%$
(D) $\dfrac{{0.1}}{{2.6}}\%$

Answer 1

Hint
Error in the calculation of a parameter changes according to the laws of error propagation as and when the quantity is modified. Error percentage is calculated by taking a fraction with the original value.

Complete step by step answer
Error is the measure of uncertainty or miscalculation in the value of a parameter. However, error in a quantity does not follow linear algebra as the parameter does. The rules of error propagation are simple, but different.
In this question, we are provided with a sphere and are required to calculate the percentage error in its volume. The values provided to us include:
Radius of the sphere $R = 2.6$
Error in the radius of the sphere$\Delta R = 0.1$
Volume of the sphere V
Error in the volume of the sphere $\Delta V$
We know that the volume of a sphere is given as:
$V = \dfrac{4}{3}\pi {R^3}$
Putting the known values in this:
$V = \dfrac{4}{3} \times 3.14 \times {(2.6)^3} = 4.18 \times 17.576$
This gives us the volume as $V = 73.6$
We know, the error for exponential powers will propagate as:
$\dfrac{{\Delta V}}{V} = 3\left( {\dfrac{{\Delta R}}{R}} \right)$ [As the volume depends on the 3rd power of the radius]
Substituting the known values in this equation gives us:
$\Delta V = 3 \times 73.6\left( {\dfrac{{0.1}}{{2.6}}} \right) = 8.49$
This is the expected error in the measurement of the volume of the sphere. Now, to find the error percentage, we have:
$\dfrac{{\Delta V}}{V} \times 100\%$
This is equal to $\dfrac{{8.49}}{{73.6}} \times 100\%$. Hence, the correct answer is option C.

Note
Errors are very natural to occur in measurements and are almost inevitable. Sometimes these are caused by the instrument itself, and sometimes by the handler of the instrument. The extent of error in the calculation of a value may drastically affect the end result. Hence, most measurements are provided with a range for error.