Question

If the error in measurement of radius of a sphere is 2%, then the error in determination of
volume of the sphere will be ......

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Answer 1

Given that:

Percentage error in radius is given as 2%

or we can say that

$\large\rm { \frac{ \Delta r}{r} \times 100 = 2 \%}$

Solution

we know that,

volume → $\large\rm { V = \frac{4}{3} \pi r^{3}}$

Percentage error in volume:

$\large\rm { \frac { \Delta V}{V} \times 100 }$

$\large\rm { = 3 \times \frac{ \Delta r}{r} \times 100}$

$\large\rm { = 3 \times 2}$

$\large\boxed{\bf { \therefore 6 \% }}$ is the required answer.