Question

Error in the measurement of radius of a sphere is $1\% $ . Find the error in the measurement of volume.

Answer 1

Hint:For finding the error in the measurement of volume, we apply the formula of volume of sphere, that is $\dfrac{4}{3}\pi {R^3}$ and put error in the measurement of radius in it to get the final answer.

Complete step-by-step solution:
Error in the measurement of radius of a sphere is given, which is $1\% $
It is also written as in the form of a delta $R$ divided by $R$.
$ \Rightarrow \dfrac{{\Delta R}}{R} \times 100 = 1\% \cdots \cdots \cdots $ equation \[\left( 1 \right)\]
Where $\Delta R$ is an error in measurement of radius of sphere and $R$ is radius of sphere and they are represented in the form of percentage.
The volume of the sphere is $V$ and radius is $R$ . Volume of a sphere will be,
$ \Rightarrow V = \dfrac{4}{3}\pi {R^3} \cdots \cdots \cdots $ equation $\left( 2 \right)$
Applying the delta operator on the above equation
$ \Rightarrow \Delta V = \dfrac{4}{3}\pi \times \Delta {R^3}$
Delta operator also works as a differentiation. So, delta of ${R^3}$ will be $3 \times {R^2} \times \Delta R$ , putting this value on the above equation we get,
$ \Rightarrow \Delta V = \dfrac{4}{3}\pi \times 3 \times {R^2} \times \Delta R \cdots \cdots \cdots $ equation $\left( 3 \right)$
Dividing the equation $\left( 3 \right)$ by equation \[\left( 2 \right)\] , we get,
$ \Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{\dfrac{4}{3}\pi \times {R^2} \times \Delta R}}{{\dfrac{4}{3}\pi \times {R^3}}}$
Now , we divide the same value in the expression and get,
$ \Rightarrow \dfrac{{\Delta V}}{V} = 3 \times \dfrac{{\Delta R}}{R}$
It is a relative error. Now we multiply the equation by $100$ to get percentage error in measurement of volume.
$ \Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 3 \times \dfrac{{\Delta R}}{R} \times 100$
From equation $\left( 1 \right)$ , we put the value of percentage error in measurement of radius and get,
\[ \Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 3 \times 1\% \]
Therefore, percentage error in measurement of volume of a sphere will be
$ \Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 3\% $
Hence, the final answer is \[3\% \] error in measurement of volume of a sphere

Note:- Measurement error is also called the observation error, it is the difference between a measured quantity and its true value. It includes random error, which is naturally occurring errors and systematic error, which is caused by a not calibrated instrument that affects all the results of measurement.