Question

If error is in measurement of radius of sphere is 1% what will be the error in measurement of volume
A) 1%
B) 1/3 %
C) 3%
D) none of these

Answer 1

Hint
In order to calculate the percentage error, we need to follow the formula of the error in question which is percent error= [experimental value – theoretical value]/theoretical value×100%
As volume of sphere is $V = \dfrac{4}{3}\pi {R^3}$, then we can write it in the form of percentage error i.e. $\dfrac{{\Delta V}}{V} \times 100 = 3\left( {\dfrac{{\Delta R}}{R} \times 100} \right)$, on substituting the values we will get the desired result.

Complete step by step answer
Here, let us consider R is radius of the sphere
And it is also given that error is in measurement of radius of sphere is 1% i.e. $\dfrac{{\Delta R}}{R} = 1\% $
We have to find the value of the error in the measurement of the volume of the sphere.
For this, we know that the volume of the sphere is $V = \dfrac{4}{3}\pi {R^3}$…………… (1)
Where, R is the radius of the sphere.
As we also know that the formula of error is percent error= [experimental value – theoretical value]/theoretical value×100%, therefore the formula of error for the equation (1) can be written as
$ \Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 3\left( {\dfrac{{\Delta R}}{R} \times 100} \right)$ ……………… (2)
Now, substitute the given values in above equation, we get
$ \Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 3\left( {1 \times 100} \right)$
$ \Rightarrow \dfrac{{\Delta V}}{V} = 3\% $
Hence, the value of the error in the measurement of the sphere is 3%
Therefore, option (C) is correct.

Note
The purpose of percent error calculation is to gauge how close a measured value is to true value. In some fields the percent error is always expressed as a positive number. In others, it is correct to have the positive or negative value. The sign kept to determine whether the recorded value consistently falls above or below to the expected value.
While solving this question, we need to be more careful with formula for finding the percentage error as it is used in a modified way. This question can also be solved by different ways as per the convenience.