Question

If there is an error of \[0.1\% \] in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.

Answer 1

Hint: We will first express the volume of the sphere in terms of its radius. Then, we will differentiate on both sides with respect to the radius. Finally, we will substitute the percentage error in radius in the formula of percentage error in a quantity to find the percentage error in the volume.

Formula used:
We will use the following formulas:
1.Volume of a sphere, \[V = \dfrac{4}{3}\pi {r^3}\]
2.Percentage error in a quantity \[x = \dfrac{{dx}}{x} \times 100\]

Complete step-by-step answer:
Let us denote the radius and volume of the sphere as \[r\] and \[V\] respectively.
We know that Volume of a sphere, \[V = \dfrac{4}{3}\pi {r^3}\] …………………….\[\left( 1 \right)\]
Let us differentiate on both sides with respect to \[r\]. Therefore, we get
\[\dfrac{{dV}}{{dr}} = \dfrac{4}{3}\pi (3{r^2})\]
Cancelling 3 on the RHS of the above equation,
\[ \Rightarrow \dfrac{{dV}}{{dr}} = 4\pi {r^2}\]
On cross multiplication, we get
  \[dV = 4\pi {r^2}dr\] …………………….\[\left( 2 \right)\]
Now we will find the percentage error in volume, using the formula Percentage error in a quantity \[x = \dfrac{{dx}}{x} \times 100\].
So, percentage error in volume \[ = \dfrac{{dV}}{V} \times 100\]
Using equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we have
\[\dfrac{{dV}}{V} = \dfrac{{4\pi {r^2}dr}}{{\dfrac{4}{3}\pi {r^3}}}\]
Cancelling the similar terms, we get
\[ \Rightarrow \dfrac{{dV}}{V} = 3 \times \dfrac{{dr}}{r}\]
Multiplying both sides by 100, we get
\[ \Rightarrow \dfrac{{dV}}{V} \times 100 = 3 \times \dfrac{{dr}}{r} \times 100\]
\[ \Rightarrow \] Percentage error in volume \[ = 3 \times \dfrac{{dr}}{r} \times 100\]…………………….\[\left( 3 \right)\]
It is given that percentage error in radius is \[0.1\% \], so
Percentage error in radius \[ = \dfrac{{dr}}{r} \times 100 = 0.1\]
Substituting this in equation \[\left( 3 \right)\], we have
Percentage error in volume \[ = 3 \times 0.1 = 0.3\]
$\therefore $ If there is an error of \[0.1\% \] in the radius of a sphere, then there is an error of \[0.3\% \] in the volume of the sphere.

Note: Whenever we are measuring any physical quantity using an instrument, then there occurs an error in the measurements. Percentage error is defined as the ratio of the least count to the measured value of the physical quantity multiplied by hundreds. We have multiplied by 100 on both sides of the equation because the given error is expressed as a percentage. We can also write percentage error in radius \[ = \dfrac{{dr}}{r} = 0.1\% = \dfrac{{0.1}}{{100}} = 0.001\].
In this case, percentage error in volume \[ = \dfrac{{dV}}{V} = 3 \times \dfrac{{dr}}{r} = 3 \times 0.001 = \dfrac{{0.3}}{{100}} = 0.3\% \].