Question

The radius of the sphere is $(4.3 \pm 0.1)cm$. The percentage error in its volume is:
A) $\dfrac{{0.1}}{{4.3}} \times 100$
B) $3 \times \dfrac{{0.1 \times 100}}{{4.3}}$
C) $\dfrac{1}{3} \times \dfrac{{0.1 \times 100}}{{4.3}}$
D) $\dfrac{1}{3} + \dfrac{{0.1 \times 100}}{{4.3}}$

Answer 1

Hint: In this question, we can use the formula of volume of the sphere. Then we can convert it in the percentage error formula by calculating the difference between the accepted value and the experimental values.

Complete step by step solution:
According to the question, the radius of the sphere is $(4.3 \pm 0.1)cm$. Here, we can write the radius of the sphere as below-
$r + \Delta r = (4.3 \pm 0.1)cm$
Where $r = 4.3cm$ and $\Delta r = \pm 0.1cm$ or $\Delta r = 0.1cm$
Now, we know that the formula of the volume $V$ of the sphere having radius $r$ is given as-
$\Rightarrow$ $V = \dfrac{4}{3}\pi {r^3}$.............(i)
So, the error in the volume of the sphere will be given as-
$\Rightarrow$ $\dfrac{{\Delta V}}{V} = 3 \times \dfrac{{\Delta r}}{r}$..................(ii)
Where $\Delta V$ and $\Delta r$ are the differences between the accepted values and experimental values of volume and the radius respectively.
And the percentage error in the volume of the sphere will be given as-
$\Rightarrow$ $\dfrac{{\Delta V}}{V} \times 100 = 3 \times \dfrac{{\Delta r}}{r} \times 100$...........(iii)
So, putting the value of $r$ and $\Delta r$ in the above equation (iii), we get-
$\Rightarrow$ $\dfrac{{\Delta V}}{V} \times 100 = 3 \times \dfrac{{0.1}}{{4.3}} \times 100$
Or we can write
$\Rightarrow$ $\dfrac{{\Delta V}}{V} \times 100 = 3 \times \dfrac{{0.1 \times 100}}{{4.3}}$
Hence, the percentage error in the volume of the sphere is $3 \times \dfrac{{0.1 \times 100}}{{4.3}}$.

Therefore, option B is correct.

Note: In this question, we have to remember that only the formula of the volume of a sphere is used to find the percentage error in the volume. We have to keep in mind that $\Delta r$ is the difference between the accepted values and experimental value radius. We have to keep in mind that the power of the radius will be multiplied with the error in the radius. The value of $\Delta r$ is also known as the least count of that apparatus which is used in the measurement of the radius of the sphere.