Explanation:
It is given that, the radius of a ball is [tex](5.2\pm 0.2)\ cm[/tex].
We need to find the percentage error in the volume of the ball. The volume of a sphere is : [tex]V=\dfrac{4}{3}\pi r^2[/tex]
The percentage error is given by :
[tex]\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}\times 100[/tex]
We have, [tex]\Delta r=0.2[/tex] and r = 5.2
So,
[tex]\dfrac{\Delta V}{V}=3\times \dfrac{0.2}{5.2}\times 100\\\\\%=11\%[/tex]
So, the percentage error in the volume of the ball is 11%
