1 Answer

Gladis · Answer 1 · 2021-07-01T22:29:39+0000

Step-by-step explanation:

It is given that, the radius of a ball is
$(5.2\pm 0.2)\ cm$ .

We need to find the percentage error in the volume of the ball. The volume of a sphere is :
$V=(4)/(3)\pi r^2$

The percentage error is given by :

$(\Delta V)/(V)=3(\Delta r)/(r)* 100$

We have,
$\Delta r=0.2$ and r = 5.2

So,

$(\Delta V)/(V)=3* (0.2)/(5.2)* 100\\\\\%=11\%$

So, the percentage error in the volume of the ball is 11%

Please log in or register to add a comment.