Question

the radius of a ball is 5.2 plus minus 0.2 CM the percentage error in the value of the ball is approximately​

Answer 1

Answer: 11.5 %

= 11 % ( approx )

Explanation:

Answer 2

Answer:

The percentage error in the volume of the ball is 11%.

Explanation:

Given that,

Radius of ball = 5.2±0.2

We need to calculate volume of the ball

Using formula of volume

$V = \dfrac{4}{3}\p r^3$

On differentiating

$\Delta V=\dfrac{4}{3}\pi\times3r^2dr$

We need to calculate the percentage error in the volume of the ball

Using formula of percentage error

$\%\ error=\dfrac{\Delta V}{V}\times100$

Put the value into the formula

$\%\ error=\dfrac{\dfrac{4}{3}\pi\times3r^2dr}{ \dfrac{4}{3}\pi r^3}\times100$

$\%\ error=3\times\dfrac{dr}{r^2}\times100$

Put the value into the formula

$\%\ error=3\times\dfrac{0.2}{5.2}\times100$

$\%\ error=11\%$

Hence, The percentage error in the volume of the ball is 11%.