Question

If the error in measuring the radius of a sphere is 2%, then the error in the measurement of volume is:

• A. 8%
• B. 6%
• C. 2%
• D. 9%.

Answer 1

The correct option is B 6%

Percentage error in radius is given as $2$% i.e. $\frac{\Delta r}{r}\times 100=2$ %

Volume of sphere $V=\frac{4\pi }{3}{r}^3$

Percentage error in volume $\frac{\Delta V}{V}\times 100=3\times \frac{\Delta r}{r}\times 100=3\times 2=6$ %