Question

If there is a possible error of 0.02 cm in the measurement of the diameter of a sphere,
then find the possible percentage error in its volume, when the radius is 10cm

Answer 1

As error in the diameter=0.02 cm
so error in the radius =0.01cm
➡️Δr=0.01
and radius=10cm.
and volume of sphere =
$4 \div 3 \times\pi \times r {}^{2} \times r$

So percentage error = 3*(Δr/r)*100
=(0.01/10)*100*3
=0.1*3
=0.3%

Answer 2

The percentage error in its volume is 0.3

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