BITSAT 2021 Paper

Let r be the radius of the sphere and Δr be the error in measuring the radius. Then, r=9‌cm and Δr=0.03‌cm
Let V be the volume of the sphere. Then,
V=

4

3

πr3
⇒

dV

dr

=4πr2
⇒ (

dV

dr

)r=9=4π×92=324π
Let ΔV be the error in V due to error Δr in r.
Then,
ΔV=

dV

dr

Δr
⇒ΔV=324π×0.03 =9.72πcm3