The radius of a spherical ball is increasing at the rate of 1m/sec. At the radius equal to 3m, the volume of the ball is increasing at the rate given by :

$30\pi \, m^3/sec$ $38\pi \, m^3/sec$ $36 \, m^3/sec$ $36\pi \, m^3/sec$

$36\pi \, m^3/sec$

The correct answer is Option (4) → $36\pi \, m^3/sec$ $\frac{dR}{dt}=1m/sec$ $v=\frac{4}{3}πR^3$ $\frac{dv}{dt}=4πR^2\frac{dR}{dt}=4πR^2$ $\left.\frac{dv}{dt}\right]_{R=3}=4π×3^2$ $=36\pi \, m^3/sec$