The edge of a cube is equal to the radius of the sphere. If the rate at which the volume of the cube is increasing is equal to 'λ., then the rate of increase of volume of the sphere, is

Let r be the radius of the sphere. Let $V_{1}$ and $V_{2}$ be

the volumes of the sphere and the cube whose edge is r. Then,

$V_{1} = \frac{4}{3} π r^{3}$ and $V_{2} = r^{3}$

$\Rightarrow \frac{d V_{1}}{d t} = 4 π r^{2} \frac{d r}{d t}$ and $\frac{d V_{2}}{d t} = 3 r^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d V_{1}}{d t} = 4 π r^{2} \frac{d r}{d t}$ and $λ = 3 r^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d V_{1}}{d t} = \frac{4 π λ}{3}$

The edge of a cube is equal to the radius of the sphere. If the rate at which the volume of the cube is increasing is equal to ' $λ$ ., then the rate of increase of volume of the sphere, is