Question

The volume of a sphere is increasing at the rate of 4π cm³/sec. The rate of increase of the radius when the volume is 288 π cm³, is

विकल्प

• 1/4

•  1/12

•  1/36

•  1/9

Answer 1

1/36

\[\text { Let r be the radius and V be the volume of the sphere at any time t. Then },\]

\[V=\frac{4}{3}\pi r^3 \]

\[ \Rightarrow \frac{4}{3}\pi r^3 =288\pi\]

\[ \Rightarrow r^3 = \frac{288 \times 3}{4}\]

\[ \Rightarrow r^3 = 216\]

\[ \Rightarrow r = 6\]

\[ \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\]

\[ \Rightarrow \frac{dV}{dt} = 4\pi \left( 6 \right)^2 \frac{dr}{dt} \]

\[ \Rightarrow 4\pi = 144\pi\frac{dr}{dt}\]

\[ \Rightarrow \frac{dr}{dt} = \frac{1}{36}\]