Question

The volume of a sphere is increasing at 3 cm³/sec. The rate at which the radius increases when radius is 2 cm, is

विकल्प

• \[\frac{3}{32\pi}cm/\sec\]

• \[\frac{3}{16\pi}cm/\sec\]

• \[\frac{3}{48\pi}cm/\sec\]

• \[\frac{1}{24\pi}cm/\sec\]

Answer 1

 \[\frac{3}{16\pi}cm/\sec\]

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

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

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

\[\Rightarrow\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}\]

\[\Rightarrow\frac{dr}{dt}=\frac{3}{4\pi \left( 2 \right)^2}\]

\[\Rightarrow\frac{dr}{dt}=\frac{3}{16\pi}cm/sec\]