Question

The rate of change of volume of a sphere with respect to its surface area when the radius is 4 cm is:

• (A) $4{\mathrm{cm}}^3/{\mathrm{cm}}^2$
• (B) $6{\mathrm{cm}}^3/{\mathrm{cm}}^2$
• (C) $2{\mathrm{cm}}^3/{\mathrm{cm}}^2$
• (D) $8{\mathrm{cm}}^3/{\mathrm{cm}}^2$

Answer 1

The Correct Option is C

Explanation:

We know that, volume of sphere $\mathrm{V}=\frac{4}{3}\pi {\mathrm{r}}^3$Surface area of sphere $\mathrm{S}=4\pi {\mathrm{r}}^2$, where $r$ is the radius of the sphere.So, rate of change of volume of sphere,$\Rightarrow \frac{\mathrm{d}\mathrm{V}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\times 3\times \pi r^2\frac{\mathrm{dr}}{\mathrm{dt}}$$\Rightarrow \frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{t}}=4\pi {\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}$    ----(1)Rate of change of surface area of sphere,$\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(4\pi r^2)=4\pi \times 2\times r\frac{\mathrm{dr}}{\mathrm{dt}}$$\Rightarrow \frac{\mathrm{dS}}{\mathrm{d}t}=8\pi \mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}$    ----(2)From equation (1) and equation (2),$\Rightarrow \frac{\mathrm{dV}}{\mathrm{dS}}=\frac{\frac{\mathrm{dV}}{\mathrm{dt}}}{\frac{\mathrm{dS}}{\mathrm{dt}}}=\frac{4\pi {\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}}{8\pi \mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}}$$\Rightarrow \frac{4\mathrm{r}}{8}=2(\because r=4\mathrm{cm})$Therefore, the rate of change of volume of a sphere with respect to its surface area is $2{\mathrm{cm}}^3/{\mathrm{cm}}^2$.Hence, the correct option is (C).