Q91 / 180 · KCET 2017 · Mathematics
The rate of change of volume of a sphere with respect to its surface area when the radius is \( 4 \)
\( \mathrm{cm} \) is
\( \mathrm{cm} \) is
- A\( 4 \mathrm{~cm}^{3} \mathrm{~cm}-2 \)
- Correct answerB\( 2 \mathrm{~cm}^{3} \mathrm{~cm}^{-2} \)
- C\( 6 \mathrm{~cm}^{3} \mathrm{~cm}^{-2} \)
- D\( 8 \mathrm{~cm}^{3} \mathrm{~cm}^{-2} \)
Solution
Given that, radius of the sphere is
\[ r=4 \]
We now that, volume sphere is given by
\[ V=\frac{4}{3} \Pi r^{2} \rightarrow(1) \]
and surface area of sphere is given by
\[ S=4 \Pi r^{2} \rightarrow(2) \]
Differentiating Eqs. (1) and (2) with respect to \( r \), we get
\[ \frac{d V}{d r}=\frac{4}{3} \times 3 \Pi r^{2}=4 \Pi r^{2} \]
\[ \begin{array}{l} \text { And } \frac{d S}{d r}=4 \times 2 \pi r=8 \pi r \\ \text { So, } \frac{d V}{d S}=\frac{d V / d r}{d S / d r}=\left[\frac{4 m r^{2}}{8 \pi r}\right]=\frac{r}{2} \\ \text { At } r=4 \mathrm{~cm}, \text { we get } \\ \left(\frac{r}{2}\right)_{r=4}=\frac{4}{2}=2 \mathrm{~cm}^{3} \mathrm{~cm}^{-2} \end{array} \]
\[ r=4 \]
We now that, volume sphere is given by
\[ V=\frac{4}{3} \Pi r^{2} \rightarrow(1) \]
and surface area of sphere is given by
\[ S=4 \Pi r^{2} \rightarrow(2) \]
Differentiating Eqs. (1) and (2) with respect to \( r \), we get
\[ \frac{d V}{d r}=\frac{4}{3} \times 3 \Pi r^{2}=4 \Pi r^{2} \]
\[ \begin{array}{l} \text { And } \frac{d S}{d r}=4 \times 2 \pi r=8 \pi r \\ \text { So, } \frac{d V}{d S}=\frac{d V / d r}{d S / d r}=\left[\frac{4 m r^{2}}{8 \pi r}\right]=\frac{r}{2} \\ \text { At } r=4 \mathrm{~cm}, \text { we get } \\ \left(\frac{r}{2}\right)_{r=4}=\frac{4}{2}=2 \mathrm{~cm}^{3} \mathrm{~cm}^{-2} \end{array} \]
Hence, the correct answer is option: 'B'.