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The radius of a sphere increases at the rate of $0.04 \mathrm{~cm} / \mathrm{sec}$. The rate of increase in the volume of that sphere with respect to its surface area, when its radius is $10 \mathrm{~cm}$ is
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The correct answer is:
5
Let $r$ be the radius of the sphere.
Given, rate of change in radius $\frac{d r}{d t}=0.04 \mathrm{~cm} / \mathrm{sec}$
Volume of sphere $(V)=\frac{4}{3} \pi r^3$
Differentiating w.r.t, $t$, we get
$$
\frac{d V}{d t}=\frac{4}{3} \pi\left(3 r^2\right) \cdot \frac{d r}{d t}
$$
Surface area of sphere $(S)=4 \pi r^2$
$$
\frac{d S}{d t}=4 \pi(2 r) \cdot \frac{d r}{d t}
$$
Eq. (i) divided by Eq. (ii), we get
$$
\begin{array}{ll}
\frac{\frac{d V}{d t}}{\frac{d S}{d t}}=\frac{4 \pi r^2 \cdot \frac{d r}{d t}}{8 \pi r \cdot \frac{d r}{d t}} & \\
\frac{d V}{d S}=\frac{r}{2}=\frac{10}{2} & {[\because r=10 \mathrm{~cm}]} \\
\frac{d V}{d S}=5 \mathrm{~cm} &
\end{array}
$$
Hence, option (d) is correct.
Given, rate of change in radius $\frac{d r}{d t}=0.04 \mathrm{~cm} / \mathrm{sec}$
Volume of sphere $(V)=\frac{4}{3} \pi r^3$
Differentiating w.r.t, $t$, we get
$$
\frac{d V}{d t}=\frac{4}{3} \pi\left(3 r^2\right) \cdot \frac{d r}{d t}
$$
Surface area of sphere $(S)=4 \pi r^2$
$$
\frac{d S}{d t}=4 \pi(2 r) \cdot \frac{d r}{d t}
$$
Eq. (i) divided by Eq. (ii), we get
$$
\begin{array}{ll}
\frac{\frac{d V}{d t}}{\frac{d S}{d t}}=\frac{4 \pi r^2 \cdot \frac{d r}{d t}}{8 \pi r \cdot \frac{d r}{d t}} & \\
\frac{d V}{d S}=\frac{r}{2}=\frac{10}{2} & {[\because r=10 \mathrm{~cm}]} \\
\frac{d V}{d S}=5 \mathrm{~cm} &
\end{array}
$$
Hence, option (d) is correct.
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