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If the volume of spherical ball is increasing at the rate of 4 \pi \mathrm{cc} / \mathrm{sec} then the rate of change of its surface area when the volume is 288 \pi \mathrm{cc} is
(A) \frac{4}{3} \pi \mathrm{cm}^{2} / \mathrm{sec}
(B) \frac{2}{3} \pi \mathrm{cm}^{2} / \mathrm{sec}
(C) 4 \pi \mathrm{cm}^{2} / \mathrm{sec}
(D) 2 \pi \mathrm{cm}^{2} / \mathrm{sec}
maths | Maths
If the volume of spherical ball is increasing at the rate of 4 \pi \mathrm{cc} / \mathrm{sec} then the rate of change of its surface area when the volume is 288 \pi \mathrm{cc} is
(A) \frac{4}{3} \pi \mathrm{cm}^{2} / \mathrm{sec}
(B) \frac{2}{3} \pi \mathrm{cm}^{2} / \mathrm{sec}
(C) 4 \pi \mathrm{cm}^{2} / \mathrm{sec}
(D) 2 \pi \mathrm{cm}^{2} / \mathrm{sec}

Correct option is

(A)

Given, volume \mathrm{V}=288 \pi \mathrm{cc}, \frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=4 \pi \mathrm{cc} / \mathrm{sec}

volume of sphere

\begin{array}{l} \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3} \\ 288 \pi=\frac{4}{3} \pi \mathrm{r}^{3} \\ \Rightarrow \mathrm{r}=6 \end{array}

\frac{\mathrm{dV}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}

4 \pi=4 \pi 6^{2} \frac{\mathrm{dr}}{\mathrm{dt}}

\Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec}

Now, s=4 \pi r^{2}
Differentiating w.r.t. t, we get
\begin{array}{l} \left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)=4 \pi(2 \mathrm{r}) \cdot \frac{\mathrm{dr}}{\mathrm{dt}} \\ =8 \pi \mathrm{r} \cdot \frac{\mathrm{dr}}{\mathrm{dt}} \end{array}
When r=6 \mathrm{~cm}
\begin{array}{l} \left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)_{\mathrm{r}=6}=8 \pi(6) \cdot \frac{1}{36} \\ =\frac{4 \pi}{3} \end{array}
Thus, the surface area of the spherical ball is increasing at the rate of \frac{4 \pi}{3} \mathrm{~cm}^{2} / \mathrm{sec}.

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