Search any question & find its solution
Question:
Answered & Verified by Expert
A football is inflated by pumping air in it. When it acquires spherical shape its radius increases at the rate of $0.02 \mathrm{~cm} / \mathrm{s}$. The rate of increase of its volume when the radius is $10 \mathrm{~cm}$ is $\pi \mathrm{cm} / \mathrm{s}$
Options:
Solution:
2989 Upvotes
Verified Answer
The correct answer is:
8
$$
\begin{array}{l}
\mathrm{v}=\frac{4}{3} \pi \mathrm{r}^{3} \\
\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{4}{3} \pi \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{r}^{3}=\frac{4}{3} \pi 3 \mathrm{r}^{2} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}
\end{array}
$$
when $\mathrm{r}=10 \mathrm{~cm}$;
$\frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi(10)^{2} \cdot(0.02)=8 \pi \mathrm{cm}^{3} / \mathrm{s}$
\begin{array}{l}
\mathrm{v}=\frac{4}{3} \pi \mathrm{r}^{3} \\
\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{4}{3} \pi \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{r}^{3}=\frac{4}{3} \pi 3 \mathrm{r}^{2} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}
\end{array}
$$
when $\mathrm{r}=10 \mathrm{~cm}$;
$\frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi(10)^{2} \cdot(0.02)=8 \pi \mathrm{cm}^{3} / \mathrm{s}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.