Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A spherical balloon is expanding. If the radius is increasing at the rate of 2 centimeters per minute, the rate at which the volume increases (in cubic centimeters per minute) when the radius is 5 , centimetres is
MathematicsStraight LinesVITEEEVITEEE 2008
Options:
  • A $10 \pi$
  • B $100 \pi$
  • C $200 \pi$
  • D $50 \pi$
Solution:
2984 Upvotes Verified Answer
The correct answer is: $200 \pi$
Let $\mathrm{r}$ and $\mathrm{V}$ be the respectively radius and volume of the balloon. Let $\mathrm{t}$ represents the time. The rate of increament in radius is $\frac{\mathrm{dr}}{\mathrm{dt}}=2 \mathrm{~cm} / \mathrm{minute} .$ The volume of the balloon is given by
$\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$
Differentiating w.r. to $\mathrm{t}$, we get
$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}\right)$
Substituting the values of and $\frac{\mathrm{dr}}{\mathrm{dt}}$, we get
$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \times 5^{2} \times 2\right)=200 \pi \mathrm{cm}^{3} /$ minute

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.