Question:
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
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
Updated On: Apr 28, 2024
- 10$\pi$
- 100$\pi$
- 200$\pi$
- 50$\pi$
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The Correct Option is C
Solution and Explanation
Let r and V be the respectively radius and volume of the balloon. Let t represents the time. The rate of increament in radius is $\frac{dr}{dt}=2$ cm/minute. The volume of the balloon is given by
$V =\frac{4}{3}\pi r^{3}$
Differentiating w.r. to t, we get
$\frac{dV}{dt}=\frac{4}{3}\pi\left(3r^{2}\frac{dr}{dt}\right)$
Substituting the values of and $\frac{dr}{dt}$ , we get
$\frac{dV}{dt}=\frac{4}{3}\pi\left(3\times5^{2}\times2\right)=200\pi cm^{3} / minute$
$V =\frac{4}{3}\pi r^{3}$
Differentiating w.r. to t, we get
$\frac{dV}{dt}=\frac{4}{3}\pi\left(3r^{2}\frac{dr}{dt}\right)$
Substituting the values of and $\frac{dr}{dt}$ , we get
$\frac{dV}{dt}=\frac{4}{3}\pi\left(3\times5^{2}\times2\right)=200\pi cm^{3} / minute$
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