Question

Air is being pumped into a spherical balloon at a constant rate such that its radius increases constantly with respect to time according to the equation $r\left(t\right)=0.5t^2+r_{\circ },$ (where $r$ is in $\text{cm},t$ is in $\text{minute}$ and $r_{\circ }$ is the initial radius of the balloon). Then the rate of change of its volume after $2\text{minute}$ is

(Assume that the initial radius of the balloon is $2\text{cm}$)

• A. $128\pi {\text{cm}}^3\text{/min}$
• B. $64\pi {\text{cm}}^3\text{/min}$
• C. $32\pi {\text{cm}}^3\text{/min}$
• D. $16\pi {\text{cm}}^3\text{/min}$

Answer 1

The correct option is A $128\pi {\text{cm}}^3\text{/min}$

Volume of the sphere whose radius $r$ is
$V\left(r\right)=\frac{4}{3}\pi r^3$
Differentiate with respect to time$\left(t\right)$
$\frac{dV}{dt}=\frac{4}{3}\pi \times 3r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}\cdots \left(1\right)$

Now, $r\left(t\right)=0.5t^2+r_{\circ }$
$\Rightarrow r\left(2\right)=0.5\times 4+2=4\text{cm}$
$\frac{dr}{dt}{|}_{t=2}=t+0=2\text{cm/min}$
Putting the values in $\left(1\right),$ we get
$\frac{dV}{dt}=128\pi {\text{cm}}^3\text{/min}$