Question

If the volume of a spherical balloon is increasing at the rate $900{\mathrm{cm}}^3/\sec$, then the rate of change of radius of balloon instant when radius is $15$ cm [in cm/sec]

• A. $\frac{22}{7}$
• B. $22$
• C. $\frac{7}{22}$
• D. None of these

Answer 1

The correct option is C

$\frac{7}{22}$

Explanation for the correct option :

Find the rate of change of radius,

The volume of the sphere is given as $V=\frac{4}{3}{\mathrm{\pi r}}^3$.

Now differentiate the equation with respect to time:

$$\begin{gathered} \frac{dV}{dt}=\frac{4}{3}\mathrm{\pi }·3{\mathrm{r}}^2\frac{d\mathrm{r}}{d\mathrm{t}} \\ \frac{d\mathrm{V}}{d\mathrm{t}}=4{\mathrm{\pi r}}^2\frac{d\mathrm{r}}{d\mathrm{t}} \end{gathered}$$

Now the rate of change of volume is $900{\mathrm{cm}}^3/\sec$, so $\frac{dV}{dt}=900$.

Then the rate of change of radius when $r=15$ cm is given as:

$\begin{array}{rcl}900& =& 4\mathrm{\pi }\times {15}^2\frac{d\mathrm{r}}{d\mathrm{t}}\\ 900& =& 900\mathrm{\pi }\frac{d\mathrm{r}}{d\mathrm{t}}\\ \frac{900}{900\mathrm{\pi }}& =& \frac{d\mathrm{r}}{d\mathrm{t}}\\ \frac{1}{\mathrm{\pi }}& =& \frac{d\mathrm{r}}{d\mathrm{t}}\end{array}$

Now taking the value of $\mathrm{\pi }$ as $\frac{22}{7}$ the rate of change is:

$$\begin{gathered} \frac{d\mathrm{r}}{d\mathrm{t}}=\frac{1}{{\displaystyle \frac{22}{7}}} \\ \frac{d\mathrm{r}}{d\mathrm{t}}=\frac{7}{22} \end{gathered}$$

Hence, the correct option is C.