JEE Main 2013 (Online) 25th AprilQuestion 1
A spherical balloon is being inflated at the rate of $35\,\text{cc/min}$. The rate of increase in the surface area (in $\text{cm}^{2}/\text{min}$) of the balloon when its diameter is $14$ cm, is :
Differentiation and its Application
Solution
Concept. Find $\frac{dr}{dt}$ from the volume rate, then the surface rate. Formulas used. (1) $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ and $\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$Step 1 - $\frac{dr}{dt}$. At $r=7$: $35=4\pi\cdot49\cdot\frac{dr}{dt}$, so $\frac{dr}{dt}=\frac{35}{196\pi}$. Step 2 - $\frac{dS}{dt}$. $8\pi\cdot7\cdot\frac{35}{196\pi}=\frac{56\cdot35}{196}=10$. Answer: (A)