JEE Main 2020 (Online) 3rd September Evening ShiftQuestion 1
If the surface area of a cube is increasing at a rate of $3.6\ \mathrm{cm}^2/\mathrm{sec}$, retaining its shape; then the rate of change of its volume (in $\mathrm{cm}^3/\mathrm{sec}$), when the length of a side of the cube is $10$ cm, is :
Differentiation and its Application
Solution
Concept. Link the rates of $S=6s^2$ and $V=s^3$ through $\frac{ds}{dt}$. Formulas used. (1) $\dfrac{dS}{dt}=12s\dfrac{ds}{dt}$, $\ \dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}$Step 1 - $\frac{ds}{dt}$. By (1), $120\frac{ds}{dt}=3.6$, so $\frac{ds}{dt}=0.03$. Step 2 - $\frac{dV}{dt}$. $3\cdot100\cdot0.03=9$. Answer: (B)