JEE Main 2020 (Online) 9th January Morning ShiftQuestion 1
A spherical iron ball of $10$ cm radius is coated with a layer of ice of uniform thickness that melts at a rate of $50\ \mathrm{cm}^3/\mathrm{min}$. When the thickness of ice is $5$ cm, then the rate (in cm/min.) at which the thickness of ice decreases, is :
Differentiation and its Application
Solution
Concept. Only the outer radius $R=10+(\text{thickness})$ changes; the rate of thickness equals $\frac{dR}{dt}$. Formulas used. (1) $V=\frac43\pi\left(R^3-10^3\right)$, so $\dfrac{dV}{dt}=4\pi R^2\dfrac{dR}{dt}$Step 1 - At thickness $5$. $R=15$. By (1), $$-50=4\pi\cdot225\cdot\frac{dR}{dt}\ \Longrightarrow\ \frac{dR}{dt}=-\frac{1}{18\pi}.$$ Step 2 - Answer. The thickness decreases at $\frac{1}{18\pi}$ cm/min. Answer: (C)