JEE PYQ: Application Of Derivatives Question 65
Question 65 - 2019 (10 Apr Shift 2)
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm$^3$/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is:
(1) $\frac{1}{18\pi}$
(2) $\frac{1}{36\pi}$
(3) $\frac{5}{6\pi}$
(4) $\frac{1}{9\pi}$
Show Answer
Answer: (1)
Solution
$V_{ice} = \frac{4}{3}\pi[(10+r)^3 - 10^3]$. $\frac{dV}{dt} = 4\pi(10+r)^2\frac{dr}{dt} = -50$. At $r = 5$: $\frac{dr}{dt} = \frac{-50}{4\pi \cdot 225} = \frac{-1}{18\pi}$.
BETA
