JEE PYQ: Application Of Derivatives Question 53
Question 53 - 2020 (09 Jan Shift 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 $50$ cm$^3$/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which of the thickness of ice decreases, is:
(1) $\frac{5}{6\pi}$
(2) $\frac{1}{54\pi}$
(3) $\frac{1}{36\pi}$
(4) $\frac{1}{18\pi}$
Show Answer
Answer: (4)
Solution
$V = \frac{4}{3}\pi(10+x)^3 - \frac{4}{3}\pi(10)^3$. $\frac{dV}{dt} = 4\pi(10+x)^2\frac{dx}{dt} = -50$. At $x = 5$: $\frac{dx}{dt} = \frac{-50}{4\pi(225)} = \frac{-1}{18\pi}$. Rate of decrease $= \frac{1}{18\pi}$.
BETA
