JEE PYQ: Application Of Derivatives Question 63
Question 63 - 2019 (09 Apr Shift 2)
A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is $\tan^{-1}\left(\frac{1}{2}\right)$. Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.) at which the level of water is rising at the instant when the depth of water in the tank is 10 m, is:
(1) $1/15\pi$
(2) $1/10\pi$
(3) $2/\pi$
(4) $1/5\pi$
Show Answer
Answer: (4)
Solution
$r = h/2$. $V = \frac{1}{3}\pi r^2 h = \frac{\pi h^3}{12}$. $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt} = 5$. At $h = 10$: $\frac{dh}{dt} = \frac{20}{100\pi} = \frac{1}{5\pi}$.
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