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Question:

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan⁻¹(1/2). Water is poured into it at a constant rate of 5 cubic meters per minute. 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 10m is?

110π

15π

115π

Solution:

tanθ = 1/2 = r/h => r = h/2
V = (1/3)πr²h = (1/3)π(h/2)²h = (1/12)πh³
dV/dt = (π/4)(3h²)dh/dt
5 = (π/4)(3(10)²)dh/dt => dh/dt = 15π