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Question:

A spherical iron ball of radius 10cm is coated with a layer of ice of uniform thickness that melts at a rate of 50cm^3/min. When the thickness of the ice is 5cm, then the rate at which the thickness (in cm/min) of the ice decreases is:

19π

118π

136π

56π

Solution:

Correct option is C. 1/18π
Given r=10, R=10+h and dV/dt = -50
V = (4/3)π(R³ - r³)
V = (4/3)π((10+h)³ - 10³)
dV/dt = 4π(10+h)² dh/dt
-50 = 4π(10+5)² dh/dt
=> dh/dt = -1/18π cm/min