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

Step-by-step solution

We know that the volume of a sphere of radius r is 4 3 πr 3 . Given the radius of the spherical iron ball is R = 10   c m and the thickness of the ice is x   c m . Volume of ice: V = 4 3 π R + x 3 - 4 3 π R 3 ⇒ V = 4 3 π 10 + x 3 - 4 3 π 10 3 Differentiating with respect to t , we get d V d t = 4 π 10 + x 2 d x d t - 0 Given d V d t = 50   c m 3 / m i n ⇒ 50 = 4 π 10 + x 2 d x d t ⇒ d x d t = 50 4 π 10 + x 2 ⇒ d x d t x = 5 = 50 4 π