A spherical iron ball 10 cm in 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 i - Mathematics MCQ AIEEE 2026 | ExamTest.live

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A spherical iron ball 10 cm in 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 at which the thickness of ice decreases, is:

Solution & Step-by-step Explanation

Let r be the radius of the iron ball (r = 10) and x be the thickness of the ice.The total radius R = r + x = 10 + x .The volume of the ice V = \frac{4}{3} π R^{3} - \frac{4}{3} π r^{3} = \frac{4}{3} π (10 + x)^{3} - \frac{4}{3} π (10)^{3} .Differentiating with respect to time t : \frac{d V}{d t} = 4 π (10 + x)^{2} \frac{d x}{d t} Given \frac{d V}{d t} = - 50 cm^{3} /min (melting implies volume decrease).When thickness x = 5 cm : - 50 = 4 π (10 + 5)^{2} \frac{d x}{d t} - 50 = 4 π (15)^{2} \frac{d x}{d t} - 50 = 4 π (225) \frac{d x}{d t} - 50 = 900 π \frac{d x}{d t} \frac{d x}{d t} = - \frac{50}{900 π} = - \frac{1}{18 π} .The rate at which thickness decreases is \frac{1}{18 π} cm/min .

Try it yourself before checking the explanation above.

A spherical iron ball 10 cm in 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 at which the thickness of ice decreases, is:

\frac{1}{36 π} cm/min

\frac{1}{18 π} cm/min

\frac{1}{54 π} cm/min

\frac{5}{6 π} cm/min

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