A spherical iron ball $10 \mathrm{~cm}$ in radius is coated with a layer of ice of uniform thickness than melts at a rate of $50 \mathrm{~cm}^3 / \mathrm{min}$. When the thickness of ice is $5 \mathrm{~cm}$, then the rate at which the thickness of ice decreases, is

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Aicharya · Accepted Answer

$ \begin{aligned} & \frac{d v}{d t}=50 \ & 4 \pi r^2 \frac{d r}{d t}=50 \ & \Rightarrow \frac{d r}{d t}=\frac{50}{4 \pi(15)^2} \quad 	ext { where } r=15 \ & =\frac{1}{16 \pi} \end{aligned} $

A spherical iron ball $10 \mathrm{~cm}$ in radius is coated with a layer of ice of uniform thickness than…

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