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

$\begin{aligned} & V=\frac{4}{3} \pi r^3 \ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=4 \pi r^2 \frac{\mathrm{d} r}{\mathrm{~d} t} \ & \Rightarrow 50=4 \pi 	imes 15^2 	imes \frac{\mathrm{d} r}{\mathrm{~d} t} \ & \Rightarrow \frac{\mathrm{d} r}{\mathrm{~d} t}=\frac{50}{4 \pi 	imes 15^2}=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\end{aligned}$

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

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