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

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Aicharya · Accepted Answer

$\begin{aligned} & v=\frac{4}{3} \pi r^3 \ & \frac{d v}{d t}=50 \quad r=(10+5)=15 \mathrm{~cm} \ & \Rightarrow \frac{d v}{d t}=4 \pi r^2 \frac{d r}{d t} \ & \Rightarrow 50=4 \pi 	imes 15 	imes 15 	imes \frac{d r}{d t} \ & \Rightarrow \frac{d r}{d t}=\frac{50}{4 \pi 	imes 15 	imes 15}=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\end{aligned}$

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

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