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