A spherical iron ball of $10 \mathrm{~cm}$ radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \mathrm{~cm}^{3} /$ min. When the thickness of ice is $5 \mathrm{~cm}$, then the rate (in $\mathrm{cm} / \mathrm{min} .)$ at which of the thickness of ice decreases, is:
Correct Option: , 4
Let the thickness of ice layer be $=x \mathrm{~cm}$
Total volume $V=\frac{4}{3} \pi(10+x)^{3}$
$\frac{d V}{d t}=4 \pi(10+x)^{2} \frac{d x}{d t}$ ......(1)
Since, it is given that
$\frac{d V}{d t}=50 \mathrm{~cm}^{3} / \mathrm{min}$.........(2)
From (1) and (2), $50=4 \pi(10+x) \frac{d x}{d t}$
$\Rightarrow 50=4 \pi(10+5)^{2} \frac{d x}{d t}[\because$ thickness of ice $x=5]$
$\Rightarrow \quad \frac{d x}{d t}=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$
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