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(C) Let the thickness of the ice be $h \; cm$. The radius of the iron ball is $r = 10 \; cm$. The total radius of the sphere (iron ball + ice) is $R =

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$\frac{1}{18 \pi}$

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(C) Let the thickness of the ice be $h \; cm$. The radius of the iron ball is $r = 10 \; cm$. The total radius of the sphere (iron ball + ice) is $R = 10 + h$. The volume of the ice layer $V$ is given by the difference between the volume of the total sphere and the volume of the iron ball: $V = \frac{4}{3} \pi (10 + h)^{3} - \frac{4}{3} \pi (10)^{3}$. Differentiating both sides with respect to time $t$: $\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3(10 + h)^{2} \cdot \frac{dh}{dt} = 4 \pi (10 + h)^{2} \frac{dh}{dt}$. Given that the ice melts at a rate of $\frac{dV}{dt} = 50 \; cm^{3}/min$,we have: $50 = 4 \pi (10 + h)^{2} \frac{dh}{dt}$. When the thickness $h = 5 \; cm$: $50 = 4 \pi (10 + 5)^{2} \frac{dh}{dt} = 4 \pi (15)^{2} \frac{dh}{dt} = 4 \pi (225) \frac{dh}{dt} = 900 \pi \frac{dh}{dt}$. Therefore,$\frac{dh}{dt} = \frac{50}{900 \pi} = \frac{1}{18 \pi} \; cm/min$.

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

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