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(D) Let the radius of the iron ball be $r_0 = 10 \ cm$ and the thickness of the ice layer be $x \ cm$. The total radius of the sphere (iron ball + ice

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the radius of the iron ball be $r_0 = 10 \ cm$ and the thickness of the ice layer be $x \ cm$. The total radius of the sphere (iron ball + ice) is $R = 10 + x \ cm$. The volume of the ice layer is $V = \frac{4}{3}\pi (10+x)^3 - \frac{4}{3}\pi (10)^3$. Since the iron ball is constant,the rate of change of the total volume is equal to the rate of change of the ice volume. $V = \frac{4}{3}\pi (10+x)^3$. Differentiating with respect to time $t$: $\frac{dV}{dt} = 4\pi (10+x)^2 \frac{dx}{dt}$. Given $\frac{dV}{dt} = -50 \ cm^3/min$ (since it melts) and $x = 5 \ cm$: $-50 = 4\pi (10+5)^2 \frac{dx}{dt}$. $-50 = 4\pi (15)^2 \frac{dx}{dt}$. $-50 = 4\pi (225) \frac{dx}{dt}$. $-50 = 900\pi \frac{dx}{dt}$. $\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \ cm/min$. The negative sign indicates a decrease in thickness. Therefore,the rate at which the thickness decreases is $\frac{1}{18\pi} \ cm/min$. Thus,option $(D)$ is correct.

$A$ 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 $5 \ cm$,then the rate (in $cm/min$) at which the thickness of ice decreases is:

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