A spherical iron ball of 10 text{ cm} radius | vedclass

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Question

(A) Let $r$ be the radius of the sphere including the ice layer.Given that the radius of the iron ball is $10 	ext{ cm}$ and the thickness of the ice

Vedclass · Accepted Answer

$\frac{1}{18 \pi} 	ext{ cm/min}$

Vedclass · Answer

(A) Let $r$ be the radius of the sphere including the ice layer.Given that the radius of the iron ball is $10 	ext{ cm}$ and the thickness of the ice is $x = 5 	ext{ cm}$,the total radius is $r = 10 + x = 15 	ext{ cm}$.The volume of the sphere is $V = \frac{4}{3} \pi r^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Since the ice melts at a rate of $50 	ext{ cm}^3/	ext{min}$,the rate of change of volume is $\frac{dV}{dt} = -50 	ext{ cm}^3/	ext{min}$.Substituting the values: $-50 = 4 \pi (15)^2 \frac{dr}{dt}$.$\frac{dr}{dt} = \frac{-50}{4 \pi 	imes 225} = \frac{-50}{900 \pi} = \frac{-1}{18 \pi} 	ext{ cm/min}$.Since $\frac{dr}{dt} = \frac{dx}{dt}$,the rate at which the thickness decreases is $\frac{1}{18 \pi} 	ext{ cm/min}$.

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

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