A spherical iron ball of radius 10 , cm is co | vedclass

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(D) Let $r = 10 \, cm$ be the radius of the iron ball and $h$ be the thickness of the ice layer.The total radius of the sphere including the ice is $R

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $r = 10 \, cm$ be the radius of the iron ball and $h$ be the thickness of the ice layer.The total radius of the sphere including the ice is $R = 10 + h$.The volume of the ice layer $V$ is given by the difference between the volume of the sphere with ice 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 $50 \, cm^3/min$,we have $\frac{dV}{dt} = -50 \, cm^3/min$.Substituting $h = 5 \, cm$ and $\frac{dV}{dt} = -50$ into the equation:$-50 = 4\pi (10 + 5)^2 \frac{dh}{dt}$$-50 = 4\pi (15)^2 \frac{dh}{dt}$$-50 = 4\pi (225) \frac{dh}{dt}$$-50 = 900\pi \frac{dh}{dt}$$\frac{dh}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \, cm/min$.The rate at which the thickness decreases is $\frac{1}{18\pi} \, cm/min$.

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

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