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

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(D) Let $r$ be the radius of the iron ball,which is $10 	ext{ cm}$. Let $x$ be the thickness of the ice layer. The total radius of the sphere (iron b

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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