A spherical chocolate ball has a layer of ice | vedclass

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(D) Let $r$ be the radius of the chocolate ball and $x$ be the thickness of the ice-cream layer. The total radius of the sphere (chocolate + ice-cream

Vedclass · Accepted Answer

$256$

Vedclass · Answer

(D) Let $r$ be the radius of the chocolate ball and $x$ be the thickness of the ice-cream layer. The total radius of the sphere (chocolate + ice-cream) is $R = r + x$.Given $x = 1 \ cm$, the total volume $V$ 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 chocolate ball is constant, $\frac{dR}{dt} = \frac{dx}{dt}$.We are given $\frac{dV}{dt} = -81 \ cm^3/min$ (as it melts) and $\frac{dx}{dt} = -\frac{1}{4\pi} \ cm/min$.Substituting these values: $-81 = 4\pi R^2 \left(-\frac{1}{4\pi}\right)$.$-81 = -R^2 \implies R^2 = 81 \implies R = 9 \ cm$.Since $R = r + x$ and $x = 1 \ cm$, we have $r = 9 - 1 = 8 \ cm$.The surface area of the chocolate ball is $4\pi r^2 = 4\pi(8)^2 = 4\pi(64) = 256\pi \ cm^2$.

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