JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Application of Derivatives

A spherical chocolate ball has a layer of ice-cream of uniform...

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of

81 \mathrm{~cm}^3 / \mathrm{min}

and the thickness of the ice-cream layer decreases at the rate of

\frac{1}{4 \pi} \mathrm{~cm} / \mathrm{min}

. The surface area (in

\mathrm{cm}^2

) of the chocolate ball (without the ice-cream layer) is : A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream

Options

A $196 \pi$
B $256 \pi$
C $225 \pi$
D $128 \pi$

Correct answer

B. $256 \pi$

Step-by-step solution

\begin{aligned} & \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & 81=4 \pi \mathrm{r}^2 \times \frac{1}{4 \pi} \\ & \mathrm{r}^2=81 \\ & \mathrm{r}=9 \end{aligned}

surface area of chocolate

=4 \pi(r-1)^2=256 \pi

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