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 cm 3 / min and the thickness of the ice-cream layer decreases at the rate of 1/4 π cm / min . The surface area (in cm 2 ) of the chocolate ball (without the ice-cream layer) is :

Mathematics · Application of Derivatives

JEE Main 2025 — 23 January, Evening Shift — Question 9

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 :

Option A:
$225 \pi$
Option B:
$128 \pi$
Option C:
$196 \pi$
Option D:
$256 \pi$
Correct

Answer: D

Step-by-step solution

$\mathrm{v}=\frac{4}{3} \pi \mathrm{r}^{3}$

$\frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}$

$81=4 \pi r^{2} \times \frac{1}{4 \pi}$

$\mathrm{r}^{2}=81$

$r=9$

surface area of chocolate $=4 \pi(r-1)^{2}=256 \pi$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives
Topic: Rate Measure using Derivatives