(D) The volume $V$ of a spherical snowball is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to time $t$,we get $\frac{dV}

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$\frac{1}{2\pi} \text{ cm/sec}$

(D) The volume $V$ of a spherical snowball is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Given that $\frac{dV}{dt} = 8 \text{ cm}^3/\text{sec}$ and $r = 2 \text{ cm}$.Substituting these values,we have $8 = 4 \pi (2)^2 \frac{dr}{dt}$.$8 = 16 \pi \frac{dr}{dt}$.Therefore,$\frac{dr}{dt} = \frac{8}{16\pi} = \frac{1}{2\pi} \text{ cm/sec}$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

Easy

$A$ spherical snowball is forming such that its volume is increasing at the rate of $8 \text{ cm}^3/\text{sec}$. Find the rate of increase of its radius when the radius is $2 \text{ cm}$.

MHT CET 2021 All MHT CET

A
$\pi \text{ cm/sec}$
B
$\frac{1}{8\pi} \text{ cm/sec}$
C
$2\pi \text{ cm/sec}$
D
$\frac{1}{2\pi} \text{ cm/sec}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Applications of Derivatives

Subtopic

Rate of Change of Quantities

Previous Year

MHT CET 2021 Paper All MHT CET years →

$A$ balloon,which always remains spherical,has a variable diameter of $ \frac{3}{2}(2x + 3) $. Find the rate of change of its volume with respect to $ x $.

Difficult

View Solution

$A$ particle moves according to the law $s=t^{3}-6t^{2}+9t+25$. Find the displacement of the particle when its velocity is zero. (in $\text{ units}$)

EasyMHT CET 2020

View Solution

If by dropping a stone in a quiet lake a wave moves in a circle at a speed of $3.5 \, cm/sec$,then the rate of increase of the enclosed circular region when the radius of the circular wave is $10 \, cm$ is ......... $cm^2/sec$. $\left( \pi = \frac{22}{7} \right)$

Medium

View Solution

$A$ particle is moving in a straight line. At time $t$, the distance of the particle from its starting point is given by $x = t^3 - 6t^2 + t$. Its acceleration will be zero at

EasyWBJEE 2009

View Solution

$A$ particle is moving along the curve $x = at^2 + bt + c$. If $ac = b^2$,then the particle would be moving with uniform

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ spherical snowball is forming such that its volume is increasing at the rate of $8 \text{ cm}^3/\text{sec}$. Find the rate of increase of its radius when the radius is $2 \text{ cm}$.

Similar Questions

$A$ balloon,which always remains spherical,has a variable diameter of $ \frac{3}{2}(2x + 3) $. Find the rate of change of its volume with respect to $ x $.

$A$ particle moves according to the law $s=t^{3}-6t^{2}+9t+25$. Find the displacement of the particle when its velocity is zero. (in $\text{ units}$)

If by dropping a stone in a quiet lake a wave moves in a circle at a speed of $3.5 \, cm/sec$,then the rate of increase of the enclosed circular region when the radius of the circular wave is $10 \, cm$ is ......... $cm^2/sec$. $\left( \pi = \frac{22}{7} \right)$

$A$ particle is moving in a straight line. At time $t$, the distance of the particle from its starting point is given by $x = t^3 - 6t^2 + t$. Its acceleration will be zero at

$A$ particle is moving along the curve $x = at^2 + bt + c$. If $ac = b^2$,then the particle would be moving with uniform

Vedclass Test Series

Exam Paper Generator

Online Exam Module