(A) The volume of a sphere is given by $V = \frac{4}{3} \Pi r^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \Pi r^2 \frac{dr}{

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

$ \frac{1}{90 \Pi} \text{ cm s}^{-1} $

(A) The volume of a sphere is given by $V = \frac{4}{3} \Pi r^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \Pi r^2 \frac{dr}{dt}$.Given $\frac{dV}{dt} = 10 \text{ cm}^3 \text{s}^{-1}$ and $r = 15 \text{ cm}$.Substituting these values: $10 = 4 \Pi (15)^2 \frac{dr}{dt}$.$\frac{dr}{dt} = \frac{10}{4 \Pi (225)} = \frac{10}{900 \Pi} = \frac{1}{90 \Pi} \text{ cm s}^{-1}$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

Medium

$A$ balloon which always remains spherical is being inflated by pumping in $10 \text{ cm}^3$ of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is $15 \text{ cm}$.

KCET 2015 All KCET

A
$ \frac{1}{90 \Pi} \text{ cm s}^{-1} $
B
$ \frac{1}{9 \Pi} \text{ cm s}^{-1} $
C
$ \frac{1}{30 \Pi} \text{ cm s}^{-1} $
D
$ \frac{1}{\Pi} \text{ cm s}^{-1} $

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

KCET 2015 Paper All KCET years →

The distance $s$ in meters covered by a body in $t$ seconds is given by $s = 3t^2 - 8t + 5$. The body will stop after

EasyMHT CET 2024

View Solution

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

DifficultJEE MAIN 2020

View Solution

In $\Delta ABC$,sides $a$ and $b$ are kept constant such that there is an error $\alpha$ in angle $C$. What is the relative error in the area?

Medium

View Solution

The rate of change of the volume of a sphere of radius $r$ with respect to its surface area is:

Medium

View Solution

The distance covered by a particle in $t$ seconds is given by $x = 3 + 8t - 4t^2$. After $1$ second, the velocity will be:

EasyWBJEE 2009

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

$A$ balloon which always remains spherical is being inflated by pumping in $10 \text{ cm}^3$ of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is $15 \text{ cm}$.

Similar Questions

The distance $s$ in meters covered by a body in $t$ seconds is given by $s = 3t^2 - 8t + 5$. The body will stop after

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

In $\Delta ABC$,sides $a$ and $b$ are kept constant such that there is an error $\alpha$ in angle $C$. What is the relative error in the area?

The rate of change of the volume of a sphere of radius $r$ with respect to its surface area is:

The distance covered by a particle in $t$ seconds is given by $x = 3 + 8t - 4t^2$. After $1$ second, the velocity will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module