(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 \ cm/sec$. At $t = 0$,the radius $r = 0$. Integrating $\frac{dr}{dt} = 5$,we get

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

(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 \ cm/sec$. At $t = 0$,the radius $r = 0$. Integrating $\frac{dr}{dt} = 5$,we get $r = 5t$. At $t = 2 \ seconds$,the radius $r = 5(2) = 10 \ cm$. The area of the circular ripple is $A = \pi r^2$. Differentiating with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$. Substituting the values $r = 10 \ cm$ and $\frac{dr}{dt} = 5 \ cm/sec$: $\frac{dA}{dt} = 2 \pi (10)(5) = 100 \pi \ cm^2/sec$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

Medium

$A$ stone is dropped into a pond. Waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of $5 \ cm/sec$. The rate at which the area increases after $2 \ seconds$ is:

MHT CET 2019 All MHT CET

A
$100 \pi \ cm^2/sec$
B
$40 \pi \ cm^2/sec$
C
$50 \pi \ cm^2/sec$
D
$25 \pi \ cm^2/sec$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Application of Derivatives

Subtopic

Rate of Change of Quantities

Previous Year

MHT CET 2019 Paper All MHT CET years →

The distance $s$ in meters covered by a particle in $t$ seconds is given by $s = 2 + 27t - t^3$. The particle will stop after covering a distance of:

EasyMHT CET 2021

View Solution

The relation between pressure $p$ and volume $V$ is given by $p V^{1/4} = \text{constant}$. If the percentage decrease in volume is $\frac{1}{2} \%$, then the percentage increase in pressure is:

DifficultAP EAMCET 2013

View Solution

$A$ particle covers a distance $x$ in time $t$ given by $x = t^3 - 9t^2 + 24t + 6$. When the acceleration of the particle is $6$,its velocity is .... units.

Medium

View Solution

Let a kind of bacteria grow following the function $f(t) = t^4$,where $t$ is given in seconds. If the rate of growth of the bacteria after $t_0$ seconds is $4000 \text{ units/second}$,then $t_0 =$

EasyAP EAMCET 2017

View Solution

If the area of a circle increases at a uniform rate,then prove that the rate of change of its perimeter varies inversely as the radius.

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

$A$ stone is dropped into a pond. Waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of $5 \ cm/sec$. The rate at which the area increases after $2 \ seconds$ is:

Similar Questions

The distance $s$ in meters covered by a particle in $t$ seconds is given by $s = 2 + 27t - t^3$. The particle will stop after covering a distance of:

The relation between pressure $p$ and volume $V$ is given by $p V^{1/4} = \text{constant}$. If the percentage decrease in volume is $\frac{1}{2} \%$, then the percentage increase in pressure is:

$A$ particle covers a distance $x$ in time $t$ given by $x = t^3 - 9t^2 + 24t + 6$. When the acceleration of the particle is $6$,its velocity is .... units.

Let a kind of bacteria grow following the function $f(t) = t^4$,where $t$ is given in seconds. If the rate of growth of the bacteria after $t_0$ seconds is $4000 \text{ units/second}$,then $t_0 =$

If the area of a circle increases at a uniform rate,then prove that the rate of change of its perimeter varies inversely as the radius.

Vedclass Test Series

Exam Paper Generator

Online Exam Module