A balloon,which always remains spherical,is b | vedclass

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

Question

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

Vedclass · Accepted Answer

$1/\pi \ cm/sec$

Vedclass · Answer

(A) The volume $V$ of a spherical balloon 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 $\frac{dV}{dt} = 900 \ cm^3/sec$ and the radius $r = 15 \ cm$,we substitute these values into the equation:$900 = 4 \pi (15)^2 \frac{dr}{dt}$.$900 = 4 \pi (225) \frac{dr}{dt}$.$900 = 900 \pi \frac{dr}{dt}$.Solving for $\frac{dr}{dt}$:$\frac{dr}{dt} = \frac{900}{900 \pi} = \frac{1}{\pi} \ cm/sec$.Thus,the radius of the balloon is increasing at a rate of $1/\pi \ cm/sec$.

$A$ balloon,which always remains spherical,is being inflated by pumping in $900 \ cm^3/sec$ of gas. Find the rate at which the radius of the balloon is increasing when the radius is $15 \ cm$.

Similar Questions

The distance $s$ (in meters) covered by a particle in $t$ seconds is given by $s = ae^t + \frac{b}{e^t}$. Then the acceleration of the particle at time $t$ is:

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

$A$ particle moves in a straight line so that its velocity at any point is given by ${v^2} = a + bx$,where $a, b \neq 0$ are constants. The acceleration is

The rate of change of the diagonal $R$ of a square with respect to its area $A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module