A balloon,which always remains spherical on i | vedclass

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

Question

(A) The volume of a sphere $(V)$ with radius $(r)$ is given by $V = \frac{4}{3} \pi r^3$. Differentiating with respect to time $(t)$,we get the rate o

Vedclass · Accepted Answer

$\frac{1}{\pi} \, cm/s$

Vedclass · Answer

(A) The volume of a sphere $(V)$ with radius $(r)$ is given by $V = \frac{4}{3} \pi r^3$. Differentiating with respect to time $(t)$,we get the rate of change of volume: $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt}$. Given that $\frac{dV}{dt} = 900 \, cm^3/s$,we substitute this into the equation: $900 = 4 \pi r^2 \frac{dr}{dt}$. Solving for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{900}{4 \pi r^2} = \frac{225}{\pi r^2}$. When the radius $r = 15 \, cm$: $\frac{dr}{dt} = \frac{225}{\pi (15)^2} = \frac{225}{225 \pi} = \frac{1}{\pi} \, cm/s$. Thus,the rate at which the radius increases is $\frac{1}{\pi} \, cm/s$.

$A$ balloon,which always remains spherical on inflation,is being inflated by pumping in $900 \, cm^3$ of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15 \, cm$.

Similar Questions

$A$ point is moving along the curve $y^3 = 27x$. The interval in which the abscissa changes at a slower rate than the ordinate is

$A$ point is moving on $y=4-2x^2$. The $x$-coordinate of the point is decreasing at the rate of $5 \text{ units/s}$. Then, the rate at which the $y$-coordinate of the point is changing when the point is at $(1, 2)$ is

The radius of a circular plate is increasing at the rate of $0.01 \text{ cm/s}$ when the radius is $12 \text{ cm}$. Then,the rate at which the area increases,is

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

If the rate of change of $x$ is more than the rate of change of $y$ on the curve $x^3 = 12y$ for $x > 0$,then $x$ lies in the interval:

Vedclass Test Series

Exam Paper Generator

Online Exam Module