Air is discharging from a large spherical bal | vedclass

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

Question

(B) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. Given, the rate of discharge of air is $\frac{dV}{dt}

Vedclass · Accepted Answer

$1 \,m^2 / min$

Vedclass · Answer

(B) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. Given, the rate of discharge of air is $\frac{dV}{dt} = -4 \,m^3 / min$ (since air is discharging, volume is decreasing). The volume of a sphere is $V = \frac{4}{3} \pi r^3$. Differentiating with respect to $t$, we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$. Substituting the given values at $r = 8 \,m$: $-4 = 4 \pi (8)^2 \frac{dr}{dt} \Rightarrow -4 = 256 \pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = -\frac{1}{64 \pi} \,m / min$. The surface area of a sphere is $S = 4 \pi r^2$. Differentiating with respect to $t$, we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$. Substituting $r = 8 \,m$ and $\frac{dr}{dt} = -\frac{1}{64 \pi} \,m / min$: $\frac{dS}{dt} = 8 \pi (8) \left( -\frac{1}{64 \pi} \right) = -1 \,m^2 / min$. Since the surface area is shrinking at a rate of $1 \,m^2 / min$, the rate of shrinking is $1 \,m^2 / min$.

Air is discharging from a large spherical balloon at the rate of $4 \,m^3 / min$. The rate at which the surface area is shrinking when the radius of the balloon is $8 \,m$, is

Similar Questions

$A$ 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 $\text{ units}$)

$A$ stone is dropped into a quiet lake and waves move in circles at a speed of $8 \,cm/sec$. At the instant when the radius of the circular wave is $12 \,cm$, how fast is the enclosed area increasing?

The volume of a sphere is increasing at the rate of $1200 \text{ cm}^3/\text{s}$. The rate of increase in its surface area when the radius is $10 \text{ cm}$ is: (in $\text{ cm}^2/\text{s}$)

$A$ spherical balloon is filled with $4500\pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72\pi$ cubic meters per minute,then the rate (in meters per minute) at which the radius of the balloon decreases $49$ minutes after the leakage began is:

The volume of a ball is increasing at the rate of $4 \pi \text{ cc/sec}$. The rate of increase of the radius,when the volume is $288 \pi \text{ cc}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module