The volume of a spherical balloon being infla | vedclass

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

Question

(A) Let the volume of the spherical balloon be $V = \frac{4}{3} \pi r^3$. Given that the rate of change of volume is constant,let $\frac{dV}{dt} = k$,

Vedclass · Accepted Answer

$(63t + 27)^{\frac{1}{3}}$

Vedclass · Answer

(A) Let the volume of the spherical balloon be $V = \frac{4}{3} \pi r^3$. Given that the rate of change of volume is constant,let $\frac{dV}{dt} = k$,where $k$ is a constant. Integrating with respect to $t$,we get $V = kt + C$. Substituting $V = \frac{4}{3} \pi r^3$,we have $\frac{4}{3} \pi r^3 = kt + C$. At $t = 0$,$r = 3$,so $\frac{4}{3} \pi (3)^3 = k(0) + C \Rightarrow C = 36 \pi$. At $t = 3$,$r = 6$,so $\frac{4}{3} \pi (6)^3 = k(3) + 36 \pi$. $288 \pi = 3k + 36 \pi \Rightarrow 3k = 252 \pi \Rightarrow k = 84 \pi$. Substituting $k$ and $C$ back into the equation: $\frac{4}{3} \pi r^3 = 84 \pi t + 36 \pi$. Dividing by $\frac{4}{3} \pi$,we get $r^3 = 63t + 27$. Therefore,$r = (63t + 27)^{\frac{1}{3}}$.

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is $3$ units and after $3$ seconds it is $6$ units,find the radius of the balloon after $t$ seconds.

Similar Questions

The decay rate of radioactive material at any time $t$ is proportional to its mass at that time. The mass is $27 \text{ grams}$ when $t=0$. After $3 \text{ hours}$,it was found that $8 \text{ grams}$ are left. Then the substance left after one more hour is

In a certain culture of bacteria,the rate of increase is proportional to the number present. It is found that the number doubles in $4$ hours. Then the number of times the bacteria are increased in $12$ hours is

If the population grows at the rate of $5 \%$ per year,then the time taken for the population to become double is (Given $\log 2=0.6912$ ) (in $years$)

$A$ spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is $3 \text{ mm}$ and $1 \text{ hour}$ later has been reduced to $2 \text{ mm}$,then the expression of radius $r$ of the raindrop at any time $t$ is (where $0 \leq t < 3$):

Water flows from the base of a rectangular tank of depth $16 \ m$. The rate of flow of the water is proportional to the square root of the depth at any time $t$. If the depth is $4 \ m$ when $t = 2 \ hours$,then after $3.5 \ hours$,the depth (in meters) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module