A spherical balloon is expanding. If at any i | vedclass

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

Question

(B) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$,where $r$ is the radius.Differentiating both sides with respect to tim

Vedclass · Accepted Answer

$\frac{2}{\sqrt{\pi}}$

Vedclass · Answer

(B) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$,where $r$ is the radius.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.According to the problem,the rate of increase of volume is $16$ times the rate of increase of the radius:$\frac{dV}{dt} = 16 \frac{dr}{dt}$.Substituting this into the derivative equation:$16 \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Assuming $\frac{dr}{dt} 
eq 0$,we can divide both sides by $\frac{dr}{dt}$:$16 = 4 \pi r^2$.$r^2 = \frac{16}{4 \pi} = \frac{4}{\pi}$.Taking the square root of both sides:$r = \sqrt{\frac{4}{\pi}} = \frac{2}{\sqrt{\pi}}$.

$A$ spherical balloon is expanding. If at any instant the rate of increase of its volume is $16$ times the rate of increase of its radius,then its radius at that instant is:

Similar Questions

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

$A$ man $2 \, m$ high walks at a uniform speed of $5 \, m/h$ away from a lamp post $6 \, m$ high. The rate at which the length of his shadow increases is

The radius of a right circular cylinder increases at the rate of $0.1 \text{ cm/min}$ and the height decreases at the rate of $0.2 \text{ cm/min}$. The rate of change of the volume of the cylinder in $\text{cm}^3\text{/min}$,when the radius is $2 \text{ cm}$ and the height is $3 \text{ cm}$,is:

$A$ man of $5 \text{ feet}$ height is walking away from a light fixed at a height of $15 \text{ feet}$ at the rate of $K \text{ miles/hour}$. If the rate of increase of his shadow is $\frac{11}{5} \text{ feet/sec}$, then $K=$ (Take $1 \text{ mile} = 5280 \text{ feet}$)

The equation of motion of a particle is $s = at^2 + bt + c$. If the displacement after $1 \text{ s}$ is $20 \text{ m}$,velocity after $2 \text{ s}$ is $30 \text{ m/s}$,and the acceleration is $10 \text{ m/s}^2$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module