The volume of a spherical ball is increasing | vedclass

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

Question

(B) Given,$\frac{dV}{dt} = 4 \pi 	ext{ cm}^3/	ext{s}$  ...$(i)$We need to find $\frac{dr}{dt}$ when $V = 288 \pi 	ext{ cm}^3$.Let $r$ be the radius

Vedclass · Accepted Answer

$\frac{1}{36}$

Vedclass · Answer

(B) Given,$\frac{dV}{dt} = 4 \pi 	ext{ cm}^3/	ext{s}$  ...$(i)$We need to find $\frac{dr}{dt}$ when $V = 288 \pi 	ext{ cm}^3$.Let $r$ be the radius of the spherical ball.The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to $t$,we get:$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting the given rate $\frac{dV}{dt} = 4 \pi$:$4 \pi = 4 \pi r^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{r^2}$  ...(ii)When $V = 288 \pi$,we have:$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow r^3 = 288 	imes \frac{3}{4} = 216$.Thus,$r = \sqrt[3]{216} = 6 	ext{ cm}$.Substituting $r = 6$ into equation (ii):$\frac{dr}{dt} = \frac{1}{6^2} = \frac{1}{36} 	ext{ cm/s}$.

The volume of a spherical ball is increasing at a rate of $4 \pi \text{ cm}^3 \text{ s}^{-1}$. The rate at which its radius increases,when its volume is $288 \pi \text{ cm}^3$,is ....... $\text{cm s}^{-1}$.

Similar Questions

$A$ particle moves along the curve $y=x^2+2x$. Then, the point on the curve such that the $x$ and $y$ coordinates of the particle change with the same rate is

The volume of a spherical balloon is increasing at the rate of $40 \ cm^3/\min$. The rate of change of the surface area of the balloon at the instant when its radius is $8 \ cm$ is ........ $cm^2/\min$.

$A$ 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:

If the sides of a triangle vary slightly in such a way that its circumradius remains constant,then $\frac{da}{\cos A} + \frac{db}{\cos B} + \frac{dc}{\cos C}$ is equal to

If the area of a circle increases at the rate of $\frac{1}{\sqrt{\pi}}$ sq. units/sec, then the rate (in units/sec) at which the perimeter of the circle changes, when perimeter is $\sqrt{\pi}$ units, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module