The vertical angle of a right circular cone i | vedclass

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

Question

(C) Let $r$ be the radius and $h$ be the height of the water level in the cone at any time $t$.The vertical angle is $60^{\circ}$,so the semi-vertical

Vedclass · Accepted Answer

$\frac{1}{9 \pi}$

Vedclass · Answer

(C) Let $r$ be the radius and $h$ be the height of the water level in the cone at any time $t$.The vertical angle is $60^{\circ}$,so the semi-vertical angle is $30^{\circ}$.From the geometry of the cone,$\frac{r}{h} = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$,which implies $h = \sqrt{3}r$.The volume $V$ of the water in the cone is given by $V = \frac{1}{3} \pi r^2 h$.Substituting $h = \sqrt{3}r$,we get $V = \frac{1}{3} \pi r^2 (\sqrt{3}r) = \frac{\sqrt{3}}{3} \pi r^3 = \frac{1}{\sqrt{3}} \pi r^3$.Differentiating with respect to $t$,we get $\frac{dV}{dt} = \frac{1}{\sqrt{3}} \pi (3r^2) \frac{dr}{dt} = \sqrt{3} \pi r^2 \frac{dr}{dt}$.Given $\frac{dV}{dt} = \frac{1}{\sqrt{3}}$,we have $\sqrt{3} \pi r^2 \frac{dr}{dt} = \frac{1}{\sqrt{3}}$,so $\frac{dr}{dt} = \frac{1}{3 \pi r^2}$.Since $h = \sqrt{3}r$,when $h = 3 	ext{ m}$,$r = \frac{3}{\sqrt{3}} = \sqrt{3} 	ext{ m}$.Substituting $r = \sqrt{3}$ into the expression for $\frac{dr}{dt}$,we get $\frac{dr}{dt} = \frac{1}{3 \pi (\sqrt{3})^2} = \frac{1}{3 \pi (3)} = \frac{1}{9 \pi} 	ext{ m/min}$.

The vertical angle of a right circular cone is $60^{\circ}$. If water is being poured into the cone at the rate of $\frac{1}{\sqrt{3}} \text{ m}^3/\text{min}$,then the rate $(\text{m/min})$ at which the radius of the water level is increasing when the height of the water level is $3 \text{ m}$ is

Similar Questions

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 ....... $\text{cm s}^{-1}$.

$A$ particle moves along the curve $y = x^{3/2}$ in the first quadrant in such a way that its distance from the origin increases at the rate of $11$ units per second. The value of $\frac{dx}{dt}$ when $x = 3$ is

The rate of change of the volume of the sphere with respect to its surface area $S$ is . . . . . . .

The maximum height is reached in $5$ seconds by a stone thrown vertically upwards and moving under the equation $10s = 10ut - 49t^2$,where $s$ is in metres and $t$ is in seconds. The value of $u$ is ........ $m/sec$.

If water is poured into a cylindrical tank of radius $3.5 \ ft$ at the rate of $1 \ ft^3/min$,then the rate at which the level of the water in the tank increases (in $ft/min$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module