The height of the cone of maximum volume insc | vedclass

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

Question

(C) Let the height of the cone be $h$ and the radius of the base of the cone be $r$. Given,the radius of the sphere is $R$. In the right-angled triang

Vedclass · Accepted Answer

$\frac{4 R}{3}$

Vedclass · Answer

(C) Let the height of the cone be $h$ and the radius of the base of the cone be $r$. Given,the radius of the sphere is $R$. In the right-angled triangle $	riangle OPB$,where $O$ is the center of the sphere,$P$ is the center of the base of the cone,and $B$ is a point on the circumference of the base of the cone: $R^2 = r^2 + (h - R)^2$ $r^2 = R^2 - (h - R)^2 = R^2 - (h^2 - 2Rh + R^2) = 2Rh - h^2$. The volume of the cone $V$ is given by: $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2Rh - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$. To find the maximum volume,differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2)$. Setting $\frac{dV}{dh} = 0$: $\frac{\pi}{3} h(4R - 3h) = 0$. Since $h 
eq 0$,we have $h = \frac{4R}{3}$. Checking the second derivative: $\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6h)$. At $h = \frac{4R}{3}$,$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6(\frac{4R}{3})) = \frac{\pi}{3} (4R - 8R) = -\frac{4\pi R}{3} Since the second derivative is negative,the volume is maximum at $h = \frac{4R}{3}$.

The height of the cone of maximum volume inscribed in a sphere of radius $R$ is

Similar Questions

What is common between chloroplasts,chromoplasts and leucoplasts?

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents

If $C$ is the midpoint of $AB$ and $P$ is any point outside $AB$,then:

$A$ and $B$ are two concentric circular loops carrying currents $i_1$ and $i_2$ as shown in the figure. If the ratio of their radii is $1:2$ and the ratio of the magnetic flux densities at the centre $O$ due to $A$ and $B$ is $1:3$,then the value of $\frac{i_1}{i_2}$ will be:

In which of the following molecules/ions are all the bonds not equal?

Vedclass Test Series

Exam Paper Generator

Online Exam Module