A solid hemisphere is mounted on a solid cyli | vedclass

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

Question

(A) Let $r$ be the common radius and $h$ be the height of the cylinder.The total surface area $S$ of the solid is the sum of the curved surface area o

Vedclass · Accepted Answer

$1: 1$

Vedclass · Answer

(A) Let $r$ be the common radius and $h$ be the height of the cylinder.The total surface area $S$ of the solid is the sum of the curved surface area of the hemisphere $(2\pi r^2)$,the curved surface area of the cylinder $(2\pi rh)$,and the base area of the cylinder $(\pi r^2)$.$S = 2\pi r^2 + 2\pi rh + \pi r^2 = 3\pi r^2 + 2\pi rh$From this,we can express $h$ in terms of $S$ and $r$:$h = \frac{S - 3\pi r^2}{2\pi r}$The volume $V$ of the solid is the sum of the volume of the hemisphere $(\frac{2}{3}\pi r^3)$ and the volume of the cylinder $(\pi r^2 h)$:$V = \pi r^2 h + \frac{2}{3}\pi r^3$Substituting the expression for $h$:$V = \pi r^2 \left( \frac{S - 3\pi r^2}{2\pi r} \right) + \frac{2}{3}\pi r^3$$V = \frac{1}{2} (Sr - 3\pi r^3) + \frac{2}{3}\pi r^3 = \frac{1}{2} Sr - \frac{3}{2}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{1}{2} Sr - \frac{5}{6}\pi r^3$To find the maximum volume,differentiate $V$ with respect to $r$ and set it to $0$:$\frac{dV}{dr} = \frac{S}{2} - \frac{5}{2}\pi r^2 = 0$$S = 5\pi r^2$Now,substitute $S = 5\pi r^2$ back into the expression for $h$:$h = \frac{5\pi r^2 - 3\pi r^2}{2\pi r} = \frac{2\pi r^2}{2\pi r} = r$Thus,the ratio of the height of the cylinder to the common radius is $h:r = 1:1$.

$A$ solid hemisphere is mounted on a solid cylinder,both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume,then the ratio of the height of the cylinder to the common radius is

Similar Questions

The function $f(x) = 2x^3 - 3x^2 - 12x + 4, x \in R$ has

The two parts of $100$ for which the sum of double of the first part and the square of the second part is minimum,are

If $f(x)=3x^3-9x^2-27x+15$,then the maximum value of $f(x)$ is $.....$

Of all the closed cylindrical cans (right circular) with a given volume of $100 \text{ cm}^3$,find the dimensions of the can that has the minimum surface area.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module