Question

A cyclindrical gas container is closed at the top and open at the bottom, if the iron plate of the top is $\frac{5}{4}$ times as thick as the plate forming the cylindrical sides, the ratio of the radius to the height of the cylinder using minimum material for the same capacity is

• A. $3:4$
• B. $5:6$
• C. $4:5$
• D. None of these

Answer 1

The correct option is C $4:5$

Let $x$ be the radius and $y$ the height of the cylindrical gas container.Also let $k$ be the thickness of the plates forming the cylindrical sides. Therefore, the thickness of the plate forming the top will be $\frac{5k}{4}$.
Capacity of the vessel $=$ vol. of cylinder
$=\pi x^{2}y=V\left(Given\right)\Rightarrow y=V/\left(\pi x^2\right)\cdots \left(i\right)$
Volume of the material used, $V_1=\left(2\pi xy\right)k+\left(\pi x^2\right)\frac{5k}{4}$
$\Rightarrow V_1=2\pi kx\left(\frac{V}{\pi x^2}\right)+\left(\pi x^2\right)\frac{5k}{4}{V}_1=\frac{2kV}{x}+\left(\frac{5\pi k}{4}\right)x^2$
$\Rightarrow \frac{d{V}_1}{dx}=-\frac{2kV}{x^2}+\left(\frac{5\pi k}{2}\right)x$
and $\frac{d^2{V}_1}{d{x}^2}=\frac{4kV}{x^3}+\frac{5\pi k}{2}>0$
$\therefore V_1$ is minimum when $\frac{d{V}_1}{dx}=0$
$\Rightarrow V=\frac{5\pi x^3}{4}$
From $\left(i\right),y=\frac{5x}{4}$
$\Rightarrow \frac{x}{y}=\frac{4}{5}$