Question

A cylindrical 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. $\frac{3}{4}$
• B. $\frac{5}{6}$
• C. $\frac{4}{5}$
• D. None of these

Answer 1

Answer: (C)

$\frac{4}{5}$

We know

Volume of cylinder $V=\pi r^{2}h$

If $k$ be the thickness of the sides then that of the top will be $\frac{5}{4}k$

$\therefore S=\left(\pi rh\right)k+\left(\pi r^2\right)\frac{5}{4}$

Or $S=2\pi r.\frac{V}{pi{r}^2}k+\frac{5}{4}\pi r^{2}k$

$=k(\frac{2V}{r}+\frac{5}{4}\pi r^2)$

We have taken into consideration the actual material used and not as in last question in which the base was made from a square sheet thus wasting the material

$\therefore \frac{dS}{dr}=k(\frac{-2V}{r^2}+\frac{5}{2}\pi r)(5\pi +\frac{5}{2})=+ve$

where $r^3=\frac{4V}{5\pi }$

Or $5\pi r^3=4\pi r^{2}h$

$\therefore \frac{r}{h}=\frac{4}{5}$