A cylindrical gas container is closed at the top and open at the bottom. If the iron plate of the top is 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 3:4 (b) 5:6 (c) 4:5 (d) none of these

Options

A. `3:4`
B. `5:6`
C. `4:5`
D. none of these

Answer

3

Explanation

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 `5k/4`
Capacity of the vessel = volume of cylinder
`=pix^(2)y` =V (Given)
`therefore y=v//(pix^(2))`
Now the volume `V_(1)` of the iron plate used for construction of the container is given by
`V_(1) =pi(x+k^(2))(y+5k//4)-pix^(2)y`
`therefore (dv_(1))/(dx)=2vk(x+k)xx(56pi)/(4v)-(1)/(x^(3))`
For maximum or minimum of `V_(1)dv_(1)//dx=0 or x =[4v//(5pi)]^(1//3)`
For this value of x `d^(2)v_(1)//dx^(2)` is + ve
Hence `V_(1)` is minimum when x =`[4V//(5pi)]^(1//3)`
Now x = `[4v//(5pi)]^(1//3)`
or `5pix^(3)=4V=4pix^(2)y or x//y =4//5`
Hence the required ratio is 4:5