The volume of the greatest cylinder which can be inscribed in a cone of height 30 cm and semi-vertical angle `30^0` is `4000pi/(sqrt(3))` (b) `400pi/3c m^3` `4000pi/(sqrt(3)c m^3` (d) none of these

Options

A. `4000 pi//3 cm^(3)`
B. `400 pi//3 cm^(3)`
C. `4000 pi//sqrt(3) cm^(3)`
D. `4000 pi//3 cm^(3)` none of these

Answer

1

Explanation

From geometry we have `(r )/(30 tan 30^(@))=(30-h)/(30)`
or `H=30-sqrt(3r)`
Now the volume of cylinder V= `pir^(2)h=pir^(2)(30-sqrt(3r))`
Now let `(dv)/(dr)=0 or pi 60r-3sqrt(3r^(2))=0 or r=(20)/sqrt(3)`
Hence `V_(max)=pi(20)/sqrt(3)^(2)(30-sqrt(3))(20)/sqrt(3)`
`=pi(400)/(3)xx10=(4000pi)/(3)`