A block is freely sliding down from a vertical height `4 m` on smooth inclined plane. The block reaches bottom of inclined plane and then it decribes vertical circle of radius `1 m` along smooth track. The ratio of normal reactions on the block while it crossing lowest point and highest point of vertical circle is

Options

A. `6 :1`
B. `5 : 1`
C. `3 : 1`
D. `5 : 2`

Answer

`(3)`

Explanation

Let `v_(2), v_(1)` be the velocities at lowest point, highest point of vertical circle.
`v_(2)^(2) = 2gh = 2g xx 4 = 8g`
`v_(2)^(2) - v_(1)^(2) = 4gr = 4g xx 1 = 4g` `therefore v_(1)^(2) = 4g`
normal reaction at lowest point , `R_(2) = (mv_(2)^(2))/(r) + mg`,
At highest point, `R_(1) = (mv_(1)^(2))/r - mg`