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A block is freely sliding down from a ve...

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

A

`6 : 1`

B

`5 : 1`

C

`3 : 1`

D

`5 : 2`

Text Solution

Verified by Experts

The correct Answer is:
C

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`
`:. 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`.
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