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A body of mass 10 kg slides a long a rou...

A body of mass 10 kg slides a long a rough horizontal surfce. The coefficient of friction is `1//sqrt(3)`. Taking `g=10m//s^(2)`. The least force which acts at an angle of `30^(@)` to the horizontal is

A

25 N

B

100 N

C

50 N

D

`(50)/(sqrt3)` N

Text Solution

Verified by Experts

The correct Answer is:
C
`mu = (1)/(sqrt3)`

`f = mu (mg - P sin theta)`
`P cos 30^(@) = (1)/(sqrt3) (mg -P sin 30^(@))`
`P "" (sqrt3)/(2) = (1)/(sqrt3) (100 - P xx (1)/(2))`
`implies (3P)/(2) = 100 - (P)/(2) implies P = 50 N`
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