A small body of mass m is released from rest at the point A on the smooth vertical track ABC. AB is a quarter circle of radius R and BC is a semi circle of radius 0.5R. Then,

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The height of the point above B where the particle crosses the line BC is 25R27

The height of the point above B where the particle crosses the line BC is 25R32

The magnitude of force applied by the particle on the track near point B is 3mg

The magnitude of force applied by the particle on the track near point B is 5mg

answer is B.

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Detailed Solution

Applying conservation of energy from A to B, mgR=12mvB2 ⇒vB=2gRAt point B, NB-mg=mvB20.5R ⇒NB=5mgThe particle leaves the track and follows a parabolic path. Assume it leaves track at a point P as shown in figure.Applying conservation of energy from B to P, 12mvB2=12mvP2+mg0.5R1+sinθ ...........(1)FBD at point P, NP+mgsinθ=mvP20.5RIf it leaves track, NP=0 ⇒mgsinθ=mvP20.5R.............(2)Solving (1) and (2) , sinθ=23 and vP=Rg3Using, equation of trajectory, y=xtanα1-xRange here, tanα=52Range=vP2sin2αg=45R27x=0.5Rcosθ=5R6After substitution, we get ⇒y=-5R96∴Distance above point B = 0.5R+0.5Rsinθ+-5R96=25R32

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