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Question

Figure shows a smooth vertical circular track of radius R. A block slides along the surface AB when it is given a velocity equal to 6gR at point A. The ratio of the force exerted by the track on the block at point A to that at point B is


A
0.25
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B
0.35
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C
0.45
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D
0.55
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Solution

The correct option is D 0.55
Free body diagram at A and B shown is figure


Solving from block's frame of reference, we have to assume a centrifugal force of mv2R.
At point A: NA+mgcos60=mv2AR
NA=m(6gR)2/Rmgcos60=5.5mg

Using energy conservation between A and B, taking B as a reference
12mv2A+mgR(1+cos60)=12mv2B
v2B=6gR+3gRv2B=9gR

At point B, NB=mg+mv2BR
NB=10mg
Therefore, NANB=5.5mg10mg=0.55

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