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Q.

Figure shows a smooth vertical circular track AB 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

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a

0.25

b

0.35

c

0.45

d

0.55

answer is D.

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Detailed Solution

NA+mgcos⁡60∘=mvA2RNA=mvA2R−mgcos⁡60∘ =6mg−0.5mg=5.5mg12mvA2+mgR1+cos⁡60∘=12mvB2vA2+2gR1+12=vB26gR+3gR=vB2vB2=9gR⇒vB=9gRNB=mg+mvB2R=10mg
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