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

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

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

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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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 $\sqrt{6 gR}$ 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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0.25

0.45

0.55

0.35

answer is D.

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Detailed Solution

$\begin{array}{l} N_{A} + mgcos 60^{\circ} = \frac{{mv}_{A}^{2}}{R} \\ N_{A} = \frac{{mv}_{A}^{2}}{R} - mgcos 60^{\circ} \\ = 6 mg - 0 .5 mg = 5 .5 mg \\ \frac{1}{2} {mv}_{A}^{2} + mgR (1 + \cos 60^{\circ}) = \frac{1}{2} {mv}_{B}^{2} \end{array}$

$\begin{array}{l} v_{A}^{2} + 2 gR (1 + \frac{1}{2}) = v_{B}^{2} \\ 6 gR + 3 gR = v_{B}^{2} \\ v_{B}^{2} = 9 gR \Rightarrow v_{B} = \sqrt{9 gR} \\ N_{B} = mg + \frac{{mv}_{B}^{2}}{R} = 10 mg \end{array}$