A bob of mass M is suspended by a masslessstring of length L . The horizontal velocityat position 4 is just sufficient to make itreach the point The angle θ at which thespeed of the bob is half of that at A, satisfiesa)theta=pi/4 b)pi/4

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Arun · Answer

If you see in the diagram, You get (1/2)* 5 mgl = (1/2)* m* (5gl/4) + mgl (1- cos&theta;) On solving Cos &theta; = - 7/8 Hence &theta; > 3&pi;4 &theta;&pi; Hope it helps

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A bob of mass M is suspended by a masslessstring of length L . The horizontal velocityat position 4 is just sufficient to make itreach the point The angle θ at which thespeed of the bob is half of that at A, satisfiesa)theta=pi/4 b)pi/4

A bob of mass M is suspended by a masslessstring of length L . The horizontal velocityat position 4 is just sufficient to make itreach the point The angle θ at which thespeed of the bob is half of that at A, satisfiesa)theta=pi/4 b)pi/4

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A bob of mass M is suspended by a masslessstring of length L . The horizontal velocityat position 4 is just sufficient to make itreach the point The angle θ at which thespeed of the bob is half of that at A, satisfiesa)theta=pi/4 b)pi/4

A bob of mass M is suspended by a masslessstring of length L . The horizontal velocityat position 4 is just sufficient to make itreach the point The angle θ at which thespeed of the bob is half of that at A, satisfiesa)theta=pi/4 b)pi/4

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