An object of mass $^{'} m^{'}$ is moving in a vertical circle of radius $^{'} R^{'}$ what is the work done on the body. it moves when from lowest point of circle to highest point of circle

π m g R

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\frac{π m g}{R^{2}}

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2 π m g R

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2 R m g

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Solution

The correct option is D $2 R m g$
Let the diameter of the circle be reference point,

Work done= $(P_{f} - P_{i})$

Work done= $M g R - (- M g R) = M g (R - (- R)) = 2 M g R$

Option $D$ is the correct answer

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An object of mass ′m′ is moving in a vertical circle of radius ′R′ what is the work done on the body. it moves when from lowest point of circle to highest point of circle

The correct option is D 2RmgLet the diameter of the circle be reference point,Work done=(Pf−Pi)Work done=MgR−(−MgR)=Mg(R−(−R))=2MgROption D is the correct answer

An object of mass $^{'} m^{'}$ is moving in a vertical circle of radius $^{'} R^{'}$ what is the work done on the body. it moves when from lowest point of circle to highest point of circle

The correct option is D $2 R m g$
Let the diameter of the circle be reference point,

Work done= $(P_{f} - P_{i})$

Work done= $M g R - (- M g R) = M g (R - (- R)) = 2 M g R$

Option $D$ is the correct answer