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A metre stick weighing 600 g, is displaced through an angle of `60^@` in vertical plane as shown. The change in its potential energy is (`g = 10 m s^(-2)`) A. 1.5 JB. 15 JC. 30 JD. 45 J

Answer» Correct Answer - A
`DeltaU=(MgL)/2 (1-cos theta)=(600xx10xx1)/(1000xx2)xx1/2`
`DeltaU=6/4=1.5 J`


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