1.

A simple pendulum of length 1 m has a bob of 200 g suspended to a fixed point. It is displaced through 60^(@) and then released. What is K.E. when its inclination is 30^(@) with the vertical ? (g = 10 ms^(-2))

Answer»

0.73 J
1 J
0.54 J
1.73 J

Solution :In rt `DeltaODB, COS 60^(@)=(l-h)/(l)impliesh=l-l cos60^(@)`

Similarly,`h-l-l cos 30^(@)`
`:.` K.E. = LOSS of P.E.
=mgh-mgh.
=`MG(h-h.)=mg[l-l cos 60^(@)-l+l cos 30^(@)]`
=`mgl[cos 30^(@)-cos 60^(@)]`
=`0.2xx10xx1[(1.732-1)/(2)]0.732 J`


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