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))`

Options

A. 0.73 J
B. 1 J
C. 0.54 J
D. 1.73 J

Answer

a

Explanation

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`