JEE Mains Model Paper 21

Given l = 1m
As the pendulum released from an angular position
T.E at this = PE + KE
= mgl [1 - cos60°] + 0
When it is at θ = 30° , let v is the velocity of bob
T.E = mgl [1 - cos30°] +

1

2

mv2
TE = TEf
mgl [1−1/] = mgl [1−

√3

2

] +

1

2

mv2
v = 2.68 m/s
= (mg sin30°) v
= (1) (10) (

1

2

) (2.68) = 13.4 W