The correct choice is (a) 107 mm
For explanation: Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m
Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,
n = 20/40 = 0.5 Hz
Let kG = Radius of gyration of the connecting rod.
We know that frequency of oscillation (n),
0.5 = 1/2πkG √gxy/l = 1/2πkG √9.81 x 0.12 x 0.12/1.25 = 0.0535/k
kG = 0.0535/0.5 = 0.107 m = 107 mm.