Two uniform solid cylinders A and B each of mass 1 kg are connected by a spring of constant `200Nm^(-1)` at their axles and are placed on a fixed wedge as shown in fig.There is no friction between cylinders and wedge. The angle made by the line AB with the horizontal, in equilibirum is

Options

A. `0^(@)`
B. `45^(@)`
C. `30^(@)`
D. `60^(@)`

Answer

C

Explanation

`mgsin60^(@)=kxcos(60^(@)-theta)` .(1)
`mgsin30^(@)=kxcos(30+theta)` .(2)
`=kxsin((pi)/(theta)-(30+theta))`
`=kxsin(60^(@)-theta)` … (3)
From (1) and (3)
`tan(60^(@)-theta)=(sin30^(@))/(sin60^(@))=(1)/(sqrt(3))impliestheta=30^(@)`