Correct Answer - D
Let rod is displaced by an angle `theta`, taking torque about edge of rod
`k_(b)bthetaxxbcostheta+k_(2)lthetaxxlcostheta=-(Id^(2)theta)/(dt^(2))`
Here, `I=(ml^(2))/(3) and theta` is small `costheta~~1`
`therefore(ml^(2)alpha)/(3)+(k_(1)b^(2)+k_(2)l^(2))theta=0`
`impliesalpha=(-3(k_(1)b^(2)+k_(2)l^(2))theta)/(ml^(2))`
`thereforeomega=sqrt((3(k_(1)b^(2)+k_(2)l^(2)))/(ml^(2)))`
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/16161153872068281384RDb2FUkQzQgzvmSz.png)