Question

A weightless rigid rod with a small iron bob at the end is hinged at point A to the wall so that it can rotate in all directions. The rod is kept in the horizontal position by a vertical inextensible string of length 20 cm, fixed at its mid point. The bob is displaced slightly, perpendicular to the plane of the rod and string. The period of small oscillations of the system is found to be $\frac{\pi X}{10}sec$, then find the value of $X$. ($g=10m/s^2)$

• A. 1
• B. 3
• C. 4
• D. 2

Answer 1

The correct option is C 4

The bob will execute SHM about a stationary axis passing through AB.
Balancing the torque due to tension $T$ and weight $mg$ of the bob about hinged point A
$T\times l=mg\times 2l$
$\Rightarrow Tension,T=2mg$
When the rod is displaced horizontally, restoring torque will only be due to tension in the string which is $T\times lsin\theta =2mglsin\theta$
Since the bob is displaced slightly, $\theta$ will be small
Thus, $Torque=2mglsin\theta \approx 2mgl\theta$
Time period for oscillation where torque is represented in terms of $\theta$ is given by $T=2\pi \sqrt{\frac{I}{C}}$
where, $I$ = Moment of inertia of bob about point A
$C=\frac{Torque}{\theta }=2mgl$
$\therefore Time,T=2\pi \sqrt{\frac{m(2l{)}^2}{2mgl}}=2\pi \sqrt{\frac{2l}{g}}$
Putting the values of $l$ and $g$ we get
Time period, $T=\frac{2\pi }{5}=\frac{4\pi }{10}s$
Therefore, x is 4.