A) \[3\pi \sqrt{\frac{2{{l}_{0}}}{3g}}\]
B) \[4\pi \sqrt{\frac{{{l}_{0}}}{g}}\]
C) \[2\pi \sqrt{\frac{{{l}_{0}}}{3g}}\]
D) \[2\pi \sqrt{\frac{2{{l}_{0}}}{3g}}\]
Correct Answer: D
Solution :
[d] : Here the rod is oscillating about an end point O. Hence, moment of inertia of rod about the point of oscillation is\[I=\frac{1}{3}ml_{0}^{2}\] Moreover, length \[l\] of the pendulum = distance from the oscillation axis to centre of mass of rod\[={{l}_{0}}/2\] \[\therefore \]Time period of oscillation \[T=2\pi \sqrt{\frac{I}{mgl}}=2\pi \sqrt{\frac{\frac{1}{3}ml_{0}^{2}}{mg\left( \frac{{{l}_{0}}}{2} \right)}},2\pi \sqrt{\frac{2{{l}_{0}}}{3g}}\]You need to login to perform this action.
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