A) \[\frac{1}{\sqrt{3}}\]
B) 1
C) \[\sqrt{\frac{4}{3}}\]
D) \[\sqrt{\frac{3}{2}}\]
Correct Answer: D
Solution :
Time period of simple pendulum is given by \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\] and time period of uniform rod in given position is given by \[{{T}_{2}}=2\pi \sqrt{\frac{inertia\text{ }factor}{spring\text{ }factor}}\] Here, inertia factor = moment of inertia of rod at one end \[=\frac{m{{l}^{2}}}{12}+\frac{+m{{l}^{2}}}{4}\] \[=\frac{m{{l}^{2}}}{3}\] Spring factor = restoring torque per unit angular displacement \[=mg\times \frac{1}{2}\frac{\sin \theta }{\theta }\] \[=mg\frac{1}{2}\] (if \[\theta \] is small) \[\therefore \] \[{{T}_{2}}=2\pi \sqrt{\frac{m{{l}^{2}}/3}{mgl/2}}=2\pi \sqrt{\frac{2}{3}\frac{l}{g}}\] Hence, \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{3}{2}}\]You need to login to perform this action.
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