question_answer

Time period of a simple pendulum of length L is \[{{T}_{1}}\] and time period of a uniform rod of the same length L pivoted one end oscillating in a vertical plane is\[{{T}_{2}},\]. Amplitude of oscillation in both the case is small. Then, \[{{T}_{1}}/{{T}_{2}}\] is:

A)  \[\sqrt{\frac{4}{3}}\]                        

B)  \[1\]

C)  \[\sqrt{\frac{3}{2}}\]                        

D)  \[\sqrt{\frac{1}{3}}\]

Correct Answer: C

Solution :

            As we know that the centre of mass of an uniform is at point situated at its half length             \[{{T}_{1}}=2\pi \sqrt{\frac{L}{g}}\] and \[{{T}_{2}}=2\pi \sqrt{\frac{1}{mg\,L/2}}\] (physical pendulum) Here, I moment of inertial \[=\frac{m{{L}^{2}}}{3}\]             \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{3}{2}}\]