THE_FREQUENCY_OF_OSCILLATION_OF_A_TORSIONAL_PENDULUM_IS?$ – McqOptions

MCQOPTIONS

1.	THE_FREQUENCY_OF_OSCILLATION_OF_A_TORSIONAL_PENDULUM_IS?$
A.	2‚âà√¨‚àö√ëk<sub>G</sub>/r ‚Äö√Ñ√∂‚àö‚Ä†‚àö‚àÇg/I
B.	r/2‚âà√¨‚àö√ëk<sub>G</sub>‚Äö√Ñ√∂‚àö‚Ä†‚àö‚àÇg/I
C.	2‚âà√¨‚àö√ëk<sub>G</sub>/r‚Äö√Ñ√∂‚àö‚Ä†‚àö‚àÇI/g
D.	r/2‚âà√¨‚àö√ëk<sub>G</sub>‚Äö√Ñ√∂‚àö‚Ä†‚àö‚àÇI/g
Answer» C. 2‚âà√¨‚àö√ëk<sub>G</sub>/r‚Äö√Ñ√∂‚àö‚Ä†‚àö‚àÇI/g

Discussion

No Comment Found

Related MCQs

In SHM, what is the phase difference between velocity and acceleration?
In an SHM the time taken to go from mean position to A/2 is the same as that from A/2 to A. True or False? Here, A is the amplitude of motion.
What is the amplitude of motion for x = 2sin(2t) + 4sin2t ?
A particle starts from the extreme position, at t = 0, in a SHM. If the time period of motion is 2s & maximum speed is 5m/s, find the equation of motion.
What is the relation between time periods of the two given SHMs.
A particle has an equation of motion given by: x = cos2wt – sin2wt. Select the correct statement regarding the same.
2 particles undergoing SHM start from the mean position and go in opposite directions. Particle 1 starts with a speed of 10m/s and particle 2 starts with a speed 0f 5m/s. If the amplitude(=10cm) is the same. At what position will they first meet?
The displacement vs time graphs of 2 SHMs are given below. Which parameter is the same for both of them?
A particle is undergoing SHM with amplitude 10cm. The maximum speed it achieves is 1m/s. Find the time it takes to reach from the mean position to half the amplitude.
A particle is initially at the centre and going towards the left. Let T be the time period of the SHM it is undergoing. What will be its position and velocity at time 3T/4, if it starts from the centre at t=0?