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A uniform rod of length 2.0 m is suspend...

A uniform rod of length 2.0 m is suspended through an end and is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately

A

1.60 s

B

1.80 s

C

2.0 s

D

2.40 s

Text Solution

Verified by Experts

The correct Answer is:
D
Time period for physical pendulum-
`T=2pisqrt((I)/((Mgl//2)))`
Here, `I=(Ml^(2))/(3),T=2pisqrt((2l)/(3g))`
`T=2pisqrt((2xx2)/(3xx9.8))=2.32sec implies T=2.4 sec`. (Approx)
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