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Time period of a simple pendulum of leng...

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 about an end and oscillating in a vertical plane is `T_(2)`. Amplitude of osciallations in both the cases is small. Then `T_(1)/T_(2)` is

A

`n_(1)=1,n_(2)=2`

B

`n_(1)=2,n_(2)=4`

C

`n_(1)=1,n_(2)=3`

D

`n_(1)=3,n_(2)=6`

Text Solution

Verified by Experts

The correct Answer is:
C
`T=(2pir_(n))/(u_(n))=(2pir_(1)n^(2))/(u_(1)//n)`
or `T prop n^(3)`
`((T_(1))/(T_(2)))=((n_(1))/(n_(2)))^(3)=8`
`:.(n_(1))/(n_(2))=2`
Hence c is the correct answer
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