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If the length of a simple pendulum is do...

If the length of a simple pendulum is doubled then the `%` change in the time period is `:`

A

50

B

`41:4`

C

25

D

100

Text Solution

Verified by Experts

The correct Answer is:
B
`(DeltaT)/(T)xx100=(1)/(2)(Deltal)/(l)xx100` is not valid as `Deltal` is not small.
`(DeltaT)/(T)xx100=(1)/(2)(Deltal)/(l)xx100`
`T_(1)=2pisqrt((l)/(g))=T_(2)=2pisqrt((2l)/(g)) %` change
`=(T_(2)-T_(1))/(T_(1))xx100=(sqrt(2)-1)xx100=41.4`
`i=[(M(Rsqrt(2))^(2))/(12)+M((R)/(sqrt(2)))^(2)]xx4+mR^(2)`
`=20kgm^(2)`.
`(4M+m)g sin theta -F=(4M+m)a.`
`F.R.=I((a)/(R))`
Solving
`a=(7g)/(24)`
`F=20alemu(4M+m)g cos 30`
`muge(5)/(12sqrt(3))`
`:. mu _(m i n)=(5)/(12sqrt(3))`
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