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If the time period of a simple pendu...

If the time period of a simple pendulum is `T = 2pi sqrt(l//g)` , then the fractional error in acceleration due to gravity is

A

`(4 pi^(2)Deltal)/(DeltaT^(2))`

B

`(Deltal)/(l) - 2(Deltal)/(T)`

C

`(Delta l)/(l)+2(Delta T)/(T)`

D

None of these

Text Solution

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The correct Answer is:
c
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Knowledge Check

  • The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is

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