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The period of oscillation of a simple pe...

The period of oscillation of a simple pendulum is given by `T=2pisqrt((l)/(g))` where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

A

`0.4%`

B

`0.1%`

C

`0.3%`

D

`0.2%`

Text Solution

Verified by Experts

The correct Answer is:
D
`(Deltag)/(g)xx100={(Deltal)/(l)+2((DeltaT)/(T))}100`
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Knowledge Check

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