1.

The period of oscillation of a simple pendulum is T=2pisqrt(L/(g)) . L is about 10 cm and is known to 1 mm accurancy. The period of oscillation is about 0.5 second. The time of 100 oscillation is measured with a wrist watch of 1 s resolution. What is the accurancy in the determination of g ?

Answer»

`3%`
`2%`
`5%`
`4%`

Solution :`(DELTAG)/(g)=(DeltaL)/(L)+2XX(DeltaT)/(T)`
In TERMS of percentage,
`(Deltag)/(g)xx100=(DeltaL)/(L)xx100+2xx(DeltaT)/(T)xx100`
Percentage error in `L=100xx(DeltaL)/(L)=100xx(0.1)/(10)=1%`
Percentage error in `T=100xx(DeltaT)/(T)=100xx(1)/(50)=2%`
Percentage error in `g=100(Deltag)/(g)=1%+2xx2%=5%`
`(c )` is CORRECT.


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