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The time period of oscillation of simple...

The time period of oscillation of simple pendulum is given by `t = 2pisqrt(l/g)` What is the accurancy in the determination of 'g' if 10cm length is known to 1mm accuracy and 0.5 s time period is measured form time of 100 oscillations with a watch of 1 sec. resolution.

Text Solution

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Here `(Deltal)/(l)=(0.1)/(10)`
`Deltat=1` second, and
time of 100 oscillations, `t=100xx0.5=50s`
`therefore" "(Deltat)/(t)=(1)/(50)`
From`" "t^(2)=4pi^(2)(l)/(g)`
`t^(2)=4pi^(2)(l)/(g)" "g=4pi^(2)(l)/(t^(2))`
`therefore" "(Deltag)/(g)=pm ((Deltal)/(l)+2(Deltat)/(t))`
`therefore" Error, "(Deltag)/(g)xx100=pm((0.1)/(10)+(2xx1)/(50))xx100=pm5%`
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