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In an experiment to determine accelerati...

In an experiment to determine acceleration due to gravity, the length of the pendulum is measured as 98 cm be a scale of least count of 1 cm. The period of swing/oscillations is measured with the help of a stop watch having a least count of 1s. The time period of 50 oscillation is found to be 98 s. Express value of g with proper error limits.

Text Solution

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As `T=2pi sqrt((l)/(g))` Now, time period of 50 oscillation is 98 s
`:.` Time period of one oscillation is `(98)/(50) = 1.96 s`
As `T=2 pi sqrt((l)/(g))` We have `g =(4 pi^(2)l)/(T^(2))=4xx(3.14)^(2)xx(0.98)/(1.96xx1.96)=10.06 m//s^(2)`
Let us find the permissible error in the measurement.
As `g=(4pi^(2)l)/(T^(2))` We have `(Deltag)/(g)=(Deltal)/(l)+(2DeltaT)/(T),Deltag=10xx((1)/(98)+(2xx1)/(98))`
(`because` Least count of meter scale is 1 cm and least count of stop watch is 1s), `Delta g = 0.3 m//s^(2)`
So, final result can be expressed as `(10.1 pm 0.3)ms^(-2)`
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