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The acceleration due to gravity is deter...

The acceleration due to gravity is determined by using a simple pendulum of length `l = (100 pm 0.1)cm`. If its time period is `T = (2 pm 0.01)s`, find the maximum percentage error in the measurement of g.

Text Solution

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The time period of oscillation of a simple pendulum is given by
`T = 2pi sqrt(l/g)`
Squaring both sides
`T^2 = 4pi//g`
`:. G = 4pi l/T^2`
Now `Deltal = 0.1, l = 100 cm, DeltaT = 0.01 s, T = 2s`
Precentage error ` = (Deltag xx100)/(g)`
` =((Deltal)/l + (2 DetlaT)/T) xx 100`
` = (0.1/100 + (2 xx 0.01)/2) xx 100`
` = (0.001 + 0.01) xx 100 = 1.1`
Prcentage error in measurement of g is 1.1%
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