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 isA. `0.1%`B. `1%`C. `0.2%`D. 0.8%`

The period of oscillation of a simple pendulum is given by `T=2pisqrt(

1.	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 isA. `0.1%`B. `1%`C. `0.2%`D. 0.8%`
Answer» Correct Answer - c `T = 2 pi sqrt(1//g)rArr T^(2) = 4 pi^(2) 1//g rArr g = (4 pi^(2)l)/(T^(2))` `Here % error in l = (1 mm)/(100cm) xx 100 = (0.1)/(100) xx 100 = 0.1%` and `% "error in" T = (0.1)/(2 xx 100) xx 100 = 0.05%` `:. % "error in" g = % "error in" l + 2(% "error in" T)`.

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