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The period of oscillation of a simple pendulum is `T = 2pisqrt((L)/(g))`. Meaured value of `L` is `20.0 cm` know to `1mm` accuracy and time for `100` oscillation of the pendulum is found to be `90 s` using a wrist watch of `1 s` resolution. The accracy in the determinetion of `g` is :A. `1%`B. `5%`C. `2%`D. `3%` |
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Answer» Correct Answer - D `T=2pisqrt((L)/(g)tog=(4pi^(2)L)/(T^(2))=(4pi^(2)Ln^(2))/(t^(2))(T=(t)/(n))` Maximum percentage error in `g` `(Deltag)/(g)xx100=(DeltaL)/(L)xx100+2(Deltat)/(t)xx100` `(0.1)/(20.0)xx100+2xx(1)/(90)xx100=2.72%=3%` |
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