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A student uses a simple pendulum of exactly `1m` length to determine `g`, the acceleration due ti gravity. He uses a stop watch with the least count of `1sec` for this and record `40 seconds` for `20` oscillations for this observation, which of the following statement `(s) is (are)` true?A. Error `DeltaT` in measuring T, the time period, is 0.05 sB. Error `DeltaT` in measuring T, the time period is, 1 sC. Percentage error in the determination of g is 5%D. Percentage error in the determination of g is 2.5 % |
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Answer» Correct Answer - a, c `T=40/20=2s` `(DeltaT)/T=(Deltat)/t=1/40` `rArr DeltaT=T/40=2/40=0.05 s` `g=(4pi^(2)Ln^(2))/t^(2)" "`where `t=nT` `(Deltag)/g=(2 Delta t)/t` `rArr %` error `=(2 Delta t)/txx100=5%` |
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