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A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations. For this observation, which of the following statements is true? |
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Answer» Error `DeltaT` in measuring T, the time PERIOD, is0.02 SECONDS. `(Deltat)/t=(1s) /(40S)=1/40` Time period , `T=(40 s)/20=2s` Error in measurement of time period `DeltaT=T xx (Deltat)/t= 2 s xx 1/40`= 0.05 s The time period of simple PENDULUM is `T=2pisqrt(l/g)` or `T^2=(4pi^2l)/g` or `g=(4pi^2l)/T^2` `therefore (Deltag)/g=(2DELTAT)/T=2xx1/40 =1/20 (because (DeltaT)/T = (Deltat)/t)` Percentage error in determination of g is `(Deltag)/g xx 100=1/20 xx 100`=5% |
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