Q.

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 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true           [2010]

1 Error ΔT in measuring T, the time period, is 0.05 seconds  
2 Error ΔT in measuring T, the time period, is 1 second  
3 Percentage error in the determination of g is 5%  
4 Percentage error in the determination of g is 2.5%  

Ans.

(1, 3)

The length of the string of simple pendulum is exactly 1 m (given), therefore the error in length Δl=0.

Further the possibility of error in measuring time is 1s in 40s as the least count of stop watch is 1s.

 Δtt=ΔTT=140

Time period T=total timeno. of oscillations=4020=2 s

 ΔTT=140  ΔT2=140  ΔT=0.05 s

Now for measuring error in g

using, T=2πlg  T2=4π2lg  g=4π2lT2

 Percentage error in g

Δgg×100=Δll×100+2ΔTT×100

 Δgg×100=0+2(140)×100=5%