A student uses simple pendulum of length 1 m and commits an error of Δl=1mm to determine g(the acceleration due to gravity). He uses a stopwatch with the least count of 1s for this and records 40 seconds for 20 oscillations. For this obervation which of the following statement(s) is(are) true?
(1) Error in ΔT in measuring the time period T is 0.05s
(2) Error in ΔT in measuring the time period T is 1s
(3) Percentage error in the determination of g is 5.1%
(4) Percentage error in the determination of g is 2.6%
ΔT/T = 1/40 And T = 2s ---why?
Δg/g ×100 = Δl/l ×100 + 2× ΔT/T ×100
Δg/g ×100 = ((10)^(-3)/1)×100 + 2×(1/40)×100
Δg/g ×100 = 5.1%
I did not get the part where relative error was calculated as least count over observation beacuse if one is taking measurment of time the only time one can make error is either at the start or at the end and that would be human error and or or leact count error(1 sec in this case) but this was what my teacher told me to learn as a formula that error equals least count over observation. But why?
I did not get the part where in the solution the value of T is taken to be 2. Why?
Also in the solution while calulating percentage error the solution takes ΔT=1 . Why?