Question

Student `I ,II , and III` perform an experiment for measuring the acceleration due to gravity `(g) ` usinf a simple pendulum. They use lengths of the pendulum and // or record time for different number of oscillations . The observations are shown in the following table . Least count for length ` = 0.1 cm`
`{:(underset(III)underset(II)underset(I)underset()underset()("Student"),underset(20.0)underset(64.0)underset(64.0)underset((cm))underset("Pendulam")("Length of "),underset(4)underset(4)underset(8)underset((n))underset("n Oscillation")("Number of"),underset(9.0)underset(16.0)underset(16.0)underset((s))underset("Period")("Time")):}`
Least count for time `= 0.1 s`.
If `E_(I), E_(II)` , and `E_(III)` are the percentage errors in `g` , i.,e., `((Delta g)/( g) xx 100)` for students I,II , and III, respectively , then
A. `E_(t) =0`
B. `E_(I)` is minimum
C. `(E_(I)=E_(II)`
D. `E_(II)` is maximum

Answer 1

The least count of legnth `Deltal = 0.1 cm`
The least count of time `Deltat = 0.1 s`
`%` error of `g = (Deltag)/(g) xx 100`
`T=2pisqrt((l)/(g))rArrg=(4pi^(2)l)/(T^(2))=(4pi^(2)l)/(t^(2)).n^(2)`
`(Deltag)/(g)=(Deltal)/(l)+2(Deltat)/(t)`
For student `I(100 xx (Deltag)/(g))_(I)=((0.1)/(64.0)+(2xx0.1)/(128.0))xx100`
`E_(I)=(0.2)/(64.0)xx100=(20)/(64)`
For student `(100 xx (Deltag)/(g))_(II) = ((0.1)/(64.0) + 2 xx (0.1)/(64.0)) xx 100`
`E_(II) = (0.3)/(64.0) xx 100 = (30)/(64)`
For student III `(100 xx (Deltag)/(g))_(III) = ((0.1)/(20.0) + 2 xx (0.1)/(36.0))xx100`
`E_(III) = ((0.1)/(20.0)+(0.1)/(18.0))xx 100 = (19)/(18)`
`E` is least .