Student I, II and III perform an experiment for measuring the acceleratioon due to gravity (g) using a simple pendulum . Theu use different lengths of the pedulum and /or record time for different number of oscillations . The observation are shoen in the table .
Least count for length =0.1cm
Least count for time =0.1s

If `E_(I)`, `E_(II)`and `E_(III)`are the percentage errors in g,i.e
`(Deltag)/(g)xx100`for students I, II, and III respectively ,

Options

A. `E_(I)=0`
B. `E_(I)` is minimum
C. `E_(I)=E_(II)`
D. `E_(II)` is maximum

Answer

B

Explanation

Time period of a simple pendulum can be written as follows :
`T = 2pi sqrt((l)/(g))`
If it takes time t to complete n oscillations then T = t/n, hence time period can be written as
`(t)/(n) = 2pi sqrt((l)/(g))`
`rArr g = (4pi^(2)l)/(t^(2))`
Differentiating and simplifying we get the following :
`(Deltag)/(g) = (Deltal)/(l) + 2 (Deltat)/(t)`
Percentage error : `E = (Deltag)/(g) xx 100`
`E = ((Delta)/(l) + 2(Delta)/(t)) xx 100`
In above equation `Deltal = 0.1` cm and `Deltat = 0.1s `
`E_(I) = ((0.1)/(64) + 2 xx (0.1)/(128)) xx 100 = 0.31 %`
`E_(II) = ((0.1)/(64) + 2xx(0.1)/(64))xx 100 = 0.46%`
`E_(III) = ((0.1)/(20) + 2xx(0.1)/(36)) xx 100 = 1.05%`