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Question

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum.They use different lenghts of the pendulum and record time for different number of oscillations.The observations are shown in the table.Least count for length = 0.1cm, Least count for time = 0.1s

StudentLength of the pendulum(cm)Number of oscillations(n)Total time for (n) oscillation(s)Time period(s)I64.08128.016.0II64.0464.016.0III20.0436.09.0

If EI,  EII and EIII are the percentage errors in g, i.e. (gg×100) for student I, II and III, respectively, t = 2πlg , then

 


  1. = 0

  2. is minimum

  3. and  = 0

  4. is maximum


Solution

The correct option is B

is minimum


Given l = 0.1cm and t = 0.1s.

T=2πLg g=4π21T2 gg=ll+2TT

(As relative error gets added when quantities are multiplied or divided)

Comparing E1 and E2

l and l will be the same but,

T1<T2 as no. of observations of student 1 is higher.
So

E1<E2

Comparing E2 and E3

L2>l3, T2>T3l2=l3, as least count is same , also T2=T3 as number of observations is same.

E2<E3 

E1<E2<E3

E1 is minimum.

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