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 = 2π√lg , then   = 0 is minimum and  = 0 is maximum

Solution

## The correct option is B is minimum Given △l = 0.1cm and △t = 0.1s. T=2π√Lg ⇒g=4π21T2 ⇒△gg=△ll+2△TT (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>T3, △l2=△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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