Simple Harmonic Motion 3 Question 4

4. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s,91s,92s and 95s. If the minimum division in the measuring clock is 1s, then the reported mean time should be

(2016 Main)

(a) (92±2s)

(c) (92±1.8s)

(b) (92±5s)

(d) (92±3s)

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Answer:

Correct Answer: 4. (a)

Solution:

  1. True value =90+91+95+924=92

Mean absolute error

=|9290|+|9291|+|9295|+|9292|4=2+1+3+04=1.5

 Value =(92±1.5)

Since, least count is 1sec

Value =(92±2s)



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