Physics And Measurement Question 189

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

Options:

A) 92±2s

B) 92±5.0s

C) 92±1.8s

D) 92±3s

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Measured time period of 100 oscillations are 90 sec, 91 sec, 95 sec and 92 sec.

Mean value of time= tm=90+91+95+924=92sec Absolute error in measurement |Δt1|=|tmt1|=2sec

|Δt2|=|tmt2|=1sec

|Δt3|=|tmt3|=3sec

|Δt4|=|tmt4|=0sec

Mean absolute error Δtmean=2+1+3+04=1.5sec

But the least count of the measuring clock is 1 sec, so it cannot measure up to 0.5 second, so we have to round it off.

So mean error will be 2 second. Hence men time (92±2sec) .



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