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Question

Physics Question on physical world

A student measures the time period of 100100 oscillations of a simple pendulum four times. The data set is 90s,91s,95s90\, s, 91\, s, 95\, s and 92s92\, s. If the minimum division in the measuring clock is 1s1 \,s, then the reported mean time should be :

A

92±2s92 \pm 2\, s

B

92±5.0s92 \pm 5.0 \,s

C

92±1.8s92 \pm 1.8 \,s

D

92±3s92 \pm 3 \,s

Answer

92±2s92 \pm 2\, s

Explanation

Solution

x=xiN=90+91+95+924=92\overline{x} = \frac{\sum x_i}{N} = \frac{90 + 91+ 95 + 92}{4} = 92
Mean deviation = xˉxiN=2+1+3+04=1.5\frac{\sum |\bar{x} - x_i|}{N} = \frac{2 + 1+ 3 +0}{4} = 1.5
L.C.=1sL.C. = 1\,s.
\therefore Required value =92±2s= 92 \pm 2\, s