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:

92 \pm 2 s

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92 \pm 5.0 s

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92 \pm 1.8 s

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92 \pm 3 s

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Solution

The correct option is A $92 \pm 2 s$
Sum of all observation= $90 + 91 + 95 + 92 = 368$
Average $= 368 / 4 = 92$

Sum of modulus errors= $2 + 1 + 3 = 6$
Average error $= 6 / 4 = 1.5$ ,rounded of $2$ .
Final answer $= (92 \pm 2) s$

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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:

The correct option is A 92±2 sSum of all observation= 90+91+95+92=368Average =368/4=92Sum of modulus errors=2+1+3=6Average error =6/4=1.5,rounded of 2.Final answer =(92±2) s

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:

The correct option is A $92 \pm 2 s$
Sum of all observation= $90 + 91 + 95 + 92 = 368$
Average $= 368 / 4 = 92$

Sum of modulus errors= $2 + 1 + 3 = 6$
Average error $= 6 / 4 = 1.5$ ,rounded of $2$ .
Final answer $= (92 \pm 2) s$