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 :

Units and Measurements PYQ — JEE Main 2016 · Shift 0 · Q63 with Solution

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Correct answer: B: 92 \PM 3 S

Given observations for time of $100$ oscillations: $90\,\text{s},\;91\,\text{s},\;95\,\text{s},\;92\,\text{s}$
Calculate the mean time $\bar t = \frac{90+91+95+92}{4} = \frac{368}{4} = 92\,\text{s}$
Estimate the random error from repeated measurements

For JEE-level reporting of repeated measurements, the mean absolute error is used: $\Delta t_{\text{mean}} = \frac{|90-92|+|91-92|+|95-92|+|92-92|}{4}$ $= \frac{2+1+3+0}{4} = \frac{6}{4} = 1.5\,\text{s}$
Instrument error

Minimum division of the clock $=1\,\text{s}$ .

The instrumental uncertainty is taken as $\pm 1\,\text{s}$ .
Total reported error

The reported uncertainty should account for both observational spread and instrument limitation. So, $\Delta t = 1.5 + 1 = 2.5\,\text{s}$

Since the error is to be reported appropriately, this is rounded to one significant figure: $\Delta t \approx 3\,\text{s}$
Reported mean time $t = 92 \pm 3\,\text{s}$
Check options
- A: $92\pm1.8\,\text{s}$ — does not include instrument error properly.
- B: $92\pm3\,\text{s}$ — correct.
- C: $92\pm2\,\text{s}$ — too small.
- D: $92\pm5.0\,\text{s}$ — too large.

Therefore, the correct option is B.