JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
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
- A \(92\pm 2\;s\)
- B \(92\pm 3\;s\)
- C \(92\pm 1.8\;s\)
- D \(92\pm 5\;s\)
Answer & Solution
Correct Answer
(A) \(92\pm 2\;s\)
Step-by-step Solution
Detailed explanation
\(\Delta T = \frac{{\left[ {\Delta {T_1}| + |\Delta {T_2}| + |\Delta {T_3}| + |\Delta {T_4}} \right]}}{4}\) \( = \frac{{2 + 1 + 3 + 0}}{4} = 1.5\) As the resolutions of measuring clock is \(1.5\) therefore the mean time should be \(92\) pm \(1.5\) but Least count is \(1\; sec\)…
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