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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2016
  • Medium
  • MCQ

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

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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 \(1\; sec\)

then answer should be \(92\, \pm \,2\;\;sec\)

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