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
Pick an option to check your answer.
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 \(1\; sec\)
\( = \frac{{2 + 1 + 3 + 0}}{4} = 1.5\)
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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