A student measures the time period of 100 oscillations of a simple...

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:

Options

A $92 \pm 1.8s$
B $92 \pm 3s$
C $92 \pm 2s$
D $92 \pm 5.0s$

Correct answer

C. $92 \pm 2s$

Step-by-step solution

T	${\text{T}}_{\text{s}}$	${\text{T}}_{\text{i}} - \text{T}$
${\text{t}}_{1}$	90	- 2
${\text{t}}_{2}$	91	- 1
${\text{t}}_{3}$	95	3
${\text{t}}_{4}$	92	0
${\text{T}}_{\text{i}}$	92	$\dfrac{Σ{\|T}_{i} - T\|}{N} = 1.5$

{T}_{r} = 9 2 \pm \text{1.5}

{T}_{r} = 9 2 \pm 2

Because least count of clock is 1s

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