Simple Harmonic Motion 3 Question 4
4. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is $90 s, 91 s, 92 s$ and $95 s$. If the minimum division in the measuring clock is $1 s$, then the reported mean time should be
(2016 Main)
(a) $(92 \pm 2 s)$
(c) $(92 \pm 1.8 s)$
(b) $(92 \pm 5 s)$
(d) $(92 \pm 3 s)$
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Answer:
Correct Answer: 4. (a)
Solution:
- True value $=\frac{90+91+95+92}{4}=92$
Mean absolute error
$$ \begin{aligned} & =\frac{|92-90|+|92-91|+|92-95|+|92-92|}{4} \\ & =\frac{2+1+3+0}{4}=1.5 \end{aligned} $$
$$ \text { Value }=(92 \pm 1.5) $$
Since, least count is $1 sec$
$\therefore \quad$ Value $=(92 \pm 2 s)$
