Question
Four persons measure the length of a rod as $20.00\text{ }cm,19.75\text{ }cm,17.01\text{ }cm$ and 18.25 cm . The relative error in the measurement of average length of the rod is :
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Solution
$\mathcal{l}_{\text{mean~}} = \frac{\mathcal{l}_{1} + \mathcal{l}_{2} + \mathcal{l}_{3} + \mathcal{l}_{4}}{4}$
$${\mathcal{l}_{\text{mean~}} = \frac{20.00 + 19.75 + 17.01 + 18.25}{4} }{= 18.75 }{\Delta\mathcal{l}_{\text{mean~}} = \frac{\left| \Delta\mathcal{l}_{1} \right| + \left| \Delta\mathcal{l}_{2} \right| + \left| \Delta\mathcal{l}_{3} \right| + \left| \Delta\mathcal{l}_{4} \right|}{4} }{= \frac{1.25 + 1 + 1.74 + 0.5}{4} = 1.12 }$$So, relative error
$$= \frac{\Delta\mathcal{l}_{\text{mean~}}}{\mathcal{l}_{\text{mean~}}} = \frac{1.12}{18.75} = 0.06$$
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