JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 20)
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier
scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm. The average of
these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The
average diameter
of the pencil should therefore be recorded as :
of the pencil should therefore be recorded as :
(5.54 $$ \pm $$ 0.07) mm
(5.5375 $$ \pm $$ 0.0740) mm
(5.5375 $$ \pm $$ 0.0739) mm
(5.538 $$ \pm $$ 0.074) mm
Explanation
Given, dav = 5.5375 mm
$$\Delta $$d = 0.07395 mm
Significant rule says that reading should has same significant figure as that of reading given.
$$ \because $$ Measured data are up to two digits after decimal.
$$ \therefore $$ 5.5375 rounded to $$ \to $$ 5.54
$$\Delta $$d = 0.07395 mm
Significant rule says that reading should has same significant figure as that of reading given.
$$ \because $$ Measured data are up to two digits after decimal.
$$ \therefore $$ 5.5375 rounded to $$ \to $$ 5.54
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