Question

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings \[5.50mm\] , \[5.55{\text{ }}mm\], \[5.45{\text{ }}mm\],\[5.65{\text{ }}mm\]. The average of these four readings is \[5.5375{\text{ }}mm\]and the standard deviation of the data is \[0.07395{\text{ }}mm\]. The average diameter of the pencil should therefore be recorded as:
(A) \[\left( {5.5375\; \pm {\text{ }}0.0739} \right)mm\;\]
(B) $(5.54 \pm 0.07)mm$
(C) $(5.5538 \pm 0.074)mm$
(D) $(5.5375 \pm 0.0740)mm$

Answer 1

Hint:In order to solve this question, we will first note down the average value of the four readings of the diameter of the pencil, and then using the general relation between average value and standard deviation, we will solve for the exact final average value of the diameter of the pencil.

Formula used:
 If ‘d’ is the average value of multiple readings of a measurement and ‘S.D’ is the standard deviation of the measurement then the average value ‘d’ will be recorded as $d' = d \pm S.D$ measured with significant figures and rounding off the digits.

Complete answer:
According to the question, we have given that out of the four readings the average value of the diameter of the pencil is given as $d = 5.5375mm$ on rounding off the digits up to two decimal places we have, $d = 5.54mm$ and the standard deviation is given to us $S.D = 0.07395mm$ on rounding off we get, $S.D = 0.07mm$

So, the average value of diameter of the pencil will be recorded as $d' = d \pm S.D$so we get,
$d' = (5.54 \pm 0.07)mm$

Hence, the correct answer is option (B) $(5.54 \pm 0.07)mm$

Note: The Vernier caliper is an instrument used to measure the linear dimensions of the objects like the inner and outer radius of the cylindrical object and it’s widely used in physics laboratories for the measurement of dimensions of the object.