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Pitch of a micrometer is 1mm and it has 100 divisions on a circular scale. There is no zero error. Thickness of a pile of 50 papers is to be found out. While measuring the thickness of a paper it is observed the linear scale reading is unknown but the $25^{th}$ division of circular scale coincides with the reference line. Possible thickness of the pile isA.15.20mmB. 23.12mmC. 21.45mmD. 12.25mm

Last updated date: 17th Sep 2024
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Hint: From the given value of pitch of the micrometer, you could find the least count of the micrometer by dividing the pitch by the given number of divisions on the circular scale. Now you could recall the expression for the reading taken by a micrometer. We will get an approximated value for the thickness of a single paper as the L.S.R is unknown. Now multiplying this value by 50 gives you the thickness of the pile.

$t=\left( L.S.R+\left( C.S.C\times L.C \right) \right)\pm Z.C$

In the question we are given the pitch of the micrometer and also the number of divisions on the circular scale. We are also said that there is no zero error and also the circular scale coincidence of the micrometer while measuring the thickness of one paper is given. We are asked to find the thickness of a pile consisting of 50 papers.
Pitch of the micrometer is given as 1mm. We know that pitch of the screw gauge is the linear distance covered by the screw head in every rotation. Let us represent pitch by P and the number of divisions on the circular scale as N then, the least count (L.C) of the micrometer is given by,
$L.C=\dfrac{P}{N}$
$P=1mm$
$N=100$
$\Rightarrow L.C=\dfrac{1mm}{100}=0.01mm$ ……………………. (1)
We know that measurement made by the micrometer is given by,
$t=\left( L.S.R+\left( C.S.C\times L.C \right) \right)\pm Z.C$
Where, t = thickness measured
C.S.C = circular scale coincidence
Z.C = zero correction
But, in the question we are said that the L.S.R is unknown and also, the zero correction is absent. We are said that 25th division of circular scale coincide with reference line, so,
$C.S.C=25$
From (1),
$C.S.C\times L.C=25\times 0.01mm$
$C.S.C\times L.C=0.25mm$
So the thickness of a single paper,
$t\approx 0.25mm$
For 50 papers the thickness of the pile will be,
${{t}_{pile}}\approx 0.25\times 50mm$
${{t}_{pile}}\approx 12.5mm$
Therefore, among the given options we could say that the possible thickness of the pile is 12.25mm.

So, the correct answer is “Option D”.

Note: While solving we have approximated the linear scale reading as zero. Though we are only told that the L.S.R is unknown, we know that the thickness of the paper will surely be less than 1mm. So, we have accordingly taken the value for L.S.R and then chose the nearest value among the given options for the thickness of the pile with 50 such papers.