Question

CASE 1. A spherometer has a least count of 0.005mm and its head scale is divided into 200 equal divisions. Let the distance between consecutive threads on the spherometer screw be ${{d}_{1}}$ mm.
CASE 2: A vernier caliper has each division on its main scale to be 1mm. On a vernier scale, 99mm is divided into 100 divisions. Let the vernier constant be ${{d}_{2}}$ mm.
CASE 3: The pitch of a screw gauge is 1mm and there are 100 divisions on its circular scale. When no article is placed in between its jaws, the zero of the circular scale coincides with the reference line. When a steel wire is placed between the jaws, two main scale divisions are clearly visible and 67 divisions on the circular scale are observed. Let the diameter of the wire be ${{d}_{3}}$mm.
The values of ${{d}_{1}},{{d}_{2}}$ and ${{d}_{3}}$ are:
A. ${{d}_{1}}=1,{{d}_{2}}=1.99,{{d}_{3}}=0.01$
B. ${{d}_{1}}=0.01,{{d}_{2}}=9.9,{{d}_{3}}=1$
C. ${{d}_{1}}=1,{{d}_{2}}=0.01,{{d}_{3}}=2.67$
D. ${{d}_{1}}=0.01,{{d}_{2}}=2.67,{{d}_{3}}=0.01$

Answer 1

Hint: Recall the expression for pitch in terms of least count and number of divisions and hence solve the first two cases. Now recall the expression for the reading taken using a screw gauge. Now by substituting the given values in the question in that expression determine the answer to case 3.

Formula used:
Expression for pitch (P),
$P=L.C\times N$
Expression for reading (d) taken using a screw gauge,
$d=\left( M.S.R+\left( C.S.C\times L.C \right) \right)\pm Z.C$

Complete step by step answer:
In the CASE 1 we are given the least count of spherometers as 0.005mm and also the number of divisions on the head scale is also given. If ${{d}_{1}}mm$ is the distance between consecutive threads on the spherometer then, we are asked to find the value of it.
We know that the distance between consecutive threads on the spherometer is called the pitch of the device. If $L.C$is the least count of the device and $N$is the number of divisions on the head scale then pitch ${{d}_{1}}$ is given by,
${{d}_{1}}=L.C\times N$
Substituting the given values,
${{d}_{1}}=0.005mm\times 200$
$\Rightarrow {{d}_{1}}=1mm$…………………. (1)
In CASE 2 we are asked to determine the vernier constant ${{d}_{2}}$mm from the given values in the question. We know that vernier constant is actually the least count of the vernier calipers. Therefore, if$M.S.D$ is the main scale division and $V.S.D$is the vernier scale division then, the vernier constant, that is, the least count, ${{d}_{2}}$is given by,
${{d}_{2}}=1M.S.D-1V.S.D$
We are given,
$1M.S.D=1mm$
$99mm\to 100div$
$\Rightarrow 1V.S.D=\dfrac{99}{100}=0.99mm$
Therefore,
${{d}_{2}}=1-0.99=0.01mm$ ……………………….. (2)
For CASE 3, from the question, pitch (P) is 1mm and number of divisions (N) is 100 divisions. Therefore, the least count (L.C) is,
$L.C=\dfrac{P}{N}=\dfrac{1mm}{100}$
$L.C=0.01mm$
Also, the error is zero hence there is no zero correction.
$Z.C=0$
Also,
The main scale reading, $M.S.R=2mm$
The circular scale coincidence, $C.S.C=67div$
The reading of the diameter of a steel wire ${{d}_{3}}$mm taken in a screw gauge is given by,
${{d}_{3}}=\left( M.S.R+\left( C.S.C\times L.C \right) \right)\pm Z.C$
Substituting the values,
${{d}_{3}}=\left( 2+\left( 67\times 0.01 \right) \right)\pm 0$
$\Rightarrow {{d}_{3}}=2.67mm$ …………………………………. (3)
From (1), (2) and (3), we get that, ${{d}_{1}}=1,{{d}_{2}}=0.01,{{d}_{3}}=2.67$

Hence, the answer to the question is option C.

Note:
Zero correction has great significance while taking the reading on a screw gauge or any other device. In the third part of the question we are told that when no article is placed in between its jaws, the zero of the circular scale coincides with the reference line, which clearly says that the zero correction is null. If the zero of the circular had been to the left of the reference line, we should have subtracted that value from the reading and if it had been to the right then we should have added the same to the reading taken.