Question

In a Vernier calliper the main scale and the Vernier scale are made up of different materials. When the room temperature increases by $\Delta {{\rm T}^0}C$ it is found that the reading of the instrument remains the same. Earlier it was observed that the front edge of the wooden rod placed for measurement crossed the ${{\text{N}}^{th}}$ main scale division and${\text{N + 2}}$ $msd$ coincided the ${{\text{2}}^{nd}}$ $vsd$. Initially, $10$ $vsd$ coincided with $9msd$. If coefficient of linear expansion of the main scale is ${\alpha _1}$​ and that of the Vernier scale is ${\alpha _2}$​ then what is the value of $\dfrac{{{\alpha _1}}}{{{\alpha _2}}}$? (Ignore the expansion of the rod on heating)

A.)$\dfrac{{1.8}}{N}$
B.)$\dfrac{{1.8}}{{(N + 2)}}$
C.)$\dfrac{{1.8}}{{(N - 2)}}$
D.)None of these

Answer 1

Hint- Here the two different scales of Vernier Calliper are made of two different materials so the effect of the temperature will be there. It is given that the reading is same then for reading to remain same ,the increase in the length of main scale till ${\text{N + 2}}$ division should be same as that of the increase in the length of Vernier up to ${{\text{2}}^{nd}}$ division.

Complete answer:
The change in the length due to thermal expansion is given as follows here,

\[\Delta l = l\alpha \Delta T\]-----equation (1)

Now we will write the change in lengths of both the scales separately,

Calculation for the Change in the length of the main scale (upto $N + 2$division)
${l_{ms}} = (N + 2)(msd)$, where ${l_{ms}} = $measurement in the main scale.

And now change in the length of the main scale,

$\Delta {l_{ms}} = (N + 2)(msd){\alpha _1}\Delta T$-----equation (2)

Now calculation for the change in the length of the Vernier scale (upto${2^{nd}}$division)

${l_{vs}} = (2)(vsd)$, where ${l_{vs}} = $measurement in the Vernier scale
And now change in the length of the Vernier scale

$\Delta {l_{vs}} = (2)(vsd){\alpha _2}\Delta T$-------equation (3)
Now on equating equation (2) and equation (3), that means equating the change in lengths of main scale and Vernier scale,

$(N + 2)(msd){\alpha _1}\Delta T = (2)(vsd){\alpha _2}\Delta T$------equation (4)

Also we see that there is given in the question that $10$ vsd coincide with $9$ msd, this gives a relation as follows,

$9msd = 10vsd$

$ \Rightarrow msd = \dfrac{{10}}{9}vsd$

So now putting this value in the equation (4)

$(N + 2)(\dfrac{{10}}{9}vsd){\alpha _1}\Delta T = (2)(vsd){\alpha _2}\Delta T$
$ \Rightarrow \dfrac{5}{9}(N + 2){\alpha _1} = {\alpha _2}$
$ \Rightarrow (N + 2)\dfrac{{{\alpha _1}}}{{{\alpha _2}}} = \dfrac{9}{5}$
$ \Rightarrow (N + 2)\dfrac{{{\alpha _1}}}{{{\alpha _2}}} = 1.8$
$ \Rightarrow \dfrac{{{\alpha _1}}}{{{\alpha _2}}} = \dfrac{{1.8}}{{(N + 2)}}$

Hence option (B) is the correct answer.

Note- Vernier calliper is a decimal measuring instrument. Here it is given that the $9$ graduations on the main scale coincide with the $10$ graduations of the Vernier scale. This implies that the Vernier’s ${10^{th}}$ graduation is $0$, not $10.$ so the difference between scales is that the $9$ main scale division is of the same length as the $10$ Vernier scale division.