Question

In a vernier calipers the main scale and the vernier scale are made up of different materials. When the room temperature increases by $\Delta T^{o}C$, 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 $N^{th}$ main scale division and $N+2$ msd coincided the $2^{nd}$ vsd. Initially, $10$ vsd coincided with $9$ msd. 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 ${\alpha }_1/{\alpha }_2$? (Ignore the expansion of the rod on heating)

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

Answer 1

The correct option is B $\frac{1.8}{(N+2)}$

For reading to remain same, the increase in length of main scale upto $N+2$ division should be same as increase in length of vernier upto $2^{nd}$ division:
The change in length is given as
$\Delta l=l\alpha \Delta T$...(i)
Change in length for Main Scale upto $N+2$ division
$l_{mainscale}=(N+2)\left(msd\right)$
$\Delta l_{mainscale}=(N+2)\left(msd\right){\alpha }_1\Delta T$...(ii)
Change in length for Vernier Scale upto $2^{nd}$ division:
$l_{vernierscale}=\left(2\right)\left(vsd\right)$
$\Delta l_{vernierscale}=\left(2\right)\left(vsd\right){\alpha }_2\Delta T$...(iii)
Equating change in lengths of main scale and vernier scale:
$(N+2)\left(msd\right){\alpha }_1\Delta T=\left(2\right)\left(vsd\right){\alpha }_2\Delta T$
using relation $9msd=10vsd$
$(N+2)\times \frac{10}{9}vsd{\alpha }_1\Delta T=2\times vsd{\alpha }_2\Delta T$
$\frac{{\alpha }_1}{{\alpha }_2}=\frac{1.8}{N+2}$