Question

Suppose that we have a vernier callipers which does not expand or contract on heating. Such a device is very useful in engineering applications where the dimensions are to be found at different temperatures. One such vernier callipers is used to measure the diameter of a rod at different temperatures as shown in table below. What will be the vernier scale reading (division coinciding) at a temperature of ${50}^{\circ }C$.

Given that $10$ vernier scale divisions coincide with $9$ main scale divisions and $1$ main scale division is equal to $1\text{mm}$.

Temperature T in ${}^{\circ }C$	0	20
Main scale reading	$2.4\text{cm}$	$2.4\text{cm}$
Vernier scale reading	$4^{th}$ division coinciding	$8^{th}$ division coinciding

Answer 1

Using vernier callipers,
Total reading $=MSR+VSD\left(LC\right)$
Given that $10$ vernier scale divisions coincide with 9 main scale division,
$10VSD=9MSD$
$1VSD=0.9MSD$
$LC=1MSD-1VSD=1MSD-0.9MSD=0.01\text{cm}$
Diameter of rod at $0^{\circ }C$ is $2.4+4\left(0.01\right)\text{cm}=2.44\text{cm}$
Diameter of rod at ${20}^{\circ }C$ is $2.4+8\left(0.01\right)\text{cm}=2.48\text{cm}$

Let $\alpha$ be coefficient of linear expansion,
Change in length as the temperature changes from $0^{\circ }$ to ${20}^{\circ }$ is
$\Delta l_1=\alpha l\Delta T_{1}4\times 0.01=\alpha \times 2.44\times 20$
$\alpha =\frac{1}{244\times 5}$
Change in length as the temperature changes from $0^{\circ }$ to ${50}^{\circ }$ is
$\Delta l_2=\alpha l\Delta T_2$
$\Delta l_2=\frac{1}{244\times 5}\times 2.44\times 50=0.1cm$

Diameter of rod at ${50}^{\circ }C$ is
$L=(2.44+0.1)\text{cm}=2.54\text{cm}$

Diameter of rod at ${50}^{\circ }C$ can be written as $2.5+VSD\left(0.01\right)\text{cm}$
$2.5+VSD\left(0.01\right)=2.54$
$VSD=4$
Therefore division coinciding at ${50}^{\circ }C$ is $4^{th}$ division