Question

A vernier caliper has no zero error, one main scale division = $1\text{mm}$ and $9$ main scale division = $10$ vernier scale divisions. A spring is held between jaws without exerting force on jaws. Main scale reading is $3.4\text{cm}$, and 5th vernier scale division coincides with a main scale division. When jaws are pressed by force of $10\text{N}$ on both sides, main scale reading remains $3.4\text{cm}$, but 1st division of vernier scale coincides with a main scale division. Now spring is removed, cut into two pieces and when inserted between jaws without any force, main scale reading = $1.1\text{cm}$ and 5th vernier scale division coincides with a main scale. If we apply force of $30\text{N}$ on both the jaws, main scale reading gets $1.0\text{cm}$ and $x$ vernier scale division coincides with a main scale division. Find ${}^{\prime }{x}^{\prime }$.

Answer 1

Least count of vernier caliper:
$LC=1MSR-1VSR$
$LC=1MSR-\frac{9}{10}MSR$
$LC=0.1MSR=0.1mm$

If $\Delta t=$ change in length due to compression of spring, we can write:
$\Delta t=34+(5\times 0.1)-34-1\times 0.1=0.4$
$k\Delta t=10$
$k\times 0.4=10$
$k=25N/mm$

Now, consider $\Delta p=$ change in length when force of $30N$ is applied.
$\Delta p=11+5\times 0.1-10-x\times 0.1=1.5-0.1x$
But, we know $k\Delta p=30$
$\Delta p=\frac{30}{k}=\frac{30}{25}=1.2mm$
$\Rightarrow 1.5-0.1x=1.2$
$\Rightarrow x=3$