Question

There are two Vernier calipers both of which have $1\text{cm}$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $\left({\text{C}}_1\right)$ has $10$ equal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other calipers $\left({\text{C}}_2\right)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure.

• A. The measured value by caliper ${\text{C}}_1$ is $2.83\text{cm}$.
• B. The measured value by caliper ${\text{C}}_1$ is $2.87\text{cm}$.
• C. The measured value by caliper ${\text{C}}_2$ is $2.83\text{cm}$.
• D. The measured value by caliper ${\text{C}}_2$ is $2.87\text{cm}$.

Answer 1

Answer: (B), (C)

The measured value by caliper ${\text{C}}_2$ is $2.83\text{cm}$.

For ${\text{C}}_1:$

$9MSD=10VSD$

$1VSD=0.9MSD=0.9\text{mm}$

$\therefore LC=1MSD-1VSD=1-0.9=0.1\text{mm}=0.01\text{cm}$

So, the reading of ${\text{C}}_1$ is given by,

$\text{Reading}=MSR+(LC\times VCR)$

$\text{Reading}=2.8+(0.01\times 7)=2.87\text{cm}$

For ${\text{C}}_2:$

$11MSD=10VSD$

$1VSD=1.1MSD=1.1\text{mm}$

$\therefore LC=1MSD-1VSD=1-1.1=-0.1\text{mm}=-0.01\text{cm}$

So, the reading of ${\text{C}}_2$ is given by,

$\text{Reading}=MSR+(LC\times VCR)$

$\text{Reading}=2.8+[0.01\times (10-7\left)\right]=2.83\text{cm}$

Hence, options $\text{(B) and (C)}$ are the correct answer.