# The least count of the main scale of a vernier callipers is 1 mm.

Question:

The least count of the main scale of a vernier callipers is $1 \mathrm{~mm}$. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the $7^{\text {th }}$ division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale betwen $3.1 \mathrm{~cm}$ and $3.2 \mathrm{~cm}$ and $4^{\text {th }}$ VSD coincides with a main scale division. The length of the cylinder is : (VSD is vernier scale division)

1. (1) $3.2 \mathrm{~cm}$

2. (2) $3.21 \mathrm{~cm}$

3. (3) $3.07 \mathrm{~cm}$

4. (4) $2.99 \mathrm{~cm}$

Correct Option: 3

Solution:

(3) L.C. of vernier callipers $=1 \mathrm{MSD}-1 \mathrm{VSD}$

$=\left(1-\frac{9}{10}\right) \times 1=0.1 \mathrm{~mm}=0.01 \mathrm{~cm}$

Here $7^{\text {th }}$ division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale.

Zero error $=7 \times 0.1=0.7 \mathrm{~mm}=0.07 \mathrm{~cm}$

Length of the cylinder $=$ measured value $-$ zero error $=(3.1+4 \times 0.01)-0.07=3.07 \mathrm{~cm}$.