Question

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is :

• A. $\frac{1}{2}$
• B. $\frac{10}{11}$
• C. $\frac{50}{11}$
• D. $\frac{5}{11}$

Answer 1

Answer: (D)

$\frac{5}{11}$

Answer 2

Given:

$10 \, \text{MSD} = 11 \, \text{VSD}$

1 VSD (Vernier Scale Division) is equivalent to:

$1 \, \text{VSD} = \frac{10}{11} \, \text{MSD}$

The least count (LC) of the Vernier caliper is given by:

$LC = 1 \, \text{MSD} - 1 \, \text{VSD}$

Substituting the values:

$LC = 1 \, \text{MSD} - \frac{10}{11} \, \text{MSD} = \frac{1}{11} \, \text{MSD}$

Given that 1 MSD corresponds to 5 units:

$LC = \frac{5 \, \text{units}}{11}$