QJEEM24Ph23d6: JEE Main 2024 Physics - Class 11 JEEM24Ph23d6 Solution: 10 divisions on the main scale of a Vernier calliper coincide…

The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is determined by the difference in the measurement of one main scale division and one Vernier scale division.

In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. This means that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale. Since each division on the main scale is 5 units, this is equal to:

$$10\text{ divisions on main scale}\times 5\text{ units per division}=50\text{ units}$$

The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions on the Vernier scale. This means that:

$$1\text{ division on the Vernier scale}=\frac{50\text{ units}}{11}$$

Therefore, the least count of the Vernier caliper, which is the difference between one division on the main scale and one division on the Vernier scale, can be calculated as follows:

$$\text{Least count}=\text{value of one main scale division}-\text{value of one Vernier scale division}$$

$$\text{Least count}=5\text{ units}-\frac{50\text{ units}}{11}$$

$$\text{Least count}=\frac{55\text{ units}}{11}-\frac{50\text{ units}}{11}$$

$$\text{Least count}=\frac{5\text{ units}}{11}$$

Thus, the least count of the Vernier caliper is

$$\frac{5\text{ units}}{11}$$

Therefore, the correct answer is:

Option A: $$\frac{5}{11}$$