JEE PYQ: Units & Measurements - Question ID 1b6c5fdf08ad (JEE Main 2025)

ID: 1b6c5fdf08adJEE Main 2025Single Correct MCQ

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm ) :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In any vernier instrument (such as a travelling microscope), the least count (LC) is the smallest length that can be measured accurately. It is defined as the difference between one main-scale division and one vernier-scale division.

Key formulas:

  • Value of one main-scale division (MSD): MSD=Total length of main scaleNumber of divisions on main scale\text{MSD} = \frac{\text{Total length of main scale}}{\text{Number of divisions on main scale}}
  • Value of one vernier-scale division (VSD): VSD=Number of main-scale divisions matchedNumber of vernier divisions×MSD\text{VSD} = \frac{\text{Number of main-scale divisions matched}}{\text{Number of vernier divisions}} \times \text{MSD}
  • Least count (LC): LC=MSDVSD\text{LC} = \text{MSD} - \text{VSD}
Step-by-Step Derivation:

Step 1: Compute the value of one main-scale division (MSD)

Given: 300 divisions on the main scale correspond to 15 cm. Therefore, MSD=15 cm300=0.05 cm\text{MSD} = \frac{15\ \text{cm}}{300} = 0.05\ \text{cm}

Step 2: Compute the value of one vernier-scale division (VSD)

The vernier scale has 25 divisions that match 24 divisions of the main scale. Since each main-scale division is 0.05 cm, 24 main-scale divisions equal 24×0.05 cm=1.2 cm.24 \times 0.05\ \text{cm} = 1.2\ \text{cm}. This 1.2 cm is covered by 25 vernier divisions, so VSD=1.2 cm25=0.048 cm.\text{VSD} = \frac{1.2\ \text{cm}}{25} = 0.048\ \text{cm}.

Step 3: Compute the least count (LC)

The least count is the difference between one main-scale division and one vernier-scale division: LC=MSDVSD=0.05 cm0.048 cm=0.002 cm.\text{LC} = \text{MSD} - \text{VSD} = 0.05\ \text{cm} - 0.048\ \text{cm} = 0.002\ \text{cm}.

Conclusion: The least count of the travelling microscope is 0.002 cm, which corresponds to option A.

Common Traps & Exam Tip:

1. Misidentifying matched divisions: Students often confuse which scale’s divisions match the other. Here, 25 vernier divisions match 24 main-scale divisions, not the other way around. 2. Incorrect unit conversion: Ensure all lengths are in the same unit (cm) before computing. 3. Arithmetic errors: Double-check the subtraction 0.050.0480.05 - 0.048 to avoid careless mistakes.

Exam Tip: Always write down the definitions of MSD, VSD, and LC explicitly before plugging in numbers. This keeps the logic clear and reduces errors.

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