JEE PYQ: Units & Measurements - Question ID 7ebbc9e74f07 (JEE Main 2021)

ID: 7ebbc9e74f07JEE Main 2021Single Correct MCQ
One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with (n - 1)th division of the main scale. The least count of the callipers in mm is :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In vernier callipers, 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.

Let:

  • aa = length of one main scale division (in cm)
  • nn = the vernier scale division that coincides with the (n1)(n-1)th main scale division
The key idea is: One vernier scale division = (n1)n×\frac{(n-1)}{n} \times one main scale division. Hence, the least count is: LC=One main scale divisionOne vernier scale divisionLC = \text{One main scale division} - \text{One vernier scale division}

Step-by-Step Derivation:

1. Express the main scale division:
One main scale division = aa cm. 2. Express the vernier scale division:
Since the nnth vernier division coincides with the (n1)(n-1)th main scale division, the total length covered by nn vernier divisions equals the length covered by (n1)(n-1) main scale divisions: n×(one vernier division)=(n1)×an \times \text{(one vernier division)} = (n-1) \times a Solving for one vernier division: One vernier division=(n1)n×a cm\text{One vernier division} = \frac{(n-1)}{n} \times a \text{ cm} 3. Calculate the least count (LC):
LC=One main scale divisionOne vernier divisionLC = \text{One main scale division} - \text{One vernier division} Substituting the values: LC=a(n1)na=(1n1n)a=an cmLC = a - \frac{(n-1)}{n} a = \left(1 - \frac{n-1}{n}\right) a = \frac{a}{n} \text{ cm} 4. Convert cm to mm:
Since 1 cm=10 mm1 \text{ cm} = 10 \text{ mm}, LC=an×10=10an mmLC = \frac{a}{n} \times 10 = \frac{10a}{n} \text{ mm}

Common Traps & Exam Tip:

  • Misidentifying coinciding divisions: Students often confuse which division coincides with which, leading to incorrect expressions for the vernier division length.
  • Unit inconsistency: Forgetting to convert the final answer from cm to mm (as the options are in mm) is a frequent error.
  • Algebraic simplification: Errors in simplifying (1n1n)\left(1 - \frac{n-1}{n}\right) can lead to incorrect options like DD.
Tip: Always verify the coincidence condition carefully and ensure unit consistency in the final answer.

Related Questions from Units & Measurements

ID: 32f827e3012fJEE Main 2026

In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and 7th 7^{\text {th }} Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has

View Solution →
ID: 4c5472dca7e2JEE Main 2026

Dimensions of universal gravitational constant (GG) in terms of Planck's constant (hh), distance (LL), mass (MM) and time (TT) are _______.

View Solution →
ID: 605eaee8ed80JEE Main 2026

The time period of a simple harmonic oscillator is T=2πkmT = 2\pi \sqrt{\frac{k}{m}}. Measured value of mass (m)(m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k)(k) is ________%.

View Solution →
ID: 5561befb0f22JEE Main 2026

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, 4th 4{ }^{\text {th }} mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and 5th 5^{\text {th }} division of vernier scale coincides with a main scale division. Measured length of cylinder is ____\_\_\_\_ mm.

(Least count of Vernier calliper =0.1 mm=0.1 \mathrm{~mm} )

View Solution →