JEE PYQ: Units & Measurements - Question ID 20d41cc1ed71 (JEE Main 2021)

ID: 20d41cc1ed71JEE Main 2021Single Correct MCQ
Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.

Reason R :

Least Count = PitchTotaldivisionsoncircularscale{{Pitch} \over {Total\,divisions\,on\,circular\,scale}}

In the light of the above statements, choose the most appropriate answer from the options given below :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In a screw gauge (or micrometer), two scales are used to measure small lengths with high precision:

  • Main scale: A linear scale marked along the axis of the screw.
  • Circular scale: A rotating scale attached to the screw, which moves along the main scale as the screw is turned.

The key definitions are:

  • Pitch: The distance moved by the screw (and hence the circular scale) along the main scale in one complete rotation of the circular scale. It is the smallest division on the main scale that the screw advances per full turn.
  • Least Count (LC): The smallest length that can be measured with the instrument. It is given by: Least Count=PitchTotal divisions on circular scale\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}} This formula arises because the circular scale subdivides the pitch into finer parts, allowing measurement of fractions of the pitch.
Step-by-Step Derivation:

Step 1: Determine the Pitch

The question states that in 5 complete rotations of the circular scale, the distance travelled on the main scale is 5 mm.

Since pitch is the distance moved per one complete rotation, we calculate: Pitch=Distance moved on main scaleNumber of rotations=5mm5=1mm\text{Pitch} = \frac{\text{Distance moved on main scale}}{\text{Number of rotations}} = \frac{5\,\text{mm}}{5} = 1\,\text{mm} So, the pitch is 1 mm.

Step 2: Use the Least Count Formula

The circular scale has 50 total divisions. Using the formula: Least Count=PitchTotal divisions on circular scale=1mm50=0.02mm\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}} = \frac{1\,\text{mm}}{50} = 0.02\,\text{mm}

Step 3: Convert Units for Comparison

The assertion claims the least count is 0.001 cm. Let's convert our result to cm: 0.02mm=0.02×0.1cm=0.002cm0.02\,\text{mm} = 0.02 \times 0.1\,\text{cm} = 0.002\,\text{cm} This is not equal to 0.001 cm. Hence, Assertion A is incorrect.

Step 4: Evaluate Reason R

Reason R states: Least Count=PitchTotal divisions on circular scale\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}} This is the correct formula for least count in a screw gauge. So, Reason R is correct.

Step 5: Match with Given Options

  • A: A is not correct but R is correct. ✅
  • B: Both A and R are correct and R explains A. ❌
  • C: A is correct but R is not correct. ❌
  • D: Both A and R are correct but R does not explain A. ❌

Since A is incorrect and R is correct, the correct option is A.

Common Traps & Exam Tip:

Trap 1: Misinterpreting Pitch
Students often confuse the total distance moved with the number of rotations. They might divide 5 mm by 50 divisions (instead of 5 rotations), leading to an incorrect pitch of 0.1 mm. Always remember: pitch is distance per one full rotation.

Trap 2: Unit Conversion Errors
The assertion gives the least count in cm, while the calculation is often done in mm. Forgetting to convert units (1 mm = 0.1 cm) can lead to incorrect comparison. Always double-check units.

Trap 3: Misapplying the Least Count Formula
Some students mistakenly use: Least Count=Total distanceTotal divisions\text{Least Count} = \frac{\text{Total distance}}{\text{Total divisions}} This is wrong. The correct formula uses pitch, not total distance.

Exam Tip:
When solving screw gauge problems, always:

  1. Find the pitch first (distance per rotation).
  2. Use the correct least count formula.
  3. Convert units consistently (preferably to mm or cm).
  4. Verify the assertion and reason separately.

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 →