JEE PYQ: Units & Measurements - Question ID 20d41cc1ed71 (JEE Main 2021)
Reason R :
Least Count =
In the light of the above statements, choose the most appropriate answer from the options given below :
Select Option
Step-by-step Explanation
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: This formula arises because the circular scale subdivides the pitch into finer parts, allowing measurement of fractions of the pitch.
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: So, the pitch is 1 mm.
Step 2: Use the Least Count Formula
The circular scale has 50 total divisions. Using the formula:
Step 3: Convert Units for Comparison
The assertion claims the least count is 0.001 cm. Let's convert our result to cm: This is not equal to 0.001 cm. Hence, Assertion A is incorrect.
Step 4: Evaluate Reason R
Reason R states: 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:
This is wrong. The correct formula uses pitch, not total distance.
Exam Tip:
When solving screw gauge problems, always:
- Find the pitch first (distance per rotation).
- Use the correct least count formula.
- Convert units consistently (preferably to mm or cm).
- Verify the assertion and reason separately.
Related Questions from Units & Measurements
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 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
Dimensions of universal gravitational constant () in terms of Planck's constant (), distance (), mass () and time () are _______.
The time period of a simple harmonic oscillator is . Measured value of mass 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 is ________%.
When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, 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 division of vernier scale coincides with a main scale division. Measured length of cylinder is mm.
(Least count of Vernier calliper )