JEE PYQ: Units & Measurements - Question ID 4ce272101871 (JEE Main 2024)

ID: 4ce272101871JEE Main 2024Single Correct MCQ

While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is 1 mm1 \mathrm{~mm} and circular scale reading is equal to 42 divisions. Pitch of screw gauge is 1 mm1 \mathrm{~mm} and it has 100 divisions on circular scale. The diameter of the wire is x50 mm\frac{x}{50} \mathrm{~mm}. The value of xx is :

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Step-by-step Explanation

Core Formula & Concept:

When measuring small lengths (like the diameter of a wire) with a screw gauge, two scales work together:

  • The main scale (linear scale) gives a coarse reading in whole millimeters.
  • The circular scale (rotating thimble) subdivides each millimeter into finer parts.

The pitch of the screw gauge is the distance the spindle advances along the main scale when the circular scale makes one complete revolution. Here, the pitch is 1 mm1 \text{ mm} and the circular scale has 100100 divisions. Therefore, the least count (smallest measurable length) is: Least count=PitchNumber of divisions on circular scale=1 mm100=0.01 mm.\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}.

The total measured diameter DD is the sum of the main scale reading and the circular scale reading multiplied by the least count: D=Main scale reading+(Circular scale reading×Least count).D = \text{Main scale reading} + (\text{Circular scale reading} \times \text{Least count}).

Step-by-Step Derivation:

Given:

  • Main scale reading = 1 mm1 \text{ mm}
  • Circular scale reading = 4242 divisions
  • Pitch = 1 mm1 \text{ mm}
  • Number of divisions on circular scale = 100100

Step 1: Calculate the least count. Least count=PitchNumber of divisions=1 mm100=0.01 mm.\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}.

Step 2: Compute the circular scale contribution. Circular scale contribution=Circular scale reading×Least count=42×0.01 mm=0.42 mm.\text{Circular scale contribution} = \text{Circular scale reading} \times \text{Least count} = 42 \times 0.01 \text{ mm} = 0.42 \text{ mm}.

Step 3: Add the main scale reading to get the total diameter. D=Main scale reading+Circular scale contribution=1 mm+0.42 mm=1.42 mm.D = \text{Main scale reading} + \text{Circular scale contribution} = 1 \text{ mm} + 0.42 \text{ mm} = 1.42 \text{ mm}.

Step 4: Express the diameter in the form x50 mm\frac{x}{50} \text{ mm}. 1.42 mm=x50 mm.1.42 \text{ mm} = \frac{x}{50} \text{ mm}. Multiply both sides by 5050: x=1.42×50=71.x = 1.42 \times 50 = 71.

Common Traps & Exam Tip:

Students often confuse the pitch with the least count. They might directly multiply the circular scale reading by the pitch (e.g., 42×1 mm=42 mm42 \times 1 \text{ mm} = 42 \text{ mm}), which is incorrect. Always remember:

  • The pitch is the distance per full revolution.
  • The least count is the pitch divided by the number of circular scale divisions.
  • The circular scale reading must be multiplied by the least count, not the pitch.

Another common mistake is miscounting the number of divisions or misinterpreting the main scale reading. Double-check the given values before performing calculations.

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