JEE PYQ: Units & Measurements - Question ID 741cfd246c0a (JEE Main 2008)

ID: 741cfd246c0aJEE Main 2008Single Correct MCQ
Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

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Step-by-step Explanation

Core Formula & Concept:

A screw gauge is a precision instrument used to measure small lengths (like the diameter of a wire) with high accuracy. It consists of two scales:

  • Main Scale (Linear Scale): Fixed scale along the length of the screw, usually marked in millimeters (mm).
  • Circular Scale (Thimble Scale): Rotating scale attached to the screw, which moves over the main scale. The number of divisions on this scale determines the least count (smallest measurable length) of the screw gauge.

The least count (LC) of the screw gauge is calculated as:

Least Count (LC)=Pitch of the screwNumber of divisions on circular scale\text{Least Count (LC)} = \frac{\text{Pitch of the screw}}{\text{Number of divisions on circular scale}}

Where the pitch is the distance moved by the screw along the main scale in one complete rotation of the circular scale.

The observed reading is the sum of the main scale reading and the circular scale reading multiplied by the least count:

Observed Reading=Main Scale Reading+(Circular Scale Reading×LC)\text{Observed Reading} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times \text{LC})

However, screw gauges often have a zero error, which is a systematic error due to misalignment when the jaws are closed. The zero error can be positive or negative:

  • Positive Zero Error: Circular scale reading is ahead of the zero mark when jaws are closed. The error is subtracted from the observed reading.
  • Negative Zero Error: Circular scale reading is behind the zero mark when jaws are closed. The error is added to the observed reading.

The corrected diameter is obtained by adjusting the observed reading for the zero error:

Corrected Diameter=Observed ReadingZero Error\text{Corrected Diameter} = \text{Observed Reading} - \text{Zero Error}

(Note: The sign of the zero error is crucial. A negative zero error of −0.03 mm means the instrument reads 0.03 mm less than the actual value, so we add 0.03 mm to correct it.)

--- Step-by-Step Derivation:

Step 1: Determine the Pitch of the Screw Gauge

Given that two full turns of the circular scale cover 1 mm on the main scale:

Pitch=Distance covered on main scaleNumber of rotations=1 mm2=0.5 mm\text{Pitch} = \frac{\text{Distance covered on main scale}}{\text{Number of rotations}} = \frac{1 \text{ mm}}{2} = 0.5 \text{ mm}

Step 2: Calculate the Least Count (LC)

The circular scale has 50 divisions. The least count is:

LC=PitchNumber of divisions=0.5 mm50=0.01 mm\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}

Step 3: Compute the Observed Reading

The student notes:

  • Main scale reading = 3 mm
  • Circular scale reading = 35 divisions

The circular scale contribution is:

35×LC=35×0.01 mm=0.35 mm35 \times \text{LC} = 35 \times 0.01 \text{ mm} = 0.35 \text{ mm}

Thus, the observed reading is:

Observed Reading=Main Scale+Circular Scale Contribution=3 mm+0.35 mm=3.35 mm\text{Observed Reading} = \text{Main Scale} + \text{Circular Scale Contribution} = 3 \text{ mm} + 0.35 \text{ mm} = 3.35 \text{ mm}

Step 4: Apply the Zero Error Correction

The screw gauge has a zero error of −0.03 mm. This means the instrument reads 0.03 mm less than the actual value. To correct this, we add the absolute value of the zero error to the observed reading:

Corrected Diameter=Observed ReadingZero Error=3.35 mm(0.03 mm)=3.35 mm+0.03 mm=3.38 mm\text{Corrected Diameter} = \text{Observed Reading} - \text{Zero Error} = 3.35 \text{ mm} - (-0.03 \text{ mm}) = 3.35 \text{ mm} + 0.03 \text{ mm} = 3.38 \text{ mm}

Step 5: Match with the Given Options

The corrected diameter is 3.38 mm, which corresponds to option D.

--- Common Traps & Exam Tip:

1. Misinterpreting the Zero Error Sign:

Many students confuse the sign of the zero error. A negative zero error means the instrument reads less than the actual value, so the correction involves adding the absolute value of the error. Conversely, a positive zero error would require subtracting the error. Always double-check the sign convention.

2. Incorrect Calculation of Least Count:

Students often miscalculate the least count by using the total distance covered in one turn instead of the pitch. Remember, the pitch is the distance moved per rotation, and the least count is the pitch divided by the number of circular scale divisions.

3. Forgetting to Multiply Circular Scale Reading by Least Count:

The circular scale reading is in divisions, not millimeters. Always multiply the circular scale reading by the least count to convert it to millimeters before adding it to the main scale reading.

4. Rounding Errors:

Avoid premature rounding during intermediate steps. For example, keeping the least count as 0.01 mm (instead of rounding to 0.01) ensures accuracy in the final result.

Exam Tip:

Always write down the formula for least count and zero error correction explicitly. This helps in avoiding sign errors and ensures clarity in calculations. Practice similar problems to build confidence in handling screw gauge measurements.

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