JEE PYQ: Units & Measurements - Question ID 7fa80afe67b0 (JEE Main 2016)

A screw gauge is a precision instrument used to measure small lengths (e.g., thickness of sheets) with high accuracy. It consists of two scales:

• Main Scale (Linear Scale): This is a fixed scale along the axis of the screw, typically marked in millimeters (mm). The reading on this scale gives the approximate measurement.
• Circular Scale (Rotating Scale): This is a rotating scale attached to the screw. The number of divisions on this scale determines the least count (smallest measurable length) of the screw gauge.

  The pitch of the screw gauge is the distance moved by the screw along the main scale for one complete rotation of the circular scale. The least count (LC) is calculated as: $\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$

  When the jaws of the screw gauge are closed (zero error condition), if the zero of the circular scale does not coincide with the main scale line, the instrument has a zero error. This error must be accounted for in measurements.

  The observed thickness of the sheet is given by: $\text{Observed Thickness} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times \text{Least Count})$ The actual thickness is then: $\text{Actual Thickness} = \text{Observed Thickness} - \text{Zero Error}$

  Step-by-Step Derivation:

  Step 1: Calculate the Least Count (LC)

  Given:

  • Pitch = 0.5 mm
  • Number of divisions on circular scale = 50
  $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}$

  Step 2: Determine the Zero Error

  When the jaws are closed (no sheet), the 45^th division of the circular scale coincides with the main scale line, and the zero of the main scale is barely visible. This implies:

  • Main Scale Reading (MSR) = 0 mm (since zero is barely visible, we take it as 0)
  • Circular Scale Reading (CSR) = 45 divisions
  The zero error is the reading when no object is placed between the jaws: $\text{Zero Error} = \text{MSR} + (\text{CSR} \times \text{LC}) = 0 + (45 \times 0.01 \text{ mm}) = 0.45 \text{ mm}$ Since the zero of the circular scale is ahead of the main scale line, the zero error is positive. This means the instrument reads 0.45 mm when it should read 0 mm.

  Step 3: Measure the Observed Thickness of the Sheet

  Given:

  • Main Scale Reading (MSR) = 0.5 mm
  • Circular Scale Reading (CSR) = 25 divisions
  The observed thickness is: $\text{Observed Thickness} = \text{MSR} + (\text{CSR} \times \text{LC}) = 0.5 \text{ mm} + (25 \times 0.01 \text{ mm}) = 0.5 \text{ mm} + 0.25 \text{ mm} = 0.75 \text{ mm}$

  Step 4: Calculate the Actual Thickness

  Since the zero error is positive, the actual thickness is less than the observed thickness: $\text{Actual Thickness} = \text{Observed Thickness} - \text{Zero Error} = 0.75 \text{ mm} - 0.45 \text{ mm} = 0.30 \text{ mm}$ Wait! This result does not match any of the options. This indicates a misinterpretation of the zero error condition.

  Re-evaluating Zero Error: The question states: "the zero of the main scale is barely visible." This implies that the main scale reading is not exactly 0 mm but very close to it. Since the 45^th division coincides with the main scale line, the zero error is the excess reading beyond the main scale zero.

  However, the main scale zero is barely visible, meaning the main scale reading is 0 mm (as no other value is given). The zero error is thus correctly calculated as +0.45 mm. But the actual thickness calculation leads to 0.30 mm, which is not among the options. This suggests that the zero error should be subtracted from the observed thickness, but the options still do not match.

  Alternative Interpretation: The zero error is the reading when the jaws are closed. If the 45^th division coincides with the main scale line, the zero error is: $\text{Zero Error} = - (50 - 45) \times \text{LC} = -5 \times 0.01 \text{ mm} = -0.05 \text{ mm}$ This is because the zero of the circular scale is 5 divisions behind the main scale line (since 45 coincides, the zero is at 50 - 45 = 5 divisions behind). A negative zero error means the instrument reads less than the actual value.

  Now, recalculate the actual thickness: $\text{Observed Thickness} = 0.5 \text{ mm} + (25 \times 0.01 \text{ mm}) = 0.75 \text{ mm}$ $\text{Actual Thickness} = \text{Observed Thickness} - \text{Zero Error} = 0.75 \text{ mm} - (-0.05 \text{ mm}) = 0.75 \text{ mm} + 0.05 \text{ mm} = 0.80 \text{ mm}$ This matches option D.

  Conclusion: The correct zero error is -0.05 mm (not +0.45 mm). The actual thickness of the sheet is 0.80 mm.

  Common Traps & Exam Tip:

  Trap 1: Misinterpreting Zero Error Sign Students often confuse whether the zero error is positive or negative. If the zero of the circular scale is ahead of the main scale line, the zero error is positive. If it is behind, the zero error is negative. In this question, the 45^th division coincides, meaning the zero is 5 divisions behind, leading to a negative zero error.

  Trap 2: Ignoring the "Barely Visible" Clue The phrase "zero of the main scale is barely visible" implies that the main scale reading is 0 mm, but the circular scale is offset. Students might mistakenly assume a non-zero main scale reading, leading to incorrect zero error calculation.

  Trap 3: Incorrect Least Count Calculation Some students divide the number of divisions by the pitch instead of the pitch by the number of divisions. Always remember: $\text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions}}$

  Exam Tip: Always sketch the screw gauge reading to visualize the zero error. For zero error:

  • If the zero of the circular scale is below the main scale line, the zero error is negative.
  • If it is above, the zero error is positive.
  In this question, since the 45^th division coincides, the zero is 5 divisions below, making the zero error negative.