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Physics Question 32 – JEE-MAIN 2026

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In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50th division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.

Recall the components of a screw gauge reading: Main Scale Reading (MSR), Circular Scale Reading (CSR), Least Count (LC), and Zero Error (ZE).

Step 1: Calculate the Least Count (LC)✦ Active

The least count of the screw gauge is calculated using the pitch and the total number of divisions on the circular scale.

LC=PitchNumber of divisions on circular scale=0.1 mm100=0.001 mm
Step 2: Determine the Zero Error (ZE)○ Expand

When the studs are in contact, the zero of the main scale coincides with the 5th division of the circular scale. This indicates a positive zero error, as the circular scale zero is effectively 'ahead' of the main scale zero.

ZE=+(Zero error division×LC)=+(5×0.001 mm)=+0.005 mm
💡 Teacher's Secret Hint

Remember that a positive zero error is subtracted from the observed reading.

Step 3: Calculate the Actual Diameter○ Expand

First, calculate the observed reading using the Main Scale Reading (MSR) and Circular Scale Reading (CSR). Then, apply the zero error correction to find the actual diameter.

OR=MSR+(CSR×LC)=5 mm+(50×0.001 mm)=5 mm+0.050 mm=5.050 mm
AR=OR−ZE=5.050 mm−0.005 mm=5.045 mm
💡 Teacher's Secret Hint

Ensure correct sign convention for zero error. Positive zero error is subtracted.

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