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Physics Question 26 – NEET-UG 2026

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One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that 4th Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be 1 cm, the actual length of the wire is :

A Vernier caliper is used for precise length measurements, and its accuracy depends on correctly determining its least count and accounting for any zero error.

🥷
Ninja Strategy•Zero Error Direction

Recognize that a negative zero error implies the actual measurement must be greater than the observed measurement, eliminating options that are less than or equal to the observed value.

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

The main scale division (MSD) is 1 mm. The number of divisions on the Vernier scale is 10. The least count is calculated as:

LC=1 MSDNumber of VSDs=1 mm10=0.1 mm
Step 2: Determine the Zero Error (ZE)○ Expand

The Vernier scale shifts to the left of zero, indicating a negative zero error. The 4th Vernier division coincides. The zero error is:

ZE=−(coinciding Vernier division×LC)=−(4×0.1 mm)=−0.4 mm
💡 Teacher's Secret Hint

Remember that a shift to the left of zero indicates a negative zero error.

Step 3: Calculate the Actual Length○ Expand

The observed length is 1 cm, which is equal to 10 mm. The actual length is given by:

Actual Length=Observed Length−Zero Error

Substituting the values:

Actual Length=10 mm−(−0.4 mm)=10 mm+0.4 mm=10.4 mm

Converting to centimeters:

Actual Length=10.4 mm=1.04 cm
💡 Teacher's Secret Hint

Pay attention to unit consistency throughout the calculation.

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