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Physics Question 25 – NEET-UG 2025

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Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at x=0.1 cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is M=5 cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is

To solve this problem, you need to understand how a Vernier caliper works, including how to calculate its least count and how to apply zero correction.

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

Given that 10 Vernier Scale Divisions (V.S.D.) are equal to 9 Main Scale Divisions (M.S.D.), we have 1 V.S.D.=910 M.S.D.. The least count (LC) of the Vernier caliper is the difference between one M.S.D. and one V.S.D.:

LC=1 M.S.D.−1 V.S.D.=1 M.S.D.−910 M.S.D.=110 M.S.D.

Given that the least division in the M.S. (i.e., 1 M.S.D.) is 0.1 cm, we can calculate the LC:

LC=110×0.1 cm=0.01 cm
Step 2: Determine the Zero Error (ZE)○ Expand

The problem states that the zero of the Vernier Scale (V.S.) is at x=0.1 cm when the jaws of the Vernier calipers are closed. This means the instrument shows a reading of 0.1 cm when it should ideally show 0 cm. This indicates a positive zero error.

ZeroError(ZE)=+0.1 cm
💡 Teacher's Secret Hint

A positive zero error means the instrument reads higher than the actual value, so the correction will be negative.

Step 3: Calculate the Observed and Corrected Diameter○ Expand

The observed reading (OR) is calculated using the Main Scale Reading (MSR), Coinciding Vernier Division (CVD), and the Least Count (LC):

ObservedReading(OR)=MSR+(CVD×LC)

Given MSR = 5 cm and CVD = 8:

OR=5 cm+(8×0.01 cm)=5 cm+0.08 cm=5.08 cm

Finally, the corrected reading (CR) is obtained by subtracting the zero error from the observed reading:

CorrectedReading(CR)=OR−ZE=5.08 cm−(+0.1 cm)=5.08 cm−0.1 cm=4.98 cm
💡 Teacher's Secret Hint

Remember to subtract a positive zero error from the observed reading to get the accurate measurement.

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