NEET2025Physics-Measurements

NEET 2025 Physics Vernier Calipers Numerical Question

Type: Numerical-numerical-Medium-Class 11

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

A

5.18 cm

B

5.08 cm

C

4.98 cm

D

5.00 cm

Correct Answer

Option C

Detailed Explanation

To solve the question regarding the measurement of the diameter of a spherical object using Vernier callipers, let's break it down step-by-step.

Given Data:

  1. Main Scale Reading (MSR): M=5cmM = 5 \, \text{cm}
  2. Number of coinciding Vernier Scale Divisions (VSD): n=8n = 8
  3. Least Count (LC) of Main Scale: 0.1cm0.1 \, \text{cm}
  4. Zero Error: The zero of the Vernier scale reads at 0.1cm0.1 \, \text{cm} when the jaws are closed.
  5. Vernier Scale Divisions (VSD) to Main Scale Divisions (MSD) relationship: 10VSD=9MSD10 \, \text{VSD} = 9 \, \text{MSD}

Step 1: Calculate the Least Count of the Vernier Scale

The least count of the Vernier calipers can be calculated using the formula:

Least Count (LC)=Value of 1 MSDValue of 1 VSD\text{Least Count (LC)} = \text{Value of 1 MSD} - \text{Value of 1 VSD}

Here, we know:

  • Value of 1 MSD = 0.1cm0.1 \, \text{cm}
  • Let's denote the value of 1 VSD as xx.

From the relationship given, we have:

10x=9(0.1)    10x=0.9    x=0.09cm10x = 9(0.1) \implies 10x = 0.9 \implies x = 0.09 \, \text{cm}

Thus, the least count of the Vernier scale is:

LC=0.1cm0.09cm=0.01cm\text{LC} = 0.1 \, \text{cm} - 0.09 \, \text{cm} = 0.01 \, \text{cm}

Step 2: Calculate the Observed Diameter

The total reading when the Vernier scale is taken into account is given by:

Total Reading=MSR+(n×LC)\text{Total Reading} = \text{MSR} + (n \times \text{LC})

Substituting the known values:

Total Reading=5cm+(8×0.01cm)=5cm+0.08cm=5.08cm\text{Total Reading} = 5 \, \text{cm} + (8 \times 0.01 \, \text{cm}) = 5 \, \text{cm} + 0.08 \, \text{cm} = 5.08 \, \text{cm}

Step 3: Correct for Zero Error

Since the zero of the Vernier scale is at 0.1cm0.1 \, \text{cm} when the jaws are closed, this indicates a positive zero error. To correct for this zero error, we subtract it from the total reading:

Corrected Reading=Total ReadingZero Error\text{Corrected Reading} = \text{Total Reading} - \text{Zero Error}

Thus:

Corrected Reading=5.08cm0.1cm=4.98cm\text{Corrected Reading} = 5.08 \, \text{cm} - 0.1 \, \text{cm} = 4.98 \, \text{cm}

Conclusion

After zero error correction, the measured diameter of the spherical object is 4.98cm4.98 \, \text{cm}.

Option Analysis

  • A) 5.18 cm: Incorrect as it does not account for the zero error.
  • B) 5.08 cm: This is the total reading before correcting for zero error; hence incorrect as a final answer.
  • C) 4.98 cm: This is the correct answer after correcting for zero error.
  • D) 5.00 cm: Incorrect as it arbitrarily presents a different value without proper justification.

Thus, the correct answer is C) 4.98 cm.

Found an issue with this question?