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JEE Advanced 2025 Paper 1, Physics Section 1 Q4: Measurements

JEE Advanced2025Paper 1Physics Section 1
Q.

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter of a tube. The measured value of is:

  1. A

    0.12 cm

  2. B

    0.11 cm

  3. C

    0.13 cm

  4. D

    0.14 cm

Solution

From Fig. 1: 10 main scale divisions equal 1 cm, so 1 MSD = 0.1 cm. Also, 10 VSD coincide with 7 MSD, giving 1 VSD = 0.07 cm. The least count is cm, and there is a zero error of cm (Vernier zero coincides with main-scale zero but seven Vernier divisions match seven main-scale divisions, so the zero correction subtracts one VSD).

From Fig. 2: main scale reading just before the Vernier zero is 0.2 cm. Applying the correction:

cm

Hence option (C).

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