JEE Advanced 2025 — Paper 1 — PHY-I solution

JEE Advanced2025Paper 1PHY-I

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

A
0.12 cm
B
0.11 cm
C
0.13 cm
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 $1 MSD - 1 VSD = 0.03$ cm, and there is a zero error of $- 0.07$ 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:

$D = 2 MSD - 1 VSD = 0.20 - 0.07 = 0.13$ cm

Hence option (C).

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