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JEE Advanced 2025 · Paper 1, Physics question 4

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

Two Vernier configurations. Main scale: 0 to 1 cm divided into 10 divisions (1 mm each), continuing beyond 1 cm. The Vernier scale has 10 divisions whose total length equals 7 main-scale divisions (7 mm), so one Vernier division is 0.7 mm. Fig. 1: the Vernier zero coincides with the main-scale zero (no zero error). Fig. 2: the Vernier zero lies between the 1 mm and 2 mm marks of the main scale, and the 1st Vernier division coincides with the 2 mm main-scale mark.
Choose one option
Answer and worked solution

Official answer: C

Least count.

  • 1 MSD = 1 mm.
  • 10 VSD span 7 MSD, so 1 VSD = 0.7 mm.
  • LC = 1 MSD − 1 VSD = 0.3 mm.

Zero error. In Fig. 1 the zeros coincide, so there is none.

Reading (Fig. 2). The vernier zero lies just past the 1 mm mark, so the main-scale reading is 1 mm. The 1st vernier division coincides with a main-scale mark.
D=1 mm+1×0.3 mm=1.3 mm=0.13 cm.D=1\ \text{mm}+1\times0.3\ \text{mm}=1.3\ \text{mm}=0.13\ \text{cm}.

Topics:Vernier callipers and screw gaugeLeast count
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