JEE Advanced 2025 Paper 1 Q04 Physics Errors & Measurements Vernier and Screw Gauge Medium

JEE Advanced 2025 Paper 1 · Q04 · Vernier and Screw Gauge

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

  1. A. 0.12 cm
  2. B. 0.11 cm
  3. C. 0.13 cm
  4. D. 0.14 cm
Reveal answer + step-by-step solution

Correct answer:C

Solution

Read the Vernier scale carefully using the figures.

Step 1 — Least count and zero error from Fig. 1. The main-scale 1 cm is divided into 10 parts (1 MSD = 0.1 cm) and 10 Vernier divisions cover 9 MSDs, so $1\text{ VSD} = 0.9\text{ MSD}$ and the least count is $\text{LC} = 1\text{ MSD}-1\text{ VSD} = 0.01\text{ cm}$. In Fig. 1 (zero configuration) the $0$ of the Vernier sits to the right of the main-scale $0$; counting Vernier divisions to the next coincidence shows a positive zero error of $+0.02$ cm.

Step 2 — Reading from Fig. 2. Main-scale reading just before the Vernier $0$ is $0.1$ cm, and the Vernier division that lines up with a main-scale mark is the 5th, contributing $5\times0.01 = 0.05$ cm. Apparent reading $= 0.1 + 0.05 = 0.15$ cm.

Step 3 — Correct for zero error: $D = 0.15 - 0.02 = 0.13$ cm. Answer: (C).

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