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:
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:
Show Hint
- \( 0.12 \text{ cm} \)
- \( 0.11 \text{ cm} \)
- \( 0.13 \text{ cm} \)
\( 0.14 \text{ cm} \)
The Correct Option is A
Solution and Explanation
Here's how to read the Vernier scale and find the correct answer:
1. Main Scale Reading: Determine the main scale reading just before the zero mark of the Vernier scale. In Figure 2, the zero of the Vernier scale lies just after the 0.1 cm mark on the main scale. Thus, main scale reading = 0.1 cm.
2. Vernier Coincidence: Find the Vernier scale division that exactly coincides with a main scale division. This is the most crucial part. In the image, the 2nd division of the Vernier scale coincides with a main scale division.
3. Least Count of Vernier Scale: The number of divisions of the main scale in 1 cm is 10. The number of divisions of the Vernier scale is 10, each division being a tenth of the centimeter. Then, \[ \text{Least Count} = \text{Value of 1 Main Scale Division} - \text{Value of 1 Vernier Scale Division} = 0.1 - 0.09 = 0.01\, \text{cm}. \]
4. Final Reading: The final reading is the main scale reading plus (Vernier coincidence \(\times\) least count). In this case, \[ 0.1\, \text{cm} + (2 \times 0.01\, \text{cm}) = 0.1\, \text{cm} + 0.02\, \text{cm} = 0.12\, \text{cm}. \] Therefore, the answer is \(\boxed{\text{(A) } 0.12\, \text{cm}}\).
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