JEE Advance - Physics (2013 - Paper 1 Offline - No. 12)
The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero
of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions
equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale
divisions. The diameter of the cylinder is
5.112 cm
5.124 cm
5.136 cm
5.148 cm
Wyjaśnienie
From the given data :
- One main scale division (MSD) is $ 5.15 \text{ cm} - 5.10 \text{ cm} = 0.05 \text{ cm} $.
- One Vernier scale division (VSD) is $ \frac{2.45 \text{ cm}}{50} = 0.049 \text{ cm} $.
The least count (LC) of the Vernier calipers is:
$$ \text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 \text{ cm} - 0.049 \text{ cm} = 0.001 \text{ cm} $$
Given the main scale reading (MSR) is 5.10 cm and the Vernier scale reading (VSR) is 24. The diameter $ D $ of the cylinder is calculated as:
$$ \begin{aligned} D & = \text{MSR} + \text{VSR} \times \text{LC} \\\\ & = 5.10 \text{ cm} + 24 \times 0.001 \text{ cm} \\\\ & = 5.124 \text{ cm} \end{aligned} $$