The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7ᵗʰ division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and 4ᵗʰ VSD coincides with a main scale division. The length of the cylinder is : (VSD is vernier scale division)

Solution

Physics sir · 4 steps

Find the least count, read the scale, then subtract the positive zero error. Do not add it.

1. Least count. 10 VSD fit into 9 MSD, so 1 VSD = 0.9 mm. The least count is 1 MSD - 1 VSD = 0.1 mm = 0.01 cm.
2. Zero error. With the jaws closed, the vernier zero sits to the right of the main scale zero, so the zero error is positive. The 7th division coincides, so the zero error is +7 × 0.01 = +0.07 cm.
3. Observed reading. The main scale reading is 3.1 cm and the 4th VSD coincides. The observed reading is 3.1 + 4 × 0.01 = 3.14 cm.

   L = 3.14 cm - 0.07 cm = 3.07 cm

4. So. A positive zero error makes every reading too large, so you subtract it. The length of the cylinder is 3.07 cm, option (c).

The idea to keep. True value = observed reading minus zero error, with the sign included. If the vernier zero sits right of the main zero, the error is positive and you subtract it.

Answer from the NTA final answer key. Solution by PrepWiser.