The one division of main scale of Vernier callipers reads 1 mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2 cm and 6ᵗʰ Vernier division coincides scale division. The diameter of the bob will be ____________ × 10⁻² cm.

Solution

Physics sir · 4 steps

Read the bob, then take off the zero error: the vernier was already 0.04 cm ahead before you started.

1. Least count. 10 vernier divisions equal 9 main scale divisions, so 1 VSD = 0.9 mm. The least count is 1 - 0.9 = 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. That is a positive zero error. The 4th division coincides, so the error is +4 × 0.01 = +0.04 cm.
3. Observed reading. The main scale reading is 4.1 cm and the 6th vernier division coincides. The observed reading is 4.1 + 6 × 0.01 = 4.16 cm.

   d = 4.16 - (+0.04) = 4.12 cm

4. So. The diameter of the bob is 4.12 cm = 412 × 10⁻² cm, so the answer is 412.

The common slip: 420. Adding the zero error instead of subtracting it gives 4.16 + 0.04 = 4.20 cm, that is 420, not 412.

The idea to keep. True reading = observed reading minus zero error, with its sign. A vernier zero to the right of the main zero means a positive error, so you subtract it.

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