The Vernier constant of Vernier callipers is 0.1 mm and it has zero error of (-0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be _________ × 10⁻² cm.

Solution

Physics sir · 3 steps

Read the callipers first, then subtract the zero error with its sign, and a negative error adds on.

1. Same units. The vernier constant is 0.1 mm = 0.01 cm. Work in centimetres throughout, since the main scale reading and the zero error are both in cm.
2. Observed reading. Reading = MSR + (coinciding division) × VC = 1.7 + 5 × 0.01 = 1.75 cm.

   Corrected = Reading - (zero error) = 1.75 - (-0.05) = 1.80 cm

3. So. The corrected diameter is 1.80 cm = 180 × 10⁻² cm, so the answer is 180.

The common slip: 170. Subtracting 0.05 cm instead of -0.05 cm gives 1.70 cm, that is 170. A negative zero error must be added back.

The idea to keep. Always subtract the zero error, keeping its sign. A negative zero error means the instrument reads too little, so the correction adds to the reading.

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