One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10 . When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that 4^ th Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be 1 cm, the actual length of the wire is :

Step-by-step solution

The least count (LC) of the Vernier calliper is given by: LC = 1 mm 10 = 0.1 mm = 0.01 cm Since the zero of the Vernier scale shifts to the left of the zero of the main scale, the instrument has a negative zero error. The zero error is calculated as: Zero error = -4 LC = -4 0.01 cm = -0.04 cm The actual length of the wire is obtained by subtracting the zero error from the measured length: Actual length = Measured length - Zero error Actual length = 1.00 cm - (-0.04 cm ) = 1.04 cm Answer: 1.04 cm