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 :
A. 1.04 cm
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
Related: Physics — Experimental Physics · All PYQ Banks