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 4th Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be 1 cm, the actual length of the wire is :
- 1.04 cm
- 0.60 cm
- 0.96 cm
- 1.00 cm
Answer
Solution
Least count LC = 1 mm10 = 0.1 mm = 0.01 cm.
When the zero of the Vernier scale lies to the left of the main scale zero, it represents negative zero error:
Negative zero error = -(10 - 4) LC = -6 0.01 cm = -0.06 cm
Actual length = Measured length - (Zero error) = 1.00 cm - (-0.06 cm) = 1.06 cm
Since 1.06 cm is not present in the options, no option is correct.