In a Vernier calliper, 1 cm on the main scale is divided into 10 equal parts. It is given that 20 divisions of the Vernier scale coincide with 19 divisions of the main scale. When the two jaws of the instrument are brought into contact, the zero of the Vernier scale lies to the left of the zero of the main scale and the 14^ th Vernier scale division coincides with a main scale division. When a small block is held bet

Step-by-step solution

First, find the value of one main scale division (MSD): 1 MSD = 1 cm 10 = 1 mm Given that 20 VSD = 19 MSD , the value of one Vernier scale division is: 1 VSD = 19 20 MSD = 0.95 mm The least count (LC) is: LC = 1 MSD - 1 VSD = 1 mm - 0.95 mm = 0.05 mm When the jaws are closed, the Vernier zero is to the left of the main scale zero, indicating a negative zero error. The 14^ th division coincides, so the zero error is: e = -(N - n) LC e = -(20 - 14) 0.05 mm = -6 0.05 mm = -0.30 mm For the measurement of the block, the