A specially designed Vernier caliper has a least count of 0.005 cm . If 20 Vernier scale divisions are equal to 19 main scale divisions, what is the number of main scale divisions present in 1 cm of the main scale?

Step-by-step solution

Let the length of one main scale division be 1 MSD and one Vernier scale division be 1 VSD . Given that 20 VSD = 19 MSD , we have: 1 VSD = 19 20 MSD The least count (LC) is given by: LC = 1 MSD - 1 VSD LC = 1 MSD - 19 20 MSD = 1 20 MSD We are given that LC = 0.005 cm . Therefore: 1 20 MSD = 0.005 cm 1 MSD = 20 0.005 cm = 0.1 cm The number of main scale divisions in 1 cm is: Number of divisions = 1 cm 0.1 cm = 10 Answer: 10