Question:
10: There are 20 divisions in 4 cm of the main scale . The vernier scale has 10 divisions . The least count of the instrument is
Answer:
The least count of the vernier callipers is given by L.C. = 1 MSD – 1 VSD
Main scale –
20 main scale divisions = 4cm
Hence, 1 MSD = (4/20) cm = 1/5 cm
Vernier scale –
- Case 1 Consider 10 VSD coincides with 5 MSD (1 cm)
Hence, 1 VSD = (5/10) MSD = 5/10 * (1/5) cm = 1/10 cm = 0.1 cm
Least count = 1 MSD – 1 VSD = 1/5 cm – 0.1 cm = 0.2 – 0.1 = 0.1 cms
- Case 2 Consider 10 VSD coincides with 9 MSD
Hence, 1 VSD = (9/10) MSD = 9/10 * (1/5) cm = 9/50 cm = 0.18 cm
Least count = 1 MSD – 1 VSD = 1/5 cm – 0.18cm = 0.2 – 0.18 = 0.02 cms
- Case 3 Consider 5 VSD coincides with 4 MSD
Hence, 1 VSD = (4/5) MSD = 4/5 * (1/5) cm = 4/25 cm = 0.16 cm
Least count = 1 MSD – 1 VSD = 1/5 cm – 0.16 cm = 0.2 – 0.16 = 0.04 cms
In the problem, if the number of divisions of MS coinciding with the number of divisions of VS has to be given to proceed to the correct answer
Not what you are looking for? Go ahead and submit the question, we will get back to you.
