Question

A vernier callipers having 1 main scale division = 0.1 cm to have a least count of 0.02 cm.If n be the number of divisions on vernier scale and m be the length of vernier scale, then.

Answer 1

Complete Question:

A vernier calipers having 1 main scale division = 0.1 cm to have a least count of 0.02 cm.If n be the number of divisions on vernier scale and m be the length of vernier scale, then:

     a. n = 10, m = 0.5 cm

     b. n = 9, m = 0.4 cm

     c. n = 10, m = 0.8 cm

     d. n = 10, m = 0.2 cm

Given:

• 1 MSD (main scale division) = 0.1 cm
• 1 LC (least count) = 0.02 cm

To find:

• Number of divisions on vernier scale (VSD) = n
• Length of vernier scale = m

Answer:

•     LC = 1 MSD - 1 VSD

         ⇒ 1 VSD = 1 MSD - LC = 0.1 cm - 0.02 cm = 0.08 cm

• 1 VSD = $\frac{length of vernier scale}{number of divisions on vernier scale} = \frac{m}{n}$

• Now checking with the options:

            a. $\frac{m}{n} = \frac{0.5}{10} = 0.05 cm$  , which is not equal to 0.08 cm

            b. $\frac{m}{n} = \frac{0.4}{9} = 0.04 cm$ , which is not equal to 0.08 cm

            c. $\frac{m}{n} = \frac{0.8}{10} = 0.08 cm$ , which is equal to the required VSD

• Hence, n = 10 and m = 8.