Consider a Vernier callipers in which each `1cm` on the main scale is divided into `8` equal divisions and a screw gauge with `100` divisions in its circular scale. In the Vernier callipers, `5` divisions of the Vernier scale coincide with `4` divisions on the scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :

Options

A. If the pitch of the screw gauge is twice the least count of the vernier callipers, the least count of screw gauge is 0.01 mm.
B. if the pitch of the screw gauge is twice the least count of the vernier callipers, the least count of screw gauge is 0.005 mm.
C. If the least count of the linear scale of the screw gauge is twice the least count of the vernier callipers, the least count of screw gauge is 0.01 mm.
D. If the least count of the linear scale of the screw gauge is twice the least count of the vernier calliper, the least count of screw gauge is 0.005 mm.

Answer

(b,c)

Explanation

Smallest division on main scale of vernier callipers
`=(1)/(8) cm = 0.125 cm`
5 division of the vernier scale
= 4 div. of main scale
`= 4xx 0.125 = 0.5 cm`
1 division of vernier scale `=(0.5)/(5) = 0.1 cm`
L.C of vernier callipers = 1 MSD - 1 VSD
=` 0.125 - 0.1 = 0.025 cm`
L.C. of screw gauge `= (pitch)/(100)`
If pitch of screw gauge is twice that L.C of vernier callipers, then L.C of screw gauge
`=(0.025xx2)/(100) = (0.05)/(100) cm = (0.5)/(100) m`
=`0.005 mm`
Hence option (b) is correct.
Also, If L.C of linear scale of screw gauge is twice the L.C. of vernier calliper, then L.C of linear scale of screw gauge
`= 2xx 0.025 = 0.05 cm`
pitch `= 2xx 0.05 = 0.1 cm =1 mm`
L.C of screw gauge `=(1mm)/(100) = 0.01m`
Hence option (c ) is correct.