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Consider a vernier callipers in which ea...

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 on its circular scale In the vernier callipers 5 divisions of the vernier scale coincide with 4 divisions on the linear scale Then
(A) If the pitch of the screw gauge is twice the leat count of the Vernier callipers the least count of the screw gaguge is `0.01mm`
(B) If the pitch of the screw gauge is twice the leat count of the Vernier callipers the least count of the screw gaguge is `0.01mm`
(C)If the pitch of the screw gauge is twice the leat count of the Vernier callipers the least count of the screw gauge is `0.01mm` (D) If the pitch of the screw gauge is twice the leat count of the Vernier callipers the least count of the screw gaguge is `0.01mm` .

A

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the 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 the 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 the screw gauge is 0.01 mm

D

If the count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

Text Solution

Verified by Experts

The correct Answer is:
B, C

1 main scale division `(M.S.D) = (1)/(8) cm`
5 vernier scale division `(V.S.D) = 4 M.S` ltbr. `1 V.S.D = (4)/(5) M.S.D`
Least count of vernier scale (L.C) = 1 M.S.D - 1 V.S.D
`= 1 M.S.D=(4)/(5) M.S.D`
`(L.C)=(1M.S.D)/(5)=(1)/(40)cm`
If the pitch of the screw gauge is twice the least count of the vernier callipers then
pitch = `2 xx L.C` of vernier scale `= (1)/(20) cm`
hence least count of screw gauge `=("Pitch")/(100) = 0.005 m`
For option C and D
Least count of linear scale of screw gauge
`= 2 xx (1)/(40)=(1)/(20)cm`
Pitch `= 2 xx (1)/(20)=(1)/(10)=1mm`
Least count of screw gauge `= (1mm)/(100)=0.01 mm`
Hence answer is (B,C)
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