Question

Consider a Vernier calipers in which each 1 cm 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 main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:

• 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 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 0.005 mm.

Answer 1

Answer: (B), (C)

The correct options are
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.

Vernier Calipers:
$1cm$ is divided into $8$ divisions, $1$ main scale division is given as $\frac{1}{8}cm$
L.C. of Vernier is given as $1M.S.D-1V.S.D$
Given 5 Vernier scale coincides with $4$ main scale divisions:
$5V.S.D=4M.S.D$
$L.C=$$1M.S.D-\frac{4}{5}M.S.D$
$L.C=$$1M.S.D-\frac{4}{5}M.S.D=\frac{1}{5}M.S.D=\frac{1}{40}cm$
Checking option A and B:
Given pitch of Screw gauge is 2 times L.C. of V.S
$p=2\times \frac{1}{40}cm=\frac{1}{20}cm$
The least count of screw gauge is $\frac{pitch}{no.ofdivisionsoncircularscale}$
No. of divisions on circular scale $=100$$=\frac{\frac{1}{20}}{100}=0.0005cm=0.005mm$
Hence, option B is correct.
Checking Option C and D:
Given least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers.
One complete rotation of the circular scale moves it by two divisions on the linear scale, so pitch is 2 times linear scale division of screw gauge,
Pitch $=2\times$ linear scale division of screw gauge $=2\times 2\times$ $L.C$ of $V.S=4\times L.C$ of $V.S$ $=4\times \frac{1}{40}=\frac{1}{10}$
The least count of screw gauge is $\frac{pitch}{no.ofdivisionsoncircularscale}$
No. of divisions on circular scale $=100$
$L.C=\frac{\frac{1}{10}}{100}=0.001cm=0.01mm$
Option C is correct .
Hence, option B, C is correct.