Let $A$ represents the least count of a vernier caliper whose 10 main scale divisions are equal in length to 12 vernier scale divisions, and one main scale division is equal to 1 mm. Say $B$ represents the least count of a screw gauge with 25 circular scale divisions and a pitch of 1 mm, $A - B$ in mm is

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Solution

Given that,
$1 M S D = 1 m m$ and,
$12 \times 1 V S D = 10 \times 1 M S D \Rightarrow 1 V S D = \frac{10}{12} \times 1 M S D = \frac{5}{6} m m$
Least count of vernier calliper ( $A$ )
$= 1 M S D - 1 V S D = 1 m m - \frac{5}{6} m m = \frac{1}{6} m m = 0.167 m m$
Least count of screw guage $(B) = 1 C S D = \frac{p i t c h}{c i r c u l a r d i v i s i o n} = \frac{1}{25} = 0.04 m m$
$A - B = 0.167 m m - 0.04 m m = 0.127 m m$

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Let A represents the least count of a vernier caliper whose 10 main scale divisions are equal in length to 12 vernier scale divisions, and one main scale division is equal to 1 mm. Say B represents the least count of a screw gauge with 25 circular scale divisions and a pitch of 1 mm, A−B in mm is

Given that, 1 MSD=1 mm and, 12×1VSD=10×1MSD⇒1 VSD=1012×1 MSD=56 mm Least count of vernier calliper (A) =1 MSD−1 VSD=1 mm−56 mm=16 mm=0.167 mm Least count of screw guage (B)=1CSD=pitchcircular division=125=0.04 mm A−B=0.167 mm−0.04 mm=0.127 mm