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On cm on the main scale of vernier callipers is divided into ten equal parts. If 20 division of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?

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20 div. of varnier scale=8div. Of main scale `rArr 1V.S..D. =((8)/(20))M.S.D.=((2)/(5))M.S.D`
Least count `=1 M.S.D.-1V.S.D.=1 M.S.D.-((2)/(5))M.S.D.=9=(1-(2)/(5))M.S.D.`
`=(3)/(5)M.S.D.=(3)/(5)xx0.1cm=0.06cm (because 1M.S.D.=-(1)/(10)cm=0.1 cm)`
Note: for objective question `L.C.=M-V=((b-a)/(b))M=((20-8)/(20))((1)/(10))cm=(3)/(50) vm=0.06cm`
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