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Q.

The least count of the main scale of a screw  gauage is 1mm. The minimum number of divisions on its circular scalre required to measure 5 μm diameter of the wire will be

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a

100

b

50

c

200

d

500

answer is C.

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Detailed Solution

detailed_solution_thumbnail

Value of M.S.D =1 mm
 L.C. =1MSD No.ofH.S.D 5×10−6=1×10−3N⇒N=10005⇒N=200

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