The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be 10-3 mm.
Answer
35
Solution
LC= pitchdivisions; p'=1.75 p, n'=0.5 n
LC'= 1.750.5 0.01=0.035 mm
=35 10-3 mm
Answer: 35