The least count of a screw guage 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

Step-by-step solution

aligned & Given least count of Screw Gauge =0.01 ~mm & L.C = ( pitch ) No. of circular turn = P N =0.01 ~mm & New pitch = P (1+0.75) N (1-0.5) = P N [ 1.75 0.5 ] & =(0.01) 3.5 & =0.035 ~mm & =35 10⁻³ ~mm aligned