A steel wire of diameter 0.5mm and Young’s modulus 2×1011N/m2 carries a load of mass M. The length of the wire with the load is 1.0m. A Vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0mm , is attached. The 10 divisions of the Vernier scale correspond to 9 divisions of the main scale. Initially, the zero of Vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2kg , the Vernier scale division which coincides with a main scale division is . Take g=10m/s2 and π=3.2 .
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Detailed Solution
LC of vernier=1-910mm=0.1mm & d=0.5mm Y=2×1011Nm-2 l=1m Elongation =e =Fl AY∆e=∆F l A YΔe=FlAY=mglπd24Y=1.2×10×1π4×5×10−42× 2×1011Δe=1.2×103.24×25×10−8×2×1011=120.8×25×2×103=1240×103=0.3mmSince the least count of Vernier caliper is 0.1mm, so 3rd division of a Vernier scale will coincide with main scale.