As we know, change in length is given by ΔL=AYFL
Therefore, strain required LΔL=AYF
Now the ratio will be, L2ΔL2L1ΔL1=F2F1=T2T1=1030=3
JEE Main 2024 — Physics Mechanics
One end of a metal wire is fixed to a ceiling and a load of 2kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1kg hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be
[Area of cross section of wire =0.005cm2,Y=2×1011Nm−2 and g=10ms−2]
Held on 1 Feb 2024 · Verified 6 Jul 2026.
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