Given, y=2×1011Nm−2 Stress (AF)=5×107Nm−2ΔV=0.02%=2×10−4 m3rΔr=? γ= strain stress ⇒strain(ℓ0Δℓ)= stress γΔV=2πrℓ0Δr−πr2Δℓ From eqns (i) and (ii) putting the value of Δℓ,ℓ0 and ΔV and solving we get rΔr=0.25×10−4
JEE Main 2013 — Physics Mechanics
A uniform wire (Young's modulus 2×1011Nm−2 ) is subjected to longitudinal tensile stress of 5×107Nm−2. If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is close to:
Held on 22 Apr 2013 · Verified 6 Jul 2026.
1.0×10−4
1.5×10−4
0.25×10−4
5×10−4
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