As we know, Y=lΔlAF
⇒F=lYAΔl
For both cases force applied is the same, therefore
(lYAΔl)1=(lYAΔl)2
⇒Δl1Δl2=A2A1×l1l2
⇒0.2Δl2=π(2.4R)2πR2×12
⇒Δl2=6.9×10−2mm
JEE Main 2023 — Physics Mechanics
A force is applied to a steel wire A, rigidly clamped at one end. As a result elongation in the wire is 0.2mm. If same force is applied to another steel wire B of double the length and a diameter 2.4 times that of the wire A, the elongation in the wire B will be (wires having uniform circular cross sections)
Held on 30 Jan 2023 · Verified 6 Jul 2026.
6.06×10–2mm
2.77×10–2mm
3.0×10–2mm
6.9×10–2mm
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