Question
A steel wire of length 2 m and cross-section 1 mm2 is stretched by a 200 N force. If Young’s modulus Y=2×1011 Pa, find the elongation in millimetres.
Solution — Step by Step
The single equation that solves 90% of elasticity MCQs:
ΔL=AYFL
A=1 mm2=10−6 m2. Convert once and never again.
ΔL=10−6×2×1011200×2=2×105400=2×10−3 m=2 mm
Final answer: ΔL=2 mm.
Why This Works
Stress-strain problems live or die by unit consistency. Once you convert mm² to m² and N stays as N, Young’s modulus in pascals slots in cleanly and elongation comes out in metres.
The formula itself is just σ=Yε rearranged: AF=YLΔL.
Alternative Method
Compute stress first: σ=F/A=200/10−6=2×108 Pa. Then strain ε=σ/Y=10−3. Elongation =ε×L=10−3×2=2 mm. Same answer, cleaner intermediates.
For “two wires connected end-to-end” PYQs, total elongation =ΔL1+ΔL2. Same force, different L, A, Y. JEE Main 2022 Shift 2 asked exactly this.
Common Mistake
Forgetting to square the radius (or convert mm² properly). Writing A=1×10−3 m2 instead of 10−6 m2 gives an answer off by a factor of 1000. Always convert area in one explicit step.