Skip to content
Question 82 of 84

Q.A wire 10 m long has a cross-sectional area 1.25 x 10^-4 m^2. It is subjected to a load of 5 kg. If Young's Modulus of the material is 4 x 10^10 Nm^-2, calculate the elongation produced in the wire. (Take g = 10 ms^-2)

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2025Subjective· 3mImportance★★★★★
98% · 82/84 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Using dL = FL/(AY), with F=50N, L=10m, A=1.25e-4 m^2, Y=4e10 Nm^-2, the elongation is 1e-4 m = 0.1 mm.

Young's Modulus is defined as:

Y = stress/strain = (F/A) / (dL/L) = FL / (A dL)

Rearranging for the elongation dL:

dL = FL / (AY)

Given values:

Load, mass m = 5 kg, so applied force F = mg = 5 x 10 = 50 N (using g = 10 ms^-2 as instructed).

Length L = 10 m.

Cross-sectional area A = 1.25 x 10^-4 m^2.

Young's Modulus Y = 4 x 10^10 Nm^-2.

Substitute:

dL = (50 x 10) / (1.25 x 10^-4 x 4 x 10^10)

dL = 500 / (5 x 10^6)

dL = 1 x 10^-4 m …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.