Skip to content
Exercises · 10.9

Q.A brass wire 1.8 m1.8\ \text{m} long at 27 ∘C27\ ^\circ\text{C} is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of −39 ∘C-39\ ^\circ\text{C}, what is the tension developed in the wire, if its diameter is 2.0 mm2.0\ \text{mm}? Coefficient of linear expansion of brass =2.0×10−5 K−1= 2.0 \times 10^{-5}\ \text{K}^{-1}; Young's modulus of brass =0.91×1011 Pa= 0.91 \times 10^{11}\ \text{Pa}.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★est
31% · 17/55 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 →

Cooling the clamped wire prevents its contraction, producing a tensile thermal stress σ=Y α ∣ΔT∣\sigma = Y\,\alpha\,|\Delta T|. The tension is F=σA=Y α ∣ΔT∣ πr2≈3.8×102 NF = \sigma A = Y\,\alpha\,|\Delta T|\,\pi r^2 \approx 3.8\times10^{2}\ \text{N}.

Why this approach works

A wire free to contract simply shortens when cooled, with no stress. Held rigidly at both ends, its length is fixed, so the contraction it would have undergone, α ∣ΔT∣\alpha\,|\Delta T|, is held in the wire as a strain. Young's modulus converts that strain into a tensile stress, and stress times cross-sectional area gives the tension.

F=Y α ∣ΔT∣  πr2F = Y\,\alpha\,|\Delta T|\;\pi r^2

Step 1 - Temperature change

ΔT=(−39)−27=−66 K,∣ΔT∣=66 K\Delta T = (-39) - 27 = -66\ \text{K}, \qquad |\Delta T| = 66\ \text{K}

Step 2 - Prevented (thermal) strain

α ∣ΔT∣=(2.0×10−5)(66)=1.32×10−3\alpha\,|\Delta T| = (2.0\times10^{-5})(66) = 1.32\times10^{-3}

Step 3 - Thermal stress

σ=Y α ∣ΔT∣=(0.91×1011)(1.32×10−3)=1.20×108 Pa\sigma = Y\,\alpha\,|\Delta T| = (0.91\times10^{11})(1.32\times10^{-3}) = 1.20\times10^{8}\ \text{Pa} …

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.