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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A structural steel cable of length 5 m and cross-sectional area 4 ×...

A structural steel cable of length 5 m5 \text{ m} and cross-sectional area 4×10−4 m24 \times 10^{-4} \text{ m}^2 is rigidly clamped at both ends between two massive pillars. The cable is initially at a temperature of 40∘C40^\circ\text{C}. If the temperature of the surroundings drops to 15∘C15^\circ\text{C}, what is the tension developed in the cable?

(Given: Young's modulus of steel =2.0×1011 N m−2= 2.0 \times 10^{11} \text{ N m}^{-2}, coefficient of linear expansion of steel = 1.2 \times 10^{-5} \, ^\circ\text{C}^{-1})

Options

  1. A1.2×105 N1.2 \times 10^5 \text{ N}
  2. B2.4×104 N2.4 \times 10^4 \text{ N}
  3. C4.8×103 N4.8 \times 10^3 \text{ N}
  4. D3.84×104 N3.84 \times 10^4 \text{ N}

Correct answer

B. 2.4×104 N2.4 \times 10^4 \text{ N}

Step-by-step solution

Given:

A=4×10−4 m2A = 4 \times 10^{-4} \text{ m}^2

ΔT=40∘C−15∘C=25∘C\Delta T = 40^\circ\text{C} - 15^\circ\text{C} = 25^\circ\text{C}

Y=2.0×1011 N m−2Y = 2.0 \times 10^{11} \text{ N m}^{-2}

\alpha = 1.2 \times 10^{-5} \, ^\circ\text{C}^{-1}

The thermal stress developed in the cable is given by YαΔTY \alpha \Delta T.

The tension FF developed in the cable is:

F=YAαΔTF = Y A \alpha \Delta T

Substituting the values:

F=(2.0×1011)×(4×10−4)×(1.2×10−5)×25F = (2.0 \times 10^{11}) \times (4 \times 10^{-4}) \times (1.2 \times 10^{-5}) \times 25

F=240×102 NF = 240 \times 10^2 \text{ N}

F=2.4×104 NF = 2.4 \times 10^4 \text{ N}

Answer: 2.4×104 N2.4 \times 10^4 \text{ N}
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