A steel rod with y = 2.0 \times 1011 Nm - 2 and \alpha = 10 - 5 ^\circ C - 1 of length 4 m and area of cross-section 10 cm2 is heated from 0 ^\circ C to 400 ^\circ C without being allowed to extend. The tension produced in the rod is x \times 105 N where the value of x is \underline\hspace2cm .

Properties of Matter PYQ — JEE Main 2021 · 1 Sep · Shift 2 · Q72 with Solution

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Correct answer: 8

Young’s modulus: $Y = 2.0 \times 10^{11}\,\text{N m}^{-2}$
Coefficient of linear expansion: $\alpha = 10^{-5}\,^{\circ}\text{C}^{-1}$
Length of rod: $L = 4\,\text{m}$
Cross-sectional area: $A = 10\,\text{cm}^2 = 10 \times 10^{-4} = 10^{-3}\,\text{m}^2$
Temperature rise: $\Delta T = 400^{\circ}\text{C}$

The rod is not allowed to expand, so thermal expansion is completely prevented.

If the rod were free, the thermal strain would be

\epsilon_{\text{thermal}} = \alpha \Delta T

So,

\epsilon_{\text{thermal}} = 10^{-5} \times 400 = 4 \times 10^{-3}

When expansion is prevented, an equal and opposite mechanical strain is produced. Hence,

\text{stress} = Y \alpha \Delta T

Substitute values:

\sigma = 2.0 \times 10^{11} \times 10^{-5} \times 400

\sigma = 2.0 \times 10^{11} \times 4 \times 10^{-3}

\sigma = 8 \times 10^{8}\,\text{N m}^{-2}

Force developed is

F = \sigma A

F = 8 \times 10^{8} \times 10^{-3}

F = 8 \times 10^{5}\,\text{N}

Thus, the force is of the form $x \times 10^5\,\text{N}$ , so

x = 8

\boxed{8}

This is actually a compressive force/stress in the rod, though the question uses the word tension.