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Properties of Solids and Liquids: Physics | JEE Main

A steel rod of length

1\text{ m}

and cross sectional area

10^{-4}\text{ m}^2

is heated from

0^{\circ}\text{C}

to

200^{\circ}\text{C}

without being allowed to extend or bend. The compressive tension produced in the rod is

\underline{\quad\quad\quad\quad} \times 10^4\text{ N}

. (Given Young's modulus of steel

= 2 \times 10^{11}\text{ Nm}^{-2}

, coefficient of linear expansion

= 10^{-5}\text{K}^{-1}

.

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Question type: Numerical
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Does the temperature change need conversion to Kelvin before subtracting?

No, a temperature difference in Celsius ( $\Delta T = 200 - 0 = 200^{\circ}\text{C}$ ) is identical in magnitude to a temperature difference in Kelvin ( $\Delta T = 200\text{ K}$ ).

Answer

The compressive force produced in the rod is $4 \times 10^4\text{ N}$ , so the value to fill in the blank is $4$ .

Why each option works or fails

Step-by-step solution

given: Length $L = 1\text{ m}$ , Area $A = 10^{-4}\text{ m}^2$ , $\Delta T = 200^{\circ}\text{C} - 0^{\circ}\text{C} = 200\text{ K}$ , $Y = 2 \times 10^{11}\text{ N/m}^2$ , $\alpha = 10^{-5}\text{ K}^{-1}$ .
find: The compressive tension (force) $F$ produced in the rod in units of $10^4\text{ N}$ .
visualise: The rod is fixed between two rigid supports. As the temperature rises, it attempts to expand by $\Delta L = L\alpha\Delta T$ . The rigid supports prevent this expansion entirely, creating an equivalent compressive strain $\frac{\Delta L}{L} = \alpha \Delta T$ and thermal stress $\sigma = Y\alpha\Delta T$ .
strategise: Thermal stress is $\sigma = Y\alpha\Delta T$ . The compressive force exerted is $F = \sigma A = Y A \alpha \Delta T$ .
execute: $F = (2 \times 10^{11}) \times (10^{-4}) \times (10^{-5}) \times 200 = 40000\text{ N} = 4 \times 10^4\text{ N}$ . Thus, the pre-factor is 4.
verify: Dimensions of $Y A \alpha \Delta T$ : $(\text{N}/\text{m}^2) \times \text{m}^2 \times \text{K}^{-1} \times \text{K} = \text{N}$ . The prefactor evaluates cleanly to 4.