Question

A steel rod of diameter $1\text{cm}$ is clamped firmly at each end when its temperature is ${25}^{\circ }\text{C}$ so that it cannot contract on cooling. The tension in the rod at $0^{\circ }\text{C}$ is approximately

$(\alpha =({10}^{-5}){}^{\circ }{\text{C}}^{-1},Y=2\times {10}^{11}{\text{Nm}}^{-2})$

• A. $4000\text{N}$
• B. $7000\text{N}$
• C. $7400\text{N}$
• D. $4700\text{N}$

Answer 1

The correct option is A $4000\text{N}$

$\text{Stress = Strain}\times Y$

Where, $Y$ is the Young's modulus of elasticity

$\text{Strain}=\frac{\Delta L}{L}=\alpha \left(\Delta \theta \right)$

Where, $\Delta \theta$ is the temperature difference

And, $\text{Sress}=\frac{\text{Tension in the rod}}{\text{Cross-sectional area}}=\frac{T}{\frac{\pi }{4}{d}^2}$

$d$ is the diameter of rod.

$\Rightarrow T=\frac{\pi }{4}\times d^2\times \text{Stress}$

$=\frac{\pi }{4}{d}^2\times Y\times \text{Strain}$

$=\frac{\pi }{4}{d}^2\times Y\times \alpha \left(\Delta \theta \right)$

$T=\frac{\pi }{4}\times ({10}^{-2}{)}^2\times 2\times {10}^{11}\times {10}^{-5}\times (25-0)$

$\Rightarrow T=3926.99\approx 4000\text{N}$

Hence, option $\left(\text{A}\right)$ is correct.