Question

A steel rod of length 1 m and area of cross-section is heated from 0 °C to 200 °C, without being allowed to extend or bend. The tension produced in the rod is

(Given : Young’s modulus of steel $= 2 \times 10^{\text{11}}\text{ N }\text{m}^{- 2}$ and coefficient of linear expansion of steel $= 10^{- 5}\ {^\circ}C^{- 1})$

• A. $\text{4 } \times \text{ 1}\text{0}^{3}\text{ N}$
• B. $\text{4 } \times \text{ 1}\text{0}^{4}\text{ N}$
• C. $\text{4 } \times \text{ 1}\text{0}^{5}\text{ N}$
• D. $\text{4 } \times \text{ 1}\text{0}^{6}\text{ N}$

Answer 1

Answer: (B)

$\text{4 } \times \text{ 1}\text{0}^{4}\text{ N}$

Answer 2

: Let $\Delta L$be increase in the length of the rod due to increase in temperature of the rod. Then

$\Delta L = L\alpha\Delta T$

Where $\alpha$is the coefficient of the linear expansion, $\Delta T$is the rise in temperature and L is the length of the rod.

$\therefore\frac{\Delta L}{L} = \alpha\Delta T$

Let the compressive tension of the rod be T and A be cross-section area. Then

$Y = \frac{T/A}{\Delta L/L}$

$\therefore T = Y\frac{\Delta L}{L}A = Y \times \alpha\Delta T \times A$ [Using (i)]

Here, $Y = 2 \times 10^{11}Nm^{- 2}$

$\alpha = 1{0^{- 5}}^{0}C^{- 1}$

$\Delta T = 200^{0}C - 0^{0}C = 200^{0}C$

$L = 1m,A = 1cm^{2} = 1 \times 10^{- 4}m^{2}$

$\therefore T = 2 \times 10^{11}Nm^{- 2} \times 1{0^{- 5}}^{0}C^{- 1} \times 200^{0}C \times 1 \times 10^{- 4}m^{2} = 4 \times 10^{4}N$