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# A steel rail of length $5\,m$ and area of cross section $40\,c{m^2}$ is prevented from expanding along its length while the temperature rises by $10\,{}^ \circ C$. If coefficient of linear expansion and young’s modulus of steel are $1.2 \times {10^{ - 5}}\,{K^{ - 1}}$ and $2 \times {10^{11}}\,N{m^{ - 2}}$ respectively, the force developed in the rail is approximately: A) $2 \times {10^7}\,N$B) $3 \times {10^{ - 5}}\,N$C) $1 \times {10^5}\,N$D) $2 \times {10^9}\,N$

Last updated date: 14th Sep 2024
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Hint: The force developed in the rail can be determined by using the young’s modulus formula, the young’s modulus is the ratio of the stress and the strain. By using the stress and the strain formula in the young’s modulus formula, the force can be determined.

Formula used:
The young’s modulus is given by,
$Y = \dfrac{\sigma }{\varepsilon }$
Where, $Y$ is the young’s modulus of the material, $\sigma$ is the stress in the material and $\varepsilon$ is the strain in the material.
The stress of the material is given by,
$\sigma = \dfrac{F}{A}$
Where, $\sigma$ is the stress, $F$ is the force and $A$ is the area.
The strain of the material is given by,
$\varepsilon = \dfrac{{\Delta l}}{l}$
Where, $\varepsilon$ is the strain, $\Delta l$ is the change in the length and $l$ is the original length.

Complete step by step solution:
Given that,
The length of the rail is, $l = 5\,m$,
The Area of the cross section is, $A = 40\,c{m^2} = 40 \times {10^{ - 4}}\,{m^2}$,
The change in temperature is, $\Delta T = 10\,{}^ \circ C$,
The coefficient of the linear expansion is, $\alpha = 1.2 \times {10^{ - 5}}\,{K^{ - 1}}$,
The young’s modulus of the material is, $Y = 2 \times {10^{11}}\,N{m^{ - 2}}$.
The relation between the change in length and the change in the temperature is given by,
$\Delta l = l \times \alpha \times \Delta T$
By rearranging the terms in the above equation, then
$\dfrac{{\Delta l}}{l} = \alpha \times \Delta T\,...............\left( 1 \right)$
Now,
The young’s modulus is given by,
$Y = \dfrac{\sigma }{\varepsilon }$
By substituting the stress and strain formula in the above equation, then the above equation is written as,
$Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}$
By substituting the equation (1) in the above equation, then
$Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\alpha \times \Delta T}}$
By rearranging the terms in the above equation, then the above equation is written as,
$Y = \dfrac{F}{{A \times \alpha \times \Delta T}}$
By rearranging the terms in the above equation, then the above equation is written as,
$F = Y \times A \times \alpha \times \Delta T$
By substituting the young’s modulus, cross sectional area, coefficient of the linear expansion and change in temperature in the above equation, then
$F = 2 \times {10^{11}} \times 40 \times {10^{ - 4}} \times 1.2 \times {10^{ - 5}} \times 10$
By multiplying the terms in the above equation, then
$F = 9600\,N$
The above equation is also written as,
$F = 0.96 \times {10^4}\,N$
Then the force is approximately equal to,
$F \simeq 1 \times {10^5}\,N$

Hence, the option (C) is the correct answer.

Note: The force of the object is directly proportional to the young’s modulus of the material, cross sectional area of the material, coefficient of the linear expansion and the change in temperature. As the young’s modulus of the material, cross sectional area of the material, coefficient of the linear expansion and the change in temperature increases, then the force also increases.