A metal bar of length \(L\) and area of cross-section \(A\) is clamped between two rigid supports. For the material of the rod, it's Young’s modulus is \(Y\) and the coefficient of linear expansion is \(\alpha.\) If the temperature of the rod is increased by \(\Delta t^{\circ} \text{C},\) the force exerted by the rod on the supports will be:
1. \(YAL\Delta t\)
2. \(YA\alpha\Delta t\)
3. \(\frac{YL\alpha\Delta t}{A}\)
4. \(Y\alpha AL\Delta t\)

The temperature of a wire of length \(1~\text{m}\) and an area of cross-section \(1~\text{cm}^2\) is increased from \(0^{\circ} \text {C}\) to \(100^{\circ} \text {C}.\) If the rod is not allowed to increase in length, the force required will be:
\((\alpha = 10^{-5}/ ^{\circ} \text {C} ~\text{and} ~Y = 10^{11} ~\text{N/m}^2)\)

A constrained steel rod of length \(l\), area of cross-section \(A\), Young's modulus \(Y\) and coefficient of linear expansion \(\alpha\)$$ is heated through \(t^{\circ}\text{C}\)$$. The work that can be performed by the rod when heated is:
1. \((YA\alpha t)(l\alpha t)\)
2. \(\frac{1}{2}(YA\alpha t)(l\alpha t)\)
3. \(\frac{1}{2}(YA\alpha t)\frac{1}{2}(l\alpha t)\)
4. \(2(YA\alpha t)(l\alpha t)\)

A brass wire \(1.8~\text m\) long at \(27^\circ \text C\) is held taut with a little tension between two rigid supports. If the wire is cooled to a temperature of \(-39^\circ \text C,\) what is the tension created in the wire?
( Assume diameter of the wire to be \(2.0~\text{mm}\) , coefficient of linear expansion of brass \(=2.0 \times10^{-5}~\text{K}^{-1},\) Young's modulus of brass\(=0.91 \times10^{11}~\text{Pa}\) )
1. \(3.8 \times 10^3~\text N\) 
2. \(3.8 \times 10^2~\text N\) 
3. \(2.9 \times 10^{-2}~\text N\) 
4. \(2.9 \times 10^{2}~\text N\) 

The pressure applied from all directions on a cube is P. The volume elasticity of the cube is β and the coefficient of volume expansion is α. How much should its temperature be raised to maintain the original volume?

1. $\frac{P}{\alpha \beta }$                                 

2. $\frac{P\alpha }{\beta }$

3. $\frac{P\beta }{\alpha }$                                 

4. $\frac{\alpha \beta }{P}$