Cylindrical bars P and Q have identical lengths and radii, but are composed of different linear elastic materials. The Young's modulus and coefficient of thermal expansion of Q are twice the corresponding values of P. Assume the bars to be perfectly bonded at the interface, and their weights to be negligible.

The bars are held between rigid supports as shown in the figure and the temperature is raised by AT. Assume that the stress in each bar is homogeneous and uniaxial. Denote the magnitudes of stress in P and Q by σ₁,and σ₂, respectively.

Which of the statement(s) given is/are CORRECT?

Correct Statements:
1. The interface between P and Q moves to the left after heating
2. ${\sigma }_1={\sigma }_2$

Step-by-Step Explanation:

1. Analysis of Stress at the Interface (Equilibrium Condition)
From the given schematic diagram, the cylindrical bars P (on the left) and Q (on the right) are connected in series between two rigid, unyielding supports. The interface is perfectly bonded.
Let:
- Length of both bars: $L_P=L_Q=L$
- Cross-sectional area of both bars (since they have identical radii): $A_P=A_Q=A$
- For bar P: Young's modulus is $E_P=E$, and the thermal expansion coefficient is ${\alpha }_P=\alpha$.
- For bar Q: Young's modulus is $E_Q=2E$, and the thermal expansion coefficient is ${\alpha }_Q=2\alpha$.

When the temperature is raised by $\Delta T$, the bars try to expand, but are constrained by the rigid supports. This induces a compressive reaction force $R$ at the boundaries.
For static equilibrium of the system:
$R_P=R_Q=R$
The uniaxial stress in each bar is defined as the reaction force divided by the cross-sectional area:
- Stress in bar P: ${\sigma }_1=\frac{R}{A}$
- Stress in bar Q: ${\sigma }_2=\frac{R}{A}$
Since the axial force $R$ and cross-sectional area $A$ are identical for both bars, the stress magnitudes are equal:
${\sigma }_1={\sigma }_2$

2. Determination of the Movement of the Interface
The net deformation of each bar consists of thermal expansion and mechanical compression:
- Change in length of bar P:
$\Delta L_P=L\alpha \Delta T-\frac{RL}{AE}$
- Change in length of bar Q:
$\Delta L_Q=L\left(2\alpha \right)\Delta T-\frac{RL}{A\left(2E\right)}=2L\alpha \Delta T-\frac{RL}{2AE}$

Since the outer supports are rigid, the total change in length must be zero:
$\Delta L_P+\Delta L_Q=0$
Substituting the expressions:
$\left(L\alpha \Delta T-\frac{RL}{AE}\right)+\left(2L\alpha \Delta T-\frac{RL}{2AE}\right)=0$
Simplifying the equation:
$3L\alpha \Delta T=\frac{3RL}{2AE}$
Solving for the reaction force $R$:
$R=2AE\alpha \Delta T$

Now, substitute the value of $R$ back into the change in length of bar P:
$\Delta L_P=L\alpha \Delta T-\frac{\left(2AE\alpha \Delta T\right)L}{AE}=L\alpha \Delta T-2L\alpha \Delta T=-L\alpha \Delta T$

Similarly, for bar Q:
$\Delta L_Q=2L\alpha \Delta T-\frac{\left(2AE\alpha \Delta T\right)L}{2AE}=2L\alpha \Delta T-L\alpha \Delta T=L\alpha \Delta T$

Because $\Delta L_P$ is negative (contraction) and $\Delta L_Q$ is positive (expansion), bar Q expands into the space of bar P. As a result, the bonded interface between P and Q moves to the left by a distance of $L\alpha \Delta T$ after heating.