Question

A rod of mass M, area of cross section A and length $\ell_0$ is connected with a spring as shown in figure. If coefficient of linear expansion of rod is $\alpha$ and initially no extension was there in the spring of spring constant k, then the stress developed in rod when its temperature is increased by $\Delta T$, is [Young's Modulus of material of the rod is Y]

• A. $\frac{k[Y\alpha\Delta T]\ell_0}{[k\ell_0+YA]}$
• B. $Y\alpha\Delta T$
• C. $[\frac{k\ell_0+YA}{k\ell_0}]Y\alpha\Delta T$
• D. $[\frac{k\ell_0+Y}{k\ell_0}]Y\alpha\Delta T$

Answer 1

Answer: (A)

$\frac{k[Y\alpha\Delta T]\ell_0}{[k\ell_0+YA]}$

Answer 2

1. Identify the physical phenomena:

The rod undergoes thermal expansion due to temperature increase and elastic compression due to the force from the spring. The spring also undergoes elastic compression.

2. Formulate the thermal expansion:

If the rod were free to expand, its increase in length due to a temperature rise $\Delta T$ would be: $\Delta \ell_{thermal} = \ell_0 \alpha \Delta T$

3. Formulate the elastic deformation:

Let $F$ be the compressive force developed in the rod and also the force exerted by the spring. Due to this compressive force $F$, the rod itself will be compressed. The compression in the rod is given by Young's Modulus formula ($Y = \frac{F/A}{\Delta \ell_{rod}/\ell_0}$): $\Delta \ell_{rod} = \frac{F \ell_0}{YA}$

The spring will also be compressed by the force $F$. The compression in the spring is given by Hooke's Law ($F = kx$): $\Delta \ell_{spring} = \frac{F}{k}$

4. Apply the constraint:

The rod and spring are connected between two fixed supports. This means the total length of the rod-spring system must remain constant. The potential expansion of the rod due to temperature increase must be accommodated by the compression of the rod itself and the compression of the spring. Therefore, the thermal expansion of the rod must be equal to the sum of the elastic compressions in the rod and the spring: $\Delta \ell_{thermal} = \Delta \ell_{rod} + \Delta \ell_{spring}$ $\ell_0 \alpha \Delta T = \frac{F \ell_0}{YA} + \frac{F}{k}$

5. Solve for the force F:

Factor out $F$: $\ell_0 \alpha \Delta T = F \left( \frac{\ell_0}{YA} + \frac{1}{k} \right)$ Combine the terms in the parenthesis: $\ell_0 \alpha \Delta T = F \left( \frac{k\ell_0 + YA}{YAk} \right)$ Now, solve for $F$: $F = \frac{YAk \ell_0 \alpha \Delta T}{k\ell_0 + YA}$

6. Calculate the stress:

Stress ($\sigma$) is defined as force per unit area: $\sigma = \frac{F}{A}$ Substitute the expression for $F$: $\sigma = \frac{1}{A} \left( \frac{YAk \ell_0 \alpha \Delta T}{k\ell_0 + YA} \right)$ $\sigma = \frac{Yk \ell_0 \alpha \Delta T}{k\ell_0 + YA}$

This matches option (A).