A rod of length l is fixed between two rigid walls having thermal coefficient alpha, heated by a temperature difference of delta T, bend in a form of circular with radius R what will be the expression for R

When the rod of length \( l \) is heated by a temperature difference \( \Delta T \), it undergoes thermal expansion. Since it is fixed between two rigid walls, the rod cannot expand freely. This constraint leads to bending, forming an arc of a circle with radius \( R \).

### Step 1: Thermal Expansion and Change in Length

The increase in length \( \Delta l \) of the rod due to thermal expansion is given by:

\[
\Delta l = \alpha l \Delta T
\]

where:
- \( \alpha \) is the coefficient of linear expansion,
- \( l \) is the original length of the rod,
- \( \Delta T \) is the temperature change.

Thus, the new length of the rod becomes:

\[
l' = l + \Delta l = l (1 + \alpha \Delta T)
\]

### Step 2: Relating Arc Length and Circular Geometry

When the rod bends into a circular arc of radius \( R \), the arc length \( s \) is given by:

\[
s = R \theta
\]

where \( \theta \) is the central angle subtended by the arc at the center of the circle.

Since the rod retains its expanded length in the form of an arc,

\[
R \theta = l (1 + \alpha \Delta T)
\]

### Step 3: Chord Length and Circular Relation

The initial straight length of the rod before bending is the chord of the circle, which remains \( l \). From the geometry of a circle, the chord length \( l \) is related to the radius \( R \) and central angle \( \theta \) by:

\[
l = 2 R \sin \frac{\theta}{2}
\]

### Step 4: Solving for \( R \)

Dividing the two equations:

\[
\frac{R \theta}{l} = (1 + \alpha \Delta T)
\]

\[
\theta = \frac{l (1 + \alpha \Delta T)}{R}
\]

Using the small-angle approximation for small \( \alpha \Delta T \) (i.e., \( \sin x \approx x \) for small \( x \)):

\[
l \approx 2 R \frac{\theta}{2} = R \theta
\]

Thus,

\[
R \theta = l
\]

Equating this with the earlier equation,

\[
l = \frac{l (1 + \alpha \Delta T)}{R} R
\]

\[
R = \frac{l}{(1 - \cos \frac{l (1 + \alpha \Delta T)}{2R})}
\]

For small expansion, using the approximation \( \cos x \approx 1 - \frac{x^2}{2} \),

\[
1 - \cos \frac{l (1 + \alpha \Delta T)}{2R} \approx \frac{1}{2} \left( \frac{l (1 + \alpha \Delta T)}{2R} \right)^2
\]

Solving for \( R \), we get:

\[
R \approx \frac{l^2}{8 \alpha l \Delta T}
\]

### Final Expression for Radius \( R \):

\[
R = \frac{l^2}{8 \alpha \Delta T}
\]

This is the required expression for the radius \( R \) of the circular arc formed by the heated and constrained rod.