A bimetallic strip is formed out of two identical strips– one of copper and the other of brass. The

Q: Why does the bimetallic strip bend instead of just expanding linearly?

Because the two strips are welded together, they cannot slide past each other. When one tries to expand more than the other, the only way to accommodate the different lengths is by curving, with the longer strip on the outside.

Q: Why do we use the mean radius for the calculations?

The mean radius represents the central axis of each strip. While the inner and outer surfaces of a single strip have slightly different lengths, the mean length gives the most accurate representation of the strip's overall thermal expansion.

Q: Is it always necessary to use the binomial approximation here?

Yes, for practical bimetallic strips, the thickness $d$ is extremely small compared to the radius of curvature $R$, and the thermal expansion term $\alpha \Delta T$ is much less than 1. Without this approximation, the exact equation is too complex and doesn't yield a simple, intuitive relationship.

Q: What happens if we cool the strip instead of heating it?

If we cool the strip, $\Delta T$ becomes negative. The metal with the higher $\alpha$ will contract more, causing the strip to bend in the opposite direction!

Q: Why do we take the absolute value $|\alpha_B - \alpha_C|$ in the options?

The radius of curvature $R$ is a physical distance and must be positive. Taking the absolute difference ensures that $R$ remains positive regardless of which metal has the higher coefficient of expansion.

Question

A bimetallic strip is formed out of two identical strips– one of copper and the other of brass. The coefficients of linear expansion of the two metals are $\alpha_C$ and $\alpha_B$. On heating, the temperature of the strip goes up by $\Delta T$ and the strip bends to form an arc of radius of curvature $R$. Then, $R$ is

Options:

(A) proportional to $\Delta T$

(B) inversely proportional to $\Delta T$

(C) proportional to $|\alpha_B - \alpha_C|$

(D) inversely proportional to $|\alpha_B - \alpha_C|$

Sigma Percentile · Accepted Answer

### The Magic of Bimetallic Strips

Imagine holding a perfectly straight strip made of two different metals—say, brass and copper—welded tightly together. This is a bimetallic strip, a simple yet ingenious device used in thermostats and thermometers worldwide.

When you heat this strip, something fascinating happens. Both metals absorb the thermal energy and attempt to expand. However, they don't expand equally! Brass has a higher coefficient of linear expansion ($\alpha_B > \alpha_C$), meaning it wants to grow longer than the copper.

Because the two metals are fused and cannot slide past one another, this internal tug-of-war forces the entire strip to bend. It forms an elegant circular arc, with the more expansive brass taking the outer, longer path, and the copper taking the inner, shorter path.

### Decoding the Geometry of Bending

To understand this mathematically, let's dive into the geometry of the bent strip. Let the interface where the two metals meet have a radius of curvature $R$.

If we assume each metal strip has a uniform thickness $d$, we can determine the radius of their central axes (the mean radius). For the outer brass strip, the mean radius is slightly larger: $R + \frac{d}{2}$. For the inner copper strip, the mean radius is slightly smaller: $R - \frac{d}{2}$.

Let the strip subtend an angle $	heta$ at the center of curvature. From basic geometry, the arc length is simply the radius multiplied by the angle. Therefore, the expanded lengths of the mean lines are:

$$l_B = \left(R + \frac{d}{2}ight)	heta$$

$$l_C = \left(R - \frac{d}{2}ight)	heta$$

### The Mathematics of Differential Expansion

We also know from the physics of thermal expansion that the final length of any object depends on its initial length $l_0$, its expansion coefficient $\alpha$, and the change in temperature $\Delta T$.

Equating the geometric lengths to the thermal expansion formulas, we get our master equations:

$$l_B = \left(R + \frac{d}{2}ight)	heta = l_0(1 + \alpha_B \Delta T)$$

$$l_C = \left(R - \frac{d}{2}ight)	heta = l_0(1 + \alpha_C \Delta T)$$

Our goal is to find the radius of curvature $R$. To do this, we need to eliminate the unknown initial length $l_0$ and the angle $	heta$. The most elegant way to achieve this is by simply dividing the two equations!

$$\frac{R + \frac{d}{2}}{R - \frac{d}{2}} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$$

### The Power of Approximations

This equation looks a bit messy, but we can simplify it using a powerful mathematical tool: the binomial approximation. First, let's rearrange the terms. Divide the numerator and denominator on the left side by $R$, and bring the denominator on the right side up with a negative exponent:

$$\frac{1 + \frac{d}{2R}}{1 - \frac{d}{2R}} = (1 + \alpha_B \Delta T)(1 + \alpha_C \Delta T)^{-1}$$

$$\left(1 + \frac{d}{2R}ight)\left(1 - \frac{d}{2R}ight)^{-1} = (1 + \alpha_B \Delta T)(1 + \alpha_C \Delta T)^{-1}$$

Here is where the physics justifies the math. In reality, the thickness of the strip $d$ is minuscule compared to the radius of curvature $R$ ($d \ll R$). Similarly, the thermal expansion term $\alpha \Delta T$ is incredibly small ($\alpha \Delta T \ll 1$).

Using the binomial theorem $(1+x)^n \approx 1+nx$ for small $x$, we can approximate the negative exponent terms:

$$\left(1 + \frac{d}{2R}ight)\left(1 + \frac{d}{2R}ight) \approx (1 + \alpha_B \Delta T)(1 - \alpha_C \Delta T)$$

### The Final Revelation

Now, let's expand both sides of our approximated equation:

$$1 + \frac{d}{R} + \frac{d^2}{4R^2} \approx 1 + (\alpha_B - \alpha_C)\Delta T - \alpha_B \alpha_C (\Delta T)^2$$

Because $d/R$ and $\alpha \Delta T$ are already very small, their squares—$\frac{d^2}{4R^2}$ and $\alpha_B \alpha_C (\Delta T)^2$—are vanishingly tiny. We can safely neglect them! The $1$ on both sides also cancels out, leaving us with a beautifully simple relation:

$$\frac{d}{R} = (\alpha_B - \alpha_C)\Delta T$$

Rearranging this to solve for $R$, we arrive at our final masterpiece:

$$R = \frac{d}{(\alpha_B - \alpha_C)\Delta T}$$

This elegant formula reveals the deep physical truths of the bimetallic strip. The radius of curvature $R$ is **inversely proportional** to the temperature change $\Delta T$. Heat it more, and it bends tighter (smaller $R$). Furthermore, $R$ is **inversely proportional** to the difference in the expansion coefficients $|\alpha_B - \alpha_C|$. A greater mismatch in the metals' desire to expand leads to a more dramatic bend!

Select Answer:

Visualized Solution

\text{Bimetallic Strip Bending}

\text{Geometry of the Bent Strip}

\text{Expanded Lengths}

\text{Dividing the Equations}

\text{Rearranging for Approximation}

\text{Binomial Approximation}

\text{Expanding and Neglecting Higher Order Terms}

\text{Final Expression for Radius}

The Sigma Insight: Thermal Expansion

The Magic of Bimetallic Strips

Decoding the Geometry of Bending

The Mathematics of Differential Expansion

The Power of Approximations

The Final Revelation

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