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NCERT Exemplar · Q26

Q.A thin rod having length L0L_0 at 0∘0^\circC and coefficient of linear expansion α\alpha has its two ends maintained at temperatures θ1\theta_1 and θ2\theta_2, respectively. Find its new length.

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When a rod's ends are held at different temperatures, each infinitesimal element expands according to its local temperature. Integrating over the linear temperature gradient gives a new length of L0[1+αθ1+θ22]L_0\left[1 + \alpha\frac{\theta_1 + \theta_2}{2}\right].

Why This Approach Works

Thermal expansion is local: every small piece of the rod expands according to the temperature at that point. When the two ends are at different temperatures, a steady-state temperature gradient develops along the rod. The key insight is that the rod doesn't expand uniformly—hotter regions expand more than cooler ones. To find the total new length, we must integrate the expansion of each infinitesimal element.

In steady state, heat flows from the hot end to the cold end, establishing a linear temperature profile (assuming constant thermal conductivity). Once we know T(x)T(x), we can compute how much each slice dxdx stretches, then sum—integrate—over the entire length.

Step-by-Step Solution

1. Establish the temperature distribution

In steady-state conduction through a uniform rod, the temperature varies linearly from one end to the other. Taking x=0x = 0 at the end held at θ1\theta_1 and x=L0x = L_0 at the end held at θ2\theta_2, the temperature at position xx is

T(x)=θ1+θ2−θ1L0 x.T(x) = \theta_1 + \frac{\theta_2 - \theta_1}{L_0}\,x.

This is the straight-line interpolation between the boundary temperatures.

2. Consider an infinitesimal element

Take a thin slice of the rod between xx and x+dxx + dx. At 0∘0^\circC its length would be dxdx. At temperature T(x)T(x), this element expands to a new length

dℓ=dx [1+α T(x)].d\ell = dx\,[1 + \alpha\,T(x)].

The factor 1+α T(x)1 + \alpha\,T(x) is the familiar linear-expansion formula applied locally.

3. Integrate to find the total new length

The new length LL is the sum of all the stretched elements:

L=∫0L0dℓ=∫0L0[1+α T(x)] dx.L = \int_0^{L_0} d\ell = \int_0^{L_0} \left[1 + \alpha\,T(x)\right]\,dx.

Substitute T(x)T(x):

L=∫0L0[1+α(θ1+θ2−θ1L0 x)]dx.L = \int_0^{L_0} \left[1 + \alpha\left(\theta_1 + \frac{\theta_2 - \theta_1}{L_0}\,x\right)\right]dx.

4. Evaluate the integral

Break it into parts:

L=∫0L0dx+α θ1∫0L0dx+α θ2−θ1L0∫0L0x dx.L = \int_0^{L_0} dx + \alpha\,\theta_1 \int_0^{L_0} dx + \alpha\,\frac{\theta_2 - \theta_1}{L_0} \int_0^{L_0} x\,dx.

Compute each:

∫0L0dx=L0,∫0L0x dx=L022.\int_0^{L_0} dx = L_0, \quad \int_0^{L_0} x\,dx = \frac{L_0^2}{2}.

So

L=L0+α θ1 L0+α θ2−θ1L0⋅L022.L = L_0 + \alpha\,\theta_1\,L_0 + \alpha\,\frac{\theta_2 - \theta_1}{L_0} \cdot \frac{L_0^2}{2}.

Simplify the last term:

α θ2−θ1L0⋅L022=α L0 θ2−θ12.\alpha\,\frac{\theta_2 - \theta_1}{L_0} \cdot \frac{L_0^2}{2} = \alpha\,L_0\,\frac{\theta_2 - \theta_1}{2}.

Combine: …

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