JEE Physics 2026 Question with Solution

Given: The composite rod consists of aluminium and steel parts of equal initial length. Total initial length is $120 \, \text{cm}$ at $30^\circ \text{C}$. The coefficients are $\alpha_{Al} = 24 \times 10^{-6}/^\circ \text{C}$ and $\alpha_{St} = 1.2 \times 10^{-5}/^\circ \text{C} = 12 \times 10^{-6}/^\circ \text{C}$.

Find: The final length of the composite rod at $100^\circ \text{C}$.

Since both rods have the same initial length,

$$L_{Al,0} = L_{St,0} = \frac{120}{2} = 60 \, \text{cm}$$

The temperature rise is

$$\Delta T = 100 - 30 = 70^\circ \text{C}$$

Using linear expansion,

$$L_f = L_0(1 + \alpha \Delta T)$$

For the aluminium rod,

$$L_{Al,f} = 60\left(1 + 24 \times 10^{-6} \times 70\right) = 60(1 + 0.00168) = 60.1008 \, \text{cm}$$

For the steel rod,

$$L_{St,f} = 60\left(1 + 12 \times 10^{-6} \times 70\right) = 60(1 + 0.00084) = 60.0504 \, \text{cm}$$

Therefore, the total final length is

$$L_{total,f} = L_{Al,f} + L_{St,f} = 60.1008 + 60.0504 = 120.1512 \, \text{cm}$$

So the closest option is $120.15 \, \text{cm}$. The correct option is C.

Given: Total initial length $L_{total,0} = 120 \, \text{cm}$, with equal parts of $60 \, \text{cm}$ each. Temperature change is $70^\circ \text{C}$.

Find: Total final length after heating.

The total expansion is the sum of the expansions of the two rods:

$$\Delta L_{total} = L_{Al,0}\alpha_{Al}\Delta T + L_{St,0}\alpha_{St}\Delta T$$

Substituting the values,

$$\Delta L_{total} = (60 \times 24 \times 10^{-6} \times 70) + (60 \times 12 \times 10^{-6} \times 70)$$ $$\Delta L_{total} = 0.1008 + 0.0504 = 0.1512 \, \text{cm}$$

Hence,

$$L_{total,f} = L_{total,0} + \Delta L_{total} = 120 + 0.1512 = 120.1512 \, \text{cm}$$

Therefore, the length of the composite rod is approximately $120.15 \, \text{cm}$, so the correct option is C.