The variation of length ( L ) with temperature ( T ) for two...

The equation for the linear expansion of a rod is given by $L = L_0 (1 + \alpha \Delta T) = L_0 + L_0 \alpha \Delta T$.

When length $L$ is plotted on the y-axis against temperature $T$ on the x-axis, the slope of the straight line is $\dfrac{dL}{dT} = L_0 \alpha$.

Since the lines for rod A and rod B are parallel, their slopes must be equal:

$L_{0A} \alpha_A = L_{0B} \alpha_B$

Rearranging to find the ratio of the coefficients of linear expansion:

$\dfrac{\alpha_A}{\alpha_B} = \dfrac{L_{0B}}{L_{0A}}$

Given the y-intercepts (initial lengths at $T = 0$) are $L_{0A} = 20 \text{ cm}$ and $L_{0B} = 50 \text{ cm}$:

$\dfrac{\alpha_A}{\alpha_B} = \dfrac{50}{20} = 2.5$

Answer: $2.5$