A meter rod made of steel is correct at 0°C and other meter rod of same material is correct at 30°C. If α = 12 × 10^–6/°C, the difference between their lengths at 20°C is

Due to rise in temperature, the first scale will expand and the due to fall in temperature, the second scale will contract.

For the first scale, $l_1=l(1+\alpha \cdot \Delta t_1)$

$\Rightarrow l_1=l[1+\alpha (20-0\left)\right]=l+l\left(20\alpha \right)\dots \dots \left(1\right)$

For the second scale, $l_2=l\left(1-\alpha \cdot \Delta t_2\right)$

$l_2=l[1+\alpha (20-30\left)\right]=l-l\left(10\alpha \right)\dots ..\left(2\right)$
$\Rightarrow l_1-l_2=30\alpha l_1=30\left(12\times {10}^{-6}\right)\left(100\right)$

$=36\times {10}^{-3}\mathrm{cm}$