A steel metre scale is calibrated to provide correct readings at...

The scale is calibrated at 20^{}^{{}^{\circ}} C, which means the physical distance between the $50 \text{ cm}$ mark and the $51 \text{ cm}$ mark is exactly $1 \text{ cm}$ at this temperature.

Let $L_0 = 1 \text{ cm}$ be the distance at T_0 = 20^{}^{{}^{\circ}} C.

When the temperature is changed to T = 10^{}^{{}^{\circ}} C, the scale undergoes thermal contraction. The new physical distance $L$ between the marks is given by:

$L = L_0 [1 + \alpha (T - T_0)]$

Substituting the given values:

$L = 1 \times [1 + 1.1 \times 10^{-5} \times (10 - 20)]$

$L = 1 \times [1 - 1.1 \times 10^{-5} \times 10]$

$L = 1 - 0.00011$

$L = 0.99989 \text{ cm}$