A metre scale made of steel gives accurate readings at 20^ ^ ^ C ....

The error in the reading of the scale due to a change in temperature is given by the formula for linear expansion:

$\Delta L = L_0 \alpha \Delta T$

Given that the maximum allowed error is $\Delta L = 0.055 \text{ mm} = 0.055 \times 10^{-3} \text{ m}$, the original length is $L_0 = 1 \text{ m}$, and the coefficient of linear expansion is \alpha = 11 \times 10^{-6} ^{}^{{}^{\circ}} C^{-1}.

Substituting the values into the formula:

$0.055 \times 10^{-3} = 1 \times 11 \times 10^{-6} \times \Delta T$

\Delta T = \dfrac{0.055 \times 10^{-3}}{11 \times 10^{-6}} = \dfrac{55 \times 10^{-6}}{11 \times 10^{-6}} = 5^{}^{{}^{\circ}} C

The scale gives accurate readings at 20^{}^{{}^{\circ}} C. Therefore, the maximum allowable temperature deviation is 5^{}^{{}^{\circ}} C above or below this temperature.

The temperature range within which the experiment can be conducted is:

T = 20^{}^{{}^{\circ}} C \pm 5^{}^{{}^{\circ}} C

This gives a range from 15^{}^{{}^{\circ}} C to 25^{}^{{}^{\circ}} C.