A liquid metal is poured in a mold cavity of size 200 mm × 200 mm × 200 mm. The cooling is uniform in all directions with NO additional compensation for shrinkage. Considering the volumetric shrinkage during solidification and solid contraction as 7% and 8%, respectively, the length of the cube edge after cooling down to the room temperature is _______ mm (rounded off to 1 decimal place).

The correct answer is 171.1 mm (or 171.1).

To find the final edge length of the cube after cooling down to room temperature, we apply the solidification shrinkage and solid contraction rates as linear reductions to the initial edge length of the mold cavity.

Given parameters:
Initial edge length of the mold cavity, $L_0=200\text{mm}$
Volumetric shrinkage rate during solidification, $S_1=7\%=0.07$
Solid contraction rate, $S_2=8\%=0.08$

First, we determine the edge length of the cube after solidification by applying the 7% reduction:
$L_1=L_0\times (1-0.07)$
$L_1=200\times 0.93=186\text{mm}$

Next, we apply the 8% solid contraction to this intermediate edge length to find the final edge length at room temperature:
$L_f=L_1\times (1-0.08)$
$L_f=186\times 0.92=171.12\text{mm}$

Rounding the value to one decimal place:
$L_f\approx 171.1\text{mm}$