A cube has side length 5 cm and modulus of rigidity 10^5 N/m ^2 ....

Given:
Side length of the cube, $L = 5$ cm $= 0.05$ m
Modulus of rigidity, $\eta = 10^5$ N/m$^2$
Force applied on the upper face, $F = 10$ N

Area of the upper face, $A = L^2 = (0.05)^2 = 2.5 \times 10^{-3}$ m$^2$

Shear stress is given by:
$\text{Stress} = \dfrac{F}{A} = \dfrac{10}{2.5 \times 10^{-3}} = 4000$ N/m$^2$

Shear strain is given by:
$\text{Strain} = \dfrac{\Delta x}{L}$

Modulus of rigidity is defined as the ratio of shear stress to shear strain:
$\eta = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{F/A}{\Delta x/L}$

Rearranging for displacement $\Delta x$:
$\Delta x = \dfrac{F \cdot L}{A \cdot \eta}$

Substituting the values:
$\Delta x = \dfrac{10 \times 0.05}{2.5 \times 10^{-3} \times 10^5}$

$\Delta x = \dfrac{0.5}{250} = 0.002$ m

Converting to millimeters:
$\Delta x = 0.002 \times 1000 = 2$ mm

Answer: $2$