A cube has side length \(5\) cm and modulus of rigidity \(10^5\) N/m\(^2\). The displacement produced by a force of \(10\) N in the upper face of cube is _____ mm.

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: \(2\)