The edge of an aluminium cube is 10 cm long. One face of the cube...

Given, side of a cube (l) = 10 cm & = 0.1 m

Area its each face (A) = l^{2} = (0.1)^{2} = 0.01 m^{2}

Load (m) = 100 kg

Tangential force acting on one face of the cube,

F = mg = 100 × 9.8 = 980 N

$\text{Shear stress acting on this face} = \dfrac{\text{F}}{\text{A}}$

$= \dfrac{\text{980}}{\text{0.01}} {N/m}^{2}$

= 9.8 × 10^{4} N/m^{2}

Shear modulus of aluminium (η)) = 25 GPa

= 25 × 10^{9} N/m^{2}

$\text{Shear modulus} \left(η\right) = \dfrac{\text{Shearing stress}}{\text{Shearing strain}}$

$\text{or shearing strain} \left(\dfrac{\text{\Delta} \text{y}}{\text{L}}\right) = \dfrac{\text{Shearing stress}}{\text{Shear modulus}}$

$\text{or} \text{\Delta} \text{y} = \dfrac{\text{Shearing stress}}{\text{Shear modulus}} \times \text{L}$

$= \dfrac{\text{9.8} \times {10}^{4}}{\text{25} \times {10}^{9}} \times \text{0.1}$

= 0.392 × 10^{-5} m

= 3.92 × 10^{-7} m