A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

To find the shearing strain of the aluminium cube, we use the definition of shearing strain.

Shearing strain ($\theta$) is defined as the ratio of the relative displacement ($\Delta x$) of the top face to the height or side length ($L$) of the cube:
$\theta =\frac{\Delta x}{L}$

From the given problem, we have:
- Side length of the cube, $L=0.1\text{m}$
- Relative displacement of the top face, $\Delta x=0.02\text{cm}=0.02\times {10}^{-2}\text{m}=2\times {10}^{-4}\text{m}$

Now, we substitute these values into the shearing strain formula:
$\theta =\frac{2\times {10}^{-4}\text{m}}{0.1\text{m}}$

Simplifying the calculation:
$\theta =\frac{2\times {10}^{-4}}{{10}^{-1}}=2\times {10}^{-3}=0.002$

Thus, the shearing strain is 0.002.