A square lead slab of side 50 cm and thickness l0 cm is subjected to a shearing force (on its narrow face) of 9 x${10}^4$ N. The lower edge is riveted to the floor. How much
will the upper edge be displaced? (Shear modulus of lead = 5.6 x ${10}^9$ N ${\mathrm{m}}^{-2}$)

The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure.

Area of the face parallel to which this force is A = 50 cm x l0 cm = 0.5 m x 0.1 m = 0.05 ${\mathrm{m}}^2$.
If $∆\mathrm{L}$ is the displacement of the upper edge of the slab due to tangential force F, then

$\mathrm{\eta } = \frac{{\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}}{{\displaystyle \frac{∆\mathrm{L}}{\mathrm{L}}}} \mathrm{or} ∆\mathrm{L} = \frac{\mathrm{FL}}{\mathrm{\eta A}}$

Substituting the given values, we get

$∆\mathrm{L} = \frac{(9\times {10}^4\mathrm{N})(0.5 \mathrm{m})}{5.6\times {10}^9 {\mathrm{Nm}}^{-2}\times 0.05 {\mathrm{m}}^2}$

       = $1.6 \times {10}^{-4} \mathrm{m} = 0.16 \mathrm{mm}$