A square aluminium (shear modulus is 25 × 10 ^ 9 Nm ^ - 2 ) slab of...

Shearing stress is given by $\dfrac{F}{A} = η\dfrac{x}{l}$

Thus, displacement is $x = \dfrac{Fl}{Aη}$

$⇒ x = \dfrac{18 \times {10}^{4} \times 60 \times {10}^{- 2}}{60 \times {10}^{- 2} \times 15 \times {10}^{- 2} \times 25 \times {10}^{9}}$

$= 48 \times {10}^{- 6}m = 48\mum$