Here, L=50 cm= `50xx10^(-2)` m,
`eta=5.6xx10^(9)Pa,F=9.0xx10^(4)` N
Area of the on which force is applied,
`A=50xx10=500sq. Cm 0.05 m^(2)`
If `DeltaL` is the displacement of the upper edge of the slab due to tangential force F applied, then
`eta=(Fla)/(DeltaL//L)`
or `(FL)/(etaA)=(9xx10^(4)xx50xx10^(-2))/(5.6xx10^(9)xx0.05)`
`implies DeltaL=1.6 xx10^(-4)` m