Correct Answer - A
The lead slab is fixed and force fixed and force is applied parallel to the narrow face as shown in the figure. Area of the face parrallel to which this force is applied is
`A = 50 cm xx 10 cm= 0.5m xx 0.1 m = 0.05m^(2)`
If `DeltaL` is the displacement of the upper edge of the slab due to tangential force F, then
`eta= (F//A)/(DeltaL//L) or DeltaL = (FL)/(etaA)`
Substituting the given value,
`Delta L = ((9xx10^(4)N)(0.5m))/(5.6xx10^(9)N m^(-2) xx 0.05m^(3))`
`=- 1.6 xx 10^(-4)m = 0.16mm`