$A$ square aluminium (shear modulus is $25 \times 10^{9} \, N m^{-2}$) slab of side $60 \, cm$ and thickness $15 \, cm$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^{4} \, N$. The lower edge is riveted to the floor. The displacement of the upper edge is $....... \mu m$.

Two slabs with square cross-sections of different materials $(1, 2)$ have equal sides $(l)$ and thicknesses $d_1$ and $d_2$ such that $d_2 = 2d_1$ and $l > d_2$. Considering the lower edges of these slabs are fixed to the floor,we apply equal shearing force on the narrow faces. The angle of deformation is $\theta_2 = 2\theta_1$. If the shear modulus of material $1$ is $4 \times 10^9 \ N/m^2$,then the shear modulus of material $2$ is $x \times 10^9 \ N/m^2$,where the value of $x$ is . . . . . . .

$A$ cylindrical rod of length $1 \ m$ and radius $4 \ cm$ is mounted vertically. It is subjected to a shear force of $10^5 \ N$ at the top. Considering infinitesimally small displacement in the upper edge,the angular displacement $\theta$ of the rod axis from its original position would be: (Shear modulus,$G = 10^{10} \ N/m^2$)

As shown in the figure,forces of $10^5\,N$ each are applied in opposite directions on the upper and lower faces of a cube of side $10\,cm$,shifting the upper face parallel to itself by $0.5\,cm$. If the side of another cube of the same material is $20\,cm$,then under similar conditions as above,the displacement will be ......... $cm$.

$A$ rubber cube of side $5 \ cm$ has one face fixed while a tangential force $1800 \ N$ is applied on its opposite face. If the modulus of rigidity of rubber is $2.4 \times 10^6 \ N \ m^{-2}$,then the lateral displacement of the strained face is $......$ . (in $mm$)

$A$ $2 \ m$ long rod of radius $1 \ cm$,which is fixed from one end,is given a twist of $0.8 \ radian$. The shear strain developed (in radian) will be $:-$