Two wires $A$ and $B$ of the same length and of the same material have the respective radii $r_1$ and $r_2$. Their one end is fixed to a rigid support,and at the other end,an equal twisting couple is applied. The ratio of the angle of twist at the end of $A$ to the angle of twist at the end of $B$ will be:

$A$ square lead slab of side $50\, cm$ and thickness $10\, cm$ is subject to a shearing force (on its narrow face) of $9.0 \times 10^{4} \, N$. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Given: Shear modulus of lead $G = 5.6 \times 10^{9} \, N/m^2$)

The strain energy stored in a body of volume $V$ due to shear strain $\phi$ is (shear modulus is $\eta$ ).

The units of modulus of rigidity are

$A$ wire of length $1 \text{ m}$ and radius $2 \text{ mm}$ is vertically clamped. The lower end is twisted through an angle of $45^{\circ}$. The angle of shear is $.....$ (in $^{\circ}$)

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 . . . . . . .