The modulus of rigidity of diamond is:

The force required to punch a square hole of side $2\,cm$ in a steel sheet of thickness $2\,mm$ is (given: shearing stress of steel sheet $= 3.5 \times 10^8\,N/m^2$)

$A$ 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 $G = 5.6 \times 10^{9} \, N/m^2$)

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: The stretching of a spring is determined by the shear modulus of the material of the spring.
Reason $(R)$: $A$ coil spring of copper has more tensile strength than a steel spring of same dimensions.
In the light of the above statements,choose the most appropriate answer from the options given below:

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 . . . . . . .

The upper end of a wire $1\,m$ long and $4\,mm$ radius is clamped. The lower end is twisted by an angle of $30^\circ$. The angle of shear is ...... $^\circ$.