$A$ slab of side $50 \text{ cm}$ and thickness $10 \text{ cm}$ is subjected to a shearing force of $10^5 \text{ N}$ on its narrow edge. If the lower edge is riveted to the floor and the upper edge is displaced by $0.2 \text{ mm}$, then the shear modulus of the material of the slab is: (in $\text{ GPa}$)

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

Steel ruptures when a shear stress of $3.5 \times 10^8 \, N \, m^{-2}$ is applied. The force needed to punch a $1 \, cm$ diameter hole in a steel sheet $0.3 \, cm$ thick is nearly:

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

The modulus of rigidity of diamond is:

$A$ rod of length $l$ and radius $r$ is joined to a rod of length $l/2$ and radius $r/2$ of the same material. The free end of the small rod is fixed to a rigid base,and the free end of the larger rod is given a twist of $\theta^\circ$. The twist angle at the joint will be: