Assume that a block of very low shear modulus is fixed on an inclined plane as shown. Due to elastic forces,it will deform. What will be the shape of the block?

$A$ rectangular metallic block of dimensions $40 \text{ mm} \times 20 \text{ mm}$ when pulled with a tension of $50 \text{ kN}$ undergoes only elastic deformation. The strain in the block is (The shear modulus of the material of the block is $40 \times 10^9 \text{ Nm}^{-2}$)

$A$ force $F$ of the same magnitude is applied tangentially on the upper and lower faces of a cube in opposite directions. The side of the cube is $L$. The upper face of the cube shifts parallel to itself by a distance $x_{1}$. If another cube of the same material but with side $2L$ is subjected to the same condition,then the displacement of the top layer is:

$A$ steel plate of face area $1 \, cm^2$ and thickness $4 \, cm$ is fixed rigidly at the lower surface. $A$ tangential force $F = 10 \, kN$ is applied on the upper surface as shown in the figure. The lateral displacement $x$ of the upper surface with respect to the lower surface is .............. $m$ (Modulus of rigidity for steel is $8 \times 10^{11} \, N/m^2$).

What is the modulus of rigidity of a liquid?

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: