$A$ steel rod is projecting out of a rigid wall. The shearing strength of steel is $345 \, MN/m^2$. The dimensions are $AB = 5 \, cm$ and $BC = BE = 2 \, cm$. The maximum load that can be put on the face $ABCD$ is .......... $kg$ (neglect bending of the rod). Take $g = 10 \, m/s^2$.

$A$ rectangular block of size $10\,cm \times 8\,cm \times 5\,cm$ is kept in three different positions $P, Q,$ and $R$ in turn as shown in the figure. In each case,the shaded area is rigidly fixed and a definite force $F$ is applied tangentially to the opposite face to deform the block. The displacement of the upper face will be

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 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 $= 3.5 \times 10^8\,N/m^2$)

$A$ horizontal aluminium rod of diameter $4 \text{ cm}$ projects $6 \text{ cm}$ from a wall. An object of mass $400 \pi \text{ kg}$ is suspended from the end of the rod. The shearing modulus of aluminium is $3.0 \times 10^{10} \text{ N/m}^2$. The vertical deflection of the end of the rod is (given $g = 10 \text{ m/s}^2$): (in $\text{ mm}$)

The modulus of rigidity of diamond is: