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

The upper end of a wire of radius $4\, mm$ and length $100\, cm$ is clamped and its other end is twisted through an angle of $30^\circ$. The angle of shear is ..... $^\circ$.

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

$A$ square aluminium (shear modulus is $25 \times 10^{9} \, N m^{-2}$) slab of side $60 \, cm$ and thickness $15 \, cm$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^{4} \, N$. The lower edge is riveted to the floor. The displacement of the upper edge is $....... \mu m$.

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:

$A$ 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 $\text{ mm}$)