$A$ wire of length $5 \,m$ is twisted through $30^{\circ}$ at its free end. If the radius of the wire is $1 \,mm$,the shearing strain in the wire is: (in $^{\circ}$)

$A$ steel wire is suspended vertically from a rigid support. When loaded with a weight in air,it extends by $l_a$ and when the weight is immersed completely in water,the extension is reduced to $l_w$. Then the relative density of the material of the weight is

$A$ wire of density $3 \times 10^3 \, kg/m^3$ requires a breaking stress of $10^6 \, N/m^2$ to break. What should be the length of the wire so that it breaks under its own weight? (Take $g = 10 \, m/s^2$)

$A$ rod of length $1.05 \; m$ having negligible mass is supported at its ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths as shown in Figure. The cross-sectional areas of wires $A$ and $B$ are $1.0 \; mm^{2}$ and $2.0 \; mm^{2}$ respectively. At what point along the rod should a mass $m$ be suspended in order to produce $(a)$ equal stresses and $(b)$ equal strains in both steel and aluminium wires?

Two blocks of masses $1 \,kg$ and $2 \,kg$ are connected by a metal wire going over a smooth pulley. The breaking stress of metal is $\frac{40}{3 \pi} \times 10^6 \,N m^{-2}$. What should be the minimum radius of the wire used if it should not break (in $mm$)? $(g = 10 \,m s^{-2})$

Read the following two statements below carefully and state,with reasons,if it is true or false.
$(a)$ The Young's modulus of rubber is greater than that of steel;
$(b)$ The stretching of a coil is determined by its shear modulus.