If the breaking stress of a wire is $3.18 \times 10^{10} \, N/m^2$,what should be its minimum radius so that the wire does not break? (Refer to the figure for the masses attached to the pulley system,assume $g = 10 \, m/s^2$)

The density and breaking stress of a wire are $6 \times 10^4 \ kg/m^3$ and $1.2 \times 10^8 \ N/m^2$ respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is $\frac{1}{3}$ of the value on the surface of Earth. The maximum length of the wire without breaking is ............ $m$ (take $g = 10 \ m/s^2$ on Earth).

$A$ solid that transmits light in the visible region and has a very low melting point possesses:

$A$ sphere of mass $4 \,kg$ is attached to one end of a steel wire of length $1 \,m$ and radius $1 \,mm$. It is whirled in a vertical circle with an angular velocity $10 \,rad \,s^{-1}$. If the sphere is at the lowest point of its path,the elongation in the wire is . . . . . . $(g=10 \,ms^{-2}, Y_{\text{steel}}=20 \times 10^{10} \,Nm^{-2})$ (in $\,mm$)

$Assertion :$ Solids are least compressible and gases are most compressible.
$Reason :$ Solids have a definite shape and volume,but gases do not have either a definite shape or a definite volume.

$A$ uniform wire of length $l$ and weight $w$ is suspended from the roof with a weight $W$ attached at the other end. The stress in the wire at a distance $l/3$ from the top is given by $(\frac{W}{A} + \gamma \frac{w}{A})$, where $A$ is the cross-sectional area of the wire. The value of $\gamma$ is . . . . . . .