The breaking stress of a wire depends upon:

A force \(F\) is needed to break a copper wire having radius \(R.\) The force needed to break a copper wire of radius \(2R\) will be:

lf \(\rho\) is the density of the material of a wire and \(\sigma\) is the breaking stress, the greatest length of the wire that can hang freely without breaking is:
1. \(\frac{2}{\rho g}\)
2. \(\frac{\rho}{\sigma g}\)
3. \(\frac{\rho g}{2 \sigma}\)
4. \(\frac{\sigma}{\rho g}\)

To break a wire, a force of \(10^6~\text{N/m}^{2}\) is required. If the density of the material is \(3\times 10^{3}~\text{kg/m}^3,\) then the length of the wire which will break by its own weight will be:
1. \(34\) m
2. \(30\) m
3. \(300\) m
4. \(3\) m

Overall changes in volume and radius of a uniform cylindrical steel wire are \(0.2\%\) and \(0.002\%\) respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is: \(\left(2.0\times 10^{11}~\text{Nm}^{-2}\right)\)
1. \(3.2\times 10^{11}~\text{Nm}^{-2}\)
2. \(3.2\times 10^{7}~\text{Nm}^{-2}\)
3. \(3.6\times 10^{9}~\text{Nm}^{-2}\)
4. \(3.9\times 10^{8}~\text{Nm}^{-2}\)

A 1000 kg lift is tied with metallic wires of maximum safe stress of 1.4 $\times$ 10⁸ N m^-2. If the maximum acceleration of the lift is 1.2 m s^-2, then the minimum diameter of the wire is:
1. 1 m 

2. 0.1 m

3. 0.01 m 

4. 0.001 m

A wire can sustain a weight of 10 kg before breaking. If the wire is cut into two equal parts, then each part can sustain a weight of:

A light rod of length \(2~\text{m}\) is suspended from the ceiling horizontally by means of two vertical wires of equal length. A weight \(W\) is hung from the light rod as shown in the figure. The rod is hung by means of a steel wire of cross-sectional area \(A_1 = 0.1~\text{cm}^2\) and brass wire of cross-sectional area\(A_2 = 0.2~\text{cm}^2\). To have equal stress in both wires, \(\frac{T_1}{T_2}?\)${\mathrm{}}_{}$