The breaking stress of a wire depends upon:

1.	material of the wire.
2.	length of the wire.
3.	radius of the wire.
4.	shape of the cross-section.

Three wires A, B, C made of the same material and radius have different lengths. The graphs in the figure show the elongation-load variation. The longest wire is
    

The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of:

 The area of cross-section of a wire of length \(1.1\) m is \(1\) mm². It is loaded with mass of \(1\) kg. If Young's modulus of copper is \(1.1\times10^{11}\) N/m², then the increase in length will be: (If $g=10\mathrm{ m}/s^2$)

In the CGS system, Young's modulus of a steel wire is 2×10¹² dyne/cm². To double the length of a wire of unit cross-section area, the force required is:
1. 4×10⁶ dynes
2. 2×10¹² dynes
3. 2×10¹² newtons
4. 2×10⁸ dynes

Steel and copper wires of the same length and area are stretched by the same weight one after the other. Young's modulus of steel and copper are \(2\times10^{11} ~\text{N/m}^2\) and  \(1.2\times10^{11}~\text{N/m}^2\). The ratio of increase in length is: 

Two wires of copper having length in the ratio of 4: 1 and radii ratio of 1: 4 are stretched by the same force. The ratio of longitudinal strain in the two will be:
1. 1: 16                                             
2. 16: 1
3. 1: 64                                             
4. 64: 1

The work done in stretching an elastic wire per unit volume is:

A \(5\) m long wire is fixed to the ceiling. A weight of \(10\) kg is hung at the lower end and is \(1\) m above the floor. The wire was elongated by \(1\) mm. The energy stored in the wire due to stretching is:
1. zero                                 
2. \(0.05\) J
3. \(100\) J                          
4. \(500\) J

The Young's modulus of a wire is Y.  If the energy per unit volume is E, then the strain will be:

1. $\sqrt{\frac{2E}{Y}}$                                         

2. $\sqrt{2EY}$

3. $EY$                                             

4. $\frac{E}{Y}$