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
       

1. A

2. B

3. C

4. All

The bulk modulus of a spherical object is B. If it is subjected to uniform pressure P, the fractional decrease in radius will be:

1. $\frac{P}{B}$

2. $\frac{B}{3P}$

3. $\frac{\mathrm{3P}}{B}$

4. $\frac{P}{3B}$

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:

1. 1:2

2. 2:1

3. 4:1

4. 1:1

 The area of cross-section of a wire of length 1.1 metre is 1 $m{m}^2.$It is loaded with mass of 1 kg. If Young's modulus of copper is $1.1\times {10}^{11}N/m^2,$ then the increase in length will be (If $g=10m/s^2$)
1. 0.01 mm               
2. 0.075 mm
3. 0.1 mm                 
4. 0.15 mm

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

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:
1. F/2                                     
2. 2F 
3. 4F
4. F/4

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 $2\times {10}^{11}N/m^2 and 1.2\times {10}^{11}N/m^2.$ The ratio of increase in length is:

1. $\frac{2}{5}$

2. $\frac{3}{5}$

3. $\frac{5}{4}$

4. $\frac{5}{2}$

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

On applying stress of $20\times {10}^8$ N/$m^2$, the length of a perfectly elastic wire is doubled. It's Young’s modulus will be:
1. $40\times {10}^{8}N/m^2$
2. $20\times {10}^{8}N/m^2$
3. $10\times {10}^{8}N/m^2$
4. $5\times {10}^{8}N/m^2$