Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of the work done in the two wires will be

(1) 1 : 2                                   

(2) 1 : 4

(3) 2 : 1                                   

(4) 1 : 1

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

If the force constant of a wire is K, the work done in increasing the length of the wire by l is:

1. $Kl/2$                                   

2. $Kl$

3. $K{l}^2/2$                                 

4. $K{l}^2$

When strain is produced in a body within elastic limit, its internal energy:
1. Remains constant                  
2. Decreases
3. Increases                               
4. None of the above

When shearing force is applied to a body, then the elastic potential energy is stored in it. On removing the force, this energy:

1. converts into kinetic energy.

2. converts into heat energy.

3. remains as potential energy.

4. None of the above

A wire is suspended by one end. At the other end a weight equivalent to 20 N force is applied. If the increase in length is 1.0 mm, the increase in energy of the wire will be

(1) 0.01 J                               

(2) 0.02 J

(3) 0.04 J                                (4) 1.00 J

The ratio of Young's modulus of the material of two wires is 2 : 3. If the same stress is applied on both, then the ratio of elastic energy per unit volume will be-

(1) 3 : 2                                   

(2) 2 : 3

(3) 3 : 4                                   

(4) 4 : 3

The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If $Y_A$ and $Y_B$ are the Young ‘s modulii of the materials, then

(1) $Y_B=2Y_A$

(2) $Y_A=Y_B$

(3) $Y_B=3Y_A$

(4) $Y_A=3Y_B$

If a spring extends by x on loading, then the energy stored by the spring is (if T is tension in the spring and k is spring constant)

(1) $\frac{T^2}{2x}$                                     

(2) $\frac{T^2}{2k}$

(3) $\frac{2x}{T^2}$                                     

(4) $\frac{2T^2}{k}$

When a force is applied on a wire of uniform cross-sectional area $3\times {10}^{-6}{m}^2$ and length 4m, the increase in length is 1 mm. Energy stored in it will be $\left(Y=2\times {10}^{11}N/m^2\right)$

1. 6250 J                               2. 0.177 J

3. 0.075 J                              4. 0.150 J