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

A stretched rubber has:

(1) increased kinetic energy.

(2) increased potential energy.

(3) decreased kinetic energy.

(4) decreased potential energy.

When load of 5kg is hung on a wire then extension of 3m takes place, then work done will be

(1) 75 joule                            

(2) 60 joule

(3) 50 joule                             

(4) 100 joule

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) $Y_A=3Y_B$

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

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

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

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

If the force constant of a wire is \(K\), the work done in increasing the length of the wire by \(l\) is:
1. \(\frac{Kl}{2}\)
2. \(Kl\)
3. \(\frac{Kl^2}{2}\)
4. \(Kl^2\)