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

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. \(Kl^2\)

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 the potential energy of a spring is V on stretching it by 2 cm, then its potential energy when it is stretched by 10 cm will be

1. V/25                                   

2. 5V

3. V/5                                    

4. 25V

The diagram shows stress v/s strain curve for the materials A and B. From the curves we infer that

1. A is brittle but B is ductile  

2. A is ductile and B is brittle

3. Both A and B are ductile      

4. Both A and B are brittle

The adjacent graph shows the extension $\left(∆l\right)$ of a wire of length 1m suspended from the top of a roof at one end with a load W connected to the other end. If the cross sectional area of the wire is ${10}^{-6}{m}^2$ calculate the young’s modulus of the material of the wire

1. $2\times {10}^{11}N/m^2$

2. $2\times {10}^{-11}N/m^2$

3. $3\times {10}^{-12}N/m^2$

4. $2\times {10}^{-13}N/m^2$

The strain-stress curves of three wires of different materials are shown in the figure. \(P\), \(Q\) and \(R\) are the elastic limits of the wires. The figure shows that:
           

One end of a uniform wire of length \(L\) and of weight \(W\) is attached rigidly to a point in the roof and a weight \(W_1\) is suspended from its lower end. If \(S\) is the area of cross-section of the wire, the stress in the wire at a height \(\frac{3L}{4}\) from its lower end is:
1. \(\frac{W_1}{S}\)
2. \(\frac{W_1+\left(\frac{W}{4}\right)}{S}\)
3. \(\frac{W_1+\left(\frac{3W}{4}\right)}{S}\)
4. \(\frac{W_1+W}{S}\)

Shearing stress causes a change in-

1.   Length                              

2.   Breadth

3.   Shape                               

4.   Volume