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

(a) $2\times {10}^{11}N/m^2$

(b) $2\times {10}^{-11}N/m^2$

(c) $3\times {10}^{-12}N/m^2$

(d) $2\times {10}^{-13}N/m^2$

The graph shows the behaviour of a length of wire in the region for which the substance obeys Hook’s law. \(P\) and \(Q\) represents:

             
1. \(P\) = applied force, \(Q\) = extension
2. \(P\) = extension, \(Q\) = applied force
3. \(P\) = extension, \(Q\) = stored elastic energy
4. \(P\) = stored elastic energy, \(Q\) = extension 

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

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

Two wires of the 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 \(5~\text{m}\) 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. \(Kl^2\)

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