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 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. \(\frac{Kl}{2}\)
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