When a piece of aluminum wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become :

1. Two times

2. Four times

3. Eight times

4. Sixteen times

Through a semiconductor, an electric current is due to drift off:

1. Free electrons

2. Free electrons and holes

3. Positive and negative ions

4. Protons

The specific resistance of all metals is most affected by :

1. Temperature

2. Pressure

3. Degree of illumination

4. Applied magnetic field

The positive temperature coefficient of resistance is for :

1. Carbon

2. Germanium

3. Copper

4. An electrolyte

The electric intensity E, current density j and specific resistance k are related to each other by the relation :

1. E = j/k

2. E = jk

3. E = k/j

4. k = jE

The resistance of a wire of uniform diameter d and length L is R. The resistance of another wire of the same material but diameter 2d and length 4L will be :

1. 2R

2. R

3. R/2

4. R/4

There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is 1 mm². If the number of free electrons per cm³ is 8.4 × 10²², then the drift velocity would be :

1. 1.0 mm/sec

2. 1.0 m/sec

3. 0.1 mm/sec

4. 0.01 mm/sec

An electric wire of length ‘I’ and area of cross-section a has a resistance R ohms. Another wire of the same material having the same length and area of cross-section 4a has a resistance of :

1. 4R

2. R/4

3. R/16

4. 16R

If \(n\), \(e\), \(\tau\) and \(m\) respectively represent the density, charge relaxation time and mass of the electron, then the resistance of a wire of length \(l\) and area of cross-section \(A\) will be:
1. \(\frac{ml}{ne^2\tau A}\)
2. \(\frac{m\tau^2A}{ne^2l}\)
3. \(\frac{ne^2\tau A}{2ml}\)
4. \(\frac{ne^2 A}{2m\tau l}\)