The current I as shown in the circuit will be:

                 

1.  10 A

2.  $\frac{20}{3}$ $\mathrm{A}$ $$

3.  $\frac{2}{3}$ $\mathrm{A}$ $$ $$

4.  $\frac{5}{3}$ $\mathrm{A}$

A meter bridge is set up to determine unknown resistance x using a standard 10 $\Omega$ resistor. The galvanometer shows the null point when the tapping key is at a 52 cm mark. End corrections are 1 cm and 2 cm respectively for end A and B. Then the value of x is:

1.  10.2 $\Omega$

2.  10.6 $\Omega$

3.  10.8 $\Omega$

3. 11.1 $\Omega$

The resistivity of iron is 1 × 10^–7 ohm – m. The resistance of iron wire of particular length and thickness is 1 ohm. If the length and the diameter of wire both are doubled, then the resistivity in ohm – m will be :

(1) 1 × 10^–7

(2) 2 × 10^–7

(3) 4 × 10^–7

(4) 8 × 10^–7

The resistivity of a wire :

Drift velocity v_d varies with the intensity of electric field as per the relation: 

1. $v_d\propto E$

2. $v_d\propto \frac{1}{E}$

3. v_d = constant

4. $v_d\propto E^2$ 

In a conductor 4 coulombs of charge flows for 2 seconds. The value of electric current will be :

(1) 4 volts

(2) 4 amperes

(3) 2 amperes

(4) 2 volts

The specific resistance of a wire is ρ, its volume is 3 m³ and its resistance is 3 ohms, then its length will be 

(1) $\sqrt{\frac{1}{\rho }}$

(2) $\frac{3}{\sqrt{\rho }}$

(3) $\frac{1}{\rho }\sqrt{3}$

(4) $\rho \sqrt{\frac{1}{3}}$ 

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

If a metallic block has no potential difference applied across it, then the mean velocity of free electron is:
(T = absolute temperature of the block)