A current of 1 mA is flowing through a copper wire. How many electrons will pass a given point in one second  $\left[\mathrm{e} = 1.6 \times {10}^{-19} \mathrm{Coulomb}\right]$

1.  $6.25 \times {10}^{19}$

2.  $6.25 \times {10}^{15}$

3.  $6.25 \times {10}^{31}$

4.  $6.25 \times {10}^8$

The drift velocity of free electrons in a conductor is \(v\) when a current \(i\) is flowing in it. If both the radius and current are doubled, then the drift velocity will be:

The current \(I\) as shown in the circuit will be:

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. \(10.8~\Omega\)
4. \(11.1~\Omega\)

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 the electric field as per the relation:
1. \(v_{d} \propto E\)
2. \(v_{d} \propto \frac{1}{E}\)
3. \(v_{d}= \text{constant}\)
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) $\rho \sqrt{\frac{1}{3}}$