A battery has emf \(4\) V and internal resistance \(r\). When this battery is connected to an external resistance of \(2\) ohm, a current of \(1\) ampere flows in the circuit. How much current will flow if the terminals of the battery are connected directly?

1.	\(1\) A	2.	\(2\) A
3.	\(4\) A	4.	Infinite

The equivalent resistance between \(A\) and \(B\) is:

1. \(3~\Omega\)
2. \(6~\Omega\)
3. \(9~\Omega\)
4. \(12~\Omega\)$$

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

1. \(10~\text{A}\)
2. \(\frac{20}{3}~\text{A}\)
3. \(\frac{2}{3}~\text{A}\)
4. \(\frac{5}{3}~\text{A}\)

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. \(10.8~\Omega\)
4. \(11.1~\Omega\)

The current through the \(5~\Omega\) resistor is:

1. \(3.2\) A                     

2. \(2.8\) A

3. \(0.8\) A                     

4. \(0.2\) A

In the circuit shown, the value of each of the resistances is \(r\). The equivalent resistance of the circuit between terminals \(A\) and \(B\) will be:
        
1. \(\frac{4r}{3}\)
2. \(\frac{3r}{2}\)
3. \(\frac{r}{3}\)
4. \(\frac{8r}{7}\)

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. \(v_{d}= \text{constant}\)
4. \(v_{d} \propto E^2\)

A wire of resistance \(R\) is divided into \(10\) equal parts. These parts are connected in parallel, the equivalent resistance of such connection will be:
1. \(0.01R\)
2. \(0.1R\)
3. \(10R\)
4. \(100R\)