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

         

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

$\mathrm{}$

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. \(100R\)

In the figure, the value of resistors to be connected between \(C\) and \(D\) so that the resistance of the entire circuit between \(A\) and \(B\) does not change with the number of elementary sets used is:

1. \(R\)
2. \(R(\sqrt{3}-1)\)
3. \(3R\)
4. \(R(\sqrt{3}+1)\)