The Wheatstone bridge shown in the figure below is balanced when the uniform slide wire \(AB\) is divided as shown. Value of the resistance \(X\) is:
          

1. \(3~\Omega\)
2. \(4~\Omega\)
3. \(2~\Omega\)
4. \(7~\Omega\)

The figure given below shows a circuit when resistances in the two arms of the meter bridge are \(5~\Omega\) and \(R\), respectively. When the resistance \(R\) is shunted with equal resistance, the new balance point is at \(1.6l_1\). The resistance \(R\) is:

A resistance of 4 Ω and a wire of length 5 metres and resistance 5 Ω are joined in series and connected to a cell of e.m.f. 10 V and internal resistance 1 Ω. A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is:

1. 1.5 V

2. 3.0 V

3. 0.67 V

4. 1.33 V

The metre bridge shown is in a balanced position with \(\frac{P}{Q} = \frac{l_1}{l_2}\). If we now interchange the position of the galvanometer and the cell, will the bridge work? If yes, what will be the balanced condition?

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. \(11.1~\Omega\)