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: 
           
1. \(10~\Omega\)
2. \(15~\Omega\)
3. \(20~\Omega\)
4. \(25~\Omega\)

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?

    
1. Yes, \(\frac{P}{Q}=\frac{l_1-l_2}{l_1+l_2}\)
2. No, no null point
3. Yes, \(\frac{P}{Q}= \frac{l_2}{l_1}\)
4. Yes, \(\frac{P}{Q}= \frac{l_1}{l_2}\)

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. \(11.1~\Omega\)