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 $\mathrm{\Omega }$

2. 4 $\mathrm{\Omega }$

3. 2 $\mathrm{\Omega }$

4. 7 $\mathrm{\Omega }$

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 $\Omega$

3. 11.1 $\Omega$

When two resistances X and Y are put in the left hand and right hand gaps in a Wheatstone meter bridge, the null point is at 60 cm. If X is shunted by a resistance equal to half of itself, then the shift in the null point will be: