The given circuit shows a uniform straight wire \(AB\) of \(40 ~\text{cm}\) length fixed at both ends. In order to get zero reading in the galvanometer \(G,\) the free end of \(J\) is to be placed from the end \(B\) at:

Three resistances \(\mathrm P\), \(\mathrm Q\), and \(\mathrm R\), each of \(2~\Omega\) and an unknown resistance \(\mathrm{S}\) form the four arms of a Wheatstone bridge circuit. When the resistance of \(6~\Omega\) is connected in parallel to \(\mathrm{S}\), the bridge gets balanced. What is the value of \(\mathrm{S}\)?

For the network shown in the figure below, the value of the current \(i\) is:

     
1. \(\frac{18V}{5}\)
2. \(\frac{5V}{9}\)
3. \(\frac{9V}{35}\)
4. \(\frac{5V}{18}\)

Five equal resistances each of resistance \(R\) are connected as shown in the figure below. A battery of \(V\) volts is connected between \(A\) and \(B\). The current flowing in \(AFCEB\) will be:

        
1. \(\frac{V}{R}\)
2. \(\frac{V}{2R}\)
3. \(\frac{2V}{R}\)
4. \(\frac{3V}{R}\)

In a Wheatstone bridge, all four arms have equal resistance \(R.\) If the resistance of the galvanometer arm is also \(R,\) the equivalent resistance of the combination is:

The resistance of each arm of the wheat stone bridge is \(10~ \Omega.\) A resistance of \(10~ \Omega\) is connected in series with a galvanometer. The equivalent resistance across the battery will be:
1.\(10~ \Omega\)
2.\(15~ \Omega\)
3. \(20~ \Omega\)
4. \(40~ \Omega\)

The net resistance of the circuit between \(A\) and \(B\) is: