Two metal wires of identical dimensions are connected in series. If \(\sigma_1~\text{and}~\sigma_2\) are the conductivities of the metal wires respectively, the effective conductivity of the combination is:

A potentiometer wire of length \(L\) and a resistance \(r\) are connected in series with a battery of EMF \(E_{0 }\) and resistance \(r_{1}\). An unknown EMF is balanced at a length l of the potentiometer wire. The EMF \(E\) will be given by:
1. \(\frac{L E_{0} r}{l r_{1}}\)
2. \(\frac{E_{0} r}{\left(\right. r + r_{1} \left.\right)} \cdot \frac{l}{L}\)
3. \(\frac{E_{0} l}{L}\)
4. \(\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}\)

A potentiometer wire has a length of \(4​​~\text{m}\) and resistance  \(8~\Omega.\)  The resistance that must be connected in series with the wire and an energy source of emf \(2~\text{V}\), so as to get a potential gradient of \(1~\text{mV}\) per cm on the wire is:
1. \(32~\Omega\)
2. \(40~\Omega\)
3. \(44~\Omega\)
4. \(48~\Omega\)

\(\mathrm{A, B}~\text{and}~\mathrm{C}\) are voltmeters of resistance \(R\), \(1.5R\) and \(3R\) respectively as shown in the figure above. When some potential difference is applied between \(\mathrm{X}\) and \(\mathrm{Y}\), the voltmeter readings are \({V}_\mathrm{A}\), \({V}_\mathrm{B}\) and \({V}_\mathrm{C}\) respectively. Then:

       

Two cities are \(150~\text{km}\) apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is \(8\) volts and the average resistance per km is \(0.5~\text{ohm}\). The power loss in the wire is:
1. \(19.2~\text{W}\)
2. \(19.2~\text{kW}\)
3. \(19.2~\text{J}\)
4. \(12.2~\text{kW}\)

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. \(25~\Omega\)

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of \(2.0~\text{V}\) and negligible internal resistance. The potentiometer wire itself is \(4~\text{m}\) long. When the resistance, \(R\), connected across the given cell, has values of (i) infinity (ii) \(9.5\), the 'balancing lengths, on the potentiometer wire, are found to be \(3~\text{m}\) and \(2.85~\text{m}\), respectively. The value of the internal resistance of the cell is (in ohm):
1. \(0.25\)
2. \(0.95\)
3. \(0.5\)
4. \(0.75\)

A wire of resistance \(4~\Omega\) is stretched to twice its original length. The resistance of a stretched wire would be: