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. The 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: