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 circuit contains an ammeter, a battery of $30~\text{V},$ and a resistance $40.8~\Omega$ all connected in series. If the ammeter has a coil of resistance $480~\Omega$ and a shunt of $20~\Omega,$ then the reading in the ammeter will be:
1. $0.5~\text{A}$
2. $0.02~\text{A}$
3. $2~\text{A}$
4. $1~\text{A}$

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. $48~\Omega$

${A, B}~\text{and}~{C}$ are voltmeters of resistance $R,$ $1.5R$ and $3R$ respectively as shown in the figure above. When some potential difference is applied between ${X}$ and ${Y},$ the voltmeter readings are ${V}_{A},$ ${V}_{B}$ and ${V}_{C}$ 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 $8~\text{volts}$ and the average resistance per $\text{km}$ 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:

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. $0.75$