The readings of the voltmeter and ammeter in the circuit shown in the diagram are respectively

The resistance of an iron wire is $10\,\Omega$ and the temperature coefficient of resistivity is $5 \times 10^{-3}\,^{\circ}C^{-1}$. At $20\,^{\circ}C$,it carries $30\,mA$ of current. Keeping the potential difference between its ends constant,the temperature of the wire is raised to $120\,^{\circ}C$. The current in milliamperes that flows in the wire is:

Each of the resistances in the network shown in the figure is equal to $R$. The resistance between the terminals $A$ and $B$ is

The readings of ammeters $A_1$ and $A_2$ will be respectively:

Two resistances of $400 \Omega$ and $800 \Omega$ are connected in series with a $6 \text{ V}$ battery of negligible internal resistance. $A$ voltmeter of resistance $10000 \Omega$ is used to measure the potential difference across the $400 \Omega$ resistor. The error in the measurement of potential difference in volts is approximately:

$A$ cell of internal resistance $r$ is connected across an external resistance $n r$. Then the ratio of the terminal voltage to the emf of the cell is