An electric current is passed through a circuit containing two wires of the same material,connected in parallel. If the lengths and radii of the wires are in the ratio of $4/3$ and $2/3$ respectively,then the ratio of the currents passing through the wires will be:

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:

The currents $i_1$ and $i_2$ through the resistors $R_1 (= 10\,\Omega)$ and $R_2 (= 30\,\Omega)$ in the circuit diagram with $E_1 = 3\,V, E_2 = 3\,V$ and $E_3 = 2\,V$ are respectively:

The reading of the ammeter in the adjoining figure will be .............. $A$.

$A$ resistor is connected to a battery of $12 \text{ V}$ emf and internal resistance $2 \text{ }\Omega$. If the current in the circuit is $0.6 \text{ A}$, the terminal voltage of the battery is: (in $\text{ V}$)

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: