The potential difference $V$ across the filament of the bulb shown in the given Wheatstone bridge varies as $V=i(2 i+1)$, where $i$ is the current in ampere through the filament of the bulb. The emf of the battery $V_a$, so that the bridge become balanced is

Three resistors of resistances $10 \Omega, 20 \Omega$ and $30 \Omega$ are connected as shown in the figure. If the points $A, B$ and $C$ are at potentials $10 \mathrm{~V}, 6 \mathrm{~V}$ and 5 V respectively, then the ratio of the magnitudes of the currents through $10 \Omega$ and $30 \Omega$ resistors is

A potentiometer balances at 44 cm when a cell of internal resistance $1 \Omega$ is in the secondary circuit. To obtains the balancing point at 40 cm , the resistance to be connected parallel to cell is