As shown in the figure, the resistance of a galvanometer G can be found by the half-deflection method. Here the resistance R₂ is adjusted such that when the key K is closed the deflection in the galvanometer becomes half of the value as compared to when K is open. Half-deflection is obtained at R₂ = 4 $\Omega$ and thus the galvanometer resistance is found to be 6 $\Omega$. In this half-deflection condition the current (in mA) through the resistor R₁ is :

Two resistances $R_{1}=X \Omega$ and $R_{2}=1 \Omega$ are connected to a wire $A B$ of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from $0.2 \mathrm{~mm}$ at $A$ to $1 \mathrm{~mm}$ at $B$. A galvanometer $(\mathrm{G})$ connected to the center of the wire, $50 \mathrm{~cm}$ from each end along its axis, shows zero deflection when $A$ and $B$ are connected to a battery. The value of $X$ is ____________.

In the following circuit $C_{1}=12 \mu F, C_{2}=C_{3}=4 \mu F$ and $C_{4}=C_{5}=2 \mu F$. The charge stored in $C_{3}$ is ____________ $\mu C$.

In order to measure the internal resistance r₁ of a cell of emf E, a meter bridge of wire resistance R₀ = 50$$\Omega$$, a resistance R₀/2, another cell of emf E/2 (internal resistance r) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at l = 72 cm, then the value of r₁ = __________ $$\Omega$$.