The resistance of the meter bridge $AB$ in given figure is $4 \mathrm{\Omega }.$ With a cell of emf $\epsilon =0.5 \mathrm{V}$ and rheostat resistance $R_h=2 \mathrm{\Omega }$ the null point is obtained at some point $J.$ When the cell is replaced by another one of emf $\epsilon ={\epsilon }_2$ the same null point $J$ is found for $R_h=6 \mathrm{\Omega }.$ The emf ${\epsilon }_2$ is

In the experimental set--up of metre bridge shown in the figure, the null point is obtained at a distance of $40 \mathrm{cm}$ from $A$. If a $10 \mathrm{\Omega }$ resistor is connected in series with $R_1,$ the null point shifts by $10 \mathrm{cm}.$ The resistance that should be connected in parallel with $\left(R_1+10\right) \mathrm{\Omega }$ such that the null point shifts back to its initial position is

In the circuit shown, the potential difference between $\mathrm{A}$ and $\mathrm{B}$ is :

In a meter bridge, the wire of length $1 \mathrm{m}$ has a non-uniform cross-section such that, the variation $\frac{dR}{dl}$ of its resistance $R$ with length $l$ is $\frac{dR}{dl}\propto \frac{1}{\sqrt{l}}.$ Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point $P.$ What is the length $AP$ ?

In the given circuit diagram, the currents, $I_1=0.3 A, I_4=0.8 A$ and $I_5=0.4 A,$ are flowing as shown. The currents $I_2, I_3$ and $I_6,$ respectively, are -