An infinite ladder network is constructed with $1 \mathrm{\Omega }$ and $2 \mathrm{\Omega }$ resistors as shown. The current through the $2 \mathrm{\Omega }$ resistor nearest to the battery is

In a metrebridge, when $R_{1}$ and $X$ are the resistances in left gap and right gap respectively, the null point is obtained at $40 \mathrm{cm}$ from the left. Now, when the resistance $R_{2}$ is in left gap and $X$ in right gap, then the null point is obtained at $60 \mathrm{cm}$ from the left. When the resistance in left gap is changed to $\left(R_{1}+R_{2}\right),$ the null point will be at

A potentiometer wire is $10 \mathrm{m}$ long and has a resistance of $2 \mathrm{\Omega } {\mathrm{m}}^{-1}$. It is connected in series with a battery of e.m.f. $3 \mathrm{V}$and a resistance of $10 \mathrm{\Omega }$. The potential gradient along the wire in $\mathrm{V} {\mathrm{m}}^{-1}$ is

A potentiometer wire has a resistivity of ${10}^9 \mathrm{\Omega } \mathrm{cm}$ and area of cross-section is $10^{-2} \mathrm{cm}^{2} .$ If current of $0.01 \mathrm{mA}$ passes through the wire, potential gradient is

A battery of e.m.f. $2 \mathrm{V}$ and internal resistance $2 \mathrm{\Omega }$ is connected to an external resistance $8 \mathrm{\Omega }$. If the length of the conductor is $4 \mathrm{m}$, then potential gradient between the two ends of the wire is

A resistance of $990 \mathrm{\Omega }$ and a cell of emf $2 \mathrm{V}$ is connected in series with a potentiometer wire having a length $2 \mathrm{m}$ and resistance $10 \mathrm{\Omega }$. The potential gradient along the wire will be

An accumulator of $5 \mathrm{V}$ is connected through a resistance of $40 \mathrm{\Omega }$ to a potentiometer wire $10 \text{m}$ long and of resistance $10 \mathrm{\Omega }$. For a cell, the null point is found at a length of $8 \mathrm{m}$ from the common terminal. The current through the wire is

The resistance per unit length of a wire is $1 \mathrm{\Omega } {\mathrm{m}}^{-1}$. A Leclanche cell of e.m.f. $1.45 \mathrm{V}$ is balanced against $2.9 \mathrm{m}$ length of potentiometer wire. The current through the wire is

The e.m.f. $E$ of the battery is balanced by potential difference across $75 \mathrm{cm}$ of a potentiometer wire. For a standard cell of e.m.f. $1.02 \mathrm{V}$, the balancing length is $50 \mathrm{cm}.$ The value of $E$ is