$A$ potentiometer wire of length $4 \text{ m}$ and resistance $5 \Omega$ is connected in series with a resistance of $992 \Omega$ and a cell of e.m.f. $4 \text{ V}$ with internal resistance $3 \Omega$. The length of $0.75 \text{ m}$ on the potentiometer wire balances the e.m.f. of: (in $\text{ mV}$)

$A$ null point is found at $200\,cm$ in a potentiometer when the cell in the secondary circuit is shunted by $5\,\Omega$. When a resistance of $15\,\Omega$ is used for shunting,the null point moves to $300\,cm$. The internal resistance of the cell is $..............\,\Omega$.

For comparing the $e.m.f.$'s of two cells with a potentiometer,a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessful?

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,the same cell will give a balance point at distance $x$. The value of $x$ is

$A$ $10 \ m$ long wire of resistance $20 \ \Omega$ is connected in series with a battery of e.m.f. $3 \ V$ and a resistance of $10 \ \Omega$. The potential gradient along the wire in $V/m$ is

In the given circuit of a potentiometer,the potential difference $E$ across $AB$ ($10\, m$ length) is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ (closed),the jockey is adjusted to touch the wire at point $J_{1}$ so that there is no deflection in the galvanometer. Now,the first battery $(E_{1})$ is replaced by the second battery $(E_{2})$ for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer then gives null deflection at $J_{2}$. The value of $\frac{E_{1}}{E_{2}}$ is $\frac{a}{b}$,where $a = \dots$ (Refer to the image for balancing lengths $l_{1}$ and $l_{2}$ from point $A$).