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

$A$ cell can be balanced against $110 \ cm$ and $100 \ cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \ \Omega$. Its internal resistance is

In the given potentiometer circuit,the balancing length for points $B$ and $C$ is $40 \, cm$. What is the balancing length for points $C$ and $D$ in $cm$?

An ideal battery of $4\, V$ and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1\, m$ and resistance $5\,\Omega$. The value of $R$,to give a potential difference of $5\, mV$ across $10\, cm$ of potentiometer wire,is: ................ $\Omega$

In the given potentiometer circuit arrangement,the balancing length $AC$ is measured to be $250 \, cm$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram,the balancing length becomes $400 \, cm$. The ratio of the emf of two cells,$\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

$A$ potentiometer wire of length $100 \ cm$ and resistance $3 \ \Omega$ is connected in series with a resistance of $8 \ \Omega$ and an accumulator of $4 \ V$ whose internal resistance is $1 \ \Omega$. $A$ cell of e.m.f. $E$ is balanced by $50 \ cm$ length of the wire. The e.m.f. of the cell is: (in $V$)