$A$ wire of length $100\, cm$ is connected to a cell of emf $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential difference of $1\, mV/cm$ is ............. $\Omega$.

$A$ 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$)

$A$ 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 $V$)

In a potentiometer,when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance,the balance is obtained at a length of $120 \ cm$ of the wire. Now,when the same cell is shunted with a $12 \ \Omega$ resistance,the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $Omega$)

When a cell of $E.M.F.$ $E_1$ is connected to a potentiometer wire,the balancing length is $l_1$. Another cell of $E.M.F.$ $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is