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

The given figure represents an arrangement of a potentiometer for the calculation of the internal resistance $(r)$ of an unknown battery $(E)$. The balance length is $70.0 \, cm$ with the key open and $60.0 \, cm$ with the key closed. $R$ is $132.40 \, \Omega$. The internal resistance $(r)$ of the unknown cell will be ....... $\Omega$ (Given $E_o > E$):-

$A$ cell balances against a length of $150 \ cm$ on a potentiometer wire when it is shunted by a resistance of $5 \ \Omega$. But when it is shunted by a resistance of $10 \ \Omega$,then the balancing length increases by $25 \ cm$. The balancing length when the cell is in an open circuit is: (in $cm$)

The length of a wire of a potentiometer is $100 \, cm$,and the $emf$ of its standard cell is $E \, volt$. It is employed to measure the $emf$ of a battery whose internal resistance is $0.5 \, \Omega$. If the balance point is obtained at $l = 30 \, cm$ from the positive end,the $emf$ of the battery is (where $i$ is the current in the potentiometer wire).

Resistance of a potentiometer wire is $2 \ \Omega/m$. $A$ cell of e.m.f. $1.5 \ V$ balances at $300 \ cm$. The current through the wire is (in $mA$)

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: ................ $\Omega$