$A$ wire,$10 \ m$ long,has a resistance of $40 \ \Omega$. It is connected in series with a resistance box of resistance $R$ and a $2 \ V$ storage cell. If the potential gradient along the wire is $0.1 \ mV/cm$,then the value of $R$ is (in $\Omega$)

$A$ potentiometer has a potential gradient of $2 \, mV/cm$. It is used to measure the potential difference across a $10 \, \Omega$ resistor. If the length of the potentiometer wire required to obtain the null point is $50 \, cm$,the current flowing through the $10 \, \Omega$ resistor is ............. $mA$.

To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)

$A$ potentiometer wire $PQ$ of $1\,m$ length is connected to a standard cell $E_{1}$. Another cell $E_{2}$ of emf $1.02\,V$ is connected with a resistance $r$ and switch $S$ (as shown in figure). With switch $S$ open,the null position is obtained at a distance of $49\,cm$ from $Q$. The potential gradient in the potentiometer wire is.......$V/cm$.

$A$ potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by

$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a 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: (in $\Omega$)