In the given figure,a battery $E$ is balanced on $55 \, cm$ length of a potentiometer wire. When a resistance of $10 \, \Omega$ is connected in parallel with the battery,it balances on $50 \, cm$ length of the potentiometer wire. The internal resistance $r$ of the battery is ............. $\Omega$.

$A$ cell in a secondary circuit gives a null deflection for $2.5 \ m$ length of a potentiometer wire having a total length of $10 \ m$. If the length of the potentiometer wire is increased by $1 \ m$ without changing the cell in the primary circuit,the new position of the null point is: (in $m$)

$A$ cell in the secondary circuit gives a null deflection for $2.5 \,m$ length of a potentiometer wire having a total length of $10 \,m$. If the length of the potentiometer wire is increased by $1 \,m$ without changing the cell in the primary circuit, the new position of the null point is: (in $m$)

$A$ potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is $k \, V/cm$ and the ammeter present in the circuit reads $1.0 \, A$ when the two-way key is switched off. The balance points,when the key between the terminals $(i)$ $1$ and $2$ and $(ii)$ $1$ and $3$ is plugged in,are found to be at lengths $l_1$ and $l_2$ respectively. The magnitudes of the resistors $R$ and $X$ in ohms are equal to:

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 $\Omega$)

In the primary circuit of a potentiometer,the current is $0.2 \ A$. The resistivity and cross-sectional area of the potentiometer wire are $4 \times 10^{-7} \ \Omega \cdot m$ and $8 \times 10^{-7} \ m^2$ respectively. The potential gradient will be ......... $V/m$.