$A$ potentiometer balances at $44 \ cm$ when a cell of internal resistance $1 \ \Omega$ is in the secondary circuit. To obtain the balancing point at $40 \ cm$,the resistance to be connected in parallel to the cell is: (in $\Omega$)

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 a potentiometer experiment,a cell is balanced at a length of $240 \ cm$. When the cell is shunted with a resistance of $2 \ \Omega$,it is balanced at a length of $120 \ cm$. The internal resistance of the cell is .......... $\Omega$.

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 ......... $V/m$.

$A$ cell of internal resistance $3 \, \Omega$ and $emf$ $10 \, V$ is connected to a uniform wire of length $500 \, cm$ and resistance $3 \, \Omega$. The potential gradient in the wire is .............. $mV/cm$.

In a potentiometer circuit, there is a cell of $e.m.f.$ $2\, V$, a resistance of $5\, \Omega$ and a wire of uniform thickness of length $1000\, cm$ and resistance $15\, \Omega$. The potential gradient in the wire is: