Balancing point of a potentiometer shifts from a length of $60 \ cm$ to $40 \ cm$ by shunting the cell with a $4 \ \Omega$ resistance. What is the internal resistance of the cell (in $\Omega$)?

The given potentiometer has its wire of resistance $10\, \Omega$. When the sliding contact is in the middle of the potentiometer wire,the potential drop across the $2\, \Omega$ resistor is -

$A$ potentiometer is connected between $A$ and $B$ and the balance point is obtained at $203.6 \, cm$. When the end of the potentiometer connected to $B$ is shifted to $C$,then the balance point is obtained at $24.6 \, cm$. If now the potentiometer be connected between $B$ and $C$,the balance point will be at ................. $cm$.

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

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 of emf $1.25 \,V$ gives a balancing length of $30 \,cm$. If the cell is replaced by another cell, the balancing length is found to be $40 \,cm$. What is the emf of the second cell?