$A$ cell is connected to a potentiometer,and the balancing length is obtained at $125 \ cm$. When a resistor of $2 \ \Omega$ is connected in parallel with the cell,the balancing length is obtained at $100 \ cm$. What is the internal resistance of the cell in $\Omega$?

Explain the comparison of the electromotive force (emf) of two cells using a potentiometer with a necessary diagram.

$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?

$A$ $6 \, V$ battery of negligible internal resistance is connected across a uniform wire $AB$ of length $1 \, m$. The positive terminal of another battery of emf $4 \, V$ and internal resistance $1 \, \Omega$ is joined to point $A$ as shown in the figure. The ammeter shows zero deflection when the jockey touches the wire at point $C$. The length $AC$ is equal to:

$A$ potentiometer wire is $4 \,m$ long and a potential difference of $3 \,V$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $100 \,cm$ of the potentiometer wire is: (in $\,V$)

$A$ cell can be balanced against $110 \, cm$ and $100 \, cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \, \Omega$. Its internal resistance is ............... $\Omega$.