$A$ potentiometer has a uniform potential gradient across it. Two cells connected in series $(i)$ to support each other and $(ii)$ to oppose each other are balanced over $6 \ m$ and $2 \ m$ respectively on the potentiometer wire. The ratio of the $e.m.f.$s of the cells is:

$A$ potentiometer wire has a length of $100 \ cm$ and is connected to a cell of $emf \ E$. It is used to measure the $emf \ E_0$ of a battery with an internal resistance of $0.5 \ \Omega$. If the balance point is obtained at a distance of $\ell = 30 \ cm$ from the positive terminal,the $emf \ E_0$ of the battery is:

In a potentiometer,when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance,the balance is obtained at a length of $120 \ cm$ of the wire. Now,when the same cell is shunted with a $12 \ \Omega$ resistance,the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

$A$ galvanometer can be used as a voltmeter by connecting a

$A$ potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery,used across the potentiometer wire,has an $EMF$ of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R$,connected across the given cell,has values of $(i)$ infinity and $(ii)$ $9.5\,\Omega$,the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m$,respectively. The value of internal resistance of the cell is ............... $\Omega$.

The figure shows a $2.0 \; V$ potentiometer used for the determination of internal resistance of a $1.5 \; V$ cell. The balance point of the cell in open circuit is $76.3 \; cm$. When a resistor of $9.5 \; \Omega$ is used in the external circuit of the cell,the balance point shifts to $64.8 \; cm$ length of the potentiometer wire. Determine the internal resistance (in $\Omega$) of the cell.