$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be:

For a cell,the balancing length is $0.60 \, m$. For another cell having an electromotive force $(emf)$ $0.1 \, V$ less than the first one,the balancing length is $0.55 \, m$. What are the $emf$ values of the two cells?

In the experiment of a potentiometer,at balance,there is no current in the

$A$ 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:

$A$ $6 \, V$ 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:

$A$ 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 $,V/cm$)?