The accurate measurement of $emf$ can be obtained using

$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs is:

$A$ potentiometer circuit shown in the figure is set up to measure the $e.m.f.$ of a cell $E$. As the point $P$ moves from $X$ to $Y$, the galvanometer $G$ shows deflection always in one direction, but the deflection decreases continuously until $Y$ is reached. In order to obtain a balance point between $X$ and $Y$, it is necessary to

The length of a potentiometer wire is $l$. $A$ cell of emf $E$ is balanced at a length $\left(\frac{l}{3}\right)$ from the positive end of the wire. If the length of the wire is increased by $\left(\frac{l}{2}\right)$,the distance at which the same cell gives the balancing point is (The cell in the primary circuit is ideal and no series resistance is present in the primary circuit.)

$A$ 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 ............... $\Omega$.

The reading of the voltmeter in the following circuit is ................ $V$.