Correct diagram for the determination of internal resistance of a primary cell by potentiometer is

A potentiometer wire $PQ$ of $1 \mathrm{m}$ length is connected to a standard cell $E_1$. Another cell $E_2$ of emf $1.02 \mathrm{V}$ is connected with a resistance $r$ and switch $S$ (as shown in figure). With switch $S$ open, the null position is obtained at a distance of $49 \mathrm{cm}$ from $\mathrm{Q}$. The potential gradient in the potentiometer wire is :

In the given potentiometer circuit arrangement, the balancing length $\mathrm{AC}$ is measured to be $250 \mathrm{cm}$. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes $400 \mathrm{cm}$. The ratio of the emf of two cells $\frac{{\epsilon }_1}{{\epsilon }_2}$ is:

Figure shows a potentiometer. Length of the potentiometer wire $AB$ is $100 \mathrm{cm}$ and its resistance is $100\Omega$. EMF of the battery $E$ is $2 \mathrm{V}$. A resistance $R$ of $50\Omega$ draws current from the potentiometer. What is the voltage across $R$ when the sliding contact $C$is at the mid-point of $AB ?$