A cell having an emf $\varepsilon $ and internal resistance $r$ is connected across a variable external resistance $R$. As the resistance $R$ is increased, the plot of potential difference $V$ across $R$ is given by-

(A)

(B)

(C)

               (D)

Hint: To solve this question, we need to draw the circuit diagram according to the given information. Then using the Ohm’s law we can find out the expression for the potential difference across the external resistor in terms of the external resistance. From there we can predict the shape of the graph.

Complete step-by-step solution:
According to the information given in the question, a cell of emf $\varepsilon $ and internal resistance $r$ is connected across a variable external resistance $R$. So we can represent this information by the circuit diagram shown below.

Let the current in the circuit be $I$. The internal resistance and the external resistance are connected in series combination with each other. So the net resistance in the circuit is
${R_N} = R + r$ ……………………...(1)
From the Ohm’s law we can write
$\varepsilon = I{R_N}$
From (1)
$\varepsilon = I\left( {R + r} \right)$
$ \Rightarrow I = \dfrac{\varepsilon }{{R + r}}$................(2)
Now, the potential difference across the resistance is given by the Ohm’s law as
$V = IR$
Substituting (2) we get
$V = \dfrac{{\varepsilon R}}{{R + r}}$..........................(3)
Since the independent variable $R$ is appearing at both the numerator and the denominator, so we have to simplify the above equation (3), which can also be written as
$V = \dfrac{{\varepsilon \left( {R + r - r} \right)}}{{R + r}}$
$V = \varepsilon - \dfrac{{\varepsilon r}}{{R + r}}$...............................(4)
So now we have the independent variable $R$ only in the denominator. Now, we substitute in (4) to get
$V\left( 0 \right) = \varepsilon - \dfrac{{\varepsilon r}}{{0 + r}}$
$ \Rightarrow V\left( 0 \right) = 0$
So for $R = 0$, we have $V = 0$. So the plot must pass through the origin. As we can see that the plots given in the options A and D do not pass through the origin, so they must be incorrect.
Therefore options A and D are incorrect.
Now, we take the limit $R \to \infty $ at both sides in (4) to get
$\mathop {\lim }\limits_{R \to \infty } V = \mathop {\lim }\limits_{R \to \infty } \left( {\varepsilon - \dfrac{{\varepsilon r}}{{R + r}}} \right)$
\[ \Rightarrow \mathop {\lim }\limits_{R \to \infty } V = \varepsilon - \varepsilon r\mathop {\lim }\limits_{R \to \infty } \left( {\dfrac{1}{{R + r}}} \right)\]
We know that $\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{x + k}}} \right) = 0$. Therefore we have
\[\mathop {\lim }\limits_{R \to \infty } V = \varepsilon - \varepsilon r\left( 0 \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{R \to \infty } V = \varepsilon \]
So the ordinate of the plot will approach the value $\varepsilon $ as the variable resistance $R$ is infinitely increased.
In the plot shown in option B, we can see this happening.

Hence, the correct answer is option B.

Note: Do not try to obtain the plot by using transformations of the graph. Although we can obtain the plot by that method also, that would take much time and also chances of mistakes are huge. So after getting the equation, always guess the plot by substituting the end point values and taking the limits.