1. What is actually being plotted?

Because the voltmeter is connected across the battery, the measured voltage is the terminal potential difference of the battery, not just the voltage across the external resistor.

For a battery of emf $E$ and internal resistance $r$, when current $I$ flows,

$V = E - Ir$

So a graph of $V$ versus $I$ must be a straight line with:

• intercept on voltage axis $= E$
• slope $= -r$

2. Read the graph

From the graph:

• At $I=0$, the voltage is about $1.5\,\text{V}$. Hence, $E = 1.5\,\text{V}$

• The graph meets the current axis near $I = 1000\,\text{mA} = 1.0\,\text{A}$, and given $V_0 \approx 0$, this means terminal voltage becomes nearly zero there.

Using $V = E - Ir$

At $V \approx 0$ and $I=1\,\text{A}$,

$0 = 1.5 - (1)r$

so,

$r = 1.5\,\Omega$

3. Check each option

Option A: The value of the resistance $R$ is $1.5\,\Omega$

From the graph we determine battery parameters $E$ and $r$, not directly the external resistance $R$. So this is incorrect.

Option B: The emf of the battery is $1.5\,\text{V}$ and its internal resistance is $1.5\,\Omega$

This matches exactly what we found:

$E = 1.5\,\text{V}, \qquad r = 1.5\,\Omega$

So this is correct.

Option C: The emf of the battery is $1.5\,\text{V}$ and the value of $R$ is $1.5\,\Omega$

Again, $R$ cannot be concluded as $1.5\,\Omega$ from this graph. So this is incorrect.

Option D: The potential difference across the battery is $1.5\,\text{V}$ when it sends a current of $1000\,\text{mA}$

At $I=1000\,\text{mA}$, the graph shows $V \approx 0$, not $1.5\,\text{V}$. So this is incorrect.

4. Final answer

The correct statement is:

$\boxed{\text{B}}$

The battery has:

$\boxed{E = 1.5\,\text{V}, \quad r = 1.5\,\Omega}$