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Current Question (ID: 16685)

Question:

$\text{A student measures the terminal potential difference } V \text{ of a cell (of emf } E \text{ and internal resistance } R) \text{ as a function of the current } I \text{ flowing through it.}$ $\text{The slope and intercept of the graph between } V \text{ and } I, \text{ respectively, is equal to:}$

Options:

1. $E \text{ and } -r$
2. $-r \text{ and } E$
3. $r \text{ and } -E$
4. $-E \text{ and } r$

Solution:

$\text{Hint: } V = \varepsilon - Ir$ $\text{Step: Find the slope and intercept of the graph between } V \text{ and } I.$ $\text{According to Ohm's law:}$ $V \propto I$ $V = IR \text{ where } I \text{ is the current and } R \text{ is the resistance present in the conductor}$ $\text{Suppose, } V \text{ is the terminal potential difference of a cell}$ $\text{And, the cell has the emf } \varepsilon$ $\text{Then, we can write:}$ $V + Ir = \varepsilon$ $\text{If } I = 0$ $\text{Then we can write, } V = \varepsilon \text{ i.e., the terminal potential difference is equal to the emf of the cell.}$ $\text{The slope will be given as:}$ \frac{dV}{dI} = -r$ $\text{Therefore, the slope and intercept of the graph between } V \text{ and } I, \text{ respectively are } -r \text{ and } E.$ $\text{Hence, option (2) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}