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Current Question (ID: 16702)

Question:

$\text{Two cells of emf } E \text{ and internal resistance } r_1 \text{ and } r_2 \text{ are connected in series through an external resistance } R. \text{ The value of } R \text{ for which the potential difference across one of the cells becomes zero will be:}$

Options:

1. $\frac{r_1 r_2}{r_1 + r_2}$
2. $r_1 + r_2$
3. $|r_1 - r_2|$
4. $\frac{r_1}{r_2}$

Solution:

$\text{Hint: Net EMF of one of the cells drops in its own internal resistance.}$ $\text{Step 1: Find the current in } r_2.$ $i_1 = \frac{E}{r_2}$ $\text{Step 2: Find the current in the circuit.}$ $i_2 = \frac{2E}{R + r_1 + r_2}$ $\text{As } \frac{E}{r_2} = i_1 = i_2, \Rightarrow \frac{2E}{R + r_1 + r_2} = \frac{E}{r_2}$ $\text{or } R = r_2 - r_1$

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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
}