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Current Question (ID: 16684)

Question:

$\text{For a cell, the terminal potential difference is } 2.2 \text{ V when the circuit is open and reduces to } 1.8 \text{ V when the cell is connected to the resistance of } R = 5 \, \Omega. \text{ The internal resistance of cell } (r) \text{ is:}$

Options:

1. $\frac{10}{9} \, \Omega$
2. $\frac{9}{10} \, \Omega$
3. $\frac{11}{9} \, \Omega$
4. $\frac{5}{9} \, \Omega$

Solution:

$\text{Hint: } V = E - Ir$ $\text{Step: Find the internal resistance of the cell } (r).$ $V = E - Ir \Rightarrow V = E - \left( \frac{E}{R+r} \right) r = \frac{ER}{(R+r)}$ $\text{From the given conditions, } E = 2.2 \text{ and when } R = 5 \text{ then, TPD } V = 1.8 \text{ V}$ $1.8 = \frac{2.2 \times 5}{5+r}$ $\Rightarrow r = \frac{10}{9} \, \Omega$ $\text{Hence, option (1) is the correct answer.}$

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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
}