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Current Question (ID: 16568)

Question:

$\text{In the circuit shown in the figure initially key } K_1 \text{ is closed and key } K_2 \text{ is open.}$ $\text{Then } K_1 \text{ is opened and } K_2 \text{ is closed (order is important).}$ $\text{[Take } Q_1 \text{ and } Q_2 \text{ as charges on } C_1 \text{ and } C_2 \text{ and } V_1 \text{ and } V_2 \text{ as voltage respectively.]}$ $\text{Then,}$

Options:

1. $\text{(a) charge on } C \text{ gets redistributed such that } V_1 = V_2$
2. $\text{(b) charge on } C \text{ gets redistributed such that } Q_1 = Q_2$
3. $\text{(c) charge on } C \text{ gets redistributed such that } C_1V_1 + C_2V_2 = C_1E$
4. $\text{(d) charge on } C \text{ gets redistributed such that } Q_1 + Q_2 = Q$

Solution:

$\text{(2) Hint: The final potential difference across the capacitors will be the same.}$ $\text{Step 1: Find the final potential difference on the capacitors.}$ $\text{The charge stored by capacitor } C_1 \text{ gets redistributed between } C_1 \text{ and } C_2 \text{ till their potentials become the same i.e., } V_2 = V_1.$ $\text{Step 2: Find the final charge on the capacitors.}$ $\text{By the law of conservation of charge, the charge stored on capacitor } C_1 \text{ when key } K_1 \text{ is closed and key } K_2 \text{ is open is equal to the sum of charges on capacitors } C_1 \text{ and } C_2 \text{ when } K_1 \text{ is opened and } K_2 \text{ is closed i.e.,}$ $C_1V_1 + C_2V_2 = C_1E.$

Import JSON File

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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
}