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Current Question (ID: 16543)

Question:

$\text{A parallel plate capacitor is connected to a battery as shown in the figure.}$ $\text{Consider two situations.}$ $\text{A: Key } K \text{ is kept closed and plates of capacitors are moved apart using the insulating handle.}$ $\text{B: Key } K \text{ is opened and plates of capacitors are moved apart using the insulating handle.}$ $\text{Choose the correct option(s):}$ $\text{a. In A, } Q \text{ remains the same but } C \text{ changes.}$ $\text{b. In B, } V \text{ remains the same but } C \text{ changes.}$ $\text{c. In A, } V \text{ remains the same and hence } Q \text{ changes.}$ $\text{d. In B, } Q \text{ remains the same and hence } V \text{ changes.}$

Options:

1. $(a) \text{ and } (b)$
2. $(a) \text{ and } (d)$
3. $(b) \text{ and } (c)$
4. $(c) \text{ and } (d)$

Solution:

$\text{Hint: The capacitance depends on the separation between the plates.}$ $\text{Step 1: Find the charge and potential in the first case.}$ $\text{When key } K \text{ is kept closed and plates of capacitors are moved apart using the insulating handle, the separation between two plates increases which in turn decreases its capacitance } \left( C = \frac{K \varepsilon_0 A}{d} \right) \text{ and hence, the charge stored decreases as } Q = CV \text{ (the potential continue to be the same as capacitor is still connected with the cell).}$ $\text{Step 2: Find the charge and potential in the second case.}$ $\text{When key } K \text{ is opened and plates of capacitors are moved apart using the insulating handle, the charge stored by the disconnected charged capacitor remains conserved and with the decreases in capacitance, the potential difference } V \text{ increases as } V = \frac{Q}{C}. $

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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
}