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Current Question (ID: 16581)

Question:

$\text{Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of } V_0 \text{ volt.}$ $\text{If one of the capacitors is completely immersed in a liquid with dielectric constant } K, \text{ the potential difference between the plates of the other capacitor will change to:}$

Options:

1. $\frac{K+1}{K} V_0$
2. $\frac{K}{K+1} V_0$
3. $\frac{K+1}{2K} V_0$
4. $\frac{2K}{K+1} V_0$

Solution:

$\text{Hint: As capacitors are connected in series so, the charge stored on each of the capacitors will be equal.}$ $\text{Step 1: Write the potential.}$ $V_0 = V_1 + V_2$ $\text{Step 2: Find } V_1 \text{ and } V_2$ $V_1 = \frac{Q}{C}$ $V_2 = \frac{Q}{KC}$ $\text{Or, } V_2 = \frac{V_1}{K}$ $\text{Step 3: Find } V_1$ $V_0 = V_1 + \frac{V_1}{K}$ $V_1 = \frac{KV_0}{1+K}$

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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
}