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Current Question (ID: 20113)

Question:

$\text{The figure shows the charge } (q) \text{ versus voltage } (V) \text{ graphs corresponding to two different combinations of the same pair of capacitors:}$ $1. \text{ In parallel: } A$ $2. \text{ In series: } B$ $\text{Based on the graphs (for } A \text{ and } B\text{), what are the capacitances of the two capacitors?}$

Options:

1. $50 \ \mu\text{F} \text{ and } 30 \ \mu\text{F}$
2. $40 \ \mu\text{F} \text{ and } 10 \ \mu\text{F}$
3. $20 \ \mu\text{F} \text{ and } 30 \ \mu\text{F}$
4. $60 \ \mu\text{F} \text{ and } 40 \ \mu\text{F}$

Solution:

$\text{Hint: Recall the combination of capacitors.}$ $\text{Step 1: Find the equivalent capacitance in series } (B) \text{ and parallel } (A) \text{ combinations.}$ $\text{For series combination:}$ $C_B = \frac{80}{10} \quad [\because q = CV]$ $\Rightarrow C_B = 8 \ \mu\text{C}$ $\Rightarrow \frac{C_1 C_2}{C_1 + C_2} = 8 \quad \ldots (1)$ $\text{For parallel combination:}$ $C_A = \frac{500}{10}$ $\Rightarrow C_A = 50 \ \mu\text{C}$ $\Rightarrow C_1 + C_2 = 50 \quad \ldots (2)$ $\text{Step 2: Find the capacitance of each capacitor.}$ $\text{From equations (1) and (2), we get:}$ $\frac{C_1 (50 - C_1)}{C_1 + 50 - C_1} = 8$ $\Rightarrow 50C_1 - C_1^2 = 400$ $\Rightarrow C_1^2 - 50C_1 + 400 = 0$ $\Rightarrow (C_1 - 10)(C_1 - 40) = 0$ $\Rightarrow C_1 = 10 \ \mu\text{C} \text{ or } 40 \ \mu\text{C}$ $\text{If } C_1 = 10 \ \mu\text{C}; \text{ then } C_2 = 40 \ \mu\text{C}$ $\text{Hence, option (2) is the correct answer.}$

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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
}