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Current Question (ID: 16544)

Question:

$\text{The figure shows charge } (q) \text{ versus voltage } (V) \text{ graph for series and parallel}$ $\text{combination of two given capacitors. The capacitances are:}$

Options:

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

Solution:

$\text{Hint: } q = CV$ $\text{Step: Find the capacitance from the slope of the given graph.}$ $\text{We know that the charge } q \text{ stored on either plate of a capacitor is directly}$ $\text{proportional to the potential difference } V \text{ across two plates given by;}$$q = CV$$C = \frac{Q}{V}$$q-V \text{ graph}$$y = mx$$\text{Here the slope of the graph will give capacitance. Now we have to determine}$ $\text{capacitance for series and parallel combinations.}$ $\text{In series combination,}$$\frac{C_1C_2}{C_1+C_2} = \frac{q}{V}$$\frac{C_1C_2}{C_1+C_2} = \frac{80}{10}$$\frac{C_1C_2}{C_1+C_2} = 8 \, \mu\text{F}$$\text{In parallel combination,}$$C_1 + C_2 = \frac{q}{V}$$C_1 + C_2 = \frac{500}{10}$$C_1 + C_2 = 50 \, \mu\text{F}$$\text{Therefore, capacitance in parallel should be } 50 \, \mu\text{F} \text{ \& capacitance in series}$ $\text{must be } 8 \, \mu\text{F}$$\text{It is only possible when } C_1 = 10 \, \mu\text{F} \text{ and } C_2 = 40 \, \mu\text{F}.$$\text{Hence, option (4) 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
}