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Current Question (ID: 20238)

Question:

$\text{A capacitor is discharging through a resistor } R. \text{ Consider in time } t_1,$ $\text{the energy stored in the capacitor reduces to half of its initial value}$ $\text{and in time } t_2, \text{ the charge stored reduces to one-eighth of its initial}$ $\text{value. The ratio } \frac{t_1}{t_2} =$

Options:

1. $\frac{1}{2}$
2. $\frac{1}{3}$
3. $\frac{1}{4}$
4. $\frac{1}{6}$

Solution:

$\text{Hint: Charge discharges exponentially.}$ $\text{Step 1: Express energy stored in the capacitor as a function of time.}$ $q = q_0 e^{-t/RC}$ $U = \frac{q^2}{2C} = \frac{q_0^2}{2C} e^{-\frac{2t}{RC}}$ $\text{Step 1: Find the value of } t_1.$ $\frac{2t_1}{RC} = \ln\left(\frac{U}{U_0}\right)$ $t_1 = \frac{RC}{2} \ln 2$

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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
}