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Current Question (ID: 20509)

Question:

$\text{In given L-R circuit connected with a D.C source of 12V, inductance is LmH and resistances is 6 } \Omega. \text{ If the emf induced in the inductor at } t = 1 \text{ms is 10V, value of L is:}$

Options:

1. $\frac{3}{\ln(1.2)}$
2. $\frac{6}{\ln(1.2)}$
3. $\frac{3}{\ln(1.8)}$
4. $\frac{6}{\ln(2.4)}$

Solution:

$\text{Hint: } i = i_0 (1 - e^{-Rt/L})$ $i = i_0 (1 - e^{-Rt/L})$ $\therefore \text{EMF} = \left| \frac{Ldi}{dt} \right| = L \left( \frac{V}{R} \right) \left( \frac{R}{L} \right) e^{-Rt/L} = V e^{-Rt/L}$ $10 = 12 \left( e^{-6/L} \right) \Rightarrow L = \frac{6}{\ln(1.2)}$

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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
}