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Current Question (ID: 20527)

Question:
$\text{A coil of inductance 1 H and resistance 100 } \Omega \text{ is connected to a battery of 6 V. Determine approximately:}$ $\text{I. The time elapsed before the current acquires half of its steady-state value.}$ $\text{II. The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on.}$ $\text{(Given: } \ln 2 = 0.693, \frac{e^{-3}}{2} = 0.25)$
Options:
  • 1. $t = 10 \text{ ms}; U = 2 \text{ mJ}$
  • 2. $t = 10 \text{ ms}; U = 1 \text{ mJ}$
  • 3. $t = 7 \text{ ms}; U = 1 \text{ mJ}$
  • 4. $t = 7 \text{ ms}; U = 2 \text{ mJ}$
Solution:
$\text{Hint: } I = I_0 \left(1 - e^{-\frac{t}{\tau}}\right); U = \frac{1}{2}LI^2$

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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
}