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Current Question (ID: 17374)

Question:

$\text{The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux } \phi \text{ linked with the primary coils is given by } \phi = \phi_0 + 4t, \text{ where } \phi \text{ is in webers, } t \text{ is time in seconds, and } \phi_0 \text{ is a constant, the output voltage across the secondary coil is:}$

Options:

1. $120 \text{ V}$
2. $220 \text{ V}$
3. $30 \text{ V}$
4. $90 \text{ V}$

Solution:

$\text{Hint: For an ideal transformer, } \frac{N_p}{N_s} = \frac{e_p}{e_s}.$ $\text{Step 1: Find the voltage across the primary coil}$ $|e_p| = \left| \frac{d\phi}{dt} \right| = 4 \text{ V}$ $\text{Step 2: Use the concept of turn ratio}$ $\frac{N_s}{N_p} = \frac{V_s}{V_p}$ $\frac{1500}{50} = \frac{V_s}{4}$ $V_s = 120 \text{ V}$

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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
}