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Current Question (ID: 20498)

Question:

$\text{A part of a complete circuit is shown in the figure. At a certain instant, the current } I \text{ is } 1 \text{ A and is decreasing at a rate of } 10^2 \text{ A s}^{-1}. \text{ The value of the potential difference } V_P - V_Q, \text{ (in volts) at that instant is:}$ $L = 50 \text{ mH} \quad I \quad 30 \text{ V} \quad R = 2 \Omega$

Options:

1. $10$
2. $25$
3. $33$
4. $53$

Solution:

$\text{Hint: } V = L \frac{dI}{dt}$ $\text{Step: Find the potential difference } V_P - V_Q.$ $\text{By applying KVL in the loop is given by;}$ $\text{Given: } \frac{dI}{dt} = 10^2 \text{ A s}^{-1}, \quad I = 1 \text{ A}$ $\Rightarrow V_P - 5 - 30 + 2 \times 1 = V_Q$ $\Rightarrow V_P - V_Q = 35 - 2 = 33 \text{ V}$ $\text{Hence, option (3) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}