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Current Question (ID: 16697)

Question:

$\text{Current through the } 2 \, \Omega \text{ resistance in the electrical network shown is:}$ $\begin{array}{c} 2 \, \text{V} \\ \begin{array}{ccc} & 10 \, \Omega & \\ 2 \, \Omega & & 20 \, \Omega \\ & & \\ \end{array} \\ 12 \, \text{V} \end{array}$

Options:

1. $\text{zero}$
2. $1 \, \text{A}$
3. $3 \, \text{A}$
4. $5 \, \text{A}$

Solution:

$\text{The current entering the battery equals the current leaving it.}$ $\text{Suppose, the current } i \text{ goes through a resistance } 2 \, \Omega \text{ then it is required that the same current must come back.}$ $\text{But there is no way for the current } i \text{ to come back, hence net current through } 2 \, \Omega \text{ resistance is zero.}$ $\text{No current will flow through resistor, because in a closed loop, total potential drop must be zero.}$ $\text{Therefore the current in } 2 \, \Omega \text{ resistor will be zero because it is not a part of any closed loop.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}