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Current Question (ID: 16513)

Question:

$\text{In a certain charge distribution, all points having zero potential can be joined by a circle } S. \text{ The points inside } S \text{ have positive potential, and points outside } S \text{ have a negative potential. A positive charge, which is free to move, is placed inside } S. \text{ What is the correct statement about } S:$

Options:

1. $\text{It will remain in equilibrium}$
2. $\text{It can move inside } S, \text{ but it cannot cross } S$
3. $\text{It must cross } S \text{ at some time}$
4. $\text{It may move, but will ultimately return to its starting point}$

Solution:

$\text{Hint: The charge moves from higher to lower potential, so it must cross the zero potential boundary } S \text{ to reach the negative potential region.}$ $\text{Explanation: A free positive charge moves from a higher (positive) potential to a lower (negative) potential. Hence, it must cross } S \text{ at some time. The positive charge inside } S \text{ experiences a force due to the electric field, which is directed from regions of higher potential to lower potential.}$ $\text{Since points inside } S \text{ have positive potential and points outside } S \text{ have negative potential, the charge will be naturally driven to move towards lower potential. Eventually, the charge must cross } S, \text{ as } S \text{ represents the zero potential boundary, and the charge will continue moving towards the negative potential region outside } S. \text{ Hence, option (3) is the correct answer.}$

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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
}