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Current Question (ID: 16499)

Question:

$\text{A positive charge } q \text{ and a negative charge } -q \text{ are placed at } x = -a \text{ and } x = +a \text{ respectively. The variation of } V \text{ along } x\text{-axis is represented by the graph:}$

Options:

1. $\text{Graph 1}$
2. $\text{Graph 2}$
3. $\text{Graph 3}$
4. $\text{Graph 4}$

Solution:

$\text{Hint: The potential at a point due to a point charge varies as } V \propto \frac{q}{r}. $ $\text{Step 1: Find the potential at } x=a, x=0, \text{ and } x=-a. $ $V_O = \frac{kq}{a} - \frac{kq}{a} = 0$ $V_{-a} = \frac{-kq}{2a} + \frac{kq}{0} = \infty$ $V_a = \frac{kq}{2a} - \frac{kq}{0} = \infty$ $\text{Step 2: From the above values of } V, \text{ we can conclude that option (1) is correct.}$

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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
}