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Current Question (ID: 20125)

Question:

$\text{A two-point charges } 4q \text{ and } -q \text{ are fixed on the x-axis at } x = -\frac{d}{2} \text{ and } x = \frac{d}{2}, \text{ respectively. If a third point charge } 'q' \text{ is taken from the origin to } x = d \text{ along the semicircle as shown in the figure, the energy of the charge will:}$

Options:

1. $\text{increased by } \frac{3q^2}{4\pi\varepsilon_0\ d}$
2. $\text{increase by } \frac{2q^2}{3\pi\varepsilon_0 d}$
3. $\text{decrease by } \frac{4q^2}{3\pi\varepsilon_0 d}$
4. $\text{decrease by } \frac{q^2}{4\pi\varepsilon_0 d}$

Solution:

$\text{Hint: Change in energy due to } -q \text{ is zero.}$ $\text{Potential of } -q \text{ is same as initial and final point of the path therefore potential due to } 4q \text{ will only change and as potential is decreasing the energy will decrease.}$ $\text{Decrease in potential energy } = q (V_i = V_f)$ $\text{Decrease in potential energy } = \left[ \frac{k4q}{d/2} - \frac{k4q}{3d/2} \right] = \frac{4q^2}{3\pi\varepsilon_0 d}$ $\text{Therefore correct answer is 3.}$

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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
}