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Current Question (ID: 16396)

Question:

$\text{Two point charges } A \text{ and } B, \text{ having charges } +Q \text{ and } -Q \text{ respectively, are placed at a certain distance apart and the force acting between them is } F. \text{ If } 25\% \text{ charge of } A \text{ is transferred to } B, \text{ then the force between the charges becomes:}$

Options:

1. $\frac{4F}{3}$
2. $F$
3. $\frac{9F}{16}$
4. $\frac{16F}{9}$

Solution:

$\text{Hint: } F = \frac{kq_1q_2}{r^2}$ $\text{Step: Find the force after the charge is transferred and use the above-obtained equation.}$ $\text{Initially, the force is given by:}$ $F = \frac{kQ^2}{d^2}$ $\text{If } 25\% \text{ of charge of } A \text{ transferred to } B \text{ then}$ $q_A = Q - \frac{Q}{4} = \frac{3Q}{4} \text{ and } q_B = -Q + \frac{Q}{4} = -\frac{3Q}{4}$ $F_1 = \frac{kq_Aq_B}{r^2}$ $F_1 = \frac{k\left(\frac{3Q}{4}\right)^2}{r^2}$ $F_1 = \frac{9}{16} \times \frac{kQ^2}{r^2}$ $\Rightarrow F_1 = \frac{9F}{16}$ $\text{Hence, option (3) is the correct answer.}$

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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
}