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Current Question (ID: 16410)

Question:

$\text{When 2 point charges } +q \text{ and } +3q, \text{ held at a distance } r \text{ from each other are released, they have an acceleration of } a \text{ and } 2a \text{ respectively.}$ $\text{When we distribute the total charge equally between them, keep them at the same distance as the original and release them, their accelerations now would be:}$

Options:

1. $2a \text{ and } 4a$
2. $\frac{4a}{3} \text{ and } \frac{8a}{3}$
3. $\frac{9a}{4} \text{ and } \frac{9a}{2}$
4. $\frac{2a}{3} \text{ and } \frac{4a}{3}$

Solution:

$\text{Initially, the forces on the charges are } F_1 = \frac{kq \cdot 3q}{r^2} \text{ and } F_2 = \frac{kq \cdot 3q}{r^2}. $ $\text{The accelerations are } a = \frac{F_1}{m_1} \text{ and } 2a = \frac{F_2}{m_2}. $ $\text{When charges are equal, } q_1 = q_2 = 2q. $ $\text{New force } F = \frac{k(2q)(2q)}{r^2} = \frac{4kq^2}{r^2}. $ $\text{New accelerations } a_1 = \frac{F}{m_1} = \frac{4a}{3} \text{ and } a_2 = \frac{F}{m_2} = \frac{8a}{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
}