Import Question JSON

Current Question (ID: 16413)

Question:

$\text{Two equal negative charges of charge } -q \text{ are fixed at the points } (0, a) \text{ and } (0, -a) \text{ on the } Y\text{-axis.}$ $\text{A positive charge } Q \text{ is released from rest at the point } (2a, 0) \text{ on the } X\text{-axis.}$ $\text{The charge } Q \text{ will:}$

Options:

1. $\text{execute simple harmonic motion about the origin.}$
2. $\text{move to the origin and remain at rest.}$
3. $\text{move to infinity.}$
4. $\text{execute oscillatory but not simple harmonic motion.}$

Solution:

$\text{Hint: Check that the net electrostatic force on charge } Q \text{ due to both charges } -q \text{ will always point towards the origin.}$ $\text{Step 1: Draw a diagram as follows, showing force vectors.}$ $\text{By symmetry of the problem the components of force on } Q \text{ due to charges at } A \text{ and } B \text{ along the } y\text{-axis will cancel each other while along the } x\text{-axis will add up and will be along with } CO. \text{ Under the action of this force charge, } Q \text{ will move towards } O. \text{ If at any time charge } Q \text{ is at a distance } x \text{ from } O. \text{ Net force on charge } Q:$ $F_{\text{net}} \Rightarrow 2F \cos \theta = 2 \frac{1}{4\pi\varepsilon_0} \frac{-qQ}{(a^2+x^2)} \times \frac{x}{(a^2+x^2)^{1/2}}$ $i.e., \ F_{\text{net}} = -\frac{1}{4\pi\varepsilon_0} \cdot \frac{2qQx}{(a^2+x^2)^{3/2}}$ $\text{As the restoring force } F_{\text{net}} \text{ is not linear, the motion will be oscillatory (with amplitude } 2a) \text{ but not simple harmonic.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}