Import Question JSON

Current Question (ID: 20060)

Question:

$\text{A fixed charge } P \text{ and a free charge } Q, \text{ both having the same charge and mass, are positioned as shown in the diagram.}$ $\text{Given that } q = 2 \mu C, \theta = 30^\circ, \text{ and } m = 20 \text{ g, what is the maximum height } h$ $\text{attained by the charge } Q \text{ in equilibrium on the smooth inclined plane?}$

Options:

1. $0.1 \text{ m}$
2. $0.3 \text{ m}$
3. $0.4 \text{ m}$
4. $0.5 \text{ m}$

Solution:

$\text{Hint: } F = \frac{k q_1 q_2}{r^2}$ $\text{Step: Find the maximum height } h \text{ attained by the charge } Q \text{ in equilibrium.}$ $\text{Given: } q = 2 \mu C, \theta = 30^\circ, m = 20 \text{ g,}$ $\text{The forces acting on the charge } Q \text{ are given by:}$ $F_1 = \frac{k q^2}{r^2}, F_2 = mg \sin \theta$ $\text{At the equilibrium condition, both forces are equal in magnitude}$ $i.e., F_1 = F_2$ $mg \sin \theta = \frac{k q^2}{r^2}$ $\Rightarrow mg \sin \theta = \frac{k q^2}{\left(\frac{h}{\sin \theta}\right)^2}$ $\Rightarrow mg \sin \theta = \frac{k q^2 \sin^2 \theta}{h^2}$ $\Rightarrow h^2 = \frac{k q^2 \sin \theta}{mg}$ $\Rightarrow h = \sqrt{\frac{k q^2 \sin \theta}{mg}}$ $\Rightarrow h = \sqrt{\frac{9 \times 10^9 \times (2 \times 10^{-6})^2 \times \frac{1}{2}}{20 \times 10^{-3} \times 10}}$ $\Rightarrow h = \sqrt{9 \times 10^{-2}} = 0.3 \text{ m}$ $\text{Hence, option (2) is the correct answer.}$

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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
}