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Current Question (ID: 20115)

Question:

$\text{In free space, a particle } A \text{ of charge } 1 \, \mu\text{C} \text{ is held fixed at a point } P. \text{ Another particle } B \text{ of the same charge and mass } 4 \, \mu\text{g} \text{ is kept at a distance of } 1 \, \text{mm from } P. \text{ If } B \text{ is released, then its velocity at a distance of } 9 \, \text{mm from } P \text{ is:}$ $\left[ \text{Take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \right]$

Options:

1. $1.0 \, \text{m/s}$
2. $3.0 \times 10^4 \, \text{m/s}$
3. $2.0 \times 10^3 \, \text{m/s}$
4. $6.2 \times 10^4 \, \text{m/s}$

Solution:

$\text{Hint: Apply conservation of energy.}$ $\text{By conservation of energy}$ $\frac{1}{2}mv^2 = kq_1q_2 \left[ \frac{1}{d_1} - \frac{1}{d_2} \right]$ $= k \times 10^{-12} \left[ 1 - \frac{1}{9} \right] \times 10^3$ $= 9 \times 10^9 \times 10^{-9} \left[ \frac{8}{9} \right] = 8$ $\frac{1}{2}mv^2 = 8$ $v = \sqrt{\frac{16}{m}} = \sqrt{\frac{16}{4 \times 10^{-9}}}$ $= 2 \times 10^4 \times \sqrt{10}$ $\approx 6 \times 10^4 \, \text{m/s}$ $\text{Of mass is considered as } m = 4 \times 10^{-6} \text{ kg then } v = 2 \times 10^3 \, \text{m/s}$

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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
}