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Current Question (ID: 20085)

Question:

$\text{A positive charge particle of } 100 \text{ mg is thrown in opposite direction to a uniform electric field of strength } 1 \times 10^5 \text{ NC}^{-1}. \text{ If the charge on the particle is } 40 \ \mu\text{C and the initial velocity is } 200 \ \text{ms}^{-1}, \text{ how much distance it will travel before coming to the rest momentarily:}$

Options:

1. $1 \ \text{m}$
2. $5 \ \text{m}$
3. $10 \ \text{m}$
4. $0.5 \ \text{m}$

Solution:

$\text{Hint: } F = qE = ma$ $m = 100 \times 10^{-3} \ \text{kg}$ $u = 200 \ \text{m/s}$ $a = \frac{qE}{m}$ $a = \frac{40 \times 10^{-6} \times 10^5}{0.1}$ $a = -40 \ \text{m/s}^2$ $v^2 = u^2 + 2as$ $0 = (200)^2 + 2(-40)s$ $s = \frac{40000}{80} = 0.5 \ \text{m}$

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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
}