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Current Question (ID: 20430)

Question:
$\text{A charged particle is moving in a uniform magnetic field } (2\hat{i} + 3\hat{j}) \text{ T. If it has an acceleration of } (\alpha \hat{i} - 4\hat{j}) \text{ m/s}^2, \text{ then the value of } \alpha \text{ will be:}$ $1.\ 3$ $2.\ 6$ $3.\ 12$ $4.\ 2$
Options:
  • 1. $3$
  • 2. $6$
  • 3. $12$
  • 4. $2$
Solution:
$\text{Hint: } \vec{F} = q(\vec{v} \times \vec{B})$ $\text{Step: Find the value of } \alpha.$ $\text{The magnetic force is always perpendicular to the velocity of the particle and is given by:}$ $\Rightarrow \vec{F}_B \cdot \vec{v} = 0 \text{ or } \vec{a}_B \cdot \vec{v} = 0$ $\Rightarrow (\alpha \hat{i} - 4\hat{j}) \cdot (2\hat{i} + 3\hat{j}) = 0$ $\Rightarrow 2\alpha - 12 = 0$ $\Rightarrow \alpha = 6$ $\text{Hence, option (2) is the correct answer.}$

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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
}