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Current Question (ID: 20354)

Question:

$\text{A charged particle carrying charge } 1 \, \mu \text{C is moving with velocity } (2\hat{i} + 3\hat{j} + 4\hat{k}) \, \text{m/s. If an external magnetic field of } (5\hat{i} + 3\hat{j} - 6\hat{k}) \times 10^{-3} \, \text{T, exists in the region where the particle is moving then the force on the particle is } \vec{F} \times 10^{-9} \, \text{N. The vector } \vec{F} \text{ is:}$

Options:

1. $-3.0\hat{i} + 3.2\hat{j} - 0.9\hat{k}$
2. $-300\hat{i} + 320\hat{j} - 90\hat{k}$
3. $-30\hat{i} + 32\hat{j} - 9\hat{k}$
4. $-0.30\hat{i} + 0.32\hat{j} - 0.09\hat{k}$

Solution:

$\text{Hint: } \vec{F} = q(\vec{v} \times \vec{B}).$ $\vec{F} = 9 \left( \vec{V} \times \vec{B} \right) \left( \text{Force on charge particle} \right)$ $\vec{V} \times \vec{B} = \begin{pmatrix} 2\hat{i} + 3\hat{j} + 4\hat{k} \end{pmatrix} \times \begin{pmatrix} 5\hat{i} + 3\hat{j} - 6\hat{k} \end{pmatrix} \times 10^{-3}$ $= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 3 & -6 \end{vmatrix} \times 10^{-3}$ $= \hat{i} \begin{vmatrix} 3 & 4 \\ 3 & -6 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 5 & -6 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 5 & 3 \end{vmatrix}$ $= \hat{i}[-18 - 12] - \hat{j}[-12 - 20] + \hat{k}[6 - 15]$ $= \hat{i}[-30] + \hat{j}[32] + \hat{k}[-9] \times 10^{-3}$ $\text{Force} = 10^{-6} \left[ -30\hat{i} + 32\hat{j} - 9\hat{k} \right] \times 10^{-3}$ $= 10^{-9} \left[ -30\hat{i} + 32\hat{j} - 9\hat{k} \right]$

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{
  "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
}