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Current Question (ID: 20433)

Question:

$\text{A velocity selector consists of electric field } \vec{E} = E\hat{k} \text{ and magnetic field } \vec{B} = B\hat{j} \text{ with } B = 12 \text{ mT. The value } E \text{ required for an electron of energy } 728 \text{ eV moving along the positive x-axis to pass undeflected is:}$ $\text{(Given, mass of electron } = 9.1 \times 10^{-31} \text{ kg)}$

Options:

1. $192 \text{ k Vm}^{-1}$
2. $192 \text{ m Vm}^{-1}$
3. $9600 \text{ k Vm}^{-1}$
4. $16 \text{ k Vm}^{-1}$

Solution:

$\text{For an electron to pass undeflected, } v = \frac{E}{B}$ $\text{Given, energy } = 728 \text{ eV} = 728 \times 1.6 \times 10^{-19} \text{ J}$ $\frac{1}{2}mv^2 = 728 \times 1.6 \times 10^{-19}$ $v^2 = \frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $v = \sqrt{\frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}$ $E = vB = \left(\sqrt{\frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}\right) \times 12 \times 10^{-3}$ $E = 192 \text{ k Vm}^{-1}$

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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
}