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Current Question (ID: 20135)

Question:

$\text{An electron with kinetic energy } K_1 \text{ enters the region between the parallel plates of a capacitor at an angle } \alpha \text{ with the plates.}$ $\text{It leaves the plates at an angle } \beta \text{ with kinetic energy } K_2. \text{ The ratio of kinetic energies } K_1 : K_2 \text{ is:}$

Options:

1. $\frac{\sin^2 \beta}{\cos^2 \alpha}$
2. $\frac{\cos^2 \beta}{\cos^2 \alpha}$
3. $\frac{\cos \beta}{\cos \alpha}$
4. $\frac{\cos \beta}{\sin \alpha}$

Solution:

$\text{Hint: The Electron does not experience any force in the } x\text{-direction, hence momentum in } x\text{-direction is conserved.}$ $\text{Step: Find the ratio of kinetic energy.}$ $\text{We know that the Electric field is always perpendicular to the plates of the capacitor.}$ $\text{Let the speed of the electron be } v_1 \text{ when it enters the gap between the plates and } v_2 \text{ when it leaves the plates.}$ $\text{As the electron only experiences force in the direction perpendicular to the plates, by the conservation of momentum in the direction parallel to the plates, we have velocity along the plate will not change.}$ $v_1 \cos \alpha = v_2 \cos \beta$ $\frac{K_1}{K_2} = \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{v_1^2}{v_2^2} = \frac{\cos^2 \beta}{\cos^2 \alpha}$ $\text{Hence, option (2) is the correct answer.}$

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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
}