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Current Question (ID: 16500)

Question:
$\text{When a proton at rest is accelerated by a potential difference } V, \text{ its speed is found to be } v. \text{ The speed of an } \alpha\text{-particle when accelerated by the same potential difference from rest will be:}$
Options:
  • 1. $v$
  • 2. $\frac{v}{\sqrt{2}}$
  • 3. $v\sqrt{2}$
  • 4. $2v$
Solution:
$qV = \frac{1}{2}mv^2$ $\text{assuming charge on Proton as } q \text{ and mass as } m, \text{ above is the equation based on conservation of energy (decrease in potential energy is equal to the increase in kinetic energy here)}$ $\text{therefore, } v_{\text{proton}} = \sqrt{\frac{2qV}{m}}$ $\text{As } \alpha \text{ particle has 2 protons and 2 neutrons, so charge = } 2q \text{ and mass = } 4m \text{ and the equation becomes}$ $2qV = \frac{1}{2} \times 4m \times v_{\alpha}^2$ $\text{So, } v_{\alpha} = \sqrt{(qV/m)}$ $\text{Therefore, } v_{\alpha} = \frac{v_{\text{proton}}}{\sqrt{2}}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}