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Current Question (ID: 20424)

Question:
$\text{A charged particle moves along a circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charged particle increases to 4 times its initial value. What will be the ratio of the new radius to the original radius of the circular path of the charged particle?}$
Options:
  • 1. $1 : 1$
  • 2. $1 : 2$
  • 3. $2 : 1$
  • 4. $1 : 4$
Solution:
$\text{Hint: } r \propto \sqrt{K.E.}$ $\text{Step: Find the ratio of the new radius to the original radius.}$ $R = \frac{mv}{qB} = \frac{\sqrt{2m(K.E)}}{qB}$ $R \propto \sqrt{KE}$ $\frac{R_1}{R_2} = \sqrt{\frac{KE_1}{KE_2}} \Rightarrow \sqrt{\frac{1}{4}} = \frac{1}{2}$ $\Rightarrow \frac{R_2}{R_1} = 2$

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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
}