Import Question JSON

Current Question (ID: 20419)

Question:

$\text{A proton and an alpha particle of the same velocity enter a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is:}$

Options:

1. $1:4$
2. $4:1$
3. $2:1$
4. $1:2$

Solution:

$\text{The radius of the circular path in a magnetic field is given by } r = \frac{mv}{qB}$. $\text{For a proton, } r_p = \frac{m_p v}{q_p B}$. $\text{For an alpha particle, } r_{\alpha} = \frac{m_{\alpha} v}{q_{\alpha} B}$. $\text{The mass of an alpha particle is approximately 4 times that of a proton, and its charge is 2 times that of a proton.}$ $\text{Therefore, } \frac{r_{\alpha}}{r_p} = \frac{4m_p}{2q_p} = 2:1.$

Import JSON File

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
}