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Current Question (ID: 20409)

Question:

$\text{A beam of protons with speed } 4 \times 10^5 \text{ ms}^{-1} \text{ enters a uniform magnetic field of } 0.3 \text{ T at an angle of } 60^\circ \text{ to the magnetic field.}$ $\text{The pitch of the resulting helical path of protons is close to:}$ $\text{(Mass of the proton } = 1.67 \times 10^{-27} \text{ kg, charge of the proton } = 1.69 \times 10^{-19} \text{ C)}$

Options:

1. $12 \text{ cm}$
2. $4 \text{ cm}$
3. $5 \text{ cm}$
4. $2 \text{ cm}$

Solution:

$\text{Hint: Pitch } = T \times v \cos \theta$ $\text{Pitch } = \frac{2\pi m}{qB} v \cos \theta$ $\text{Pitch } = \frac{2(3.14)(1.67 \times 10^{-27}) \times 4 \times 10^5 \times \cos\ 60}{(1.69 \times 10^{-19})(0.3)}$ $\text{Pitch } = 0.04 \text{ m} = 4 \text{ cm}$

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
}