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Current Question (ID: 20457)

Question:
$\text{A proton and a deuteron } (q = +e, \ m = 2.0u) \text{ having same kinetic energies enter a region of uniform magnetic field } \vec{B}. \text{ The ratio of the radius } r_d \text{ of deuteron path to the radius of } r_p \text{ of the proton path is:}$
Options:
  • 1. $1:2$
  • 2. $\frac{\sqrt{2}}{1}$
  • 3. $1:1$
  • 4. $\frac{\sqrt{1}}{2}$
Solution:
$\text{Given that the kinetic energies are the same, we have:}$ $\frac{1}{2} m_p v_p^2 = \frac{1}{2} m_d v_d^2$ $\Rightarrow m_p v_p^2 = m_d v_d^2$ $\Rightarrow \frac{v_d}{v_p} = \sqrt{\frac{m_p}{m_d}}$ $\text{The radius of the path in a magnetic field is given by } r = \frac{mv}{qB}$ $\Rightarrow \frac{r_d}{r_p} = \frac{m_d v_d}{m_p v_p}$ $= \frac{m_d}{m_p} \times \frac{v_d}{v_p}$ $= \frac{2}{1} \times \frac{1}{\sqrt{2}}$ $= \sqrt{2}$

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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
}