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Current Question (ID: 19311)

Question:
$\text{A stationary particle breaks into two parts of masses } m_\text{A} \text{ and } m_\text{B}$ $\text{which move with velocities } v_\text{A} \text{ and } v_\text{B} \text{ respectively. The ratio of their}$ $\text{kinetic energies } (K_\text{B} : K_\text{A}) \text{ is:}$
Options:
  • 1. $m_\text{B} : m_\text{A}$
  • 2. $1 : 1$
  • 3. $m_\text{B}v_\text{B} : m_\text{A}v_\text{A}$
  • 4. $v_\text{B} : v_\text{A}$
Solution:
$\text{Since the particle is stationary initially, the total momentum is zero.}$ $\text{Thus, } m_\text{A}v_\text{A} = m_\text{B}v_\text{B}.$ $\text{The kinetic energy ratio is given by } \frac{K_\text{B}}{K_\text{A}} = \frac{\frac{1}{2}m_\text{B}v_\text{B}^2}{\frac{1}{2}m_\text{A}v_\text{A}^2} = \frac{m_\text{B}v_\text{B}^2}{m_\text{A}v_\text{A}^2}.$ $\text{Using the momentum relation, } v_\text{B} = \frac{m_\text{A}}{m_\text{B}}v_\text{A}, \text{ we get } \frac{K_\text{B}}{K_\text{A}} = \left(\frac{m_\text{A}}{m_\text{B}}\right)^2.$ $\text{Thus, the ratio is } v_\text{B} : v_\text{A}.$

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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
}