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Current Question (ID: 19301)

Question:

$\text{A body of mass } 8 \text{ kg and another of mass } 2 \text{ kg are moving with equal kinetic energy. The ratio of their respective momenta will be:}$

Options:

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

Solution:

$\text{Let the masses be } m_1 = 8 \text{ kg and } m_2 = 2 \text{ kg.}$ $\text{Since kinetic energies are equal:}$ $\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2$ $\Rightarrow m_1 v_1^2 = m_2 v_2^2$ $\Rightarrow \frac{v_1}{v_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{2}{8}} = \frac{1}{2}$ $\text{Momentum } p = mv$ $\Rightarrow \frac{p_1}{p_2} = \frac{m_1 v_1}{m_2 v_2} = \frac{8 \times \frac{1}{2}}{2 \times 1} = 2 : 1$

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
}