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Current Question (ID: 19248)

Question:

$\text{Two solids } A \text{ and } B \text{ of mass } 1 \text{ kg and } 2 \text{ kg respectively are moving with equal linear momentum. The ratio of their kinetic energies } (KE)_A : (KE)_B \text{ will be:}$

Options:

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

Solution:

$\text{Hint: } K = \frac{P^2}{2m}$ $\text{Step: Find the ratio of their kinetic energies.}$ $\text{Kinetic energy is given by:}$ $K = \frac{P^2}{2m}$ $\text{Since, linear momentum is same for both the solids.}$ $\Rightarrow K \propto \frac{1}{m}$ $\text{Therefore;}$ $\Rightarrow \frac{K_A}{K_B} = \frac{m_B}{m_A} = \frac{2}{1}$ $\text{Hence, option (2) is the correct answer.}$

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
}