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Current Question (ID: 10297)

Question:

$\text{The figure below shows two identical particles 1 and 2, each of mass } m\text{, moving in opposite directions with the same speed } v \text{ along parallel lines. At a particular instant, } r_1 \text{ and } r_2 \text{ are their respective position vectors drawn from point } A\text{, which is in the plane of the parallel lines.}$ $\text{Consider the following statements.}$

Options:

1. $\text{(a), (c) only}$
2. $\text{(a), (d) only}$
3. $\text{(b), (d) only}$
4. $\text{(b), (c) only}$

Solution:

$\text{Hint: In angular momentum, only perpendicular distance is considered.}$ $\text{Step 1: Find the angular momentum of the particle 1.}$ $\text{The angular momentum } L \text{ of a particle with to origin is to } L = r \times p \text{ where } r \text{ is the position vector of the particle and } p \text{ is the linear momentum. The direction of } L \text{ is perpendicular to } dr \text{ and } p \text{ by the right-hand rule.}$ $\text{For particle 1, } L_1 = r_1 \times mv \text{ is out of the plane of the perpendicular to } r_1 \text{ and } v\text{).}$ $\text{Step 2: Find the angular momentum of the system.}$ $\text{Similarly } L_2 = r_2 \times m(-v) \text{ is into the plane of perpendicular to } r_2 \text{ and } p\text{. Hence, total angular momentum}$ $L = L_1 + L_2 = r_1 \times mv + (-r_2 \times mv)$ $|L| = (mvd)_1 - (mvd)_2$ $(d_2 > d_1)\text{total angular momentum will be inward}$ $L = mv(d_2 - d_1) \otimes$ $\text{Hence, option (2) is the correct answer.}$

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
}