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Current Question (ID: 10177)

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.} \text{At a particular instant, } r_1 \text{ and } r_2 \text{ are their respective position vectors drawn from point A, which is in the plane of the parallel lines.} \text{Consider the following statements:} \text{(a) angular momentum } l_1 \text{ of particle 1 about A is } l_1 = mv(d_1) \hat{k} \text{(b) angular momentum } l_2 \text{ of particle 2 about A is } l_2 = mv(r_2) \hat{k} \text{(c) total angular momentum of the system about A is } l = mv(r_1 + r_2) \hat{k} \text{(d) total angular momentum of the system about A is } l = mv(d_2 - d_1) \hat{k} \text{Choose the correct option from the given ones:}

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 particle 1.} \text{The angular momentum } L \text{ of a particle with respect to origin is } 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 both } r \text{ and } p \text{ by the right-hand rule.} \text{For particle 1, } L_1 = r_1 \times mv \text{ (out of the plane, 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{ (into the plane, perpendicular to } r_2 \text{ and } v\text{). Hence, total angular momentum:} L = L_1 + L_2 = r_1 \times mv + r_2 \times m(-v) |L| = (mvd_1) - (mvd_2) \text{Since } d_2 > d_1\text{, the total angular momentum will be inward:} L = mv(d_2 - d_1) \hat{k}_{\text{inward}} \text{Hence, option (d) is the correct answer.}

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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
}