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Current Question (ID: 19318)

Question:

$\text{A particle of mass } m \text{ is moving along side of a square of side } 'a', \text{ with a uniform speed } v \text{ in the x-y plane as shown in the figure:}$ $\text{Which of the following statements is false for the angular momentum } \vec{L} \text{ about the origin?}$

Options:

1. $\vec{L} = -\frac{mvR}{\sqrt{2}} \hat{k} \text{ when the particle is moving from } A \text{ to } B.$
2. $\vec{L} = mv \left[ \frac{R}{\sqrt{2}} + a \right] \hat{k} \text{ when the particle is moving from } C \text{ to } D.$
3. $\vec{L} = mv \left[ \frac{R}{\sqrt{2}} + a \right] \hat{k} \text{ when the particle is moving from } B \text{ to } C.$
4. $\vec{L} = \frac{mvR}{\sqrt{2}} \hat{k} \text{ when the particle is moving from } D \text{ to } A.$

Solution:

$\text{Hint: } \vec{L} = m(\vec{r} \times \vec{v})$ $\text{Ans is [2, 4]}$ $\text{Sol. When particle move along AB}$ $\vec{L} = mvr_\perp \left(-\hat{k}\right)$ $\vec{L} = mg \frac{R}{\sqrt{2}} \left(-\hat{k}\right)$ $\text{when particle move along C to D}$ $\vec{L} = mvr_\perp \left(\hat{k}\right)$ $\vec{L} = mv \left[ \frac{R}{\sqrt{2}} + a \right] \left(\hat{k}\right)$ $\text{when particle move from B to C}$ $\vec{L} = mvr_\perp \left(\hat{k}\right) = mv \left(\frac{R}{\sqrt{2}} + a\right) \left(\hat{k}\right)$ $\text{when particle move from D to A}$ $\vec{L} = mvr_\perp \left(-\hat{k}\right) = mv \frac{R}{\sqrt{2}} \left(-\hat{k}\right)$ $\text{so (2) and (4) are false}$

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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
}