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Current Question (ID: 10180)

Question:
$\text{A particle of mass } m \text{ moves in the } XY \text{ plane with a velocity of } v \text{ along the straight line } AB\text{. If the angular momentum of the particle about the origin } O \text{ is } L_A \text{ when it is at } A \text{ and } L_B \text{ when it is at } B\text{, then:}$
Options:
  • 1. $L_A > L_B$
  • 2. $L_A = L_B$
  • 3. $\text{The relationship between } L_A \text{ and } L_B \text{ depends upon the slope of the line } AB\text{.}$
  • 4. $L_A < L_B$
Solution:
$\text{Angular momentum = Linear momentum × perpendicular distance of the line of action of linear momentum about the origin}$ $L_A = P_A \times d$ $\text{And, } L_B = P_B \times d$ $\text{As linear momentum are equal;}$ $\text{i.e., } P_A = P_B = mV$ $\text{therefore } L_A = L_B\text{.}$

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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
}