Import Question JSON

Current Question (ID: 19313)

Question:

$\text{A bob of mass } m \text{ attached to an inextensible string of length } l \text{ is suspended from a vertical support.}$ $\text{The bob rotates in a horizontal circle with an angular speed } \omega \text{ rad/s about the vertical.}$ $\text{About the point of suspension:}$

Options:

1. $\text{angular momentum changes in magnitude but not in direction.}$
2. $\text{angular momentum changes in direction but not in magnitude.}$
3. $\text{angular momentum changes in both direction and magnitude.}$
4. $\text{angular momentum is conserved.}$

Solution:

$\text{Hint: } \vec{L} = m(\vec{r} \times \vec{v})$ $\text{Step 1: Draw the free-body diagram.}$ $\text{Step 2: Find the torque about the point of suspension.}$ $\text{There are two forces acting on the bob. One is tension in the string}$ $\text{and the other is the weight of the bob.}$ $\text{The torque due to tension force will be zero because } r_\perp = 0 \text{ but the}$ $\text{torque due to weight } (Mg) \text{ is non-zero because the bob is moving}$ $\text{in circular motion so } r_\perp \text{ will be non-zero. There is non-zero torque}$ $\text{acting about the point of suspension.}$ $\text{Step 3: Find the change in angular momentum.}$ $\text{The magnitude of the angular momentum is } \vec{L} = m(\vec{r} \times \vec{v}). \text{ The}$ $\text{magnitude will remain the same but the direction will change}$ $\text{because } \vec{r}_\perp \text{ of the bob changes its direction continuously in the}$ $\text{circular motion of the bob and } \vec{r}_\perp \text{ is perpendicular to the velocity}$ $\text{vector. So, the direction of angular momentum changes.}$ $\text{Hence, option (2) is the correct answer.}$

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
}