Import Question JSON

Current Question (ID: 9802)

Question:

$\text{A point mass } m \text{ is suspended from a light thread of length } l\text{, fixed at } O\text{, and is whirled in a horizontal circle at a constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are:}$

Options:

1. $\text{Tension } T \text{ (upward and leftward), centrifugal force } F \text{ (leftward), and weight } W \text{ (downward)}$
2. $\text{Tension } T \text{ (upward and leftward) and weight } W \text{ (downward)}$
3. $\text{Tension } T \text{ (upward and leftward), centrifugal force } F \text{ (rightward), and weight } W \text{ (downward)}$
4. $\text{Tension } T \text{ (upward and leftward), centrifugal forces } F \text{ (both leftward and rightward), and weight } W \text{ (downward)}$

Solution:

$\text{From the perspective of an observer stationary with respect to the mass (non-inertial reference frame), we must include pseudo forces.}$ $\text{The forces acting on the mass are:}$ $T = \text{tension in the thread (acting along the thread toward the fixed point)}$ $W = \text{weight of the mass (acting vertically downward)}$ $F = \text{centrifugal force (acting horizontally outward, away from the center of circular motion)}$ $\text{Since the mass is whirling in a horizontal circle, the centrifugal force acts outward from the center of the circular path, which corresponds to the rightward direction in the diagram.}$ $\text{Therefore, the correct force diagram shows tension } T \text{ acting upward and leftward along the thread, weight } W \text{ acting downward, and centrifugal force } F \text{ acting horizontally to the right (outward).}$

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
}