Import Question JSON

Current Question (ID: 9664)

Question:

$\text{Which one of the following is not true?}$

Options:

1. $\text{The net acceleration of a particle in a circular motion is always along the radius of the circle towards the centre.}$
2. $\text{The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.}$
3. $\text{The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.}$
4. $\text{None of the above.}$

Solution:

$\text{Hint: In a circular motion, the particle can have radial as well as tangential acceleration.}$ $\text{Explanation: The net acceleration of a particle in a circular motion is always in the direction of the resultant of tangential and centripetal acceleration.}$ $\text{Centripetal acceleration (always directed towards the centre along the radius). Tangential acceleration (if the particle's speed is changing, this is along the tangent to the path). In uniform circular motion, the net acceleration is only centripetal. However, in non-uniform circular motion, the particle can have a tangential acceleration component.}$ $\text{The direction is along the radius only if tangential acceleration is zero.}$ $\text{Therefore, the net acceleration is not always along the radius. Hence, option (1) is the correct answer.}$

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
}