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Current Question (ID: 11255)

Question:

$\text{A particle of mass } m \text{ is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time?}$

Options:

1. $x(t) \text{ shows a curve that increases to a peak and then decreases back toward zero}$
2. $x(t) \text{ shows oscillatory motion with positive and negative values}$
3. $x(t) \text{ shows sinusoidal oscillation starting from zero}$
4. $x(t) \text{ shows cosine-like oscillation starting from maximum amplitude}$ (Correct)

Solution:

$\text{Hint: From the graph, } U(x) \propto x^2$ $\text{Step 1: Identify the type of motion.}$ $\text{By looking at this graph, we can say that it represents the potential energy vs displacement graph of an SHM.}$ $\text{We know that potential energy in SHM = } \frac{1}{2}Kx^2\text{, Force on the particle is given by: } F = -\frac{dV}{dx} = -2Kx\text{. The acceleration of the particle is given by,}$ $a = \frac{F}{m} = -\frac{2K}{m}x \Rightarrow a \propto -x \text{ ...(1)}$ $\text{The equation (1) represents that the motion of the particle is SHM.}$ $\text{Step 2: Find the graph of the motion.}$ $\text{As the particle is released from the rest from the extreme position i.e., at } t = 0, x(t) = A\text{. Therefore, } x(t) = A\cos(\omega t)$ $\text{Hence, option (4) is the correct answer.}$

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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
}