Import Question JSON

Current Question (ID: 9774)

Question:

$\text{A block is placed on a rough horizontal plane. A time dependent horizontal force, } F = kt, \text{ acts on the block. The acceleration time graph of the block is:}$

Options:

1. $\text{Graph 1: Exponential increase in acceleration with time}$
2. $\text{Graph 2: Zero acceleration initially, then linear increase}$
3. $\text{Graph 3: Constant acceleration initially, then linear increase}$
4. $\text{Graph 4: Constant acceleration throughout}$

Solution:

\text{Analysis of the motion:} \text{When } F < f_L \text{, body does not move i.e. at rest i.e. acceleration is zero} \text{When } F > f_L: \quad kt - \mu mg = ma \text{Initially, when the applied force } F = kt \text{ is less than the limiting friction } f_L\text{, the block remains at rest and acceleration is zero.} \text{Once the applied force exceeds the limiting friction, the block starts moving with kinetic friction acting on it.} \text{The net force becomes } F - f_k = kt - \mu mg = ma \text{Therefore, acceleration } a = \frac{kt - \mu mg}{m} = \frac{k}{m}t - \mu g \text{This shows that after the block starts moving, acceleration increases linearly with time.} \text{Graph 2 correctly shows zero acceleration initially (when block is at rest) followed by linear increase in acceleration.}

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
}