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Current Question (ID: 19256)

Question:
$\text{A particle of mass } 500 \text{ g is moving in a straight line with velocity } v = bx^{5/2}. \text{ The work done by the net force during its displacement from } x = 0 \text{ to } x = 4 \text{ m is:}$ $\text{(Take } b = 0.25 \text{ m}^{-3/2} \text{ s}^{-1})$
Options:
  • 1. $2 \text{ J}$
  • 2. $4 \text{ J}$
  • 3. $8 \text{ J}$
  • 4. $16 \text{ J}$
Solution:
$\text{Hint: Use the work-energy theorem.}$ $\text{Step: Find the work done by the net force.}$ $\text{According to the work-energy theorem;} \text{Work done by all forces } = \Delta KE$ $\Rightarrow W = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2$ $\Rightarrow W = \frac{1}{2} \times 0.5 \times (0.25)^2 \times (4)^5 - 0$ $\Rightarrow W = 16 \text{ J}$

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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
}