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Current Question (ID: 19240)

Question:
$\text{A block of mass } 1.9 \text{ kg is at rest at the edge of a table, of height } 1 \text{ m.}$ $\text{A bullet of mass } 0.1 \text{ kg collides with the block and sticks to it.}$ $\text{If the velocity of the bullet is } 20 \text{ m/s in the horizontal direction just before the collision}$ $\text{then the kinetic energy just before the combined system strikes the floor, is:}$ $[\text{Take } g = 10 \text{ m/s}^2.]$ $\text{Assume there is no rotational motion and loss of energy after the collision is negligible.}$
Options:
  • 1. $21 \text{ J}$
  • 2. $23 \text{ J}$
  • 3. $20 \text{ J}$
  • 4. $19 \text{ J}$
Solution:
$\text{Hint: Apply the conservation of energy.}$ $p_i = p_f = 0.1 \times 20 = 2v$ $\therefore v = 1 \text{ m/s}$ $KE_f = mgh + \frac{1}{2}mv^2 = 213$

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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
}