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Current Question (ID: 19292)

Question:

$\text{A particle experiences a variable force } \vec{F} = (4x \hat{i} + 3y^2 \hat{j}) \text{ in a horizontal x-y plane. Assume distance in meters and force is in newton. If the particle moves from point } (1, 2) \text{ to point } (2, 3) \text{ in the x-y plane, the kinetic energy changes by:}$

Options:

1. $50.0 \text{ J}$
2. $12.5 \text{ J}$
3. $25.0 \text{ J}$
4. $0$

Solution:

$\text{The work done by the force is given by the line integral of } \vec{F} \text{ along the path.}$ $\text{Work done } W = \int_{(1,2)}^{(2,3)} (4x \hat{i} + 3y^2 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$ $= \int_{1}^{2} 4x \, dx + \int_{2}^{3} 3y^2 \, dy$ $= \left[ 2x^2 \right]_{1}^{2} + \left[ y^3 \right]_{2}^{3}$ $= (2 \times 2^2 - 2 \times 1^2) + (3^3 - 2^3)$ $= (8 - 2) + (27 - 8)$ $= 6 + 19 = 25 \text{ J}$ $\text{Thus, the kinetic energy changes by } 25.0 \text{ J.}$

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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
}