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Current Question (ID: 9841)

Question:

$\text{A force } \vec{F} = -k(y\hat{i} + x\hat{j}) \text{ (where k is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is:}$

Options:

1. $-2ka^2$
2. $2ka^2$
3. $-ka^2$
4. $ka^2$

Solution:

$\text{While moving from (0,0) to (a,0):}$ $\text{Along the positive x-axis, y = 0}$ $\therefore \vec{F} = -kx\hat{j}$ $\text{i.e. the force is in negative y-direction while the displacement is in positive x-direction.}$ $\therefore W_1 = 0 \text{ because the force is perpendicular to the displacement.}$ $\text{Then particle moves from (a, 0) to (a, a) along a line parallel to y-axis (x = +a). During this motion: } \vec{F} = -k(y\hat{i} + a\hat{j})$ $\text{The first component of force, } -ky\hat{i} \text{ will not contribute any work because this component is along negative x-direction } (-\hat{i}) \text{ while the displacement is in the positive y-direction. The second component of force i.e. } -ka\hat{j} \text{ will perform negative work.}$ $\therefore W_2 = (-ka\hat{j}) \cdot (a\hat{j}) = (-ka)(a) = -ka^2$ $\text{So the net work done on the particle, } W = W_1 + W_2$ $= 0 + (-ka^2) = -ka^2$

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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
}