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Current Question (ID: 9832)

Question:
$\text{The position of a particle } (x) \text{ varies with time } (t) \text{ as } x = (t - 2)^2, \text{ where } x \text{ is in meters and } t \text{ is in seconds. Calculate the work done during } t = 0 \text{ to } t = 4 \text{ s if the mass of the particle is 100 g.}$
Options:
  • 1. $0.4 \text{ J}$
  • 2. $0.2 \text{ J}$
  • 3. $0.8 \text{ J}$
  • 4. $\text{zero}$
Solution:
\text{Given: } x = (t - 2)^2, \text{ mass } m = 100 \text{ g} = 0.1 \text{ kg} \text{To find velocity, differentiate position with respect to time:} v = \frac{dx}{dt} = 2(t - 2) \text{At } t = 0: \quad v_0 = 2(0 - 2) = -4 \text{ m/s} \text{At } t = 4: \quad v_4 = 2(4 - 2) = 4 \text{ m/s} \text{Initial kinetic energy:} KE_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(0.1)(16) = 0.8 \text{ J} \text{Final kinetic energy:} KE_4 = \frac{1}{2}mv_4^2 = \frac{1}{2}(0.1)(16) = 0.8 \text{ J} \text{Work done: } \Delta KE = KE_4 - KE_0 = 0.8 - 0.8 = 0 \text{ J} \text{The work done is zero because initial and final kinetic energies are equal.}

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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
}