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Current Question (ID: 9900)

Question:

$\text{The power supplied to a particle of mass 2 kg varies with time as } P = \frac{3t^2}{2} \text{ Watt, where t is in seconds. If the velocity of a particle at t = 0 is v = 0, then the velocity of the particle at t = 2 s will be:}$

Options:

1. $1 \text{ m/s}$
2. $4 \text{ m/s}$
3. $2 \text{ m/s}$
4. $2\sqrt{2} \text{ m/s}$

Solution:

$\text{Hint: } P = mv\frac{dv}{dt}$ $\text{Step 1: Use the formula for power.}$ $P = Fv$ $= mav$ $= mv\frac{dv}{dt}$ $\text{Step 2: Find velocity using integration.}$ $P = \frac{3t^2}{2} = 2v\frac{dv}{dt}$ $\int_0^t \frac{3t^2}{4} dt = \int_0^v v dv$ $\frac{t^3}{4} = \frac{v^2}{2} \Rightarrow v = \frac{t^{3/2}}{\sqrt{2}} = 2 \text{ m/s}$

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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
}