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Current Question (ID: 8829)

Question:
$\text{The displacement } x \text{ of a particle varies with time } t \text{ as } x = ae^{-\alpha t} + be^{\beta t}, \text{ where } a, b, \alpha, \text{ and } \beta \text{ are positive constants. The velocity of the particle will:}$
Options:
  • 1. $\text{be independent of } \alpha \text{ and } \beta.$
  • 2. $\text{go on increasing with time.}$
  • 3. $\text{drop to zero when } \alpha = \beta.$
  • 4. $\text{go on decreasing with time.}$
Solution:
$\text{Hint: } v = \frac{dx}{dt}$ $\text{Step: Find the velocity of the particle.}$ $x = ae^{-\alpha t} + be^{\beta t}$ $v = \frac{dx}{dt} = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$ $\text{To check the nature of the velocity of the particle, we perform one more derivative of the velocity as a function of time.}$ $\text{The derivative of the velocity is given by,}$ $\frac{dv}{dt} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$ $\therefore \frac{dv}{dt} > 0 \text{ (Always)}$ $\text{Therefore, } v \text{ is the increasing function of } t.$ $\text{Hence, option (2) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}