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Current Question (ID: 8888)

Question:
$\text{A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to } v(x) = \beta x^{-2n} \text{ where } \beta \text{ and } n \text{ are constants and } x \text{ is the position of the particle. The acceleration of the particle as a function of } x \text{ is given by:}$
Options:
  • 1. $-2n\beta^2 x^{-2n-1}$
  • 2. $-2n\beta^2 x^{-4n-1}$
  • 3. $-2\beta^2 x^{-2n+1}$
  • 4. $-2n\beta^2 x^{-4n+1}$
Solution:
$\text{Hint: } a = \frac{dv}{dt}$ $\text{Step: Find the acceleration of the particle as a function of } x.$ $\text{Given: } v = \beta x^{-2n}$ $\text{The acceleration of the particle is given by;}$ $a = \frac{dv}{dt} = \frac{dx}{dt} \times \frac{dv}{dx}$ $\Rightarrow a = v \frac{dv}{dx} = \beta x^{-2n}(-2n\beta x^{-2n-1})$ $\Rightarrow a = -2n\beta^2 x^{-4n-1}$ $\text{Hence, option (2) is the correct answer.}$

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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
}