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Current Question (ID: 8892)

Question:
$\text{A point moves in a straight line under the retardation } av^2\text{. If the initial velocity is } u\text{, the distance covered in } t \text{ seconds is:}$
Options:
  • 1. $(aut)$
  • 2. $\frac{1}{a}\ln(aut)$
  • 3. $\frac{1}{a}\ln(1 + aut)$
  • 4. $a\ln(aut)$
Solution:
$\text{Hint: Retardation} = -\frac{dv}{dt}$ $\text{Step: Find the distance covered by the point in } t \text{ seconds.}$ $\text{Retardation} = -\frac{dv}{dt}$ $\Rightarrow -\frac{dv}{dt} = av^2$ $\text{Separating variables:}$ $\Rightarrow \int_u^v \frac{dv}{v^2} = -\int_0^t a\,dt$ $\text{Integrating both sides:}$ $\left[\frac{-1}{v}\right]_u^v = -at$ $\frac{1}{v} - \frac{1}{u} = at$ $\text{Solving for velocity:}$ $v = \frac{u}{1 + uat}$ $\text{Since } v = \frac{dx}{dt}\text{:}$ $\int ds = \int \left(\frac{u}{1 + uat}\right) dt$ $s = \frac{1}{a}\log_e(1 + uat)$ $\text{Therefore, the distance covered by the point in } t \text{ seconds is}$ $s = \frac{1}{a}\log_e(1 + uat)$

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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
}