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Current Question (ID: 19139)

Question:

$\text{A small ball of mass } m \text{ is thrown upward with velocity } u \text{ from the ground.}$ $\text{The ball experiences a resistive force } mkv^2 \text{ where } v \text{ is its speed.}$ $\text{The maximum height attained by the ball is:}$

Options:

1. $\frac{1}{2k} \ln \left( 1 + \frac{ku^2}{g} \right)$
2. $\frac{1}{2k} \tan^{-1} \frac{ku^2}{g}$
3. $\frac{1}{k} \ln \left( 1 + \frac{ku^2}{2g} \right)$
4. $\frac{1}{k} \tan^{-1} \frac{ku^2}{2g}$

Solution:

$\text{Hint: } a = v \frac{dv}{dx}$ $\vec{F} = mkv^2 - mg$ $\vec{a} = \frac{\vec{F}}{m} = -\left[ kv^2 + g \right]$ $\Rightarrow b. \frac{dv}{dh} = -\left[ kv^2 + g \right]$ $\Rightarrow \int_u^0 \frac{v \cdot dv}{kv^2 + g} = -\int_0^H dh$ $\Rightarrow \frac{1}{2K} \ln \left[ kv^2 + g \right]_u^0 = -H$ $\Rightarrow \frac{1}{2K} \ln \left[ \frac{ku^2}{g} + 1 \right] = H$

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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
}