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Current Question (ID: 9634)

Question:

$\text{An object of mass } m \text{ is projected from the ground with a momentum } p \text{ at such an angle that its maximum height is } \frac{1}{4}\text{th of its horizontal range. Its minimum kinetic energy in its path will be:}$

Options:

1. $\frac{p^2}{8m}$
2. $\frac{p^2}{4m}$
3. $\frac{3p^2}{4m}$
4. $\frac{p^2}{m}$

Solution:

$\text{Hint: } H_{\text{max}} = \frac{R}{4}$ $\text{Step 1: Find the angle of the projection.}$ $\text{Given, } H_{\text{max}} = \frac{R}{4}$ $\Rightarrow \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin 2\theta}{4g}$ $\Rightarrow \sin^2 \theta = \sin \theta \cdot \cos \theta$ $\Rightarrow \tan \theta = 1 \Rightarrow \theta = 45°$ $\text{Step 2: Find the minimum kinetic energy at the highest point.}$ $\text{Kinetic energy will be minimum at the highest point.}$ $\Rightarrow K_{\text{min}} = \frac{p_x^2}{2m} = \left(\frac{p^2}{2m} \cos^2 \theta\right) = \left(\frac{p^2}{2m} \cos^2(45)°\right)$ $\text{Hence, option (2) is the correct answer.}$

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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
}