Import Question JSON

Current Question (ID: 9607)

Question:

$\text{A stone projected with a velocity } u \text{ at an angle } \theta \text{ with the horizontal reaches maximum height } H_1\text{. When it is projected with velocity } u \text{ at an angle } \left(\frac{\pi}{2} - \theta\right) \text{ with the horizontal, it reaches maximum height } H_2\text{. The relation between the horizontal range of the projectile } R \text{ and } H_1 \text{ and } H_2 \text{ is:}$

Options:

1. $R = 4\sqrt{H_1 H_2}$
2. $R = 4(H_1 - H_2)$
3. $R = 4(H_1 + H_2)$
4. $R = \frac{H_1^2}{H_2^2}$

Solution:

$\text{Hint: } H = \frac{u^2 \sin^2 \theta}{2g}$ $\text{Step: Find the relation between the horizontal range of the projectile } R \text{ and } H_1 \text{ & } H_2\text{.}$ $\text{The maximum height at an angle } \theta \text{ and } 90 - \theta \text{ is given as:}$ $H_1 = \frac{u^2 \sin^2 \theta}{2g} \text{ and } H_2 = \frac{u^2 \sin^2(90-\theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$ $H_1 H_2 = \frac{u^2 \sin^2 \theta}{2g} \times \frac{u^2 \cos^2 \theta}{2g} = \frac{(u^2 \sin 2\theta)^2}{16g^2} = \frac{R^2}{16}$ $R = \frac{u^2 \sin 2\theta}{g}$ $\text{Therefore, the relation between the horizontal range of the projectile } R \text{ and } H_1 \text{ & } H_2 \text{ is } R = 4\sqrt{H_1 H_2}\text{.}$ $\text{Hence, option (1) is the correct answer.}$

Import JSON File

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
}