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Current Question (ID: 19121)

Question:

$\text{Two projectiles are thrown with same initial velocity making an angle of } 45^\circ \text{ and } 30^\circ \text{ with the horizontal respectively. The ratio of their respective ranges will be}$

Options:

1. $1 : \sqrt{2}$
2. $\sqrt{2} : 1$
3. $2 : \sqrt{3}$
4. $\sqrt{3} : 2$

Solution:

$\text{The range of a projectile is given by } R = \frac{u^2 \sin 2\theta}{g}.$ $\text{For } \theta = 45^\circ, R_1 = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g}.$ $\text{For } \theta = 30^\circ, R_2 = \frac{u^2 \sin 60^\circ}{g} = \frac{u^2 \times \frac{\sqrt{3}}{2}}{g} = \frac{u^2 \sqrt{3}}{2g}.$ $\text{The ratio of the ranges is } \frac{R_1}{R_2} = \frac{\frac{u^2}{g}}{\frac{u^2 \sqrt{3}}{2g}} = \frac{2}{\sqrt{3}}.$

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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
}