Import Question JSON

Current Question (ID: 9643)

Question:

$\text{Given below are two statements:}$ $\textbf{Assertion (A):} \text{ The maximum height of a projectile is always 25% of the maximum range.}$ $\textbf{Reason (R):} \text{ For maximum height, the projectile should be projected at } 90°.$

Options:

1. $\text{Both (A) and (R) are True and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are True but (R) is not the correct explanation of (A).}$
3. $\text{(A) is True and (R) is False.}$
4. $\text{(A) is False and (R) is True.}$

Solution:

\text{Hint: } H_{\text{max}} = \frac{v^2 \sin^2 \theta}{2g} \text{ and } R = \frac{v^2 \sin 2\theta}{g} \text{Step 1: Find the ratio of maximum height and range.} \frac{H_{\text{max}}}{R} = \frac{\frac{v^2 \sin^2 \theta}{2g}}{\frac{v^2 \sin 2\theta}{g}} = \frac{v^2 \sin^2 \theta}{2g} \times \frac{g}{v^2 \sin 2\theta} = \frac{\sin^2 \theta}{2 \sin 2\theta} = \frac{\sin^2 \theta}{4 \sin \theta \cos \theta} = \frac{\tan \theta}{4} \text{Hence it is only true if } \theta = 45°. \text{ So Assertion is incorrect.} \text{Step 2: Find the angle for maximum height.} \text{As } H_{\text{max}} = \frac{v^2 \sin^2 \theta}{2g} \text{So, height will be maximum at } \theta = 90°. \text{ So Reason is correct.}

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
}