Import Question JSON

Current Question (ID: 9752)

Question:

$\text{The figure shows a rod of length 5 m. Its ends, } A \text{ and } B\text{, are restrained to moving in horizontal and vertical guides. When the end } A \text{ is 3 m above } O\text{, it moves at 4 m/s. The velocity of end } B \text{ at that instant is:}$

Options:

1. $2 \text{ m/s}$
2. $3 \text{ m/s}$
3. $4 \text{ m/s}$
4. $0.20 \text{ m/s}$

Solution:

\text{Given information:} \text{- Rod length = 5 m} \text{- End A is 3 m above O} \text{- Velocity of end A: } v_A = 4 \text{ m/s (downward)} \text{From the geometry of triangle AOB:} \text{Using Pythagorean theorem: } OA^2 + OB^2 = AB^2 3^2 + OB^2 = 5^2 9 + OB^2 = 25 OB = 4 \text{ m} \text{Therefore:} \sin\theta = \frac{OA}{AB} = \frac{3}{5} \cos\theta = \frac{OB}{AB} = \frac{4}{5} \text{For the rod constraint (constant length):} \text{The component of velocity along the rod must be equal for both ends.} \text{Along the rod: } v_A \sin\theta = v_B \cos\theta \text{Substituting the values:} 4 \times \frac{3}{5} = v_B \times \frac{4}{5} \frac{12}{5} = v_B \times \frac{4}{5} v_B = \frac{12}{5} \times \frac{5}{4} = 3 \text{ m/s} \text{Alternatively: } v_B = v_A \tan\theta = 4 \times \frac{3}{4} = 3 \text{ m/s}

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
}