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Current Question (ID: 9659)

Question:

$\text{The magnitude of vector } \vec{A} \text{ is constant but it is changing its direction continuously. The angle between } \vec{A} \text{ and } \frac{d\vec{A}}{dt} \text{ is:}$

Options:

1. $180°$
2. $120°$
3. $90°$
4. $0°$

Solution:

\text{Hint: Draw a vector diagram of A and dA/dt.} \text{Step 1: Draw the vector diagram.} \text{A' = A + dA} \text{|A'| = |A|} \text{Step 2: Find the magnitude of A.} \text{|A'| = sqrt(A² + (dA)² + 2A × dA × cos θ)} \text{Since |A'| = |A| = A:} \text{A² = A² + (dA)² + 2A × dA × cos θ} \text{0 = (dA)² + 2A × dA × cos θ} \text{For small dA: 2A × dA × cos θ = 0} \text{Therefore cos θ = 0, so θ = 90°} \text{The angle between A and dA/dt is 90°.} \text{Hence, option (3) 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
}