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Current Question (ID: 9754)

Question:

$\text{A rigid rod is placed against the wall as shown in the figure. When the velocity at its lower end is } 10 \text{ ms}^{-1} \text{ and its base makes an angle } \alpha = 60^\circ \text{ with horizontal, then the vertical velocity of its end B (in ms}^{-1}\text{) will be:}$

Options:

1. $10\sqrt{3}$
2. $\frac{10}{\sqrt{3}}$
3. $5\sqrt{3}$
4. $\frac{5}{\sqrt{3}}$

Solution:

\text{Given: The rod maintains constant length, so } x^2 + y^2 = l^2 = \text{constant} \text{Differentiating both sides with respect to time:} 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \text{From the given conditions:} \frac{dx}{dt} = -V_A = -10 \text{ m/s} \frac{dy}{dt} = V_B \text{ and } \frac{y}{x} = \tan\alpha = \tan 60° = \sqrt{3} \text{Substituting into the constraint equation:} 2x(-10) + 2y(V_B) = 0 -20x + 2yV_B = 0 V_B = \frac{20x}{2y} = \frac{10x}{y} \text{Since } \frac{y}{x} = \sqrt{3}\text{, we have } \frac{x}{y} = \frac{1}{\sqrt{3}} \text{Therefore:} V_B = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}} \text{ m/s}

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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
}