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Current Question (ID: 10296)

Question:

$\text{A rod is falling down with constant velocity } V_0 \text{ as shown. It makes contact with hinge A and rotates around it. The angular velocity of the rod just after the moment when it comes in contact with hinge A is:}$

Options:

1. $\frac{2V_0}{3L}$
2. $\frac{3V_0}{2L}$
3. $\frac{V_0}{L}$
4. $\frac{2V_0}{5L}$

Solution:

$\text{Hint: As net torque is zero so angular Momentum is conserved}$ $\text{Step 1: Calculate the value of angular velocity using conservation of angular momentum.}$ $L_i = L_f$ $\Rightarrow \frac{L}{2} \times (Mv) = I\omega$ $\Rightarrow \frac{MvL}{2} = \frac{ML^2\omega}{3}$ $\Rightarrow \omega = \frac{3v}{2L}$ $\text{Therefore, the angular velocity is } \frac{3V_0}{2L}$

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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
}