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Current Question (ID: 9812)

Question:
$\text{A bucket full of water tied with the help of a 2 m long string performs a vertical circular motion. The minimum angular velocity of the bucket at the uppermost point so that water will not fall will be:}$
Options:
  • 1. $2\sqrt{5} \text{ rad/s}$
  • 2. $\sqrt{5} \text{ rad/s}$
  • 3. $5 \text{ rad/s}$
  • 4. $10 \text{ rad/s}$
Solution:
$\text{Hint: At the highest point, the normal reaction is zero.}$ $\text{Step 1: Draw FBD for water in the bucket at the highest point.}$ $\text{The forces acting on the water are: angular velocity } \omega\text{, normal force } N\text{, and weight } mg$ $\text{Step 2: Identify the condition for minimum angular velocity.}$ $N + mg = m\omega^2 l \text{ (As, } N = 0\text{)}$ $mg = m\omega_{\text{min}}^2 l$ $\omega_{\text{min}} = \sqrt{\frac{g}{l}}$ $= \sqrt{\frac{10}{2}} = \sqrt{5} \text{ rad/s}$

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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
}