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Current Question (ID: 9801)

Question:
$\text{A block of mass 10 kg is in contact with the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder, which is vertical and rotating about its axis, will be:}$ $(g = 10 \text{ m/s}^2)$
Options:
  • 1. $10\pi \text{ rad/s}$
  • 2. $\sqrt{10\pi} \text{ rad/s}$
  • 3. $\frac{10}{2\pi} \text{ rad/s}$
  • 4. $10 \text{ rad/s}$
Solution:
\text{Hint: The friction force balances the weight of the block.} \text{Step 1: Find the normal reaction of the block.} \text{The net force on the body towards the center is } F_c = mr\omega^2 \text{The normal reaction of the block is given by: } N = mr\omega^2 \text{Step 2: Find the minimum angular velocity.} \text{The frictional force is given by: } f = \mu N \text{Now, the block is in rest in the vertical direction.} f = \mu N \text{Substitute the known values we get: } mg = \mu mr\omega^2 \omega = \sqrt{\frac{g}{\mu r}} \omega = \sqrt{\frac{10}{0.1 \times 1}} \omega = 10 \text{ rad/s} \text{Hence, option (4) is the correct answer.}

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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
}