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Current Question (ID: 9803)

Question:

$\text{A tube of length } L \text{ is filled completely with an incompressible liquid of mass } M \text{ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity } \omega\text{. The force exerted by the liquid at the other end is:}$

Options:

1. $\frac{ML\omega^2}{2}$
2. $\frac{ML^2\omega}{2}$
3. $ML\omega^2$
4. $\frac{ML^2\omega^2}{2}$

Solution:

\text{Hint: There will be a pseudo force acting on the liquid.} \text{Step 1: Find the pseudo force on a small element of the liquid.} \text{Let the length of a small element of the tube be } dx. \text{Mass of this element, } dm = \frac{M}{L}dx \text{where M is the mass of filled liquid and L is the length of the tube.} \text{Force on this element, } dF = dm \times x\omega^2 \text{Step 2: Find the net pseudo force acting on the liquid.} \int_0^F dF = \frac{M}{L}\omega^2 \int_0^L x \, dx F = \frac{M\omega^2 L}{2}

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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
}