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Current Question (ID: 19572)

Question:

$\text{A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is } 5 \text{ cm and the angular speed of rotation is } \omega \text{ rads}^{-1}. \text{ The difference in the height, } h \text{ (in cm) of liquid at the centre of vessel and at the side will be:}$

Options:

1. $\frac{25\omega^2}{2g}$
2. $\frac{2\omega^2}{5g}$
3. $\frac{5\omega^2}{2g}$
4. $\frac{2\omega^2}{25g}$

Solution:

$\text{Hint: } P = P_0 + h \rho g$ $\text{Step: Find the value of } h.$ $\text{The change in pressure due to height is given by:}$ $\Delta P = \rho g h$ $\text{The change in pressure due to the rotation of the cylindrical vessel is given by:}$ $\Delta P = \frac{\rho \omega^2 r^2}{2}$ $\text{The pressure at the point } C \text{ along } BC \text{ is given by:}$ $P_C = P_0 + \rho g h \quad \cdots (1)$ $\text{The pressure at the point } C \text{ along } AC \text{ is given by:}$ $P_C = P_0 + \frac{\rho \omega^2 r^2}{2} \quad \cdots (2)$ $\text{From the equations } (1) \text{ and } (2) \text{ we get:}$ $P_0 + \rho g h = P_0 + \frac{\rho \omega^2 r^2}{2}$ $\Rightarrow h = \frac{\omega^2 r^2}{2g}$ $\Rightarrow h = \frac{25\omega^2}{2g}$ $\text{Hence, option } (1) \text{ is the correct answer.}$

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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
}