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Current Question (ID: 19576)

Question:

$\text{Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density } d. \text{ The area of the base of both vessels is } S \text{ but the height of liquid in one vessel is } x_1 \text{ and in the other, } x_2. \text{ When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:}$

Options:

1. $\frac{3}{4} gdS(x_2 - x_1)^2$
2. $\frac{1}{4} gdS(x_2 - x_1)^2$
3. $gdS(x_2^2 - x_1^2)$
4. $gdS(x_2 - x_1)^2$

Solution:

$\text{Hint: Apply conservation of volume.}$ $U_i = (\rho S x_1) g \cdot \frac{x_1}{2} + (\rho S x_2) g \cdot \frac{x_2}{2}$ $U_f = (\rho S x_f) g \cdot \frac{x_f}{2} \times 2$ $\text{By volume conservation}$ $S x_1 + S x_2 = S (2 x_f)$ $x_f = \frac{x_1 + x_2}{2}$ $\Delta U = \rho S g \left[ \left( \frac{x_1^2}{2} + \frac{x_2^2}{2} \right) - x_f^2 \right]$ $= \rho S g \left[ \frac{x_1^2}{2} + \frac{x_2^2}{2} - \left( \frac{x_1 + x_2}{2} \right)^2 \right]$ $= \rho S g \left[ \frac{x_1^2}{2} + \frac{x_2^2}{2} - x_1 x_2 \right]$ $= \frac{\rho S g}{4} (x_1 - x_2)^2$

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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
}