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Current Question (ID: 10446)

Question:

Two non-mixing liquids of densities $\rho$ and $n\rho$ ($n > 1$) are put in a container. The height of each liquid is $h$. A solid cylinder floats with its axis vertical and length $pL(p < 1)$ in the denser liquid. The density of the cylinder is $d$. The density $d$ is equal to:

Options:

1. $[2 + (n + 1)p]\rho$
2. $[2 + (n - 1)p]\rho$
3. $[1 + (n - 1)p]\rho$
4. $[1 + (n + 1)p]\rho$

Solution:

\text{Weight of cylinder} = \text{upthrust}_1 + \text{upthrust}_2 A \cdot L \cdot d \cdot g = (1-p) \cdot L \cdot A \cdot \rho \cdot g + (p \cdot L \cdot A) \cdot n \cdot \rho \cdot g \text{Therefore: } d = (1-p) \cdot \rho + p \cdot n \cdot \rho = \rho - p \cdot \rho + p \cdot n \cdot \rho = \rho + p \cdot (n-1) \cdot \rho = \rho \cdot [1 + p \cdot (n-1)] = [1 + (n-1) \cdot p] \cdot \rho

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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
}