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Current Question (ID: 19618)

Question:

$\text{A sphere of relative density } \sigma \text{ and diameter } D \text{ has concentric cavity of diameter } d. \text{ The ratio of } D/d, \text{ if it just floats on water in a tank is:}$

Options:

1. $\left( \frac{\sigma - 1}{\sigma} \right)^{1/3}$
2. $\left( \frac{\sigma}{\sigma - 1} \right)^{1/3}$
3. $\left( \frac{\sigma - 2}{\sigma + 2} \right)^{1/3}$
4. $\left( \frac{\sigma + 1}{\sigma - 1} \right)^{1/3}$

Solution:

$\text{For the sphere to just float, the weight of the sphere must equal the buoyant force.}$ $\text{Let the density of water be } \rho_w. \text{ Then, } \sigma \rho_w V = \rho_w V_{\text{displaced}}.$ $\text{The volume of the sphere is } V = \frac{\pi}{6} D^3 \text{ and the volume of the cavity is } V_c = \frac{\pi}{6} d^3.$ $\text{Thus, } \sigma \left( \frac{\pi}{6} D^3 - \frac{\pi}{6} d^3 \right) = \frac{\pi}{6} D^3.$ $\text{Solving for } D/d, \text{ we get } D/d = \left( \frac{\sigma}{\sigma - 1} \right)^{1/3}.$

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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
}