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Current Question (ID: 10451)

Question:
$\text{A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3. For this sphere to float with its entire volume submerged underwater, the ratio of mass of concrete to mass of sawdust will be:}$
Options:
  • 1. $8$
  • 2. $4$
  • 3. $3$
  • 4. $\text{Zero}$
Solution:
\text{Given: Specific gravities } \rho_1 \text{ (concrete) and } \rho_2 \text{ (sawdust)} \text{By flotation principle:} \frac{4}{3}\pi(R^3 - r^3)\rho_1 g + \frac{4}{3}\pi r^3 \rho_2 g = \frac{4}{3}\pi R^3 g \text{Simplifying:} R^3(\rho_1 - 1) = r^3(\rho_1 - \rho_2) \frac{R^3}{r^3} = \frac{\rho_1 - \rho_2}{\rho_1 - 1} \text{Mass ratio:} \frac{\text{Mass concrete}}{\text{Mass sawdust}} = \frac{1 - \rho_2}{\rho_1 - 1} \times \frac{\rho_1}{\rho_2} \text{Substituting values:} = \frac{1 - 0.3}{2.4 - 1} \times \frac{2.4}{0.3} = \frac{0.7}{1.4} \times 8 = 4

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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
}