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Current Question (ID: 10431)

Question:
$\text{An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm}^3 \text{ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be: (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm}^3\text{)}$
Options:
  • 1. $350 \text{ cm}^3$
  • 2. $300 \text{ cm}^3$
  • 3. $250 \text{ cm}^3$
  • 4. $22 \text{ cm}^3$
Solution:
$\text{According to Boyle's law, pressure and volume are inversely proportional to each other i.e., } P \propto \frac{1}{V}$ $\Rightarrow P_1V_1 = P_2V_2$ $\text{The diagram shows the bell at depth h with initial conditions } (P_1, V_1) \text{ and at surface with final conditions } (P_2, V_2)$ $\Rightarrow (P_0 + h\rho_wg)V_1 = P_0V_2$ $\Rightarrow V_2 = \left(1 + \frac{h\rho_wg}{P_0}\right)V_1$ $\Rightarrow V_2 = \left(1 + \frac{47.6 \times 1 \times 1000 \times 10}{0.7 \times 13.6 \times 1000 \times 10}\right)V_1$ $\Rightarrow V_2 = (1 + 5) \times 50 \text{ cm}^3 = 300 \text{ cm}^3$ $\text{As } P_2 = P_0 = 70 \text{ cm of Hg} = 0.7 \times 13.6 \times 1000 \times 10$

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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
}