Import Question JSON

Current Question (ID: 11120)

Question:

$\text{When a large bubble rises from the bottom of a lake to the surface, its radius doubles. The atmospheric pressure is equal to that of a column of water of height } H \text{. The depth of the lake is:}$

Options:

1. $H$
2. $2H$
3. $7H$ (Correct)
4. $8H$

Solution:

$\text{Hint: The amount of gas and the temperature inside the gas remains constant. Step 1: According to the question-} P_0 = \rho g H \text{ and the volume of a sphere is given by } V = \frac{4}{3} \pi r^3 \text{. Since the radius doubles, the new volume will be } V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi 8r^3 = 8V_1 \text{. Step 2: Calculate the depth. As the temperature and amount of gas are constant, we can apply Boyle's Law: } P_1 V_1 = P_0 V_2 \text{. We know that } P_1 = P_0 + \rho g d \text{ and } V_2 = 8V_1 \text{. Substituting these values into Boyle's Law equation, we get:} (P_0 + \rho g d) V_1 = P_0 (8V_1) \text{. Dividing both sides by } V_1 \text{, we get:} P_0 + \rho g d = 8P_0 \text{. Rearranging the terms, we have:} \rho g d = 8P_0 - P_0 \text{, which simplifies to } \rho g d = 7P_0 \text{. Since } P_0 = \rho g H \text{, we can substitute this into the equation to find the depth: } \rho g d = 7 (\rho g H) \text{. Dividing both sides by } \rho g \text{ gives us the depth of the lake:} d = 7H \text{.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}