Import Question JSON

Current Question (ID: 10516)

Question:

$\text{If pressure at half the depth of a lake is equal to }\frac{2}{3}^{\text{rd}}\text{ the pressure at the bottom of the lake, then the depth of the lake is:}$

Options:

1. $\text{10 m}$
2. $\text{20 m}$
3. $\text{60 m}$
4. $\text{30 m}$

Solution:

\text{Pressure at half depth: } P_0 + \frac{h \rho g}{2} \text{Pressure at bottom: } P_0 + h \rho g \text{Given condition: } P_0 + \frac{h \rho g}{2} = \frac{2}{3}(P_0 + h \rho g) \text{Solving: } 3P_0 + \frac{3h \rho g}{2} = 2P_0 + 2h \rho g \text{Therefore: } h = \frac{2P_0}{\rho g} = \frac{2 \times 10^5}{10^3 \times 10} = 20 \text{ m}

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
}