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Current Question (ID: 10482)

Question:

$\text{Fluid flows through a pipe in a horizontal plane. The fluid has a relative density of 0.9 and its velocity at P (where cross-section area is A) is 4 m/s. What is its pressure at Q? (where cross-section area is }\frac{A}{4}\text{, The pressure at P is }2.8 \times 10^5\text{ Pa)}$

Options:

1. $2.36 \times 10^5\text{ Pa}$
2. $3.88 \times 10^5\text{ Pa}$
3. $2.08 \times 10^5\text{ Pa}$
4. $1.72 \times 10^5\text{ Pa}$

Solution:

\text{1. Calculate fluid density: } \rho = 0.9 \times 1000 \text{ kg/m}^3 = 900 \text{ kg/m}^3 \text{2. Use Continuity Equation to find } v_Q\text{: } A \times 4 = \frac{A}{4} \times v_Q \Rightarrow v_Q = 16 \text{ m/s} \text{3. Apply Bernoulli's Equation (horizontal pipe): } P_P + \frac{1}{2}\rho v_P^2 = P_Q + \frac{1}{2}\rho v_Q^2 2.8 \times 10^5 + \frac{1}{2}(900)(4^2) = P_Q + \frac{1}{2}(900)(16^2) 2.8 \times 10^5 + 7200 = P_Q + 115200 P_Q = 287200 - 115200 = 172000 \text{ Pa} = 1.72 \times 10^5 \text{ Pa} \text{The final answer is } 1.72 \times 10^5 \text{ Pa}

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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
}