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Current Question (ID: 10464)

Question:
$\text{The pressure of water in a water pipe when tap is opened and closed are respectively } 3 \times 10^5 \text{ Nm}^{-2} \text{ and } 3.5 \times 10^5 \text{ Nm}^{-2}. \text{ With open tap, the velocity of water flowing is:}$
Options:
  • 1. $10 \text{ ms}^{-1}$
  • 2. $5 \text{ ms}^{-1}$
  • 3. $20 \text{ ms}^{-1}$
  • 4. $15 \text{ ms}^{-1}$
Solution:
$\text{Hint: Use Bernoulli's theorem}$ $\text{Step 1: Identify the given values}$ $P_1 = 3 \times 10^5 \text{ N/m}^2 \text{ (pressure when tap is open)}$ $P_2 = 3.5 \times 10^5 \text{ N/m}^2 \text{ (pressure when tap is closed)}$ $\text{Step 2: Apply Bernoulli's theorem}$ $\text{For the same horizontal pipe with same height level:}$ $E_{\text{open}} = E_{\text{closed}}$ $P_1 + \frac{1}{2}\rho v^2 = P_2$ $\text{where } v = 0 \text{ when tap is closed}$ $3 \times 10^5 + \frac{1}{2} \times 10^3 \times v^2 = 3.5 \times 10^5$ $\frac{1}{2} \times 10^3 \times v^2 = 0.5 \times 10^5$ $500 v^2 = 50000$ $v^2 = 100$ $v = 10 \text{ m/s}$

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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
}