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Current Question (ID: 10468)

Question:

$\text{A small hole of an area of cross-section } 2 \text{ mm}^2 \text{ is present near the bottom of a fully filled open tank of height } 2 \text{ m. Taking } (g = 10 \text{ m/s}^2), \text{ the rate of flow of water through the open hole would be nearly:}$

Options:

1. $6.4 \times 10^{-6} \text{ m}^3/\text{s}$
2. $12.6 \times 10^{-6} \text{ m}^3/\text{s}$
3. $8.9 \times 10^{-6} \text{ m}^3/\text{s}$
4. $2.23 \times 10^{-6} \text{ m}^3/\text{s}$

Solution:

$\text{Using Torricelli's law for velocity of efflux:}$ $v = \sqrt{2gh}$ $\text{where } h = 2 \text{ m (height of water column)} \text{ and } g = 10 \text{ m/s}^2$ $v = \sqrt{2 \times 10 \times 2} = \sqrt{40} = 2\sqrt{10} \text{ m/s}$ $\text{Rate of flow is given by:}$ $\text{Rate of flow} = A \times v$ $\text{where } A = 2 \text{ mm}^2 = 2 \times 10^{-6} \text{ m}^2$ $\text{Rate of flow} = 2 \times 10^{-6} \times 2\sqrt{10}$ $= 4\sqrt{10} \times 10^{-6} \text{ m}^3/\text{s}$ $= 4 \times 3.16 \times 10^{-6} \text{ m}^3/\text{s}$ $= 12.6 \times 10^{-6} \text{ m}^3/\text{s}$

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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
}