Import Question JSON

Current Question (ID: 10436)

Question:

The heart of a man pumps $5 \text{ L}$ of blood through the arteries per minute at a pressure of $150 \text{ mm}$ of mercury. If the density of mercury is $13.6 \times 10^3 \text{ kg/m}^3$ and $g = 10 \text{ m/s}^2$, then the power of heart in watt is:

Options:

1. $1.70$
2. $2.35$
3. $3.0$
4. $1.50$

Solution:

\text{Given, Pressure } = 150 \text{ mm of Hg} \text{This pressure needs to be converted into meters of Hg.} h = 150 \text{ mm} = 0.15 \text{ m} \text{Also, the volume pumping rate is given as } 5 \text{ L per minute.} 1 \text{ L} = 10^{-3} \text{ m}^3 \text{Volume per minute } = 5 \times 10^{-3} \text{ m}^3 \text{Time } = 1 \text{ minute} = 60 \text{ s} \text{Pumping rate of the heart of a man } = \frac{dV}{dt} = \frac{5 \times 10^{-3}}{60} \text{ m}^3\text{/s} \text{The power of the heart can be calculated as the rate of doing work.} \text{Power of the heart } = \frac{dW}{dt} = \frac{d}{dt}(mgh) = \frac{d}{dt}(\rho V g h) = \rho g h \frac{dV}{dt} \text{where } \rho \text{ is the density of mercury and } h \text{ is the height of the mercury.} \rho = 13.6 \times 10^3 \text{ kg/m}^3 g = 10 \text{ m/s}^2 h = 0.15 \text{ m} \frac{dV}{dt} = \frac{5 \times 10^{-3}}{60} \text{ m}^3\text{/s} \text{Power of the heart } = (13.6 \times 10^3 \text{ kg/m}^3) \times (10 \text{ m/s}^2) \times (0.15 \text{ m}) \times \left(\frac{5 \times 10^{-3} \text{ m}^3\text{/s}}{60}\right) = 1.70 \text{ W} \text{Power of the heart } = 1.70 \text{ W}

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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
}