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Current Question (ID: 10459)

Question:

$\text{The cylindrical tube of a spray pump has radius } R\text{, one end of which has } n \text{ fine holes, each of radius } r\text{. If the speed of the liquid in the tube is } v\text{, then the speed of ejection of the liquid through the holes will be:}$

Options:

1. $\frac{vR^2}{n^2r^2}$
2. $\frac{vR^2}{nr^2}$
3. $\frac{vR^2}{n^3r^2}$
4. $\frac{v^2R}{nr}$

Solution:

$\text{According to the equation of continuity } Av = \text{constant}\text{, where } A \text{ is the area of the cylindrical tube and } v \text{ is the velocity of liquid in a tube.}$ $\text{Volume inflow rate = Volume outflow rate}$ $\pi R^2 v = n\pi r^2 v_1$ $v_1 = \frac{R^2 v}{nr^2}$ $\text{Explanation:}$ $\text{The continuity equation states that for an incompressible fluid, the mass flow rate must be conserved.}$ $\text{Area of cylindrical tube = } \pi R^2$ $\text{Area of each hole = } \pi r^2$ $\text{Total area of } n \text{ holes = } n\pi r^2$ $\text{Applying continuity equation:}$ $\text{(Area of tube)} \times \text{(velocity in tube)} = \text{(Total area of holes)} \times \text{(velocity through holes)}$ $\pi R^2 \times v = n\pi r^2 \times v_1$ $\text{Solving for } v_1\text{:}$ $v_1 = \frac{\pi R^2 v}{n\pi r^2} = \frac{R^2 v}{nr^2} = \frac{vR^2}{nr^2}$

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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
}