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Current Question (ID: 19583)

Question:

$\text{If } \rho \text{ is the density and } \eta \text{ is the coefficient of viscosity of fluid which flows with a speed } v \text{ in the pipe of diameter } d, \text{ the correct formula for the Reynolds number } R_e \text{ is:}$

Options:

1. $R_e = \frac{\eta d}{\rho v}$
2. $R_e = \frac{\rho v}{\eta d}$
3. $R_e = \frac{\rho v d}{\eta}$
4. $R_e = \frac{\eta}{\rho v d}$

Solution:

$\text{Hint: Reynold's number is given by } R_e = \frac{\rho v D}{\eta}$ $\text{Explanation: Reynolds number is a dimensionless quantity that is used to determine the type of flow pattern as laminar or turbulent while flowing through a pipe.}$ $\text{Reynolds number is defined by the ratio of inertial forces to that of viscous forces.}$ $\text{Reynold's number is given by } R_e = \frac{\rho v D}{\eta}.$

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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
}