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Current Question (ID: 8741)

Question:

$\text{If dimensions of critical velocity } v_c \text{ of a liquid flowing through a tube are expressed as } \eta^x \rho^y r^z\text{, where } \eta\text{, } \rho \text{ and } r \text{ are the coefficient of viscosity of the liquid, the density of the liquid, and the radius of the tube respectively, then the values of } x\text{, } y\text{, and } z\text{, respectively, will be:}$

Options:

1. $1, -1, -1$
2. $-1, -1, 1$
3. $-1, -1, -1$
4. $1, 1, 1$

Solution:

$\text{Hint: Recall the principle of homogeneity of dimensions.}$ $\text{Step: Find the values of } x\text{, } y \text{ and } z\text{.}$ $\text{The principle of homogeneity of dimensions states that an equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equation are the same.}$ $\text{Given, the critical velocity of the liquid flowing through a tube is expressed as:}$ $v_c \propto \eta^x \rho^y r^z$ $\text{The dimensions of the coefficient of viscosity of the liquid are given by; } \eta = [ML^{-1}T^{-1}]$ $\text{The dimensions of the density of liquid are given by; } \rho = [ML^{-3}]$ $\text{The dimensions of the radius of a tube are given by; } r = [L]$ $\text{The dimensions of the critical velocity of the liquid are given by;}$ $v_c = [M^0LT^{-1}]$ $\Rightarrow [M^0L^1T^{-1}] = [ML^{-1}T^{-1}]^x \cdot [ML^{-3}]^y \cdot [L]^z$ $\Rightarrow [M^0L^1T^{-1}] = [M^{x+y}L^{-x-3y+z}T^{-x}]$ $\text{Comparing exponents of } M\text{, } L \text{ and } T\text{, we get;}$ $x + y = 0, -x - 3y + z = 1, -x = -1$ $x = 1, y = -1, z = -1$ $\text{Hence, option (1) is the correct answer.}$

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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
}