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Current Question (ID: 18917)

Question:

$\text{If speed } v, \text{ area } A \text{ and force } F \text{ are chosen as fundamental units, then the dimension of Young's modulus will be:}$

Options:

1. $[FA^{-1}v^0]$
2. $[FA^2v^{-1}]$
3. $[FA^2v^{-3}]$
4. $[FA^2v^{-2}]$

Solution:

$\text{Hint: Dimensions of Young's modulus and stress are the same.}$ $Y = F^x A^y V^z$ $M^1L^{-1}T^{-2} = [MLT^{-2}] \times [L^2]^y [LT^{-1}]^z$ $M^1L^{-1}T^{-2} = [M^x][L]^{x+2y+z}[T]^{-2x-z}$ $\text{Comparing power of } ML \text{ and } T$ $x = 1$ $x + 2y + z = -1 \quad \ldots \ldots \ldots \ldots (2)$ $-2x - z = -2 \quad \ldots \ldots \ldots \ldots (3)$ $\text{After solving}$ $x = 1$ $y = -1$ $z = 0$ $Y = FA^{-1}V^0$

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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
}