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Current Question (ID: 8735)

Question:

$\text{If energy (E), velocity (v) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:}$

Options:

1. $[Ev^{-2}T^{-1}]$
2. $[Ev^{-1}T^{-2}]$
3. $[Ev^{-2}T^{-2}]$
4. $[E^{-2}v^{-1}T^{-3}]$

Solution:

$\text{Hint: } S = \frac{F}{L}$ $\text{Step: Find the dimensional formula of surface tension in terms of E, v, and T.}$ $S = \frac{F}{L}$ $= \frac{F \times L}{L \times L}$ $\Rightarrow S = \frac{E}{L^2} = \frac{E}{\frac{L^2}{T^2} \times T^2}$ $\text{Since } L = vT$ (distance = velocity × time), we have $L^2 = v^2T^2$ $\Rightarrow [S] = \frac{[E]}{[v^2][T^2]}$ $\text{Therefore, the dimensional formula of surface tension in terms of E, v, and T is } [Ev^{-2}T^{-2}]\text{.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}