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Current Question (ID: 12056)

Question:

In the relation $x=cos(\omega t+kx)$, the dimensions of $\omega$ are

Options:

1. $[M^0LT]$
2. $[M^0L^{-1}T^0]$
3. $[M^0L^0T^{-1}]$
4. $[M^0LT^{-1}]$

Solution:

According to the principle of dimensional homogeneity, the argument of a trigonometric function must be dimensionless. Therefore, the dimension of $(\omega t+kx)$ is $[M^0L^0T^0]$. This implies that the dimension of $\omega t$ is also $[M^0L^0T^0]$. So, $[\omega] = \frac{[M^0L^0T^0]}{[t]} = \frac{[M^0L^0T^0]}{[T]} = [M^0L^0T^{-1}]$.

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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
}