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Current Question (ID: 18991)

Question:

$\text{An expression of energy density is given by } u = \frac{\alpha}{\beta} \sin \left( \frac{\alpha x}{kt} \right), \text{ where } \alpha, \beta \text{ are constants, } x \text{ is displacement, } k \text{ is Boltzmann constant and } t \text{ is the temperature. The dimensions of } \beta \text{ will be:}$

Options:

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

Solution:

$\text{The expression for energy density is given as } u = \frac{\alpha}{\beta} \sin \left( \frac{\alpha x}{kt} \right).$ $\text{Since } u \text{ is energy density, its dimensions are } [M L^{-1} T^{-2}].$ $\text{The sine function is dimensionless, so } \frac{\alpha x}{kt} \text{ must be dimensionless.}$ $\text{Thus, } \alpha \text{ has dimensions of } [M L^2 T^{-2}].$ $\text{Therefore, } \beta \text{ must have dimensions of } [M^0 L^2 T^0] \text{ to satisfy the equation.}$

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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
}