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Current Question (ID: 18918)

Question:

$\text{If momentum } (P), \text{ area } (A), \text{ and time } (T) \text{ are taken as the fundamental physical quantities, then the dimensional formula of energy in terms of } P, A, \text{ and } T \text{ is:}$

Options:

1. $[PA^{-1}T^{-2}]$
2. $[PA^{1/2}T^{-1}]$
3. $[P^2AT^{-2}]$
4. $[P^{1/2}AT^{-1}]$

Solution:

$\text{Hint: Dimensions of energy } = ML^2T^{-2}$ $\text{Step: Find the dimensional formula for energy.}$ $[E] = [P]^x[A]^y[T]^z$ $\Rightarrow [ML^2T^{-2}] = [MLT^{-1}]^x[L^2]^y[T]^z$ $\Rightarrow [ML^2T^{-2}] = [M^xL^{x+2y}T^{-x+z}]$ $\text{Compare the powers of physical quantities:}$ $x = 1, x + 2y = 2 \text{ and } z - x = -2$ $\Rightarrow x = 1, y = 1/2 \text{ and } z = -1$ $\text{Therefore;}$ $[E] = [P]^1[A]^{1/2}[T]^{-1}$ $\text{Hence, option (2) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}