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Current Question (ID: 18941)

Question:

$\text{Column-I lists a few physical quantities and Column-II lists their dimensions. Choose the correct option matching two lists correctly.}$ $\begin{array}{|c|c|} \hline \text{Column-I} & \text{Column-II} \\ \hline \text{(P) pressure gradient} & \text{(A) } M^1L^2T^{-2} \\ \text{(Q) energy density} & \text{(B) } M^1L^1T^{-1} \\ \text{(R) torque} & \text{(C) } M^1L^{-2}T^{-2} \\ \text{(S) impulse} & \text{(D) } M^1L^{-1}T^{-2} \\ \hline \end{array}$

Options:

1. $(\text{P} - \text{C}), (\text{Q} - \text{A}), (\text{R} - \text{B}), (\text{S} - \text{D})$
2. $(\text{P} - \text{A}), (\text{Q} - \text{C}), (\text{R} - \text{B}), (\text{S} - \text{D})$
3. $(\text{P} - \text{A}), (\text{Q} - \text{B}), (\text{R} - \text{S}), (\text{S} - \text{C})$
4. $(\text{P} - \text{A}), (\text{Q} - \text{C}), (\text{R} - \text{B}), (\text{S} - \text{D})$

Solution:

$\text{Hint: Apply dimensional analysis.}$ $\text{Step: Find the accurate matches.}$ $\text{The dimension of pressure gradient is given by:}$ $\Rightarrow \left[ \frac{dP}{dz} \right] = \left[ \frac{ML^{-1}T^{-2}}{L} \right] = \left[ ML^{-2}T^{-2} \right]$ $\text{The dimension of energy density is given by:}$ $\Rightarrow \left[ \frac{dU}{dV} \right] = \left[ \frac{ML^2T^{-2}}{L^3} \right] = \left[ ML^{-1}T^{-2} \right]$ $\text{The dimension of torque is given by:}$ $\Rightarrow [F] \times [r] = [MLT^{-2}] \times [L] = [ML^2T^{-2}]$ $\text{The dimension of impulse is given by:}$ $\Rightarrow [F][t] = [MLT^{-2}][T] = [MLT^{-1}]$ $\text{Hence, option (2) is the correct answer.}$

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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
}