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Current Question (ID: 18926)

Question:

$\text{Match List-I with List-II.}$ $\text{List-I}$ \begin{array}{ll} (a) & h \text{ (Planck's constant)} \\ (b) & E \text{ (kinetic energy)} \\ (c) & V \text{ (electric potential)} \\ (d) & p \text{ (linear momentum)} \end{array}$ $\text{List-II}$ \begin{array}{ll} (i) & [MLT^{-1}] \\ (ii) & [ML^2T^{-1}] \\ (iii) & [ML^2T^{-2}] \\ (iv) & [ML^2A^{-1}T^{-3}] \end{array}$ $\text{Choose from the given option from the given ones:}$

Options:

1. $(a) \rightarrow (iii), (b) \rightarrow (iv), (c) \rightarrow (ii), (d) \rightarrow (i)$
2. $(a) \rightarrow (ii), (b) \rightarrow (iii), (c) \rightarrow (iv), (d) \rightarrow (i)$
3. $(a) \rightarrow (i), (b) \rightarrow (ii), (c) \rightarrow (iv), (d) \rightarrow (iii)$
4. $(a) \rightarrow (iii), (b) \rightarrow (ii), (c) \rightarrow (iv), (d) \rightarrow (i)$

Solution:

$\text{Hint: Recall the respective dimensions.}$ $\text{Step: Find the dimension of the given quantities.}$ $\text{Using Planck's equation, the dimension of } h \text{ (Planck's constant) is given by:}$ $\frac{hc}{\lambda} = E \Rightarrow [h] = \left[ \frac{E\lambda}{c} \right] \Rightarrow [h] = [ML^2T^{-1}]$ $\text{The dimension of kinetic energy is given by:}$ $[E] = [ML^2T^{-2}]$ $\text{Similarly, the dimension of potential is given by:}$ $[V] = \left[ \frac{W}{Q} \right] = \left[ \frac{ML^2T^{-2}}{AT} \right] = [ML^2A^{-1}T^{-3}]$ $\text{Hence, option (2) is the correct answer.}$

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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
}