Import Question JSON

Current Question (ID: 20621)

Question:

$\text{Correctly match the two lists.}$ $\text{List-I:}$ \begin{array}{ll} (a) \ \text{Gauss's law (electrostatics)} & (b) \ \text{Ampere's circuital law} \\ (c) \ \text{Gauss's law (magnetism)} & (d) \ \text{Faraday's law of induction} \end{array} $ $\text{List-II:}$ \begin{array}{ll} (P) \ \vec{\mathbf{B}} \cdot d\vec{\mathbf{A}} = 0 & (Q) \ \vec{\mathbf{B}} \cdot d\vec{\mathbf{l}} = \mu_0 i_{\text{enclosed}} \\ (R) \ \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \frac{q_{\text{enclosed}}}{\varepsilon_0} & (S) \ \varepsilon = - \frac{d\phi_B}{dt} \end{array}$

Options:

1. $(a) \rightarrow (R), (b) \rightarrow (Q), (c) \rightarrow (S), (d) \rightarrow (P)$
2. $(a) \rightarrow (R), (b) \rightarrow (Q), (c) \rightarrow (P), (d) \rightarrow (S)$
3. $(a) \rightarrow (R), (b) \rightarrow (S), (c) \rightarrow (Q), (d) \rightarrow (P)$
4. $(a) \rightarrow (R), (b) \rightarrow (S), (c) \rightarrow (P), (d) \rightarrow (Q)$

Solution:

$\text{Hint: } \phi = \oint \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \frac{q_{\text{in}}}{\varepsilon_0}$ $\text{Step: Identify the correct match.}$ $\text{Gauss's law (electrostatics): } \phi = \oint \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \frac{q_{\text{enclosed}}}{\varepsilon_0} \ \text{Therefore, the correct match is } a \rightarrow R$ $\text{Ampere's circuital law: } \oint \vec{\mathbf{B}} \cdot d\vec{\mathbf{l}} = \mu_0 i_{\text{enclosed}} \ \text{Therefore, the correct match is } b \rightarrow Q$ $\text{Gauss's law (magnetism): } \phi = \oint \vec{\mathbf{B}} \cdot d\vec{\mathbf{A}} = 0 \ \text{Therefore, the correct match is } c \rightarrow P$ $\text{Faraday's law of electromagnetic induction: } \varepsilon = - \frac{d\phi_B}{dt} \ \text{Therefore, the correct match is } d \rightarrow S$ $\text{Therefore, the correct match is } a \rightarrow R; b \rightarrow Q; c \rightarrow P; d \rightarrow S$ $\text{Hence, option (2) is the correct answer.}$

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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
}