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Current Question (ID: 20482)

Question:

$\text{Match List-I with List-II}$ $\text{List - I}$ \begin{array}{|c|c|} \hline \text{A} & Y = \text{magnetic susceptibility} \\ & X = \text{magnetizing field} \\ \hline \text{B} & Y = \text{magnetic field} \\ & X = \text{distance from centre of a current carrying wire for } x < a \text{ (where } a = \text{radius of wire)} \\ \hline \text{C} & Y = \text{magnetic field} \\ & X = \text{distance from centre of a current carrying wire for } x > a \text{ (where } a = \text{radius of wire)} \\ \hline \text{D} & Y = \text{magnetic field inside solenoid} \\ & X = \text{distance from centre} \\ \hline \end{array}$ $\text{List - II}$ \begin{array}{|c|c|} \hline \text{I} & \text{Graph: Linear increase} \\ \hline \text{II} & \text{Graph: Decrease then constant} \\ \hline \text{III} & \text{Graph: Constant} \\ \hline \text{IV} & \text{Graph: Exponential decay} \\ \hline \end{array}$ $\text{Choose the correct answer from the options given below:}$

Options:

1. $(\text{A}-(\text{III}), \text{B}-(\text{I}), \text{C}-(\text{IV}), \text{D}-(\text{II}))$
2. $(\text{A}-(\text{III}), \text{B}-(\text{IV}), \text{C}-(\text{I}), \text{D}-(\text{II}))$
3. $(\text{A}-(\text{I}), \text{B}-(\text{III}), \text{C}-(\text{II}), \text{D}-(\text{IV}))$
4. $(\text{A}-(\text{IV}), \text{B}-(\text{I}), \text{C}-(\text{III}), \text{D}-(\text{II}))$

Solution:

$\text{The correct matches are:}$ $\text{A}-(\text{III}): \text{Magnetic susceptibility vs. magnetizing field is constant.}$ $\text{B}-(\text{I}): \text{Magnetic field inside a wire increases linearly with distance for } x < a.$ $\text{C}-(\text{IV}): \text{Magnetic field outside a wire decreases exponentially with distance for } x > a.$ $\text{D}-(\text{II}): \text{Magnetic field inside a solenoid is constant.}$

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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
}