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Current Question (ID: 16709)

Question:

$\text{Kirchhoff's junction rule is a reflection of:}$ $\text{(a) conservation of the current density vector.}$ $\text{(b) conservation of charge.}$ $\text{(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.}$ $\text{(d) the fact that there is no accumulation of charges at a junction.}$ $\text{Which of the above statements are correct?}$

Options:

1. $(\text{b) and (c)})$
2. $(\text{a) and (c)})$
3. $(\text{b) and (d)})$
4. $(\text{c) and (d)})$

Solution:

$(3) \text{ Hint: Use the concept of Kirchhoff's current law.}$ $\text{Kirchhoff's junction rule is also known as Kirchhoff's current law which states that the algebraic sum of the currents flowing towards any point in an electric network is zero i.e., charges are conserved in an electric network.}$ $\text{So, Kirchhoff's junction rule is the reflection of the conservation of charges.}$

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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
}