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Current Question (ID: 16459)

Question:

$\text{A hollow cylinder has a charge } q \text{ coulomb within it (at the geometrical centre).}$ $\text{If } \phi \text{ is the electric flux in units of Volt-meter associated with the curved surface } B, \text{ the flux linked with the plane surface } A \text{ in units of volt-meter will be:}$

Options:

1. $\frac{1}{2} \left( \frac{q}{\varepsilon_0} - \phi \right)$
2. $\frac{q}{2\varepsilon_0}$
3. $\frac{\phi}{3}$
4. $\frac{q}{\varepsilon_0} - \phi$

Solution:

$\text{Hint: } \phi_{\text{total}} = \frac{q}{\varepsilon_0}$ $\text{Step: Find the flux linked with the plane surface } A \text{ in units of volt-meter.}$ $\text{Gauss's law states that the net electric flux through any closed surface is}$ $\text{equal to the net charge inside the surface divided by } \varepsilon_0.$ $\phi_{\text{total}} = \frac{q}{\varepsilon_0}$ $\text{Let electric flux be linked with surfaces } A, B \text{ and } C \text{ are } \phi_A, \phi_B \text{ and } \phi_C,$ $\text{respectively. That is given by;}$ $\phi_{\text{total}} = \phi_A + \phi_B + \phi_C$ $\text{Since } \phi_C = \phi_A$ $\text{Therefore } 2\phi_A + \phi_B = \phi_{\text{total}} = \frac{q}{\varepsilon_0}$ $\text{or } \phi_A = \frac{1}{2} \left( \frac{q}{\varepsilon_0} - \phi_B \right)$ $\text{But } \phi_B = \phi \text{ (given)}$ $\text{Thus, } \phi_A = \frac{1}{2} \left( \frac{q}{\varepsilon_0} - \phi \right)$ $\text{Hence, option (1) is the correct answer.}$

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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
}