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Current Question (ID: 16461)

Question:

$\text{Refer to the arrangement of charges in the figure and a Gaussian surface of a radius } R \text{ with } Q \text{ at the centre. Then:}$ $\text{(a) total flux through the surface of the sphere is } \frac{-Q}{\varepsilon_0}. $ $\text{(b) field on the surface of the sphere is } \frac{-Q}{4\pi\varepsilon_0 R^2}. $ $\text{(c) flux through the surface of the sphere due to } 5Q \text{ is zero.}$ $\text{(d) field on the surface of the sphere due to } -2Q \text{ is the same everywhere.}$ $\text{Choose the correct statement(s):}$

Options:

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

Solution:

$\text{Hint: Use Gauss' Law.}$ $\text{Step 1: Find the net flux passing through the surface.}$ $\text{Gauss' law states that the total electric flux of an enclosed surface is given by } \frac{q}{\varepsilon_0} \text{ where } q \text{ is the charge enclosed by the surface. Thus, from the figure,}$ $\text{The total charge inside the surface is } Q - 2Q = -Q.$ $\text{The total flux through the surface of the sphere } = \frac{-Q}{\varepsilon_0}.$ $\text{Step 2: Find the flux due to the } 5Q \text{ charge.}$ $\text{Now, consider the charge } 5Q.$ $\text{The charge } 5Q \text{ lies outside the surface, thus it makes no contribution to electric flux through the given surface.}$ $\text{Therefore, the flux through the surface of the sphere due to } 5Q \text{ is zero.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}