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Current Question (ID: 16501)

Question:

$\text{Three concentric spherical shells have radii } a, b, \text{ and } c \,(a < b < c) \text{ and have surface charge densities } \sigma, -\sigma, \text{ and } \sigma \text{ respectively.}$ $\text{If } V_A, V_B \text{ and } V_C \text{ denote the potential of the three shells, and } c = a + b, \text{ it can be concluded that:}$

Options:

1. $V_C = V_A \neq V_B$
2. $V_C = V_B \neq V_A$
3. $V_C \neq V_B \neq V_A$
4. $V_C = V_B = V_A$

Solution:

$\text{(1) Electric potential inside a shell of radius } R \text{ having charge } Q = \frac{kQ}{R}$ $\text{The electric potential is the same at the surface of the shell also.}$ $\text{For the outside point at a distance } r \text{ from the centre, the electric potential,}$ $V_{\text{outside}} = \frac{kQ}{r}$ $V_A = \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi a^2}{a} - \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi b^2}{b} + \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi c^2}{c} = \frac{\sigma}{\varepsilon_0} (a - b + c) = \frac{\sigma}{\varepsilon_0} (2a)$ $V_B = \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi a^2}{b} - \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi b^2}{b} + \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi c^2}{c} = \frac{\sigma}{\varepsilon_0} \left( \frac{a^2}{b} - b + c \right) = \frac{\sigma}{\varepsilon_0} \left( \frac{a^2}{b} + a \right)$ $V_C = \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi a^2}{c} - \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi b^2}{c} + \frac{1}{4\pi\varepsilon_0} \frac{\sigma 4\pi c^2}{c} = \frac{\sigma}{\varepsilon_0} \left( \frac{a^2}{c} - \frac{b^2}{c} + c \right) = \frac{\sigma}{\varepsilon_0} (2a)$

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{
  "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
}