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Current Question (ID: 20102)

Question:

$\text{A uniformly charged solid sphere of radius } R \text{ has potential } V_0$ $\text{(measured with respect to } \infty) \text{ on its surface. For this sphere, the}$ $\text{equipotential surfaces with potentials } \frac{3V_0}{2}, \frac{5V_0}{4}, \frac{3V_0}{4} \text{ and } \frac{V_0}{4} \text{ have}$ $\text{radius } R_1, R_2, R_3 \text{ and } R_4 \text{ respectively. Then:}$

Options:

1. $R_1 = 0 \text{ and } R_2 > (R_4 - R_3)$
2. $R_1 \neq 0 \text{ and } (R_2 - R_1) > (R_4 - R_3)$
3. $R_1 = 0 \text{ and } R_2 < (R_4 - R_3)$
4. $2R < R_4$

Solution:

$\text{Hint: } V_0 = \frac{kq}{R}$ $\text{Given } V_0 = \frac{kq}{R} \Rightarrow kq = V_0 R$ $\text{For } r < R, \ V = \frac{kq}{2R^3} [3R^2 - r^2]$ $V = \frac{3}{2} \frac{kq}{R} - \frac{kqr^2}{2R^3}$ $\text{At } r = R_1, \ V = \frac{3V_0}{2}$ $\frac{3V_0}{2} = \frac{3V_0}{2} - \frac{V_0 R_1^2}{2R^2}$ $\Rightarrow R_1 = 0$ $\text{At } r = R_2, \ V = \frac{5V_0}{4}$ $\frac{5V_0}{4} = \frac{3V_0}{2} - \frac{V_0 R_2^2}{2R^2}$ $\Rightarrow R_2 = \frac{R}{\sqrt{2}}$ $\text{For } r > R, \ V < V_0$ $\text{and given as } V = \frac{kq}{r} = \frac{V_0 R}{r}$ $\text{At } r = R_3, \ V = \frac{3V_0}{4}$ $\frac{3V_0}{4} = \frac{V_0 R}{R_3}$ $R_3 = \frac{4}{3} R$ $\text{At } r = R_4, \ V = \frac{V_0}{4}$ $\frac{V_0}{4} = \frac{V_0 R}{R_4}$ $R_4 = 4R$ $\therefore \ 2R < R_4$

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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
}