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Current Question (ID: 20121)

Question:

$\text{A charge } Q \text{ is distributed over two concentric conducting thin spherical shells radii } r \text{ and } R \ (R > r).$ $\text{If the surface charge densities on the two shells are equal, the electric potential at the common centre is:}$

Options:

1. $\frac{1}{4\pi\varepsilon_0} \frac{(R+2r)Q}{2(R^2+r^2)}$
2. $\frac{1}{4\pi\varepsilon_0} \frac{(R+r)Q}{(R^2+r^2)}$
3. $\frac{1}{4\pi\varepsilon_0} \frac{(R^2+r^2)Q}{(R+r)}$
4. $\frac{1}{4\pi\varepsilon_0} \frac{(2R+r)Q}{(R^2+r^2)}$

Solution:

$\text{Hint: } V = \frac{kQ}{R}.$ $\text{Let the charges on inner and outer spheres are } Q_1 \text{ and } Q_2.$ $\sigma = \frac{Q_1}{4\pi r^2} = \frac{Q_2}{4\pi R^2} \Rightarrow \frac{Q_1}{Q_2} = \frac{r^2}{R^2}$ $Q_1 + Q_2 = Q \Rightarrow \frac{Q_2 r^2}{R^2} + Q_2 = Q$ $\Rightarrow Q_2 = \frac{QR^2}{(r^2+R^2)}$ $Q_1 = \frac{r^2}{R^2} \cdot \frac{QR^2}{(R^2+r^2)} = \frac{Qr^2}{(R^2+r^2)}$ $\text{Potential at centre } 'O' = \frac{kQ_1}{r} + \frac{kQ_2}{R}$ $= k \left[ \frac{Qr^2}{R(R^2+r^2)} + \frac{QR^2}{R(R^2+r^2)} \right]$ $= \frac{kQ(r+R)}{(R^2+r^2)} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{(R^2+r^2)}$

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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
}