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Current Question (ID: 16595)

Question:

$\text{A metallic sphere of capacitance } C_1, \text{ charged to electric potential } V_1 \text{ is connected by a metal wire to another metallic sphere of capacitance } C_2 \text{ charged to electric potential } V_2. \text{ The amount of heat produced in connecting the wire during the process is:}$

Options:

1. $\frac{C_1C_2}{2(C_1+C_2)}(V_1+V_2)^2$
2. $\frac{C_1C_2}{2(C_1+C_2)}(V_1-V_2)^2$
3. $\frac{C_1C_2}{C_1+C_2}(V_1-V_2)^2$
4. $\text{zero}$

Solution:

$\text{(2)}$ $\text{The initial charge on the spheres:}$ $Q_1 = C_1 V_1 \quad \& \quad Q_2 = C_2 V_2$ $\text{Initial potential energy of the spheres—}$ $U_1 = \frac{1}{2} C_1 V_1^2 \quad \text{and,} \quad U_2 = \frac{1}{2} C_2 V_2^2$ $\text{After connecting, their common potential:}$ $\Rightarrow V_1' = V_2' = \frac{(C_1 V_1 + C_2 V_2)}{(C_1 + C_2)}$ $\text{Final Potential energy of the spheres—}$ $U_f = \frac{1}{2} (C_1 + C_2) V'^2 = \frac{1}{2} \frac{(C_1 V_1 + C_2 V_2)^2}{(C_1 + C_2)}$ $\text{Heat produced = Loss in potential energy}$ $= \text{Initial potential energy – Final potential energy}$ $(U_i - U_f)$ $= \left[ \frac{1}{2} (C_1 V_1^2 + C_2 V_2^2) - \frac{1}{2} \frac{(C_1 V_1 + C_2 V_2)^2}{(C_1 + C_2)} \right]$ $= \frac{1}{2} \left[ \frac{(C_1 + C_2)(C_1 V_1^2 + C_2 V_2^2) - (C_1 V_1 + C_2 V_2)^2}{(C_1 + C_2)} \right]$ $= \frac{1}{2} \left[ \frac{C_1^2 V_1^2 + C_1 C_2 (V_1^2 + V_2^2) + C_2^2 V_2^2 - C_1^2 V_1^2 - C_2^2 V_2^2 - 2 C_1 C_2 V_1 V_2}{(C_1 + C_2)} \right]$ $= \frac{1}{(C_1 + C_2)} \left[ C_1 C_2 V_1^2 + C_1 C_2 V_2^2 - 2 C_1 C_2 V_1 V_2 \right]$ $= \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 - V_2)^2$

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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
}