Import Question JSON

Current Question (ID: 20137)

Question:
$\text{Twenty-seven identical spherical drops of mercury are each maintained at a potential of } 10 \text{ V.}$ $\text{If all these drops coalesce to form a single large spherical drop,}$ $\text{then the potential energy of this larger drop will be how many times that of one of the smaller drops?}$
Options:
  • 1. $143$
  • 2. $243$
  • 3. $348$
  • 4. $564$
Solution:
$\text{Hint: } V = \frac{kQ}{R}$ $\text{Step 1: Find the radius of the bigger drop.}$ $\frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \pi r^3$ $\Rightarrow R = 3r$ $\text{Step 2: Find the potential energy of the smaller drop.}$ $U = \frac{kQ^2}{2r}$ $\text{Step 3: Find the potential energy of the bigger drop.}$ $U' = \frac{k(nQ)^2}{2R}$ $= \frac{k(27Q)^2}{2 \times 3r}$ $= \frac{2 \times 3r}{729kQ}$ $= \frac{6r}{729} \times 2U$ $= \frac{729}{6} \times 2U \left[ \therefore \frac{kQ^2}{r} = 2U \right]$ $U' = 243U$ $\text{Hence, option (2) is the correct answer.}$

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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
}