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Current Question (ID: 19592)

Question:

$\text{A drop of mercury is divided into } 125 \text{ drops of equal radius } 10^{-3} \text{ m each.}$ $\text{If the surface tension of mercury is equal to } 0.45 \text{ N/m.}$ $\text{The magnitude of change in surface energy is equal to nearly:}$

Options:

1. $1.14 \times 10^{-4} \text{ J}$
2. $7.06 \times 10^{-4} \text{ J}$
3. $8.47 \times 10^{-4} \text{ J}$
4. $5.65 \times 10^{-4} \text{ J}$

Solution:

$\text{Hint: } U = T \times A$ $\text{Step 1: Find the radius of the bigger drop.}$ $\text{Let the radius of the bigger drop be } R \text{ so,}$ $\frac{4}{3} \pi R^3 = 125 \times \frac{4}{3} \pi (10^{-3})^3$ $R = 5 \times 10^{-3} \text{ m}$ $\text{Step 2: Find the magnitude of change in surface energy.}$ $\text{The initial surface energy of the liquid is given by;}$ $U_i = 4 \pi R^2 T$ $\Rightarrow U_i = 4 \pi (5 \times 10^{-3})^2 \times 0.45 = 1.41 \times 10^{-4} \text{ J}$ $\text{The final surface energy of the liquid is given by;}$ $U_f = 125 \times 4 \pi r^2 T$ $\Rightarrow U_f = 500 \times \pi (10^{-3})^2 \times 0.45 = 7.06 \times 10^{-4} \text{ J}$ $\text{The change in surface energy of the liquid is given by;}$ $\Delta U = U_f - U_i = (7.06 - 1.41) \times 10^{-4} = 5.65 \times 10^{-4} \text{ J}$ $\text{Hence, option (4) is the correct answer.}$

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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
}