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Current Question (ID: 10503)

Question:

$\text{There is a horizontal film of soap solution. On it, a thread is placed in the form of a loop. The film is pierced inside the loop and the thread becomes a circular loop of radius } R. \text{ If the surface tension of the loop is } T, \text{ then what will be the tension in the thread?}$

Options:

1. $\pi R^2 / T$
2. $\pi R^2 T$
3. $2\pi RT$
4. $2RT$

Solution:

\text{Suppose tension in thread is F, then for small part } \Delta l \text{ of thread:} \text{From the force diagram, we can see that the vertical components of tension balance the upward force due to surface tension.} \text{For a small element of length } \Delta l \text{ subtending angle } \Delta \theta \text{ at the center:} \Delta l = R\Delta\theta \text{ and } 2F\sin\left(\frac{\Delta\theta}{2}\right) = 2T \Delta l = 2TR\Delta\theta \text{Therefore: } F = \frac{TR\Delta\theta}{\sin\left(\frac{\Delta\theta}{2}\right)} \text{For small angles: } \sin\left(\frac{\Delta\theta}{2}\right) \approx \frac{\Delta\theta}{2} \text{Therefore: } F = \frac{TR\Delta\theta}{\frac{\Delta\theta}{2}} = 2TR \text{Hence, the tension in the thread is } F = 2TR

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}