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Current Question (ID: 20335)

Question:

$\text{Two long current carrying thin wires, both with current } I, \text{ are held by insulating threads of length } L \text{ and are in equilibrium as shown in the figure, with threads making an angle } \theta \text{ with the vertical. The mass per unit length of wires is } \lambda, \text{ then the value of } I \text{ is:}$ $\left( g = \text{gravitational acceleration} \right)$

Options:

1. $\sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}$
2. $2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}$
3. $2 \sqrt{\frac{\pi g L}{\mu_0}} \tan \theta$
4. $2 \sqrt{\frac{\pi \lambda g L}{\mu_0}} \tan \theta$

Solution:

$\text{Hint: Balance horizontal and vertical forces.}$ $\text{Consider the equilibrium of length } 'l' \text{ of each wire}$ $T \cos \theta = mg \quad \text{(i)}$ $T \sin \theta = F \quad \text{(ii)}$ $\text{Divide eq (ii) with (i)}$ $\tan \theta = \frac{F}{mg}$ $\tan \theta = \frac{\mu_0 I^2}{2 \pi \lambda g}$ $x = 2L \frac{\sin \theta}{\mu_0 I^2}$ $\tan \theta = \frac{2 \pi x mg}{\mu_0 I^2}$ $I^2 = \frac{2 \pi (2L \sin \theta) mg}{4 \pi L \lambda g \sin^2 \theta}$ $I^2 = \frac{\pi \lambda g L}{\mu_0 \cos \theta}$ $I = 2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}$

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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
}