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Current Question (ID: 20063)

Question:

$\text{A ring has a uniformly distributed charge of } 2\pi \text{ C and radius of } 3 \text{ cm.}$ $\text{A charge of } 10^{-6} \text{ C is placed at the centre of the ring.}$ $\text{Then the tension developed in the ring is:}$

Options:

1. $10^{13} \text{ N}$
2. $10^{5} \text{ N}$
3. $10^{7} \text{ N}$
4. $10^{11} \text{ N}$

Solution:

$\Rightarrow dF = T d\theta$ $\Rightarrow T d\theta = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q \cdot \lambda R d\theta}{R^2}$ $\Rightarrow T = \frac{1}{4\pi \varepsilon_0} \frac{q \lambda}{R}$ $= 9 \times 10^9 \times 10^{-6} \times \frac{2\pi}{2\pi R^2}$ $= \frac{9 \times 10^3}{9} \times 10^4 \text{ N}$ $= 10^7 \text{ N}$

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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
}