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Current Question (ID: 20037)

Question:

$\text{Two identical conducting spheres, each with negligible volume, carry initial charges of } 2.1 \text{ nC and } -0.1 \text{ nC, \ respectively.}$ $\text{The spheres are brought into contact, allowing charge to redistribute, and are then separated by a distance of } 0.5 \text{ m.}$ $\text{The electrostatic force acting between the spheres is:}$ $\left( 4\pi\varepsilon_0 = \frac{1}{9 \times 10^9} \text{ in SI units} \right)$

Options:

1. $36 \times 10^{-7} \text{ N}$
2. $36 \times 10^{-9} \text{ N}$
3. $18 \times 10^{-7} \text{ N}$
4. $18 \times 10^{-9} \text{ N}$

Solution:

$\text{Hint: The charge gets equally distributed when both of them are in contact.}$ $\text{Step: Find the electrostatic force acting between the spheres.}$ $\text{When the spheres are brought into contact, the charges redistribute equally because the spheres are identical.}$ $\text{The total charge is given by:}$ $\Rightarrow q_{\text{total}} = q_1 + q_2 = (2.1 - 0.1) \text{ nC} = 2.0 \text{ nC}$ $\text{After contact, the charge on each sphere is given by:}$ $\Rightarrow q_{\text{final}} = \frac{q_{\text{total}}}{2} = \frac{2.0}{2} \text{ nC} = 1.0 \text{ nC} = 1.0 \times 10^{-9} \text{ C}$ $\text{The spheres are separated by a distance } r = 0.5 \text{ m.}$ $\text{The electrostatic force between the spheres is given by Coulomb's law:}$ $\Rightarrow F = \frac{k q_1 q_2}{r^2}$ $\Rightarrow F = \frac{9 \times 10^9 (1.0 \times 10^{-9})(1.0 \times 10^{-9})}{(0.5)^2}$ $\Rightarrow F = \frac{9 \times 10^{-9} \times 10^{-18}}{0.25}$ $\Rightarrow F = \frac{9 \times 10^{-9}}{0.25} = 36 \times 10^{-9} \text{ N}$ $\text{Hence, option (2) is the correct answer.}$

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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
}