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Current Question (ID: 16619)

Question:

$\text{A wire made of a metal of uniform resistivity } 1.0 \times 10^{-6} \, \Omega\text{-m is } 2.0 \, \text{m long}$ $\text{and has a diameter of } 2.0 \times 10^{-3} \, \text{m. What is the electrical resistance of this}$ $\text{length of the wire?}$

Options:

1. $\frac{2.0 \times 10^{-3}}{\pi} \, \Omega$
2. $5.0 \times 10^{-13} \pi \, \Omega$
3. $\frac{1.0}{2\pi} \, \Omega$
4. $\frac{2.0}{\pi} \, \Omega$

Solution:

$\text{Hint: } R = \rho \frac{l}{A}$ $\text{Step: Find the resistance of the wire.}$ $\text{The resistance of the wire is given by;}$ $R = \rho \frac{l}{A}$ $\Rightarrow R = 1.0 \times 10^{-6} \times \frac{2.0}{\pi \times 1.0 \times 10^{-6}} = \frac{2.0}{\pi} \, \Omega$ $\text{Hence, option (4) is the correct answer.}$

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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
}