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Current Question (ID: 16621)

Question:

$\text{A wire of resistance } 4 \, \Omega \text{ is stretched to twice its original length. The resistance of a stretched wire would be:}$

Options:

1. $4 \, \Omega$
2. $8 \, \Omega$
3. $16 \, \Omega$
4. $2 \, \Omega$

Solution:

$\text{R} \propto \frac{L}{A}$ $\text{If the length of the wire is doubled, then the area will be } \frac{1}{2} \text{ of the original area.}$ $\text{Hence,}$ $\frac{R_n}{R} = \frac{2L}{L} \cdot \frac{A}{A/2}$ $R_n = 4R = 16 \, \Omega$

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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
}