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Current Question (ID: 20214)

Question:
$\text{A metal wire of resistance } 3 \, \Omega \text{ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends are joined to make a circle. If two points on this circle make an angle } 60^\circ \text{ at the centre, the equivalent resistance between these two points will be:}$
Options:
  • 1. $\frac{5}{2} \, \Omega$
  • 2. $\frac{5}{3} \, \Omega$
  • 3. $\frac{7}{2} \, \Omega$
  • 4. $\frac{12}{5} \, \Omega$
Solution:
$\text{Hint: } R \propto l$ $R = 3$ $\text{When length becomes double}$ $R = 12 \, \Omega \left( R = \frac{Pl}{A} = \frac{Pl^2}{V} \right)$ $\frac{R}{R_{\text{Eff}}} = \frac{l}{6}$ $= \frac{10}{6}$ $= \frac{5}{3}$

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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
}