Import Question JSON

Current Question (ID: 20237)

Question:

$\text{The variation of applied potential, } V \text{ and the current, } I \text{ flowing through a given wire is shown in the figure.}$ $\text{The length of the wire is } 31.4 \text{ cm.}$ $\text{The diameter of the wire is measured as } 2.4 \text{ cm.}$ $\text{The resistivity of the given wire is measured as } x \times 10^{-3} \, \Omega\text{-cm.}$ $\text{The value of } x \text{ is:}$ $\text{(take } \pi = 3.14)$

Options:

1. 146
2. 142
3. 140
4. 144

Solution:

$\text{Hint: } R = \frac{\rho l}{A}$ $\text{Step: Find the resistivity of the wire.}$ $\text{On a voltage-current (} V-I \text{) graph, the slope of the line represents the resistance } \left( R = \frac{V}{I} \right).$ $\text{The slope of the given graph is given by:}$ $R = \tan 45^\circ = 1 \, \Omega$ $\text{The resistivity of the wire is given by:}$ $\rho = \frac{RA}{l}$ $\Rightarrow \rho = \frac{1 \times \pi (1.2)^2}{31.4} \quad \left[ r = \frac{2.4}{2} = 1.2 \text{ cm} \right]$ $\Rightarrow \rho = \frac{3.14 \times 1.44}{31.4} = 0.144 \, \Omega\text{-cm}$ $\Rightarrow \rho = 144 \times 10^{-3} \, \Omega\text{-m} = x \times 10^{-3} \, \Omega\text{-m}$ $\text{Therefore, the value of } x \text{ is } 144.$ $\text{Hence, option (4) is the correct answer.}$

Import JSON File

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
}