Import Question JSON

Current Question (ID: 10396)

Question:

$\text{Steel and copper wires of the same length and area are stretched by the same weight one after the other. Young's modulus of steel and copper are } 2 \times 10^{11} \text{ N/m}^2 \text{ and } 1.2 \times 10^{11} \text{ N/m}^2\text{. The ratio of increase in length is:}$

Options:

1. $\frac{2}{5}$
2. $\frac{3}{5}$
3. $\frac{5}{4}$
4. $\frac{5}{2}$

Solution:

$l = \frac{FL}{AY} \Rightarrow \frac{l_s}{l_{cu}} = \frac{Y_{cu}}{Y_s} \text{ (F, L and A are constant)}$ $\therefore \frac{l_s}{l_{cu}} = \frac{1.2 \times 10^{11}}{2 \times 10^{11}} = \frac{3}{5}$

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
}