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Current Question (ID: 10381)

Question:
A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$. It is stretched by an amount $x$. The work done is:
Options:
  • 1. $\frac{YxA}{2L}$
  • 2. $\frac{Yx^2A}{L}$
  • 3. $\frac{Yx^2A}{2L}$
  • 4. $\frac{2Yx^2A}{L}$
Solution:
$\text{Work} = F_{\text{avg}} \times \text{displacement}$ $\Rightarrow W = \frac{F}{2} \times x$ $\text{Now,}$ $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L}$ $\Rightarrow F = \frac{YAx}{L}$ $\text{So, work done,}$ $W = \frac{1}{2} \left( \frac{YAx}{L} \right) \times x$ $= \frac{YAx^2}{2L}$ $\text{Alternate Method:}$ $\text{Work done} = \Delta U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$ $\text{or, } W = \frac{1}{2} \times Y \times (\text{strain})^2 \times \text{volume} \ldots \ldots (i)$ $\text{strain} = \frac{x}{L} \text{ and Volume} = A \times L$ $\text{Putting the values of strain and volume in equation (i):}$ $W = \frac{1}{2} \times Y \times \frac{x^2}{L^2} \times AL$ $\Rightarrow W = \frac{YAx^2}{2L}$

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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
}