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Current Question (ID: 19539)

Question:

$\text{The elastic behaviour of a material for linear stress and linear strain is captured in the graph below. The energy density, for a linear strain of } 5 \times 10^{-4} \text{ is:}$ $\text{(assume that the material is elastic up to the linear strain of } 5 \times 10^{-4})$

Options:

1. $15 \text{ kJ/m}^3$
2. $20 \text{ kJ/m}^3$
3. $25 \text{ kJ/m}^3$
4. $30 \text{ kJ/m}^3$

Solution:

$\text{Hint: Energy density} = \frac{1}{2} (\text{stress} \times \text{strain})$ $\text{Step 1: Find the stress developed in the wire.}$ $\text{Stress} = Y (\text{Strain})$ $Y = \frac{1}{\text{slope}}$ $\text{Stress} = \left( \frac{20}{1 \times 10^{-10}} \right) \times (5 \times 10^{-4})$ $\text{Step 2: Find the energy density of the material.}$ $\text{Energy density} = \frac{1}{2} (\text{stress} \times \text{strain})$ $= \frac{1}{2} \times 20 \times 5 \times 10^6 \times 5 \times 10^{-4}$ $= 25 \times 10^3$ $= 25 \text{ kJ/m}^3$ $\text{Hence, option (3) is the correct answer.}$

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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
}