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Current Question (ID: 10367)

Question:

$\text{The Young's modulus of a wire is numerically equal to the stress at a point when:}$

Options:

1. $\text{The strain produced in the wire is equal to unity.}$
2. $\text{The length of the wire gets doubled.}$
3. $\text{The length increases by 100\%.}$
4. $\text{All of these.}$

Solution:

\text{Hint: } Y = \frac{\text{stress}}{\text{strain}} \text{Step 1: Use formula for Young's modulus} Y = \frac{\text{Stress}}{\text{Strain}} \text{Strain} = 1 = \frac{\Delta l}{l} \Delta l = l \text{Step 2: Check any 2 options} \text{When strain = 1, Young's modulus Y = stress (numerically equal)} \text{When length gets doubled: } \Delta l = l\text{, so strain} = \frac{\Delta l}{l} = \frac{l}{l} = 1 \text{When length increases by 100\%: } \Delta l = l\text{, so strain} = \frac{\Delta l}{l} = \frac{l}{l} = 1 \text{All three conditions (options 1, 2, and 3) represent the same physical situation where strain = 1}

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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
}