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Current Question (ID: 10346)

Question:

$\text{To break a wire, a force of } 10^6 \text{ N/m}^2 \text{ is required. If the density of the material is } 3 \times 10^3 \text{ kg/m}^3\text{, then the length of the wire which will break by its own weight will be:}$

Options:

1. $34 \text{ m}$
2. $30 \text{ m}$
3. $300 \text{ m}$
4. $3 \text{ m}$

Solution:

\text{Hint: Stress} = 10^6 \text{ N/m}^2 \text{Step 1: Identify the location of maximum stress.} \text{The maximum stress occurs at the topmost point of the wire where it supports the entire weight of the wire below.} \text{Stress} = \frac{\text{Weight}}{A} \text{Step 2: Calculate stress at the topmost point.} \sigma = \frac{mg}{A} \text{For a wire of length } l \text{ and cross-sectional area } A: \text{Mass} = \rho \cdot A \cdot l \text{Weight} = mg = \rho \cdot A \cdot l \cdot g \text{Therefore: } \sigma = \frac{\rho \cdot A \cdot l \cdot g}{A} = \rho \cdot l \cdot g \text{Step 3: Equate } \sigma \text{ to the breaking stress.} \frac{(\rho A l) g}{A} = 10^6 \rho \cdot l \cdot g = 10^6 l = \frac{10^6}{\rho \cdot g} = \frac{10^6}{3 \times 10^3 \times 10} = \frac{10^6}{3 \times 10^4} = \frac{10^2}{3} = 33.33 \text{ m} \approx 34 \text{ m}

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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
}