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Current Question (ID: 19542)

Question:

$\text{A uniform heavy rod of mass } 20 \text{ kg, cross-sectional area of } 0.4 \text{ m}^2 \text{ and length of } 20 \text{ m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is:}$ $\text{(Given: Young's modulus } Y = 2 \times 10^{11} \text{ N-m}^{-2} \text{ and } g = 10 \text{ ms}^{-2})$

Options:

1. $12 \times 10^{-9} \text{ m}$
2. $30 \times 10^{-9} \text{ m}$
3. $25 \times 10^{-9} \text{ m}$
4. $35 \times 10^{-9} \text{ m}$

Solution:

$\text{Hint: } Y = \frac{T}{A} \frac{dx}{dy}$ $\text{Step: Find the elongation in the rod due to its own weight.}$ $\text{The elongation in the rod due to its own weight is given by:}$ $\Delta L = \frac{mgL}{2AY}$ $= \frac{20 \times 10 \times 20}{0.4 \times 2 \times 2 \times 10^{11}}$ $= 25 \times 10^{-9} \text{ m}$

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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
}