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Current Question (ID: 19546)

Question:

$\text{A uniform rod of mass } 10 \text{ kg and length } 6 \text{ m is hanged from the ceiling as shown in the figure.}$ $\text{Given the area of the cross-section of rod } 3 \text{ mm}^2 \text{ and Young's modulus is } 2 \times 10^{11} \text{ N/m}^2.$ $\text{The extension in the rod's length is:}$ $\text{(Take } g = 10 \text{ m/s}^2)$

Options:

1. $1 \text{ mm}$
2. $0.5 \text{ mm}$
3. $0.25 \text{ mm}$
4. $1.2 \text{ mm}$

Solution:

$\text{Hint: } Y = \frac{F/A}{\Delta L/L}$ $\text{Step: Find the extension in length of a rod.}$ $\text{The Young's modulus is given by: } Y = \frac{F/A}{\Delta L/L}$ $\text{The formula for the extension (} \Delta L \text{) in a rod due to its own weight is given by: } \Delta L = \left( \frac{mgL}{2AY} \right)$ $\text{This formula accounts for the fact that the weight is distributed uniformly along the rod, so the average force acting at the middle of the rod is considered.}$ $\text{Substitute the values:}$ $M = 10 \text{ kg}, g = 10 \text{ m/s}^2, L = 6 \text{ m}, A = 3 \times 10^{-6} \text{ m}^2, \text{ and } Y = 2 \times 10^{11} \text{ N/m}^2.$ $\text{Substituting these into the formula: } \Delta L = \frac{10 \times 10 \times 6}{2 \times 3 \times 10^{-6} \times 2 \times 10^{11}}$ $\text{Simplifying:}$ $\Delta L = \frac{600}{1.2 \times 10^6} = 0.0005 \text{ m} = 0.5 \text{ mm}$ $\text{Hence, option (2) is the correct answer.}$

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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
}