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Current Question (ID: 10402)

Question:

$\text{A uniform cylinder rod of length } L, \text{ cross-sectional area } A \text{ and Young's modulus } Y \text{ is acted upon by the forces, as shown in the figure. The elongation of the rod is:}$

Options:

1. $\frac{3FL}{5AY}$
2. $\frac{2FL}{5AY}$
3. $\frac{2FL}{8AY}$
4. $\frac{8FL}{3AY}$

Solution:

\text{To find the total elongation, we divide the rod into two sections:} \text{1. Left Section (length } 2L/3\text{):} T_1 = 3F \Delta L_1 = \frac{3F \times 2L/3}{AY} = \frac{2FL}{AY} \text{2. Right Section (length } L/3\text{):} T_2 = 2F \Delta L_2 = \frac{2F \times L/3}{AY} = \frac{2FL}{3AY} \text{Total elongation:} \Delta L_{\text{total}} = \frac{2FL}{AY} + \frac{2FL}{3AY} = \frac{8FL}{3AY}

Import JSON File

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
}