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Current Question (ID: 19551)

Question:
$\text{Two forces } F_1 \text{ and } F_2 \text{ are applied on two rods } P \text{ and } Q \text{ of the same materials such that elongation in rods are same.}$ $\text{If the ratio of their radii is } x : y \text{ and the ratio of length is } m : n, \text{ then the ratio of forces } F_1 : F_2 \text{ is:}$
Options:
  • 1. $\left( \frac{y}{x} \right)^2 \frac{n}{m}$
  • 2. $\left( \frac{x}{y} \right)^2 \frac{n}{m}$
  • 3. $\left( \frac{x}{y} \right)^2 \frac{m}{n}$
  • 4. $\left( \frac{y}{x} \right)^2 \frac{m}{n}$
Solution:
$\text{Hint: Young's modulus } Y = \frac{F \cdot L}{A \cdot \Delta L}$ $\Delta L_1 = \frac{F_1 L_1}{Y A_1}, \Delta L_2 = \frac{F_2 L_2}{Y A_2}$ $\frac{F_1}{F_2} = \frac{A_1}{A_2} \times \frac{L_2}{L_1} = \left( \frac{r_1}{r_2} \right)^2 \left( \frac{L_2}{L_1} \right) = \frac{x^2}{y^2} \cdot \frac{n}{m}$

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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
}