Import Question JSON

Current Question (ID: 10364)

Question:

$\text{If the ratio of lengths, radii, and Young's modulus of steel and brass wires in the figure are } a, b \text{ and } c \text{ respectively, then the corresponding ratio of increase in their lengths will be:}$

Options:

1. $\frac{2a^2c}{b}$
2. $\frac{3a}{2b^2c}$
3. $\frac{2ac}{b^2}$
4. $\frac{3c}{2ab^2}$

Solution:

\text{Given:} \text{Force on steel wire } F_s = 3Mg \text{ (supports } M \text{ and } 2M\text{)} \text{Force on brass wire } F_b = 2Mg \text{ (supports } 2M\text{)} \text{Ratio of lengths: } \frac{L_s}{L_b} = a \text{Ratio of radii: } \frac{r_s}{r_b} = b \text{Ratio of Young's moduli: } \frac{Y_s}{Y_b} = c \text{The formula for elongation is } \Delta L = \frac{FL}{\pi r^2 Y} \text{For steel: } \Delta L_s = \frac{3MgL_s}{\pi r_s^2 Y_s} \text{For brass: } \Delta L_b = \frac{2MgL_b}{\pi r_b^2 Y_b} \text{The ratio } \frac{\Delta L_s}{\Delta L_b} \text{ is:} \frac{\Delta L_s}{\Delta L_b} = \frac{3MgL_s}{\pi r_s^2 Y_s} \times \frac{\pi r_b^2 Y_b}{2MgL_b} = \frac{3}{2} \times \frac{L_s}{L_b} \times \frac{r_b^2}{r_s^2} \times \frac{Y_b}{Y_s} \text{Substituting the given ratios:} \frac{\Delta L_s}{\Delta L_b} = \frac{3}{2} \times a \times \frac{1}{b^2} \times \frac{1}{c} = \frac{3a}{2b^2c}

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
}