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Current Question (ID: 19534)

Question:

$\text{In an experiment, brass and steel wires of length } 1 \text{ m each with areas of cross section } 1 \text{ mm}^2 \text{ are used.}$ $\text{The wires are connected in series and one end of the combined wire is connected to a rigid support and the other end is subjected to elongation.}$ $\text{The stress required to produce a net elongation of } 0.2 \text{ mm is,}$ $\text{[Given, the Young's Modulus for steel and brass are, respectively, } 120 \times 10^9 \text{ N/m}^2 \text{ and } 60 \times 10^9 \text{ N/m}^2]}$

Options:

1. $4.0 \times 10^6 \text{ N/m}^2$
2. $1.2 \times 10^6 \text{ N/m}^2$
3. $1.8 \times 10^6 \text{ N/m}^2$
4. $8 \times 10^6 \text{ N/m}^2$

Solution:

$\text{Hint: } k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$ $k_1 = \frac{Y_1 A_1}{l_1} = \frac{120 \times 10^9 \cdot A}{l} = 120 \times 10^9 \cdot \frac{A}{l}$ $k_2 = 60 \times 10^9 \cdot \frac{A}{l}$ $k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$ $F = k_{eq} \left( x \right)$ $\frac{F}{A} = \frac{k_{eq}}{A} \left( x \right) = 8 \times 10^6$ $\text{Solution gives this answer which is not matching with any option}$

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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
}