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Current Question (ID: 19547)

Question:

$\text{Two blocks, one with a mass of } 2 \text{ kg and the other with a mass of } 1.14 \text{ kg, are suspended by steel and brass wires, respectively, as shown in the figure. Given Young's moduli for steel and brass as } 2 \times 10^{11} \text{ N/m}^2 \text{ and } 1 \times 10^{11} \text{ N/m}^2 \text{ respectively, what is the change in the length for the steel wire?}$

Options:

1. $3.2 \ \mu \text{m}$
2. $1.6 \ \mu \text{m}$
3. $0.8 \ \mu \text{m}$
4. $4.8 \ \mu \text{m}$

Solution:

$\text{Hint: } Y = \frac{F/A}{\Delta l/l}$ $\text{Step 1: Find the stress in steel wire.}$ $\text{From the free body diagram,}$ $T = 3.14g \ \text{N}$ $\text{Stress} = \frac{T}{A}$ $= \frac{3.14 \times 10}{3.14 \times (0.5 \times 10^{-2})^2}$ $\text{Step 2: Find the change in length for the steel wire.}$ $\Delta l = L \left( \frac{\text{Stress}}{Y_{\text{steel}}} \right)$ $= 1.6 \times \frac{3.14 \times 10}{3.14 (0.5 \times 10^{-2})^2 \times 2 \times 10^{11}}$ $= 3.2 \times 10^{-6} \ \text{m}$ $\text{Hence, option (1) is the correct answer.}$

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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
}