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Current Question (ID: 10369)

Question:

\text{Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded, as shown in the figure.} \text{The unloaded length of the steel wire is 1.5 m and that of the brass wire is 1.0 m.} \text{The elongation of the steel wire will be:} \text{(Given that Young's modulus of the steel, } Y_S = 2 \times 10^{11} \text{ Pa} \text{and Young's modulus of brass, } Y_B = 1 \times 10^{11} \text{ Pa)}

Options:

1. $1.5 \times 10^{-4} \text{ m}$
2. $0.5 \times 10^{-4} \text{ m}$
3. $3.5 \times 10^{-4} \text{ m}$
4. $2.5 \times 10^{-4} \text{ m}$

Solution:

\text{Hint: The elongation in the wire depends on the length and Young's modulus of the wire.} \text{Step 1: Write the formula for the extension in the wire.} \text{Use the formula: } \Delta L = \frac{FL}{AY} \text{where:} \Delta L = \text{elongation in the steel wire} A = \text{Area of the wire} = \pi r^2 A = \pi \left(\frac{0.25}{2}\right)^2 = \pi \times (0.125)^2 \text{ cm}^2 \text{Step 2: Find the elongation in the steel wire.} Y_S = 2 \times 10^{11} \text{ Pa} \text{For the steel wire:} \text{Force on steel wire} = (4.0 + 6.0) \times 9.8 = 98 \text{ N} \text{Length of steel wire} = 1.5 \text{ m} \text{Area} = \pi \times (0.125 \times 10^{-2})^2 A = \pi \times 1.5625 \times 10^{-6} \text{ m}^2 \text{Elongation in steel wire:} \Delta L = \frac{98 \times 1.5}{\pi \times 1.5625 \times 10^{-6} \times 2 \times 10^{11}} \Delta L = \frac{147}{\pi \times 1.5625 \times 2 \times 10^{5}} \Delta L = \frac{147}{\pi \times 3.125 \times 10^{5}} \Delta L = \frac{147}{9.817 \times 10^{5}} = 1.5 \times 10^{-4} \text{ m}

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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
}