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Current Question (ID: 10403)

Question:

$\text{Overall changes in volume and radius of a uniform cylindrical steel wire are 0.2\% and 0.002\% respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is:}$ $(2.0 \times 10^{11} \text{ Nm}^{-2})$

Options:

1. $3.2 \times 10^{11} \text{ Nm}^{-2}$
2. $3.2 \times 10^{7} \text{ Nm}^{-2}$
3. $3.6 \times 10^{9} \text{ Nm}^{-2}$
4. $3.9 \times 10^{8} \text{ Nm}^{-2}$

Solution:

\text{To determine the longitudinal tensile stress, we first find the longitudinal strain } \left(\frac{\Delta L}{L}\right) \text{ using the given volume change } \left(\frac{\Delta V}{V} = 0.002\right) \text{ and radius change } \left(\frac{\Delta r}{r} = -0.00002 \text{ for a decrease under tension}\right). \text{The relationship is: } \frac{\Delta V}{V} = \frac{2\Delta r}{r} + \frac{\Delta L}{L} \frac{\Delta L}{L} = 0.002 - 2(-0.00002) = 0.002 + 0.00004 = 0.00204 \text{Using Young's Modulus } Y = 2.0 \times 10^{11} \text{ N/m}^2, \text{ the longitudinal tensile stress } \sigma \text{ is:} \sigma = Y \times \frac{\Delta L}{L} = (2.0 \times 10^{11}) \times (0.00204) = 4.08 \times 10^8 \text{ N/m}^2 \text{The closest option is } \mathbf{4.} \text{ } 3.9 \times 10^8 \text{ N/m}^2

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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
}