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Current Question (ID: 19621)

Question:

$\text{The pressure that has to be applied to the ends of a steel wire of length } 10 \text{ cm to keep its length constant when its temperature is raised by } 100^\circ \text{C is:}$ $\text{(Young's modulus of steel is } 2 \times 10^{11} \text{ Nm}^{-2} \text{ and coefficient of thermal expansion is } 1.1 \times 10^{-5} \text{ K}^{-1})$

Options:

1. $2.2 \times 10^9 \text{ Pa}$
2. $2.2 \times 10^7 \text{ Pa}$
3. $2.2 \times 10^6 \text{ Pa}$
4. $2.2 \times 10^8 \text{ Pa}$

Solution:

$\text{Hint: The thermal stress developed in wire } = Y \alpha \Delta T$ $\text{Step: Find the thermal stress developed in the wire.}$ $\text{Given:}$ $l = 10 \text{ cm}, \Delta T = 100^\circ \text{C}, Y = 2 \times 10^{11} \text{ N/m}^2, \alpha = 1.1 \times 10^{-5} \text{ K}^{-1}$ $\text{The thermal stress developed in the steel wire is given by:}$ $\sigma_{\text{thermal}} = Y \alpha \Delta T$ $\Rightarrow \sigma_{\text{thermal}} = 2 \times 10^{11} \times 1.1 \times 10^{-5} \times 100$ $\Rightarrow \sigma_{\text{thermal}} = 2.2 \times 10^8 \text{ N/m}^2$ $\text{Since the pressure developed in the wire is equal to the thermal stress developed.}$ $\text{Therefore, the pressure applied to the ends of the wire is } 2.2 \times 10^8 \text{ Pa.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}