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Current Question (ID: 19626)

Question:

$\text{A uniform cylindrical rod of length } L \text{ and radius } r, \text{ is made from a material whose Young's modulus of Elasticity equals } Y. \text{ When this rod is heated by temperature } T \text{ and simultaneously subjected to a net longitudinal compressional force } F, \text{ its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to:}$

Options:

1. $\frac{3F}{\pi r^2 Y T}$
2. $\frac{6F}{\pi r^2 Y T}$
3. $\frac{F}{\pi r^2 Y T}$
4. $\frac{9F}{\pi r^2 Y T}$

Solution:

$\text{Hint: } \Delta l = l \alpha \Delta T$ $\text{Step 1: Find the value of compressive and thermal strain.}$ $\text{As the length of the rod remains unchanged,}$ $\text{The strain caused by compressive forces is equal and opposite to the thermal strain.}$ $\text{Now, compressive strain is obtained by using the formula for Young's modulus we get;} \Rightarrow Y = \frac{F \times l}{A \times \Delta l}$ $\text{The compressive strain is given by;} \frac{\Delta l}{l} = \frac{F}{AY} = \frac{F}{\pi Y r^2} \quad \cdots (1)$ $\text{Also, thermal strain in the rod is obtained by using the formula for expansion in the rod is given by;} \Delta l = l \alpha \Delta T \Rightarrow \text{Thermal strain, } \frac{\Delta l}{l} = \alpha \Delta T \quad \cdots (2)$ $\text{Step 2: Find the coefficient of volume expansion, of the material.}$ $\text{From the equations } (1) \text{ and } (2), \text{ we get;} \frac{F}{\pi r^2 Y} = \alpha T \quad [\therefore \Delta T = T] \Rightarrow \alpha = \frac{F}{\pi r^2 Y T}$ $\text{Therefore, the coefficient of volumetric expansion of the rod is given by;} \Rightarrow \gamma = 3 \alpha = \frac{3F}{\pi r^2 Y T}$ $\text{Hence, option } (1) \text{ is the correct answer.}$

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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
}