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Current Question (ID: 10413)

Question:

$\text{The density of metal at normal pressure is } \rho. \text{ Its density when it is subjected to an excess pressure } P \text{ is } \rho'. \text{ If } B \text{ is the bulk modulus of the metal, the ratio } \frac{\rho'}{\rho} \text{ is:}$

Options:

1. $\frac{1}{1-\frac{P}{B}}$
2. $1 + \frac{P}{B}$
3. $\frac{1}{1-\frac{B}{P}}$
4. $2 + \frac{P}{B}$

Solution:

\text{The density of a metal at normal pressure is } \rho\text{. When subjected to an excess pressure } P\text{, its density becomes } \rho'\text{. Given that } B \text{ is the bulk modulus of the metal, we need to find the ratio } \frac{\rho'}{\rho}\text{.} \text{The bulk modulus } B \text{ is defined as: } B = -\frac{\text{Stress}}{\text{Volumetric Strain}} = -\frac{P}{\Delta V/V} \text{Since density } \rho = \frac{m}{V} \text{ (mass/volume), and mass } m \text{ remains constant, we have } V = \frac{m}{\rho}\text{.} \text{The change in volume } \Delta V = V' - V = \frac{m}{\rho'} - \frac{m}{\rho} = m\left(\frac{1}{\rho'} - \frac{1}{\rho}\right)\text{.} \text{The volumetric strain } \frac{\Delta V}{V} = \frac{m\left(\frac{1}{\rho'} - \frac{1}{\rho}\right)}{m/\rho} = \frac{\rho - \rho'}{\rho'}\text{.} \text{Substitute this into the bulk modulus equation: } B = -\frac{P}{\frac{\rho - \rho'}{\rho'}} = \frac{P\rho'}{\rho - \rho'} \text{Rearrange to solve for } \frac{\rho'}{\rho}\text{: } \rho - \rho' = \frac{P\rho'}{B} \rho = \rho' + \frac{P\rho'}{B} = \rho'\left(1 + \frac{P}{B}\right) \frac{\rho'}{\rho} = \frac{1}{1 + \frac{P}{B}}

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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
}