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Current Question (ID: 10371)

Question:
$\text{The volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of } 7.0 \times 10^6 \text{ Pa is:}$ $\text{(Bulk modulus of copper is } 140 \times 10^9 \text{ Pa.)}$
Options:
  • 1. $3.1 \times 10^{-2} \text{ m}^3$
  • 2. $9.1 \times 10^{-3} \text{ cm}^3$
  • 3. $5.0 \times 10^{-2} \text{ cm}^3$
  • 4. $7.9 \times 10^{-2} \text{ cm}^3$
Solution:
$\text{Hint: Apply the concept of the bulk modulus.}$ $\text{Step 1: Calculate the volume contraction of a solid copper cube by using the equation of Bulk modulus.}$ $\text{Bulk modulus, } K = \frac{\text{Volumetric stress}}{\text{Volumetric strain}}$ $= \frac{P}{\Delta V/V} = \frac{PV}{\Delta V}$ $\Rightarrow \Delta V = \frac{PV}{K} = \frac{7 \times 10^6}{140 \times 10^9} \left[\frac{10}{100}\right]^3$ $= \frac{1}{2} \times \frac{10^6}{10^{13}}$ $= 5 \times 10^{-8} \text{ m}^3$ $= 5 \times 10^{-2} \text{ cm}^3$

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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
}