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Current Question (ID: 10383)

Question:
For an ideal liquid: (a) The bulk modulus is infinite. (b) The bulk modulus is zero. (c) The shear modulus is infinite. (d) The shear modulus is zero. Choose the correct option from the options given below:
Options:
  • 1. (a) and (d) only
  • 2. (b) and (d) only
  • 3. (b) and (c) only
  • 4. (c) and (d) only
Solution:
An ideal fluid is incompressible and provides zero resistance to shear forces. **Step 1: Find the Bulk modulus of the liquid.** As an ideal liquid is not compressible, the change in volume, $\Delta V = 0$. The Bulk modulus ($B$) of the liquid is given by: $B = \frac{\text{Stress}}{\text{Volume strain}} = \frac{F/A}{\Delta V/V}$ Since $\Delta V = 0$, the denominator $\Delta V/V = 0$. Therefore, the Bulk modulus $B = \frac{F/A}{0} = \infty$. Alternatively, the compressibility ($K$) is the reciprocal of the Bulk modulus: $K = \frac{1}{B}$. Since $B = \infty$, $K = \frac{1}{\infty} = 0$. So, statement (a) is correct: The bulk modulus is infinite. **Step 2: Find the shear modulus of the liquid.** An ideal fluid provides zero resistance to shear forces. This means there is no tangential (viscous) force in the case of an ideal fluid. The shear modulus ($G$) is given by: $G = \frac{\text{Shear Stress}}{\text{Shear Strain}}$ Since the shear stress (tangential force per unit area) is zero for an ideal fluid, the shear modulus $G = \frac{0}{\text{Shear Strain}} = 0$. So, statement (d) is correct: The shear modulus is zero. Therefore, the correct statements are (a) and (d).

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{
  "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
}