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Current Question (ID: 19530)

Question:

$\text{A solid sphere of radius } r \text{ made of a soft material of bulk modulus } K$ $\text{is surrounded by a liquid in a cylindrical container. A massless piston}$ $\text{of area } a \text{ floats on the surface of the liquid, covering the entire cross-section}$ $\text{of the cylindrical container. When a mass } m \text{ is placed on the}$ $\text{surface of the piston to compress the liquid, the fractional decrement}$ $\text{in the radius of the sphere, } \left( \frac{dr}{r} \right) \text{ is:}$

Options:

1. $\frac{Ka}{mg}$
2. $\frac{Ka}{3mg}$
3. $\frac{mg}{3Ka}$
4. $\frac{mg}{Ka}$

Solution:

$\text{Hint: } K = -\frac{dP}{(dV/V)}$ $\text{Step 1: Find the pressure acting on the sphere.}$ $P = \frac{\text{Force}}{\text{Area}} \Rightarrow P = \frac{mg}{a}$ $\text{Step 2: Find fractional decrease in volume.}$ $\text{As we know, } K = -\frac{dP}{(dV/V)}$ $\Rightarrow \frac{dV}{V} = -\frac{mg}{Ka}$ $\text{Step 3: Find fractional decrease in radius.}$ $\frac{dV}{V} = 3 \left( \frac{dr}{r} \right)$ $\Rightarrow -\frac{mg}{Ka} = 3 \left( \frac{dr}{r} \right)$ $\Rightarrow \frac{dr}{r} = -\frac{mg}{3Ka}$ $\text{Therefore, the fractional decrement in the radius of the sphere is}$ $\frac{mg}{3Ka}. \text{ Hence, option (3) is the correct answer.}$

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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
}