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Current Question (ID: 10489)

Question:

$\text{Two small spherical metal balls, having equal masses, are made from materials of densities } \rho_1 \text{ and } \rho_2 \text{ such that } \rho_1 = 8\rho_2 \text{ and having radii of 1 mm and 2 mm, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals } \eta \text{ and whose density is } 0.1\rho_2. \text{ The ratio of their terminal velocities would be:}$

Options:

1. $\frac{79}{72}$
2. $\frac{19}{36}$
3. $\frac{39}{72}$
4. $\frac{79}{36}$

Solution:

$\text{The terminal velocity formula for a sphere in a viscous medium is:}$ $v_T = \frac{2r^2g}{9\eta}[\sigma - \rho]$ $\text{For sphere with density } \rho_1:$ $v_{T1} = \frac{2r_1^2g}{9\eta}[8\rho_2 - 0.1\rho_2] \quad \ldots(1)$ $\text{For other sphere:}$ $v_{T2} = \frac{2(2r_1)^2g}{9\eta}[\rho_2 - 0.1\rho_2] \quad \ldots(2)$ $\text{From (1) and (2):}$ $\frac{v_{T1}}{v_{T2}} = \frac{1}{4} \times \frac{7.9}{0.9} = \frac{79}{36}$

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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
}