Import Question JSON

Current Question (ID: 19593)

Question:

$\text{A ball with mass } m, \text{ radius } r, \text{ and density } \rho \text{ is dropped into a liquid with density } \rho_0. \text{ After a period of motion, the speed of the ball becomes constant, equal to } v_0. \text{ The coefficient of viscosity of the liquid is:}$

Options:

1. $\frac{mg}{6 \pi r v_0} \left( 1 - \frac{\rho_0}{\rho} \right)$
2. $\frac{mg}{6 \pi r v_0} \left( 1 + \frac{\rho_0}{\rho} \right)$
3. $\frac{mg}{3 \pi r v_0} \left( 1 + \frac{\rho_0}{\rho} \right)$
4. $\frac{mg}{3 \pi r v_0} \left( 1 - \frac{\rho_0}{\rho} \right)$

Solution:

$\text{Hint: viscous force = effective weight of the ball}$ $\text{Step: Calculate the coefficient of viscosity of the liquid.}$ $\text{When the speed of the ball becomes constant}$ $v = v_0 \text{ (terminal velocity) then, } a = 0$ $\text{The forces acting on the ball are given by:}$ $mg \text{ (vertically downward), } 6 \pi \eta r v_0, \rho_0 V g \text{ (vertically upward)}$ $\text{By writing the equation of motion, we get:}$ $mg - 6 \pi \eta r v_0 - \rho_0 V g = 0 \quad [\text{where, } V = \text{volume of the ball} = \frac{m}{\rho}]$ $6 \pi \eta r v_0 = mg - \rho_0 \left( \frac{m}{\rho} \right) g$ $\Rightarrow \eta = \frac{mg}{6 \pi r v_0} \left( 1 - \frac{\rho_0}{\rho} \right)$ $\text{Hence, option (1) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}