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Current Question (ID: 19587)

Question:

$\text{A small spherical ball of radius } 0.1 \text{ mm and density } 10^4 \text{ kg-m}^{-3} \text{ falls freely under gravity through a distance of } h \text{ before entering a tank of water. If after entering the water, the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water, then the value of } h \text{ will be:}$ $\text{(given } g = 10 \text{ m/s}^2, \text{ the viscosity of water } = 1.0 \times 10^{-5} \text{ N-sm}^{-2})$

Options:

1. $15 \text{ m}$
2. $25 \text{ m}$
3. $20 \text{ m}$
4. $10 \text{ m}$

Solution:

$\text{Hint: } H = \frac{v^2}{2g}$ $\text{Step 1: Find the terminal velocity of the ball.}$ $\text{The terminal velocity of the ball is given by:}$ $v_c = \frac{2r^2(\rho - \rho_0)g}{9\eta}$ $\text{Substitute the values we get:}$ $v_c = \frac{2 \times (10^{-4})^2 \times (10^4 - 10^3) \times 10}{9 \times 1 \times 10^{-5}} = 20 \text{ m/s}$ $\text{Step 2: Find the height } h$ $\text{Since the ball velocity didn't change, Speed after falling through height } h \text{ should be equal to terminal velocity.}$ $h = \frac{v^2}{2g}$ $\text{Substitute the values we get:}$ $h = \frac{20^2}{2 \times 10} = 20 \text{ m}$ $\text{Hence, option (3) is the correct answer.}$

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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
}