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Current Question (ID: 19164)

Question:

$\text{A ball of mass } 2 \text{ kg is dropped from a height of } 9.8 \text{ m and rebounds to a height of } 4.9 \text{ m.}$ $\text{If it remains in contact with the ground for } 0.2 \text{ s, the average force on the ball exerted by the ground is:}$ $\text{(Take } g = 9.8 \text{ m/s}^2)$

Options:

1. $98(\sqrt{2} + 1) \text{ N}$
2. $49(\sqrt{2} + 1) \text{ N}$
3. $98(\sqrt{2} - 1) \text{ N}$
4. $49(\sqrt{2} - 1) \text{ N}$

Solution:

$\text{Hint: Impulse = change in momentum}$ $\text{Step 1: Find the velocity of the ball when it hits the ground.}$ $\text{When ball hits the ground, velocity of ball,}$ $v_{\text{initial}} = \sqrt{2g \cdot 9.8} \text{ m/s}$ $\text{Step 2: Find the velocity of the ball after hitting the ground.}$ $\text{The velocity of the ball after hitting the ground,}$ $v_{\text{final}} = \sqrt{2g \cdot 4.9}$ $\text{Step 3: Find the average force on the ball exerted by the ground.}$ $F \Delta t = m (v_{\text{final}} - v_{\text{initial}})$ $F (0.2) = 2 \times \left( \sqrt{2 \times 9.8 \times 9.8} - (-\sqrt{2 \times 9.8 \times 4.9}) \right)$ $F = 98(\sqrt{2} + 1) \text{ N}$

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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
}