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Current Question (ID: 19084)

Question:
$\text{A helicopter rises from rest on the ground vertically upwards with a constant acceleration } g. \text{ A food packet is dropped from the helicopter when it is a height } h. \text{ The time taken by the packet to reach the ground is close to } [g \text{ is the acceleration due to gravity}]:$
Options:
  • 1. $t = \sqrt{\frac{2h}{3g}}$
  • 2. $t = 1.8 \sqrt{\frac{h}{g}}$
  • 3. $t = 3.4 \left( \sqrt{\frac{h}{g}} \right)$
  • 4. $t = \frac{2}{3} \sqrt{\left( \frac{h}{g} \right)}$
Solution:
$\text{Hint: } v^2 = u^2 + 2as$ $V^2 = u^2 + 2as$ $V^2 = 2gh + 2gh$ $V = \sqrt{4gh}$ $\Rightarrow \sqrt{4gh} = \sqrt{2gh} + gt$ $\Rightarrow t = \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} \Rightarrow 3.4 \sqrt{\frac{h}{g}}$

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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
}