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Current Question (ID: 19017)

Question:

$\text{From a tower of height } H, \text{ a particle is thrown vertically upwards with a speed } u. \text{ The time taken by the particle, to hit the ground, is } n \text{ times that taken by it to reach the highest point of its path. The relation between } H, u \text{ and } n \text{ is:}$

Options:

1. $gH = (n - 2)^2 u^2$
2. $2gH = nu^2(n - 2)$
3. $gH = (n - 2)u^2$
4. $2gH = n^2u^2$

Solution:

$\text{Hint: } s = ut + \frac{1}{2}at^2$ $\text{Let time taken up to maximum height and when it hit ground is } t_2$ $\text{So } t_1 = \frac{u}{g}$ $\text{by using } S = ut + \frac{1}{2}at^2$ $-H = ut_2 - \frac{1}{2}gt_2^2$ $t_2 = \frac{u + \sqrt{u^2 + 2gH}}{g}$ $\text{Given } \frac{t_2}{t_1} = n$ $\frac{u + \sqrt{u^2 + 2gH}}{g} = n \frac{u}{g}$ $u^2 + 2gH = (n - 1)^2 u^2$ $\text{or } 2gH = nu^2(n - 2)$

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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
}