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Current Question (ID: 8864)

Question:

$\text{A stone dropped from a building of height } h \text{ and reaches the earth after } t \text{ seconds. From the same building, if two stones are thrown (one upwards and other downwards) with the same velocity } u \text{ and they reach the earth surface after } t_1 \text{ and } t_2 \text{ seconds respectively, then:}$

Options:

1. $t = t_1 - t_2$
2. $t = \frac{t_1 + t_2}{2}$
3. $t = \sqrt{t_1 t_2}$
4. $t = t_1^2 t_2^2$

Solution:

\text{Hint: } h = ut + \frac{1}{2}gt^2 \text{Step: Find the relation between the time } t, t_1, t_2. \text{If a stone is dropped from a height } h \text{, then:} h = \frac{1}{2}gt^2 \quad \text{...(i)} \text{If a stone is thrown upward with velocity } u \text{, then:} h = -ut_1 + \frac{1}{2}gt_1^2 \quad \text{...(ii)} \text{If a stone is thrown downward with velocity } u \text{, then:} h = ut_2 + \frac{1}{2}gt_2^2 \quad \text{...(iii)} \text{From equations (i), (ii) and (iii) we get:} -ut_1 + \frac{1}{2}gt_1^2 = \frac{1}{2}gt^2 \quad \text{...(iv)} ut_2 + \frac{1}{2}gt_2^2 = \frac{1}{2}gt^2 \quad \text{...(v)} \text{Dividing equation (iv) by equation (v):} \frac{-ut_1 + \frac{1}{2}gt_1^2}{ut_2 + \frac{1}{2}gt_2^2} = \frac{\frac{1}{2}gt^2}{\frac{1}{2}gt^2} = 1 \text{This gives us: } \frac{-ut_1}{ut_2} = \frac{\frac{1}{2}g(t^2 - t_1^2)}{\frac{1}{2}g(t^2 - t_2^2)} \text{Simplifying: } -\frac{t_1}{t_2} = \frac{t^2 - t_1^2}{t^2 - t_2^2} \text{By solving this equation, we get: } t = \sqrt{t_1 t_2} \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
}