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Current Question (ID: 8867)

Question:

$\text{Four marbles are dropped from the top of a tower one after the other at a one-second interval. The first one reaches the ground after 4 seconds. When the first one reaches the ground the distances between the first and second, the second and third, and the third and fourth will be, respectively:}$

Options:

1. $35 \text{ m, } 25 \text{ m, and } 15 \text{ m}$
2. $30 \text{ m, } 20 \text{ m, and } 10 \text{ m}$
3. $20 \text{ m, } 10 \text{ m, and } 5 \text{ m}$
4. $40 \text{ m, } 30 \text{ m, and } 20 \text{ m}$

Solution:

\text{Hint: } h = ut + \frac{1}{2}gt^2 \text{Step: Find the distances between the first and second, the second and third, and the third and fourth seconds.} \text{For the first marble: } h_1 = \frac{1}{2}g \times 16 = 8g \text{For the second marble: } h_2 = \frac{1}{2}g \times 9 = 4.5g \text{For the third marble: } h_3 = \frac{1}{2}g \times 4 = 2g \text{For the fourth marble: } h_4 = \frac{1}{2}g \times 1 = 0.5g \text{Calculating the distances:} h_1 - h_2 = 8g - 4.5g = 3.5g = 35 \text{ m} h_2 - h_3 = 4.5g - 2g = 2.5g = 25 \text{ m} h_3 - h_4 = 2g - 0.5g = 1.5g = 15 \text{ m} \text{Therefore, the distances between the first and second, the second and third,} \text{and the third and fourth seconds are 35 m, 25 m, and 15 m, respectively.} \text{Hence, option (1) is the correct answer.}

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}