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Current Question (ID: 8857)

Question:

$\text{A stone falls freely under gravity. It covers distances } h_1, h_2 \text{ and } h_3 \text{ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between } h_1, h_2 \text{ and } h_3 \text{ is:}$

Options:

1. $h_1 = \frac{h_2}{3} = \frac{h_3}{5}$
2. $h_2 = 3h_1 \text{ and } h_3 = 3h_2$
3. $h_1 = h_2 = h_3$
4. $h_1 = 2h_2 = 3h_3$

Solution:

$\text{Hint: } s = ut + \frac{1}{2}gt^2$ $\text{Step 1: Find the distance covered in first 5 seconds, } h_1 = ut + \frac{1}{2}gt^2$ $h_1 = 0 + \frac{1}{2}g(5)^2$ $h_1 = \frac{25g}{2}$ $\text{Distance covered in first 10 seconds} = 0 + \frac{1}{2}g(10)^2$ $\text{Distance covered between 5s and 10s} = \frac{100g}{2} - \frac{25g}{2} = \frac{75g}{2}$ $h_2 = \frac{75g}{2}$ $\text{Step 2: Distance covered in first 15 seconds} = 0 + \frac{1}{2}g(15)^2$ $\text{Distance covered between 10s and 15s} = \frac{225g}{2} - \frac{100g}{2} = \frac{125g}{2}$ $h_3 = \frac{125g}{2}$ $\text{Therefore, the relation between height is } h_1 = \frac{h_2}{3} = \frac{h_3}{5}$ $\text{Hence, option (1) is the correct answer.}$

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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
}