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Current Question (ID: 8862)

Question:

$\text{A body starting from rest moves with uniform acceleration on a horizontal surface. The body covers 3 consecutive equal distances from the beginning in time } t_1, t_2, \text{ and } t_3 \text{ seconds. The ratio of } t_1 : t_2 : t_3 \text{ is:}$

Options:

1. $1 : 2 : 3$
2. $1 : \sqrt{2} : \sqrt{3}$
3. $1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$
4. $\sqrt{3} : \sqrt{2} : 1$

Solution:

$\text{Hint: } s = ut + \frac{1}{2}at^2$ $\text{Step 1: Find the distance covered by the body in the time } t_1, t_2 \text{ and } t_3\text{.}$ $\text{As the body starts from rest then initial velocity of the body is zero i.e., } u = 0$ $\text{The distance covered by the body for the time interval } t_1\text{.}$ $s_1 = \frac{1}{2}at_1^2 \text{...(1)}$ $\text{The distance covered by the body during } t_2\text{.}$ $s_2 = \frac{1}{2}a(t_1 + t_2)^2 - \frac{1}{2}at_1^2$ $s_2 = \frac{1}{2}a(2t_1t_2 + t_2^2) \text{...(2)}$ $\text{The distance covered by the body during } t_3\text{.}$ $s_3 = \frac{1}{2}a(t_1 + t_2 + t_3)^2 - \frac{1}{2}a(t_1 + t_2)^2$ $s_3 = \frac{1}{2}a(2(t_1 + t_2)t_3 + t_3^2) \text{...(3)}$ $\text{Step 2: Find the ratio of } t_1 : t_2 : t_3\text{.}$ $\text{As the distance covered body during the time interval } t_1, t_2, t_3 \text{ are same i.e., } s_1 = s_2 = s_3 = s$ $t_1^2 = 2t_1t_2 + t_2^2 \text{...(4)}$ $t_1^2 = 2(t_1 + t_2)t_3 + t_3^2 \text{...(5)}$ $\text{By solving quadratic equations (4) and (5) we get;}$ $t_1 : t_2 : t_3 = 1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$ $\text{Hence, option (3) is the correct answer.}$

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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
}