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Current Question (ID: 19106)

Question:

$\text{A particle is projected at an angle of } 30^\circ \text{ with the horizontal. It is observed that the particle reaches the same height at } 3 \text{ s and } 5 \text{ s after projection. The initial projection speed is:}$ $\left( \text{take } g = 10 \text{ m/s}^2 \right)$

Options:

1. $40 \text{ m/s}$
2. $50 \text{ m/s}$
3. $80 \text{ m/s}$
4. $60 \text{ m/s}$

Solution:

$\text{Hint: } y = ut + \frac{1}{2} at^2$ $\text{Step: Find the speed of the projection.}$ $\text{The height of the particle at } t = 3 \text{ s and } t = 5 \text{ s are the same as shown in the figure below;}$ $\text{Let } u \text{ be the speed of projection, and the vertical component of the speed is given by; } u_y = u \sin(30^\circ) = \frac{u}{2}.$ $\text{By using the kinematic equation } y = ut + \frac{1}{2} at^2.$ $\text{The height at } t = 3 \text{ s is given as;}$ $h_1 = \frac{3u}{2} - \frac{1}{2} 10 (3^2) = \frac{3u}{2} - 45 \quad \cdots (1)$ $\text{The height at } t = 5 \text{ s is given as;}$ $h_2 = \frac{5u}{2} - \frac{1}{2} 10 (5^2) = \frac{5u}{2} - 125 \quad \cdots (2)$ $\text{Since the heights are equal, from (1) and (2), we get;}$ $\Rightarrow h_1 = h_2$ $\Rightarrow \frac{3u}{2} - 45 = \frac{5u}{2} - 125$ $\Rightarrow \frac{5u}{2} - \frac{3u}{2} = 125 - 45$ $\Rightarrow u = 80 \text{ m/s}$ $\text{Hence, option (3) is the correct answer.}$

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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
}