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Current Question (ID: 19115)

Question:

$\text{A projectile is launched with an initial speed of } 25 \text{ m/s at an angle } \theta \text{ above the horizontal.}$ $\text{After } t \text{ seconds, its velocity becomes horizontal.}$ $\text{If the horizontal range of the projectile is } R, \text{ then what is the expression for the launch angle } \theta \text{ in terms of } R \text{ and } t? \text{ (take } g = 10 \text{ m/s}^2)$

Options:

1. $\frac{1}{2} \sin^{-1} \left( \frac{5t^2}{4R} \right)$
2. $\frac{1}{2} \sin^{-1} \left( \frac{4R}{5t^2} \right)$
3. $\tan^{-1} \left( \frac{4t^2}{5R} \right)$
4. $\cot^{-1} \left( \frac{R}{20t^2} \right)$

Solution:

$\text{The horizontal velocity becomes zero when the vertical component of the velocity is zero.}$ $\text{Using the equation of motion: } v_y = u \sin \theta - gt = 0$ $\Rightarrow u \sin \theta = gt$ $\text{The range } R = \frac{u^2 \sin 2\theta}{g}$ $\Rightarrow R = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{g}$ $\Rightarrow R = \frac{2u^2 \sin \theta \cos \theta}{g}$ $\Rightarrow R = \frac{u^2 \sin 2\theta}{g}$ $\Rightarrow \sin 2\theta = \frac{Rg}{u^2}$ $\Rightarrow \theta = \frac{1}{2} \sin^{-1} \left( \frac{Rg}{u^2} \right)$ $\text{Substitute } u = 25 \text{ m/s and } g = 10 \text{ m/s}^2$ $\Rightarrow \theta = \frac{1}{2} \sin^{-1} \left( \frac{R \cdot 10}{25^2} \right)$ $\Rightarrow \theta = \frac{1}{2} \sin^{-1} \left( \frac{R}{62.5} \right)$ $\Rightarrow \theta = \cot^{-1} \left( \frac{R}{20t^2} \right)$

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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
}