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Current Question (ID: 19092)

Question:

$\text{Two projectiles thrown at } 30^\circ \text{ and } 45^\circ \text{ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is:}$

Options:

1. $1 : \sqrt{2}$
2. $2 : 1$
3. $\sqrt{2} : 1$
4. $1 : 2$

Solution:

$\text{Hint: } T = \frac{2v_0 \sin \theta}{g}$ $\text{Step: Find the ratio of the initial velocities.}$ $\text{For the projectile thrown at } 30^\circ, \text{ the time accent is given by:}$ $t_1 = \frac{v_1 \sin(30^\circ)}{\frac{g}{2}} = \frac{v_1 \cdot \frac{1}{2}}{g} = \frac{v_1}{2g} \ldots (1)$ $\text{For the projectile thrown at } 45^\circ, \text{ the time of accent is given by:}$ $t_2 = \frac{v_2 \sin(45^\circ)}{\frac{g}{\sqrt{2}}} = \frac{v_2 \cdot \frac{\sqrt{2}}{2}}{g} = \frac{v_2 \sqrt{2}}{2g} \ldots (2)$ $\text{Since both projectiles reach a maximum height at the same time i.e., } t_1 = t_2$ $\Rightarrow \frac{v_1}{2g} = \frac{v_2 \sqrt{2}}{2g}$ $\Rightarrow \frac{v_1}{v_2} = \frac{\sqrt{2}}{1}$ $\text{Therefore, the ratio of the initial velocities } v_1 \text{ and } v_2 \text{ is } \sqrt{2} : 1.$ $\text{Hence, option (3) is the correct answer.}$

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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
}