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Current Question (ID: 8872)

Question:

$\text{A particle is projected upwards. The times corresponding to height } h \text{ while ascending and while descending are } t_1 \text{ and } t_2 \text{ respectively. The velocity of projection will be:}$

Options:

1. $gt_1$
2. $gt_2$
3. $g(t_1 + t_2)$
4. $\frac{g(t_1 + t_2)}{2}$

Solution:

$\text{Hint: } h = ut - \frac{1}{2}gt^2$ $\text{Step: Find the velocity of the projection.}$ $h = ut - \frac{1}{2}gt^2$ $\Rightarrow gt^2 - 2ut + 2h = 0$ $\text{The solution of the quadratic equation is given by:}$ $\Rightarrow t = \frac{2u \pm \sqrt{4u^2 - 8gh}}{2g} = \frac{u \pm \sqrt{u^2 - 2gh}}{g}$ $t_1 = \frac{u + \sqrt{u^2 - 2gh}}{g}$ $t_2 = \frac{u - \sqrt{u^2 - 2gh}}{g}$ $\Rightarrow \frac{2u}{g} = t_1 + t_2 \Rightarrow u = \frac{g(t_1 + t_2)}{2} \text{ m/s}$ $\text{Therefore, the velocity of the projection is } \frac{g(t_1 + t_2)}{2}\text{.}$ $\text{Hence, option (4) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}