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Current Question (ID: 19025)

Question:

$\text{A ball is thrown upward with an initial velocity } v_0 \text{ from the surface of the earth.}$ $\text{The motion of the ball is affected by a drag force equal to } myv^2 \text{ (where } m \text{ is mass of the ball, } v \text{ is its instantaneous velocity and } y \text{ is a constant).}$ $\text{The time taken by the ball to rise to its zenith (maximum height) is:}$

Options:

1. $\frac{1}{\sqrt{yg}} \tan^{-1}\left(\sqrt{\frac{y}{g}} v_0\right)$
2. $\frac{1}{\sqrt{2yg}} \tan^{-1}\left(\sqrt{\frac{2y}{g}} v_0\right)$
3. $\frac{1}{\sqrt{yg}} \sin^{-1}\left(\sqrt{\frac{y}{g}} v_0\right)$
4. $\frac{1}{\sqrt{yg}} \ln\left(1 + \sqrt{\frac{y}{g}} v_0\right)$

Solution:

$\text{Hint: } a = \frac{dv}{dt}$ $a = -\left(g + pv^2\right) = \frac{dv}{dt}$ $\int_{V_0}^{V} \frac{dv}{g + pv^2} = -\int_{0}^{t} dt$ $\text{By integrating}$ $t = \frac{1}{\sqrt{gy}} \tan^{-1}\left[\sqrt{\frac{y}{g}} V_0\right]$

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
}