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Current Question (ID: 19466)

Question:

$\text{A body is projected vertically upwards from the surface of the earth with a velocity equal to one-third of escape velocity. The maximum height attained by the body will be:}$ $\text{(take the radius of the Earth } R = 6400 \text{ km and } g = 10 \text{ m/s}^2 )$

Options:

1. $800 \text{ km}$
2. $1600 \text{ km}$
3. $2133 \text{ km}$
4. $4800 \text{ km}$

Solution:

$\text{By applying the conservation of mechanical energy at point } A \text{ and } B \text{ we get;} \newline (KE_i + U_i)_A = (KE_f + U_f)_B$ $\Rightarrow \frac{1}{2} m \left( \frac{v_e}{3} \right)^2 - \frac{GMm}{R} = 0 - \frac{GMm}{(R+h)}$ $\Rightarrow \frac{1}{2} m \left( \frac{2GM}{9R} \right) = -\frac{GMm}{R} = -\frac{GMm}{R+h}$ $\Rightarrow \frac{1}{9R} - \frac{1}{R} = -\frac{1}{R+h}$ $\Rightarrow \frac{8}{9R} = \frac{1}{R+h}$ $\Rightarrow 8R + 8h = 9R$ $\Rightarrow h = \frac{R}{8} = \frac{6400}{8} = 800 \text{ km}$ $\text{Hence, option (1) is the correct answer.}$

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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
}