Import Question JSON

Current Question (ID: 19439)

Question:

$\text{The energy required to take a satellite to a height } h \text{ above the Earth's surface (radius of Earth } R = 6.4 \times 10^3 \text{ km) is } E_1 \text{ and kinetic energy required for the satellite to be in a circular orbit at this height is } E_2. \text{ The value of } h \text{ for which } E_1 \text{ and } E_2 \text{ are equal is:}$

Options:

1. $1.6 \times 10^3 \text{ km}$
2. $3.2 \times 10^3 \text{ km}$
3. $6.4 \times 10^3 \text{ km}$
4. $1.28 \times 10^4 \text{ km}$

Solution:

$\text{Hint: } v = \sqrt{\frac{GM}{(R+h)}}$ $\text{Step 1: Find the energy required to take a satellite to a height } 'h' \text{ above the Earth's surface.}$ $E_1 = U_f - U_i$ $\Rightarrow E_1 = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right)$ $\Rightarrow E_1 = \frac{GMmh}{R(R+h)}$ $\text{Step 2: Find the kinetic energy required for the satellite to be in a circular orbit at this height } 'h'.$ $F_g = F_c$ $\Rightarrow \frac{GMm}{(R+h)^2} = \frac{mv^2}{(R+h)}$ $\Rightarrow \frac{1}{2}mv^2 = \frac{GMm}{2(R+h)}$ $\Rightarrow E_2 = \frac{GMm}{2(R+h)}$ $\text{Step 3: Find the value of } h.$ $E_1 = E_2$ $\Rightarrow \frac{GMmh}{R(R+h)} = \frac{GMm}{2(R+h)}$ $\Rightarrow h = \frac{R}{2}$ $\Rightarrow h = \frac{6.4 \times 10^3}{2}$ $\Rightarrow h = 3.2 \times 10^3 \text{ km}$ $\text{Hence, option (2) is the correct answer.}$

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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
}