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Current Question (ID: 19465)

Question:

$\text{The percentage decrease in the weight of a rocket, when it is taken to a height of } 32 \text{ km above the surface of the Earth will be:}$ $\text{(radius of the Earth } R = 6400 \text{ km)}$

Options:

1. $1\%$
2. $3\%$
3. $4\%$
4. $0.5\%$

Solution:

$\text{Hint: } g = \frac{GM}{R^2}$ $\text{Step: Find the percentage decrease in the weight of the rocket.}$ $\text{The weight of the rocket is given by:}$ $\Rightarrow W_1 = mg$ $\text{The weight of the rocket at height } h \text{ given by:}$ $\Rightarrow W_2 = mg' = mg \left(1 - \frac{2h}{R}\right)$ $\text{The percentage decrease in weight is given by:}$ $\text{Percentage decrease in weight} = \frac{W_2 - W_1}{W_1} \times 100\%$ $\Rightarrow \text{Percentage decrease in weight} = \frac{mg \left(1 - \frac{2h}{R}\right) - mg}{mg} \times 100\%$ $\Rightarrow \text{Percentage decrease in weight} = -\frac{2h}{R}$ $\Rightarrow \text{Percentage decrease in weight} = -\frac{2 \times 32}{6400}$ $\Rightarrow \text{Percentage decrease in weight} = -1\%$ $\text{Therefore, the percentage decrease in the weight of the rocket is } 1\%.$ $\text{Hence, option (1) is the correct answer.}$

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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
}