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Current Question (ID: 19445)

Question:

$\text{The value of the acceleration due to gravity is } g_1 \text{ at a height } h = \frac{R}{2} \ (R = \text{radius of the earth}) \text{ from the surface of the earth. It is again equal to } g_1 \text{ at a depth } d \text{ below the surface of the earth. The ratio } \frac{d}{R} \text{ equals:}$

Options:

1. $\frac{7}{9}$
2. $\frac{4}{9}$
3. $\frac{1}{3}$
4. $\frac{5}{9}$

Solution:

$g_1 = \frac{GM}{\left(R + \frac{R}{2}\right)^2} \quad \cdots (1)$ $g_2 = \frac{GM(R-d)}{R^3} \quad \cdots (2)$ $g_1 = g_2$ $\frac{GM}{\left(\frac{3R}{2}\right)^2} = \frac{GM(R-d)}{R^3}$ $\Rightarrow \frac{4}{9} = \frac{(R-d)}{R}$ $4R = 9R - 9d$ $5R = 9d \Rightarrow \frac{d}{R} = \frac{5}{9}$

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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
}