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Current Question (ID: 19916)

Question:

$\text{The length of a seconds pendulum at a height } h = 2R \text{ from the earth's surface will be:}$ $\text{(Given } R = \text{The radius of the earth and acceleration due to gravity at the surface of the earth } g = \pi^2 \text{ m/s}^2)$

Options:

1. $\frac{2}{9} \text{ m}$
2. $\frac{4}{9} \text{ m}$
3. $\frac{8}{9} \text{ m}$
4. $\frac{1}{9} \text{ m}$

Solution:

$\text{Hint: } T = 2\pi \sqrt{\frac{l}{g}}$ $\text{Step: Find the length of the pendulum.}$ $\text{The time period of the pendulum is given by;}$ $T = 2\pi \sqrt{\frac{L}{g'}}$ $\text{At a height } h \text{ from the Earth's surface, the gravitational acceleration } g' \text{ can be expressed as:}$ $g' = \frac{g}{\left(1+\frac{h}{R}\right)^2}$ $\text{where } R \text{ is the radius of the Earth. Given } h = 2R:$ $g' = \frac{g}{\left(1+\frac{2R}{R}\right)^2} = \frac{g}{(1+2)^2} = \frac{g}{3^2} = \frac{g}{9}$ $g' = \frac{GM}{9R^2}$ $\Rightarrow g' = \frac{\pi^2}{9} \quad [g = \pi^2]$ $\text{The time period of the oscillation is given by;}$ $T = 2\pi \sqrt{\frac{L}{g'}}$ $\Rightarrow 2 = 2\pi \sqrt{\frac{L}{\pi^2} \times 9} \quad [T = 2 \text{ s}]$ $\Rightarrow 1 = \pi \sqrt{L} \times \frac{3}{\pi}$ $\Rightarrow L = \frac{1}{9} \text{ m}$ $\text{Hence, option (4) is the correct answer.}$

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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
}