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Current Question (ID: 19526)

Question:

$\text{Two planets } A \text{ and } B \text{ having masses } m_1 \text{ and } m_2 \text{ move around the sun in circular orbits of } r_1 \text{ and } r_2 \text{ radii respectively.}$ $\text{If angular momentum of } A \text{ is } L \text{ and that of } B \text{ is } 3L, \text{ the ratio of time period } (T_A/T_B) \text{ is:}$

Options:

1. $\frac{1}{27} \left( \frac{m_2}{m_1} \right)^3$
2. $\left( \frac{r_1}{r_2} \right)^3$
3. $\left( \frac{r_1}{r_2} \right)^{\frac{3}{2}}$
4. $27 \left( \frac{m_1}{m_2} \right)^3$

Solution:

$\text{Given: } L_A = L, \ L_B = 3L$ $\text{Angular momentum } L = mvr = m \cdot \omega r^2$ $\text{For planet } A: L_A = m_1 \cdot \omega_1 r_1^2 = L$ $\text{For planet } B: L_B = m_2 \cdot \omega_2 r_2^2 = 3L$ $\text{Dividing the two equations: } \frac{m_1 \cdot \omega_1 r_1^2}{m_2 \cdot \omega_2 r_2^2} = \frac{1}{3}$ $\text{Using } \omega = \frac{2\pi}{T}, \text{ we get: } \frac{m_1 \cdot \frac{2\pi}{T_A} r_1^2}{m_2 \cdot \frac{2\pi}{T_B} r_2^2} = \frac{1}{3}$ $\Rightarrow \frac{m_1 r_1^2 T_B}{m_2 r_2^2 T_A} = \frac{1}{3}$ $\Rightarrow \frac{T_A}{T_B} = \frac{m_1 r_1^2}{3 m_2 r_2^2}$ $\text{Using Kepler's third law: } T^2 \propto r^3, \text{ we find: } \frac{T_A}{T_B} = \left( \frac{r_1}{r_2} \right)^{\frac{3}{2}}$

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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
}