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Current Question (ID: 19448)

Question:

$\text{Two planets have masses } M \text{ and } 16M \text{ and their radii are } a \text{ and } 2a, \text{ respectively.}$ $\text{The separation between the centres of the planets is } 10a. \text{ A body of mass } m \text{ is fired from the surface of the larger planet}$ $\text{towards the smaller planet along the line joining their centres.}$ $\text{For the body to be able to reach at the surface of smaller planet, the}$ $\text{minimum firing speed needed is:}$

Options:

1. $\sqrt{\frac{GM^2}{ma}}$
2. $\frac{3}{2} \sqrt{\frac{5GM}{a}}$
3. $4 \sqrt{\frac{GM}{a}}$
4. $4 \sqrt{\frac{GM}{a}}$

Solution:

$\text{Hint: Apply the conservation of energy.}$ $\frac{GM}{x^2} = \frac{G(16M)}{(10a-x)^2}$ $\frac{1}{x} = \frac{4}{(10a-x)^2}$ $x = \frac{2a}{\text{COME}}$ $\frac{-GMm}{8a} - \frac{G(16M)m}{2a} + KE = \frac{-GMm}{2a} - \frac{G(16M)m}{8a}$ $KE = GMm \left[ \frac{1}{8a} + \frac{16}{2a} - \frac{1}{2a} - \frac{16}{8a} \right]$ $KE = GMm \left[ \frac{1+64-4-16}{8a} \right]$ $\frac{1}{2} mv^2 = GMm \left[ \frac{45}{8a} \right]$ $v = \sqrt{\frac{90 \cdot GM}{8a}}$ $v = \frac{3}{2} \sqrt{\frac{5GM}{a}}$

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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
}