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Current Question (ID: 9894)

Question:

$\text{A body is thrown vertically up with a certain initial velocity. The potential and the kinetic energy of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of the potential and the kinetic energies of the body when it crosses at the same point will be:}$

Options:

1. $1:1$
2. $1:4$
3. $1:7$
4. $1:8$

Solution:

\text{When the body is thrown with velocity } v\text{:} TE = KE_{\text{initial}} = \frac{1}{2}m v^2 = K \text{At the point, when KE and PE are equal:} KE + PE = TE \Rightarrow KE = PE = \frac{K}{2} \text{When the body is thrown with velocity } 2v\text{:} TE = KE_{\text{initial}} = \frac{1}{2}m (2v)^2 = 4 \times \frac{1}{2}m v^2 = 4K \text{PE at the same point will be same } = \frac{K}{2} \text{KE at that point } = 4K - \frac{K}{2} = \frac{7K}{2} \text{Ratio of PE to KE } = 1:7

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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
}