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Current Question (ID: 19299)

Question:

$\text{A body of mass } M \text{ at rest explodes into three pieces, in the ratio of masses } 1 : 1 : 2. \text{ Two smaller pieces fly off perpendicular to each other with velocities of } 30 \text{ ms}^{-1} \text{ and } 40 \text{ ms}^{-1} \text{ respectively. The velocity of the third piece will be:}$

Options:

1. $15 \text{ ms}^{-1}$
2. $25 \text{ ms}^{-1}$
3. $35 \text{ ms}^{-1}$
4. $50 \text{ ms}^{-1}$

Solution:

$\text{Using conservation of momentum:}$ $\text{Let the masses of the pieces be } m, m, \text{ and } 2m.$ $\text{Initial momentum } = 0.$ $\text{Final momentum: } m \times 30 \text{ ms}^{-1} \text{ in one direction and } m \times 40 \text{ ms}^{-1} \text{ perpendicular to it.}$ $\text{Resultant momentum of these two pieces: } \sqrt{(m \times 30)^2 + (m \times 40)^2} = m \times 50.$ $\text{Momentum of third piece } = 2m \times v = m \times 50.$ $v = 25 \text{ ms}^{-1}.$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}