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Current Question (ID: 19241)

Question:

$\text{Blocks of masses } m, 2m, 4m \text{ and } 8m \text{ are arranged in a line on a frictionless floor.}$ $\text{Another block of mass } m, \text{ moving with speed } v \text{ along the same line (see figure) collides with mass } m \text{ in perfectly inelastic manner.}$ $\text{All the subsequent collisions are also perfectly inelastic.}$ $\text{By the time the last block of mass } 8m \text{ starts moving, the total energy loss is } p\% \text{ of the original energy.}$ $\text{Value of } 'p' \text{ is close to:}$

Options:

1. 37
2. 94
3. 87
4. 77

Solution:

$\text{Hint: Apply the conservation of momentum.}$ $\text{All collisions are perfectly inelastic, so after the final collision, all blocks are moving together.}$ $\text{So let the final velocity be } v', \text{ so on applying momentum conservation: } mv = 16m v' \Rightarrow v' = \frac{v}{16}$ $\text{Now initial energy } E_1 = \frac{1}{2} mv^2$ $\text{Final energy: } E_f = \frac{1}{2} \times 16 \times m \times \left( \frac{v}{16} \right)^2 \Rightarrow E_f = \frac{1}{2} m \frac{v^2}{16}$ $\text{Energy loss: } E_i - E_f$ $\Rightarrow \frac{1}{2} mv^2 - \frac{1}{2} m \frac{v^2}{16}$ $\Rightarrow \frac{1}{2} mv^2 \left[ 1 - \frac{1}{16} \right] \Rightarrow \frac{1}{2} mv^2 \left[ \frac{15}{16} \right]$ $\%p = \frac{\text{Energy loss}}{\text{Original energy}} \times 100$ $= \frac{\frac{1}{2} mv^2 \left[ \frac{15}{16} \right]}{\frac{1}{2} mv^2} \times 100 = 93.75\%$ $\Rightarrow \text{Value of } P \text{ is close to } 94.$

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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
}