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Current Question (ID: 19532)

Question:

$\text{A boy's catapult is made of rubber cord which is } 42 \text{ cm long, with } 6 \text{ mm diameter of a cross-section and of negligible mass. The boy keeps a stone weighing } 0.02 \text{ kg on it and stretches the cord by } 20 \text{ cm by applying a constant force. When released, the stone flies off with a velocity of } 20 \text{ ms}^{-1}. \text{ Neglect the change in the area of cross-section of the cord while stretched. The Young's modulus of rubber is closest to:}$

Options:

1. $10^3 \text{ Nm}^{-2}$
2. $10^4 \text{ Nm}^{-2}$
3. $10^6 \text{ Nm}^{-2}$
4. $10^8 \text{ Nm}^{-2}$

Solution:

$\text{Hint: The energy stored in the cord is equal to the kinetic energy of the stone.}$ $\text{Step: Find Young's modulus of the rubber.}$ $\text{The potential energy stored by a stretched rubber cord is as follows:}$ $\frac{1}{2} Y \left( \frac{\Delta L}{L} \right)^2 AL = \frac{1}{2} Y \left( \frac{\Delta L}{L} \right)^2 \pi \left( \frac{D}{2} \right)^2 \ldots (1)$ $\text{The stone receives kinetic energy from the potential energy stored in the stretched cord.}$ $\text{The kinetic energy of stone is given by:}$ $\Rightarrow K.E = \frac{1}{2} mv^2 \ldots (2)$ $\text{Because energy is always conserved, solve equations (1) and (2) we get;}$ $\Rightarrow \frac{1}{2} Y \left( \frac{\Delta L}{L} \right)^2 \pi \left( \frac{D}{2} \right)^2 = \frac{1}{2} mv^2$ $\text{Substitute the given values we get;}$ $\frac{1}{2} Y \frac{0.2 \times 0.2 \times \pi \times 0.006 \times 0.006}{0.42 \times 4} = \frac{1}{2} \times 0.02 \times 20 \times 20$ $\Rightarrow Y \approx 10^6 \text{ N/m}^2$ $\text{Hence, option (3) is the correct answer.}$

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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
}