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Current Question (ID: 19556)

Question:

$\text{A wire of length } L \text{ and radius } r \text{ is clamped rigidly at one end. When the other end of the wire is pulled by a force } F, \text{ its length increases by } 5 \text{ cm.}$ $\text{Another wire of the same material of length } 4L \text{ and radius } 4r \text{ is pulled by a force } 4F \text{ under the same conditions. The increase in length of this wire is:}$

Options:

1. $3 \text{ cm}$
2. $5 \text{ cm}$
3. $10 \text{ cm}$
4. $6 \text{ cm}$

Solution:

$\text{Hint: } Y = \frac{F \cdot L}{A \cdot \Delta L}$ $\text{Step: Find the increase in the length of the first wire.}$ $\Rightarrow \Delta L = \frac{F \cdot L}{A \cdot Y} = 5 \text{ cm}$ $\text{Similarly, for second wire.}$ $\Rightarrow \Delta L' = \frac{(4F)(A \cdot L)}{(16A)Y} = 5 \text{ cm}$ $\text{Therefore, the stretch in both the wires is same.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}