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Current Question (ID: 10365)

Question:

$\text{Two wires are made of the same material and have the same volume. The first wire has a cross-sectional area } A \text{ and the second wire has a cross-sectional area } 3A. \text{ If the length of the first wire is increased by } \Delta l \text{ on applying a force } F, \text{ how much force is needed to stretch the second wire by the same amount?}$

Options:

1. $9F$
2. $6F$
3. $4F$
4. $F$

Solution:

\text{As we know Young's modulus is } Y = \frac{Fl}{A\Delta l} \text{Since, } V = Al \text{ so } l = \frac{V}{A} \text{Therefore, } F = \frac{YA\Delta l}{l} = \frac{YA^2\Delta l}{V} \text{For the same volume and the same amount of extension:} F \propto A^2 \frac{F_1}{F_2} = \left(\frac{A_1}{A_2}\right)^2 \frac{F_1}{F_2} = \left(\frac{A}{3A}\right)^2 = \frac{1}{9} F_2 = 9F

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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
}