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Current Question (ID: 19533)

Question:

$\text{Young's moduli of two wires } A \text{ and } B \text{ are in the ratio } 10 : 4.$ $\text{Wire } A \text{ is } 2 \text{ m long and has radius } R.$ $\text{Wire } B \text{ is } 1.6 \text{ m long and has radius } 2 \text{ mm}.$ $\text{If the two wires stretch by the same length for a given load,}$ $\text{then the value of } R \text{ is close to:}$

Options:

1. $\sqrt{2} \text{ mm}$
2. $\frac{1}{\sqrt{2}} \text{ mm}$
3. $2\sqrt{2} \text{ mm}$
4. $2 \text{ mm}$

Solution:

$\text{Hint: } Y = \text{Stress / Strain}$ $\text{Step 1: Find the deformation in the wire.}$ $\frac{F}{A} = Y \left( \frac{\Delta l}{l} \right)$ $\Rightarrow \Delta l = \frac{Fl}{AY}$ $\text{Step 2: Equate deformation for wires } A \text{ and } B \text{ and find the value of } R.$ $\Delta l_1 = \Delta l_2$ $\Rightarrow \frac{l_1}{A_1Y_1} = \frac{l_2}{A_2Y_2}$ $\Rightarrow A_1 = \left( \frac{l_1}{l_2} \right) \times \left( \frac{Y_2}{Y_1} \right) \times A_2$ $\Rightarrow \pi R^2 = \frac{5}{4} \times \frac{4}{10} \times \pi (2)^2$ $\Rightarrow R = \sqrt{2} \text{ mm}$ $\text{Hence, option (1) is the correct answer.}$

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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
}