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Current Question (ID: 19543)

Question:

$\text{A steel wire of length } 3.2 \text{ m } (Y_S = 2.0 \times 10^{11} \text{ N m}^{-2}) \text{ and a copper wire of length } 4.4 \text{ m } (Y_C = 1.1 \times 10^{11} \text{ N m}^{-2}), \text{ both having a radius } 1.4 \text{ mm, are connected end to end.}$ $\text{When a load is applied, the net stretch of the combined wires is found to be } 1.4 \text{ mm.}$ $\text{The magnitude of the load applied, in Newtons, will be: } \left( \text{use, } \pi = \frac{22}{7} \right)$

Options:

1. 360
2. 180
3. 1080
4. 154

Solution:

$\text{Hint: } \Delta L = \frac{FL}{AY}$ $\text{Step: Find the load applied on both wires.}$ $\text{The elongation of the wire is given by;}$ $\Delta L = \frac{FL}{AY}$ $\text{The net elongation in the steel and copper wire (both wires are connected end to end) is given as;}$ $\Delta L_1 + \Delta L_2 = 1.4 \times 10^{-3} \text{ m}$ $\Rightarrow \frac{FL_1}{AY_1} + \frac{FL_2}{AY_2} = 1.4 \times 10^{-3} \text{ m}$ $\Rightarrow F = \frac{L_1}{Y_1} + \frac{L_2}{Y_2} = \frac{A \Delta L}{\frac{L_1}{Y_1} + \frac{L_2}{Y_2}}$ $\Rightarrow F = \pi \times (1.4 \times 10^{-3})^2 \times 1.4 \times 10^{-3} \times \frac{1}{\frac{3.2}{2 \times 10^{11}} + \frac{4.4}{1.1 \times 10^{11}}}$ $\Rightarrow F = \frac{22}{7} \times \left( \frac{1.96 \times 10^{-6} \times 1.4 \times 10^{-3}}{1.6 \times 10^{-11} + 4 \times 10^{-11}} \right) = \frac{22 \times 2.744 \times 10^{-9}}{7 \times 5.6 \times 10^{-11}}$ $\Rightarrow F = 1.54 \times 10^2 = 154 \text{ N}$ $\text{Hence, option (4) is the correct answer.}$

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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
}