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Current Question (ID: 10347)

Question:

$\text{A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire } A\text{) and aluminium (wire } B\text{) of equal lengths as shown in the figure. The cross-sectional areas of wires } A \text{ and } B \text{ are } 1.0 \text{ mm}^2 \text{ and } 2.0 \text{ mm}^2 \text{ respectively. At what point along the rod should a mass m be suspended in order to produce equal stresses in both steel and aluminium wires?}$

Options:

1. $0.7 \text{ m from wire } A$
2. $0.07 \text{ m from wire } A$
3. $7.0 \text{ m from wire } A$
4. $0.007 \text{ m from wire } A$

Solution:

$\text{Hint: The rod is in translational as well as in rotational equilibrium.}$ $\text{Step 1: Find the ratio of forces on each wire.}$ $\text{As stress in the wire } = \frac{\text{Force}}{\text{Area}}$ $\text{If the two stresses on the wires are equal then,}$ $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ $\Rightarrow \frac{F_1}{F_2} = \frac{A_1}{A_2} = \frac{1}{2}$ $\text{Step 2: Applying rotational equilibrium.}$ $F_1 y = F_2 (1.05 - y)$ $\frac{F_1}{F_2} = \frac{1.05 - y}{y} = \frac{1}{2}$ $\Rightarrow y = 0.7 \text{ m}$

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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
}