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Current Question (ID: 19557)

Question:

$\text{A square aluminium (shear modulus is } 25 \times 10^9 \text{ Nm}^{-2}\text{) slab of side } 60 \text{ cm and thickness of } 15 \text{ cm is subjected to a shearing force (on its narrow face) of } 18.0 \times 10^4 \text{ N. The lower edge is riveted to the floor. The displacement of the upper edge is:}$ $1.\ 30\ \mu\text{m}$ $2.\ 48\ \mu\text{m}$ $3.\ 16\ \mu\text{m}$ $4.\ 64\ \mu\text{m}$

Options:

1. $30\ \mu\text{m}$
2. $48\ \mu\text{m}$
3. $16\ \mu\text{m}$
4. $64\ \mu\text{m}$

Solution:

$\text{Hint: } \eta = \frac{F/A}{\Delta L/L}$ $\text{Step: Find the displacement of the upper edge.}$ $\text{The displacement of the upper edge is given by:}$ $\Rightarrow \frac{F}{A} = \eta \frac{x}{l} \Rightarrow \frac{Fl}{A\eta} = x$ $\text{Substitute the given values we get:}$ $\Rightarrow x = \frac{18 \times 10^4 \times 60 \times 10^{-2}}{60 \times 10^{-2} \times 15 \times 10^{-2} \times 25 \times 10^9}$ $\Rightarrow x = 48 \times 10^{-6} \text{ m} = 48\ \mu\text{m}$ $\text{Hence, option (2) is the correct answer.}$

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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
}