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Current Question (ID: 10374)

Question:

A square lead slab of side $50 \text{ cm}$ and thickness $10 \text{ cm}$ is subject to a shearing force (on its narrow face) of $9.0 \times 10^4 \text{ N}$. The lower edge is riveted to the floor as shown in the figure below. How much will the upper edge be displaced? (Shear modulus of lead $= 5.6 \times 10^9 \text{ Nm}^{-2}$)

Options:

1. $0.16 \text{ mm}$
2. $1.8 \text{ mm}$
3. $18 \text{ mm}$
4. $16 \text{ mm}$

Solution:

\text{Hint: Displacement, } \Delta x = (\text{Stress} \times \text{L})/(\text{Shear modulus}). \text{Step 1: Find stress applied on the face of the slab.} \text{The lead slab is fixed and the force is applied parallel to the narrow face as shown in the figure below. The area of the face parallel to which this force is applied is:} $A = 50 \text{ cm} \times 10 \text{ cm}$ $= 0.5 \text{ m} \times 0.1 \text{ m}$ $= 0.05 \text{ m}^2$ \text{Therefore, the stress applied } = \frac{F}{\text{Area of face}} = \frac{9 \times (10)^4 \text{ N}}{0.05 \text{ m}^2}$ $= 1.8 \times (10)^6 \text{ Nm}^{-2}$ \text{Step 2: Find displacement, } \Delta x. \text{We know that shearing strain } = (\Delta x/L) = \text{Stress } / \text{Shear modulus} \text{Therefore the displacement } \Delta x = (\text{Stress} \times \text{L})/\text{Shear modulus} $= \frac{(1.8 \times (10)^6 \text{ Nm}^{-2} \times 0.5 \text{ m})}{(5.6 \times (10)^9 \text{ Nm}^{-2})}$ $= 1.6 \times (10)^{-4} \text{ m} = 0.16 \text{ mm}

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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
}