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The edges of an aluminum cube are 10 cm ...

The edges of an aluminum cube are `10 cm` long. One face of the cube is firmly fixed to a vertical wall. A mass of `100 kg` is then attached to the opposite face of the cube. Shear modulus of aluminum is `25 xx 10^(9)` Pa, the vertical deflection in the face to which mass is attached is

A

`4xx10^(-4)m`

B

`4xx10^(-7)m`

C

`25xx10^(-6)m`

D

`6x10^(-7)m`

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The correct Answer is:
To find the vertical deflection \( x \) in the face of the aluminum cube when a mass is attached, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Values:** - Edge length of the cube, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \) - Mass attached, \( m = 100 \, \text{kg} \) - Shear modulus of aluminum, \( \eta = 25 \times 10^9 \, \text{Pa} \) 2. **Calculate the Force (Weight) Acting on the Cube:** \[ F = m \cdot g \] where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \). \[ F = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 981 \, \text{N} \] 3. **Calculate the Area of Cross Section of the Cube:** Since the cube has a square face, the area \( A \) is given by: \[ A = L^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \] 4. **Use the Shear Modulus Formula:** The shear modulus \( \eta \) is defined as: \[ \eta = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F/A}{\tan(\theta)} \] For small angles, \( \tan(\theta) \approx \frac{x}{L} \), where \( x \) is the vertical deflection. Therefore: \[ \eta = \frac{F}{A} \cdot \frac{L}{x} \] 5. **Rearranging for Vertical Deflection \( x \):** Rearranging the equation gives: \[ x = \frac{F \cdot L}{A \cdot \eta} \] 6. **Substituting the Values:** \[ x = \frac{981 \, \text{N} \cdot 0.1 \, \text{m}}{0.01 \, \text{m}^2 \cdot 25 \times 10^9 \, \text{Pa}} \] \[ x = \frac{98.1 \, \text{N m}}{0.01 \, \text{m}^2 \cdot 25 \times 10^9 \, \text{Pa}} = \frac{98.1}{250000000} = 3.924 \times 10^{-7} \, \text{m} \] 7. **Final Answer:** The vertical deflection \( x \) is approximately: \[ x \approx 3.924 \times 10^{-7} \, \text{m} \approx 4 \times 10^{-7} \, \text{m} \]

To find the vertical deflection \( x \) in the face of the aluminum cube when a mass is attached, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Values:** - Edge length of the cube, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \) - Mass attached, \( m = 100 \, \text{kg} \) - Shear modulus of aluminum, \( \eta = 25 \times 10^9 \, \text{Pa} \) ...
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