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Question: The edge of an aluminum cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall....

The edge of an aluminum cube is 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. The vertical deflection of this face of the cube. The vertical deflection of this is (Shear modulus of aluminium = 25 GPa,g = 10 ms−2ms^{- 2})

A

4 × 10−5 m4\ \times \text{ 1}\text{0}^{- 5}\text{ m}

B

4 × 10−6 m4\ \times \text{ 1}\text{0}^{- 6}\text{ m}

C

4 × 10−7 m4\ \times \text{ 1}\text{0}^{- 7}\text{ m}

D

4 × 10−8 m4\ \times \text{ 1}\text{0}^{- 8}\text{ m}

Answer

4 × 10−7 m4\ \times \text{ 1}\text{0}^{- 7}\text{ m}

Explanation

Solution

: Shear modulus,η=F/AΔL=FLηA\eta = \frac{F/A}{\Delta L} = \frac{FL}{\eta A}

Here, F=(100kg)(10ms−2)=1000NF = (100kg)(10ms^{- 2}) = 1000N

A=(10cm×10cm)=100cm2=100×10−4m2A = (10cm \times 10cm) = 100cm^{2} = 100 \times 10^{- 4}m^{2}

η=25GPa=25×109Pa=25×109Nm−2\eta = 25GPa = 25 \times 10^{9}Pa = 25 \times 10^{9}Nm^{- 2}

L=10cm=10×10−2mL = 10cm = 10 \times 10^{- 2}m

Substituting the given values, we get

ΔL=(1000N)(10×10−2m)(25×109Nm−2)(100×10−4m2)=4×10−7m\Delta L = \frac{(1000N)(10 \times 10^{- 2}m)}{(25 \times 10^{9}Nm^{- 2})(100 \times 10^{- 4}m^{2})} = 4 \times 10^{- 7}m