A metal cube of side length 8.0cm has its upper surface displaced with respect to the bottom by 0.10 mm when tangential force of 4×104N is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is

Rigidity modulus η=F/Atan⁡θtan⁡θ≃θ≃0.10×10−38×10−2=1800∴η=FAθ=4×104×8008×8×10−1×1=5×109N/m2

A metal cube of side length 8.0cm has its upper surface displaced with respect to the bottom by 0.10 mm when tangential force of 4×104N is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is

Rigidity modulus η=F/Atan⁡θtan⁡θ≃θ≃0.10×10−38×10−2=1800∴η=FAθ=4×104×8008×8×10−1×1=5×109N/m2

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A metal cube of side length 8.0cm has its upper surface displaced with respect to the bottom by 0.10 mm when tangential force of $4 \times 10^{4} N$ is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is

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$108 {Nm}^{- 2}$

$5 \times 10^{9} {Nm}^{- 2}$

$8 \times 10^{9} {Nm}^{- 2}$

$4 \times 10^{9} {Nm}^{- 2}$

answer is B.

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Detailed Solution

Rigidity modulus $η = \frac{F / A}{\tan θ}$

$\tan θ ≃ θ ≃ \frac{0.10 \times 10^{- 3}}{8 \times 10^{- 2}} = \frac{1}{800}$

$∴ η = \frac{F}{A θ} = \frac{4 \times 10^{4} \times 800}{8 \times 8 \times 10^{- 1} \times 1} = 5 \times 10^{9} N / m^{2}$

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A metal cube of side length 8.0cm has its upper surface displaced with respect to the bottom by 0.10 mm when tangential force of 4×104N is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is

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A metal cube of side length 8.0cm has its upper surface displaced with respect to the bottom by 0.10 mm when tangential force of $4 \times 10^{4} N$ is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is