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

ℓ=8.0cm=8×10−2cmΔx=10−4cm;f=4×104N;A=64×10−4m2η=fA≜x⇒4×10464×10−4×8×10−210−4η=109×10×12=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

ℓ=8.0cm=8×10−2cmΔx=10−4cm;f=4×104N;A=64×10−4m2η=fA≜x⇒4×10464×10−4×8×10−210−4η=109×10×12=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 × 10⁴N is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is

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4 x 10⁹ Nm^-2

5 x 10⁹ Nm^-2

8 x 10⁹ Nm^-2

10⁸ Nm^-2

answer is A.

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Detailed Solution

$ℓ = 8.0 cm = 8 \times 10^{- 2} cm$

$Δ x = 10^{- 4} cm; f = 4 \times 10^{4} N; A = 64 \times 10^{- 4} m^{2}$
$η = \frac{\frac{f}{A}}{≜ x} \Rightarrow \frac{4 \times 10^{4}}{64 \times 10^{- 4}} \times \frac{8 \times 10^{- 2}}{10^{- 4}}$
$η = 10^{9} \times 10 \times \frac{1}{2} = 5 \times 10^{9} N / m^{2}$