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 = 8.0 cm = 8 x 10-2 cm∆x = 10-4 cm ; F = 4 x 104 N ; A = 64 x 10-4 m2η=FA∆xl⇒4 x 10464 x 10-4 x 8 x 10-210-4η=109 x 10 x 12=5 x 109

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 = 8.0 cm = 8 x 10-2 cm∆x = 10-4 cm ; F = 4 x 104 N ; A = 64 x 10-4 m2η=FA∆xl⇒4 x 10464 x 10-4 x 8 x 10-210-4η=109 x 10 x 12=5 x 109

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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 × 10⁹ Nm^-2

5 × 10⁹ Nm^-2

8 × 10⁹ Nm^-2

10⁸ Nm^-2

answer is B.

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Detailed Solution

l = 8.0 cm = 8 x 10^-2 cm
$∆$ x = 10^-4 cm ; F = 4 x 10⁴ N ; A = 64 x 10^-4 m²
$η = \frac{\frac{F}{A}}{\frac{∆ x}{l}} \Rightarrow \frac{4 x 10^{4}}{64 x 10^{- 4}} x \frac{8 x 10^{- 2}}{10^{- 4}}$

$η = 10^{9} x 10 x \frac{1}{2} = 5 x 10^{9}$

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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 × 10⁴N is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is