Question

A 5.0cm cube of substance has its upper surface displaced by 0.65cm by a tangential force of 0.25N Calculate the modulus of rigidity of the substance... ​

Answer 1

Modulus of rigidity is "η" = 769 N/m^2

Explanation:

Given data:

Area under shear =  5 cm x 5 cm = 25 cm^2 = 25 x 10^-2 m^2

Height of the block "h" = 5 cm = 5 x 10^-2 m

Displacement of top face = x = 0.65 cm = 0.65 x 10^-2 m = 6.5 x 10^-3 m

Shearing force = 0.25 N

Modulus of rigidity "η" = ?

Modulus of rigidity "η" = Fh/Ax

"η" = (0.25 x 5 x 10^-2) / (25 x 10^-4 x 6.5 x 10^-3)

"η" = 769 N/m^2

Thus modulus of rigidity is "η" = 769 N/m^2

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