Question

A 2 cm cube of some substance has its upper face displayed by 0.15 cm due to a tangential force of 0.3 N while keeping the lower face fixed. Calculate the rigidity modulus of the substance.

Answer 1

Given,
displacement in upper face = 0.15cm
distance between the layers = 2cm
Tangential force, F = 0.3N
Cross sectional area of cube = (edge length)² =( 2 × 10^-2)² = 4 × 10^-4 m²

so, $tan\theta$ = displacement in upper face/distance between the layers.

= 1.5/2

Now, rigidity modulus of the substance , $\eta=\frac{F}{A\times tan\theta$

= 0.3/{4 × 10^-4 × 1.5/2 }

= 0.3 × 10⁴/(2 × 1.5)

= 10⁴ N/m²

Answer 2

Explanation:

A 2 cm cube of some substance has its upper face displayed by 0.15 cm due to a tangential force of 0.3 N while keeping the lower face fixed. Calculate the rigidity modulus of the substance.