Question

A cube of side 40 mm has its upper face displaced
by 0.1 mm by a tangential force of 8 KN. The
shearing modulus of cube is
(1) 2 x 10°N m-2
(2) 4 x 109 N m-2
(3) 8 x 10°Nm-2
4) 16 x 109 Nm-2​

Answer 1

Answer:2 × 10^9 Nm^–2

Explanation:

Answer 2

answer : option (1) 2 × 10^9 N/m²

explanation : shear modulus = shear stress/shear strain.

given, side length of cube , l = 40mm

change in length due to tangential force, ∆l = 0.1 mm

so, shear strain = ∆l/l = 0.1/40 = 1/400

shear stress = force applied/base area

= 8kN/(40mm)²

= 8 × 10³ N/(40 × 10^-3 m)²

= 8000/(0.04)² N/m²

= 8000/(0.0016)

= 5 × 10^6 N/m²

now, shear modulus = shear stress/shear strain

= (5 × 10^6)/(1/400)

= 5 × 400 × 10^6 N/m²

= 2000 × 10^6 N/m²

= 2 × 10^9 N/m²

hence, shear stress of cube is 2 × 10^9 N/m²