Two slabs of different materials have square cross-sections with equal side ℓ and thicknesses d₁ and d₂, where d₂ = 2d₁. Equal shearing forces are applied on their narrow faces. Their deformation angles satisfy θ₂ = 2θ₁. If the shear modulus of material 1 is 4×10⁹ N/m² and that of material 2 is x×10⁹ N/m², find x.
Physics · Properties of solid and fluid · propertes of solid stress ,strain and modulus of elasticity
JEE Mains 2025 4 April Morning Shift

Question

Two slabs of different materials have square cross-sections with equal side ℓ and thicknesses d₁ and d₂, where d₂ = 2d₁. Equal shearing forces are applied on their narrow faces. Their deformation angles satisfy θ₂ = 2θ₁. If the shear modulus of material 1 is 4×10⁹ N/m² and that of material 2 is x×10⁹ N/m², find x.

Physics · Properties of solid and fluid · propertes of solid stress ,strain and modulus of elasticity

JEE Mains 2025 4 April Morning Shift

G1 Physics · Accepted Answer

Shear modulus is G = shear stress/shear strain. For the narrow face, area is A = ℓd, so shear stress = F/(ℓd). For small deformation, shear strain ≈ θ. Thus, G = F/(ℓdθ). For the two slabs, G₂/G₁ = (F₂/F₁)(ℓ₁/ℓ₂)(d₁/d₂)(θ₁/θ₂). Equal force and equal side length give F₂/F₁ = 1, ℓ₁/ℓ₂ = 1. Also, d₁/d₂ = 1/2, θ₁/θ₂ = 1/2. Therefore, G₂/G₁ = (1/2)(1/2) = 1/4. So, G₂ = (1/4)(4×10⁹) = 1×10⁹ N/m². Hence x = 1.