Two slabs with square cross section of different materials (1, 2) with equal sides (l$l$) and thickness d₁$d_{1}$ and d₂$d_{2}$ such that d₂ = 2d₁$d_{2} = 2d_{1}$ and l > d₂$l > d_{2}$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is θ₂ = 2θ₁$\theta_{2} = 2\theta_{1}$. If the shear moduli of material 1 is 4 × 10⁹~N/m²$4 \times 10^{9}\mathrm{~N/m}^{2}$, then shear moduli of material 2 is x × 10⁹~N/m²$x \times 10^{9}\mathrm{~N/m}^{2}$, where value of x is

Related Formula

Core Logic

The change in pressure Δ P$\Delta P$ is the hydrostatic pressure at depth h$h$.

Step 2: Calculation

Given values: ρ = 10³ kg/m³$\rho = 10^3\mathrm{\,kg/m^3}$ g = 10 m/s²$g = 10\mathrm{\,m/s^2}$ β = 9 × 10⁸ N/m²$\beta = 9 \times 10^8\mathrm{\,N/m^2}$ (Δ V)/(V) = -0.02% = -(0.02)/(100)$\frac{\Delta V}{V} = -0.02\% = -\frac{0.02}{100}$

Substitute into the equation:

Chapter Mix

Class 11 Physics: Mechanical Properties Of Solids Class 11 Physics: Mechanical Properties Of Fluids