### Related Formula Shear modulus definition: eta = fracsigmatheta = fracFA cdot theta$\eta = \frac{\sigma}{\theta} = \frac{F}{A \cdot \theta}$ where A = l cdot d$A = l \cdot d$ is the shearing face surface cross-area block configuration. ### Core Logic We are given that: * theta_2 = 2theta_1$\theta_2 = 2\theta_1$ * Shearing forces are identical (F_1 = F_2 = F$F_1 = F_2 = F$). * Cross section area parameter tracking: A_1 = l cdot d_1$A_1 = l \cdot d_1$, A_2 = l cdot d_2 = l cdot (2d_1) = 2A_1$A_2 = l \cdot d_2 = l \cdot (2d_1) = 2A_1$.

### Step 1: Link Modulus Terms Express deformation angles: theta_1 = fracFl cdot d_1 cdot eta_1$\theta_1 = \frac{F}{l \cdot d_1 \cdot \eta_1}$ theta_2 = fracFl cdot d_2 cdot eta_2 = fracFl cdot (2d_1) cdot eta_2$\theta_2 = \frac{F}{l \cdot d_2 \cdot \eta_2} = \frac{F}{l \cdot (2d_1) \cdot \eta_2}$ Substitute these into the relation theta_2 = 2theta_1$\theta_2 = 2\theta_1$: fracF2l d_1 eta_2 = 2 cdot left( fracFl d_1 eta_1 ight)$\frac{F}{2l d_1 \eta_2} = 2 \cdot \left( \frac{F}{l d_1 \eta_1} ight)$ frac12eta_2 = frac2eta_1 implies 4eta_2 = eta_1 implies eta_2 = fraceta_14$\frac{1}{2\eta_2} = \frac{2}{\eta_1} implies 4\eta_2 = \eta_1 implies \eta_2 = \frac{\eta_1}{4}$ ### Step 2: Solve for x Given eta_1 = 4 times 10^9mathrm~N/m^2$\eta_1 = 4 \times 10^{9}\mathrm{~N/m}^2$: eta_2 = frac4 times 10^94 = 1 times 10^9mathrm~N/m^2$\eta_2 = \frac{4 \times 10^9}{4} = 1 \times 10^{9}\mathrm{~N/m}^2$ Thus, x = 1$x = 1$. ### Pattern Recognition Be careful when identifying the dimensions of the shear face area. Doubling the slab width doubles the base support area resisting deformation, reducing stress maps under equivalent lateral loading vectors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids