1. **Problem Statement:** We need to choose suitable standard format bricks and mortar for the solid walls supporting the elevated tank. Given data includes brick wall unit weight $2300\ \mathrm{kg/m^3}$, dimensions, and loads. 2. **Calculate the slenderness ratio (SR):** $$SR = \frac{h_{eff}}{t_{eff}} = \frac{4000 \times 0.75}{350} = 8.57 \angle 27$$ Since SR is about 8.57, it is considered okay. 3. **Calculate ultimate load from tank and content per meter width:** $$\text{Ultimate load} = 250\ \mathrm{kN/m^2} \times 3\ \mathrm{m} = 750\ \mathrm{kN/m}$$ 4. **Calculate load on brick wall:** Brick wall volume per meter width: thickness $0.35\ \mathrm{m}$, height $4\ \mathrm{m}$ $$\text{Load on wall} = 2300 \times 9.81 \times 10^{-3} \times (0.35 \times 4) \times 1.4 = 44.22\ \mathrm{kN/m}$$ 5. **Calculate load on concrete slab:** Assuming slab load $24\ \mathrm{kN/m^2}$ over area $3 \times 0.2 \mathrm{m}^2$ $$\text{Load on slab} = 24 \times 1.4 \times (3 \times 0.2) = 20.16\ \mathrm{kN/m}$$ 6. **Calculate total load on wall:** $$\text{Total load} = 750 + 44.22 + 20.16 = 814.4\ \mathrm{kN/m}$$ 7. **Use interpolation for factor values from table:** Given values interpolate: $$= 0.88 - (0.88 - 0.66) \times \frac{0.167 - 0.1}{0.2 - 0.1} = 0.88 - 0.22 \times 0.87 = 0.73$$ 8. **Form inequality and solve for $f_k$:** $$814.4 \leq 0.73 \times 350 \times \frac{f_k}{3.5}$$ Simplify: $$814.4 \leq 73 f_k$$ $$f_k \geq \frac{814.4}{73} = 11.15\ \mathrm{N/mm^2}$$ 9. **Conclusion:** Since typical brick strength $f_k = 35\ \mathrm{N/mm^2}$ is greater than required 11.15, the chosen bricks and mortar are suitable for supporting the tank. **Final answer:** The standard bricks with characteristic strength $f_k = 35\ \mathrm{N/mm^2}$ and mortar mix with normal manufacturing and construction controls are adequate for the 350 mm thick brick walls supporting the tank.