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solutionGivene=0.49 and G_{s}=2.68( for sand )as it is dry state so the unit weigh of the dry soil\gamma_{d}=\frac{\gamma_{w} G}{1+e}=\frac{9.81(2.68)}{1+0.49}=\frac{17.644 K N}{m^{3}}for saturated \gamma_{\text {sat }}=\frac{\gamma_{w}(G+e)}{1+e}=\frac{9.81(2.68+0.49)}{1+0.49}=\frac{20.87 K N}{m^{3}}for clay e=0.9 and G_{s}=2.75 as water table is present so the degree of saturation will be 1\gamma_{s a t}=\frac{\gamma_{w}(G+e)}{1+e}=\frac{9.81(2.75+0.9)}{1+0.9}=\frac{18.84 K N}{m^{3}}The effective stress at the bottom of the clay layer\sigma^{\prime}=17.644(4)+(18.84-9.81)(3)=97.66 K p alet us assume the water table is rides to h so there is decrease in effective stress by14Kpa consider the effective stress at bottom of clay layer\begin{array}{l}97.66-14=17.644(4-h)+(20.87-9.81)(h)+(18.84-9.81) 3 \\83.66=97.666-6.584 h\end{array}h=2.127 \mathrm{mts} height form the water tableif the water table is at ground level the effective stress at bottom of clay us\sigma^{\prime}=(20.87-9.81) 4+(18.84-9.81) 3=71.33 \mathrm{Kpa} Thank you  please feel free to comment in case of any queries   ...