Please answer a$,$b & c

$13\mathrm{.}20$ A $15m$ thick normally consolidated clay layer is shown in Figure $13\mathrm{.}63.$ The liquid

limit and plastic limit of the soil are $39$ and $20,$ respectively.

Figure $13\mathrm{.}63$

An illustration has a horizontal rectangular block, which is divided in to three blocks

horizontally. The first horizontal block at the top is filled with tiny dots represents dry

sand, gamma equals $16$ kilo Newtons per meter cubed with vertical distance of $3$

meters. The second horizontal block in the middle represents the clay layer,

gamma$\_$sat equals $20\mathrm{.}5$ kilo newtons per meter cubed; $LL$ equals $39$; P L equals $20,$

which has a vertical distance of $15$ meters. The dry sand layer and the clay layer are

separated by a dashed line representing the groundwater table. The third horizontal

block at the bottom filled with small irregular particles represents the rock layer.

Part a

The effective stress at a depth of $11m$ below the ground surface, sigma' $=$

$k\frac{N}{m^2}$

$($Express your answer with one decimal place$)$

The undrained shear strength using Eq$.(13\mathrm{.}53),$ Cu$($vst$)=$

$($Express your answer with one decimal place$)$

Part b

Using Eq$.(13\mathrm{.}72),$ the correction factor, lambda $=$

$($Express your answer with three decimal places$)$

Using Eq$.(13\mathrm{.}71),$ the design value of the undrained shear strength, Cu$($design$)=$

$k\frac{N}{m^2}$

$($Express your answer with one decimal place$)$

Part c

Using Eq$.(13\mathrm{.}54),$ the new undrained cohesion $($shear strength$),$ Cu$($ocr$)=$