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RSMSSB JE Civil Degree Oct 2016 Official Paper

Option 2 : 2.67m

Construction Materials 1

2002

20 Questions
40 Marks
25 Mins

__Concept:__

Maximum height of vertical cut is given by:

\({{\rm{H}}_{\rm{c}}} = \frac{{4{\rm{\;}}{{\rm{c}}_{\rm{u}}}}}{{{\rm{\gamma \;}}\sqrt {{\rm{ka}}} }}\)

γ = γsat

For pure clay ɸ = 0°

\({{\rm{k}}_a} = \frac{{1 - {\rm{sin\;}}\phi }}{{1 + {\rm{sin\;}}\phi }}\)

__Solution:__

\({K_a} = \frac{{1 - \sin 0^\circ }}{{1 + \sin 0^\circ }} = 1\)

∴ \(Z = \frac{{4C}}{{\gamma \sqrt {{K_a}} }} = \frac{{4 \times 14}}{{21 \times \sqrt 1 }} = 2.67\;m\)