SSC JE Civil Engineering 23-Sep-2019 Shift 1 Solved Paper

Neutral axis depth for a singly reinforced beam is the axis at which the stresses are zero and it is situated at the centre of gravity of the section.
The depth of neutral axis for a singly reinforced beam is calculated by equilibrium of tensile and compressive forces as shown in figure below: Where,
b = Width of section
d = Effective depth of beam
${\mathit{A}}_{\mathrm{st}}$= Area of steel reinforcement
${\mathit{x}}_{\mathit{u}}$ = Depth of neutral axis
C = Total compression
T = Total tension
Total tension (T) = 0.87${\mathrm{f}}_{\mathit{y}}.{\mathit{A}}_{\mathrm{st}}$
Total compression $\left(\mathit{C}\right)=0.36{\mathrm{f}}_{\mathrm{ck}}{\mathit{x}}_{\mathit{u}}.\mathit{b}$
at 0.42${\mathit{x}}_{\mathit{u}}$ from the top extreme fibre (as per IS 456:2000)
For equilibrium of forces
Total Tension = Total compression
0.87${\mathrm{f}}_{\mathit{y}}.{\mathit{A}}_{\mathrm{st}}$= 0.36 ${\mathrm{f}}_{\mathrm{ck}}{\mathit{x}}_{\mathit{u}}.\mathit{b}$
${\mathit{x}}_{\mathit{u}}=0.87{\mathrm{f}}_{\mathit{y}}{\mathit{A}}_{\mathrm{st}}∕0.36{\mathrm{f}}_{\mathrm{ckbd}}$
Hence option(a) is correct