Correct Answer - Option 2 : 13 kNm/m

**Concept:**

Moment of resistance for under reinforced beam is given by:

**MOR = 0.87× f**_{y }× Ast(d - 0.42Xu)

Where, X_{u} = Depth of neutral axis, A_{st} = Area of reinforcement

Depth of neutral axis is calculated as:

Compression force = Tension force

**0.36 f**_{ck} b X_{u} = 0.87 f_{y} A_{st}

**Calculation:**

**Given,**

d = 100 mm, c/c spacing between bar = 200 mm

M20 and Fe415

Taking b = 1000 mm for slab ,Calculating A_{st,}

A_{st} = \(\left ( \dfrac{1000.A_{\phi }}{\rm Spacing} \right )\)

\(A_{st}= \dfrac{1000× \frac{\pi }{4}× 10^{2}}{200}\) = 393 mm^{2}

We know that,

**Depth of Neutral Axis (x**_{u}):

\(X_{u}= \left ( \dfrac{0.87f_{y}.A_{st}}{0.36f_{ck}.b} \right )\)

\(X_{u}= \left ( \dfrac{0.87× 415× 393}{0.36× 20 × 1000} \right )\)= 19.6 mm

MOR is given as:

M_{u} = 0.87 f_{y}.A_{st}.(d - 0.42X_{u})

M_{u} = 0.87×415× 393 (100- 0.42 × 19.6)

M_{u }= 141892.65 × 91.768 = 13021204.7 N-mm

M_{u =} **13.0 kN-m **