The correct answer is A. 14 cm2.
The formula for calculating the area of steel required is:
$A = \frac{wL}{bd}f_y$
where:
- $A$ is the area of steel required (in cm2)
- $w$ is the uniformly distributed load (in kg/m)
- $l$ is the length of the beam (in m)
- $b$ is the effective depth of the beam (in cm)
- $d$ is the lever arm factor (in dimensionless)
- $f_y$ is the permissible tensile stress of steel (in kg/cm2)
In this case, we have:
- $w = 2400 \frac{kg}{m}$
- $l = 6 m$
- $b = 50 cm = 0.5 m$
- $d = 0.85$
- $f_y = 1400 \frac{kg}{cm^2}$
Substituting these values into the formula, we get:
$A = \frac{(2400 \frac{kg}{m})(6 m)}{(0.5 m)(0.85)(1400 \frac{kg}{cm^2})} = 14 cm^2$
Therefore, the area of steel required is 14 cm2.