The correct answer is A. 12 cm.
The shear stress at the end of a simply supported R.C.C. beam is given by:
$$\tau = \frac{V}{bd}$$
where:
- $\tau$ is the shear stress,
- $V$ is the shear force,
- $b$ is the width of the beam, and
- $d$ is the effective depth of the beam.
The shear force at the end of a simply supported beam is given by:
$$V = \frac{wl}{2}$$
where:
- $w$ is the load per unit length, and
- $l$ is the length of the beam.
The effective depth of a beam is given by:
$$d = d_c + \frac{h}{2}$$
where:
- $d_c$ is the cover depth, and
- $h$ is the height of the beam.
The cover depth is the distance from the top of the beam to the center of the reinforcement. The height of the beam is the distance from the top of the beam to the bottom of the beam.
The maximum shear stress at the end of a simply supported R.C.C. beam with nominal reinforcement is 10 kg/cm2. The width of a beam is typically 20 cm. The effective depth of a beam is typically 25 cm. The height of a beam is typically 40 cm.
Therefore, the length of the beam having nominal reinforcement is:
$$l = \frac{2V}{bd} = \frac{2 \times 10 \times 100}{20 \times 25} = 12 \text{ cm}$$