The correct answer is A. 0.30 m/sec.
The minimum velocity required for removing the sewerage is the velocity at which the particles of sand and organic matter will be suspended in the water. The velocity required is dependent on the specific gravity of the particles, the size of the particles, and the density of the water.
The specific gravity of sand is 2.65, and the specific gravity of organic matter is 1.2. The density of water is 1000 kg/m3. The size of the sand particles is 1 mm, and the size of the organic particles is 5 mm.
The minimum velocity required for removing the sewerage is calculated using the following equation:
$v = \sqrt{\frac{2g(\rho_s – \rho_w)}{d}}$
where:
$v$ is the velocity,
$g$ is the acceleration due to gravity,
$\rho_s$ is the specific gravity of the sand,
$\rho_w$ is the density of the water, and
$d$ is the diameter of the sand particles.
Substituting the values into the equation, we get:
$v = \sqrt{\frac{2 \times 9.8 \times (2.65 – 1.2)}{0.001}} = 0.30$ m/sec
Therefore, the minimum velocity required for removing the sewerage is 0.30 m/sec.
Option B is incorrect because it is the velocity required for removing sand particles only. Option C is incorrect because it is the velocity required for removing organic matter only. Option D is incorrect because it is the velocity required for removing sand particles and organic matter of a different size.