This question was previously asked in

ESE Civil 2014 Paper 2: Official Paper

Option 1 : 0.018 cm / sec

ST 1: Structural Analysis

1523

20 Questions
60 Marks
25 Mins

__Concept:__

Settling velocity of particles when Reynolds no less than 0.5 is given by Stocks Law:

\(V = \frac{{\left( {G - 1} \right){\gamma _w}{d^2}}}{{18\;\mu }}\)

Where,

G is the Specific Gravity

d is the particle size

μ is the dynamic viscosity of water.

__Calculation:__

Given,

G = 2.65; Re < 0.5; d = 2 × 10^{-3} cm

Kinematic viscosity = 2 × 10^{-2} cm^{2}/sec

Let density of water is 1000 kg/m^{3}

Dynamic viscosity of water, μ = 2 × 10^{-2} × 10^{-4} × 1000 = 2 × 10^{-3} kg/m-sec

The settling velocity is

\(V = \frac{{\left( {G - 1} \right){\gamma _w}{d^2}}}{{18\;\mu }}\)

\(V = \frac{{\left( {2.65 \;- \;1} \right)\; \times \;9810\;{{\left( {2\; \times\; {{10}^{ - 3}}\; \times\; {{10}^{ - 2}}} \right)}^2}}}{{18\; \times\; 2\; \times \;{{10}^{ - 3}}}}\)

**V = 0.000179 m/sec = 0.0179 cm/sec**