Consider the following geometric solids.
2 answers:
Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given
The surface area to volume ratio for right circular cylinder is given
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
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The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 ×
friction factor = 0. 52 ×
friction factor = 0. 52 × 0. 55
friction factor
b. When V = 3mls
Friction factor = 0. 52 ×
Friction factor = 0. 52 ×
Friction factor = 0. 52 × 0. 185
Friction factor
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × ×
×
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = ×
×
×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
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