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.
Explanation:
s = ut + 1/2 a t^2
200 = 0 * 6 + 1/2 * a * (6)^2
200 = 1/2 * a * 36
200 = 18 a
a = 200/18
a= 11.1m/sec^2
v = u + at
v = 0 + 11.1 * 6
v = 66.6m/s
hope it helps you
1) Current in each bulb: 0.1 A
The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

And so, the current through the circuit is (using Ohm's law):

And since the two bulbs are connected in series, the current through each bulb is the same.
2) 4 W and 8 W
The power dissipated by each bulb is given by the formula:

where I is the current and R is the resistance.
For the first bulb:

For the second bulb:

3) 12 W
The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

M1 v1 = (m1 + m2)v2.
All of the exponents should be lowered to the bottom right of the letters.