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Colloidal Particles and their size 
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#1
Jul1714, 10:18 AM

P: 26

Suppose we have a cube, of length 1 metre. It is cut in all the three directions so that 8 cubes, each having 0.5 m as its length. Then, these cubes are again subdivided in the same manner to get cubes with length 0.25m and so on.
HOW MANY OF THESE SUCCESSIVE SUBDIVISIONS ARE REQUIRED BEFORE THE SIZE OF THE CUBES IS REDUCED TO THE SIZE OF COLLOIDAL PARTICLES, which is 100 nm?? 


#2
Jul1814, 02:30 AM

Sci Advisor
P: 3,593

Come on, we all know that all indians are brilliant mathematicians, so show us at least your intent of solution.



#3
Jul1814, 05:54 AM

P: 26

Ok, then, what I did was,
Let original length be 'l', Now one subdivision reduces the length to half of its original value, and 2, to one fourth. So, 'n' subdivisions will lead to reduction of the length to ( 1/2)^n. Let new length be 'a'. So, a = (1/2)^n *l. in the given case, a = 10^7 *l which means, 10^7 = (1/2)^n taking log, n log 2 = 7 log 10 Which gives n = 23.253. So, am I right?? 


#4
Jul1814, 06:32 AM

Sci Advisor
P: 3,593

Colloidal Particles and their size
This seems to be correct. It is helpful to have some orders of magnitude in mind. The powers of 2 we all know from informatics: ##2^{10}=1024\approx 10^3## and ##2^3=8\approx 10## so ##10^7=10^3*10^3*10\approx 10^{23}##.



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