IIT JEE , AIEEE Forum - Mathematics , Algebra

broken

In a sphere of radius R, wat is the maximum no. of spheres of radius R/9 that can be placed??

plz tell soln also ...as i m not getting the right answer ..

shreyasnivas

728?

see obvoiusly the total volume occupied by the smaller spheres will be LESS than the big one.

so just equate the volumes of the big sphere, and "x" numbers of small spheres.then find the number of spheres..this number is "x".

the answer you want is one less than this number..

(the max number can theoretically be any number less than "x" so tell me if the ans is not right .... but i guarantee that the number is less than "x".) "x" in this case is 729.

shreyas

broken

its not correct....actually it was given by my teacher and i hv asked him... he said 728 is incorrect....the only hint i have is we have to arrange it layer by layer by visualising 3d and count..it

shreyasnivas

wow thats real tough.. i think almost impossible.. uless u use some stuff from solid state..

just see if the volume occupied by the smaller spheres is 74% of the total volume of big sphere..(ask your sir if you can first)
for ccp laticce this is the maximum packing fraction..

broken

i m also thinking.....if sum one gets it plz do reply

shreyasnivas

539 i think.i used the packing fraction 74%

rate me if correct!

fused_bulb

yess....this is quite tough...

see..we cant use the solid state method here as there is a

constraint in the radius of the inner balls ( R/ 9 )....

According 2 me....this question shud b regarded as a gud question

if and only if there is a proper proof ( dat is we dont have to like

literally count the number of balls as suggested above...)

but if there is a proper mathematical proof...then yesss...its a

question worth spending time on....

so...all i want 2 ask is does this have a proper proof....???

shreyasnivas

i cant see why the radius of the inner balls matters.. unless the radius ratios are off target(i cannot remember the radius ratio to be used for ccp).. also, 74% was the packing fraction in the case of a cube.. this is a sphere. apart from these 2 flaws, my suggestion is correct.. i think.

fused_bulb

no....suppose u make the radius of inner balls smaller and

smaller...let it approach zero...then packing efficiency approaches

100 percent..just visualize the condition..

devesh_l2k007

packing fraction is 74 percent in a vube not a sphere
It shld be higher in this
Req great solving aptitude
Ans less than 729 n greater that 539 from my side

aravindh_ramaswamy

729 spheres of radius r/9 an be placed

coz 4/3pi r^3=x*4/3pi(r/9)^3

broken

@aravindh u have to fill it with solid spheres so u cant utilize all the space inside....it cant b 729!!

debabratanag99

The question is pretty simple one.................

See it has been asked for the maximum number of spheres ..............

So there lies the clue.............Maximum no. of spheres can only be there in a body of maximum DENSITY...!!!!!!!............And in this world the maximum density till date noticed is 0.7405 that is for Hexagonal packing or CCP. By density here I mean Packing Density which is commonly known as packing fraction............

Though it is theoritically proved by Kepler that maximum density can be 0.778......still it is a conjecture and has not been established in real world............

So let us not assume Kepler conjecture to be true.........at this moment.......

So maximum packing density = 0.7405
Let number of spheres of radius R/9 = n

Thus , 4/3 * (pie) * R^3 * 0.7405 = n * 4/3 * (pie) * (R/9)^3

Therefore, n = 9^3 * 0.7405
Therefore, n = 539.82.........

Thus we take n = 539 AND NOT

Ask & Discuss Questions with Community & Experts