Latest TCS Aptitude Question SOLUTION: Cubes c1,c2,c3.....and Spheres s1,s2,s3,....are defined in the following way. -s1 has radius 4cm. -For each n>0, Cn is inscribed in Sn and Sn+1 is inscribed in Cn(ie c1 is inscri

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Cubes c1,c2,c3.....and Spheres s1,s2,s3,....are defined in the following way.
-s1 has radius 4cm.
-For each n>0, Cn is inscribed in Sn and Sn+1 is inscribed in Cn(ie c1 is inscribed in s1,s2 is inscribed in c1,c2 is inscribed in s2 and so on)
Let Vn be the sum of the volumes of the first n cubes c1,c2,c3....Cn.Then as n->infinity Vn approaches.

options i)384(1+3root3)/13 2)64(1+3root3)/13 3) 128(1+3root3)/13 4) infinity , ie Vn is unbounded.

Asked In TCS (11 years ago)

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PANKAJ  (11 years ago)

c is the right answer.

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Vinay Sipani   (11 years ago)

There is no option here
Ans is 256(1 + 3sqrt3)/13

In a sphere of radius R , length of cube inscribed = 2R / sqrt3
=> V1 = (2R / sqrt3)^3

Radius of sphere inscribed in this cube = R / sqrt3
cube inside S2 = 2R / 3
=> V2 = (2R / 3)^3 and so on

Therefore V1 + V2 + . . . infinity
= (2R)^3 [ (1/sqrt)^3 + (1/3)^3 + (1/3sqrt3)^3 + . . . Infinity ]

Sum of infinite terms = first term / ( 1 - common ratio)
= 1 / (3sqrt3 - 1) = (1 + 3sqrt3) / 13

=> V = (2R)^3 ( 1 + 3sqrt3) / (13 x 2)
Here R = 4 cm
=> V = 256(1 + 3sqrt3) / 13

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unnam sai priya  (11 years ago)

ans is 256(1+3root3)/13
for cube of side a inserted in a sphere radius r
relation b/w a && r is a=2r/root3
for sphere of radius x inserted in a cube of side b
relation b/w b && x is x=b/2
by using above relations sides of the cube c1,c2,c3,...... are in gp with initial value 2r/root3 and common ratio 1/root3

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