Five identical incompressible spheres of radius 1 unit are stacked in a pyramidal form as shown in the figure. The height of the structure is

The Correct Answer is \(2+ 2\sqrt{2/3}\)

Concept:

The four spheres A, B, C, and D form a regular tetrahedron. The side of this tetrahedron connects the centers of all pairs of spheres, and so, are all equal to  2r

Explanation:

Let  E  be the center of the base  ΔBCD  that lies exactly below the top of the pyramid  A

Then, you can see that: ΔBEC  is isosceles.BE= CE = b

The perpendicular from  E to  BC ( EF ) bisects  BC

BF = FC = r

The fact that  BE bisects  ∠CBD (=  60º) tells you the relationship between  b and  r
∠CBE = 30º

b × cos(30∘) = r
b = 2r/√3

Using these, the height of the tetrahedron  h can be calculated thus:

AE is perpendicular to the base  ΔBCD
 So,  ΔAEB is right-angled with,

AB² = BE² + AE
\(h=2\times √{(2/3)} \times r\)

Finally, to calculate the total height, you need to see that:

The base of the tetrahedron itself is at a height  r from the floor, and
The top of the structure is at height r  from the center of the top sphere.

Therefore, the total height H,

H = r + h + r
2r × (1 + √2/3)

According to the question putting the value of r as 1.

H = 2(1) x (1+\(\sqrt{2/3}\))

H = 2 + 2\(\sqrt{2/3}\)z2=R1+(R1+R2)2−2R21−−−−−−−−