What is the relations with small sphere with large sphere?

Question

Into a regular glass tank, 8 small steel balls are placed. Then the tank is filled with water so that the top of the water is level with the very top of the 8 steel balls.The 8 balls are removed and replaced with 1 larger Steel Ball (into the same tank with the same water, none is lost).The larger Steel Ball is submerged in the water to exactly half of its diameter.

Which is true:

a.) The larger steel ball is exactly twice the diameter of the smaller ones.

b.) The diameter of the larger steel ball is less than twice the diameter of the smaller ones.

c.) The diameter of the larger steel ball is greater than twice that of the smaller ones.

d.) Cannot be determined.

Volume of Sphere = 4/3( Pi) r3

Pi = 3.141592653589793 (approx)

Answer 1

small ball = radius = r

A = base area of tank

V (water) in tank = A[2r] - 8[4pi/3] r^3

---------------------------------

large ball = radius = R

V1 (water) in tank now = A[R] - 2pi/3] [R]^3

given V = V1 >> no water loss

A[2r] - 8[4pi/3] r^3 = A[R] - 2pi/3] [R]^3

--------------------------------------...

A(R - 2r) + (2pi/3)[R^3 - 2 (2r)^3] =0

(R - 2r) { A + (2pi/3)[R^3 - 2 (2r)^3](R - 2r)} =0

2 options>> for equation to hold

either R = 2r >>>> will see later

(2pi/3)[R^3 - 2 (2r)^3](R - 2r)} = - A

(2pi/3)[R^3 - 2 (2r)^3](R - 2r)} < 0

[R^3 - 2 (2r)^3] <0

R^3 < 2 (2r)^3

R < (2)^1/3 [2r]

R < 1.26 [dia of small]

2R < 2.51[dia of small]

[dia of large] < 2.51 [dia of small]

[dia of large] < [2.51/2] [twice the dia of small]

[dia of large] < [twice the dia of small]

draw your conclusion >> as clear as water???