A cylinder of dia 4cm is filled with water & 300 lead balls increases its level by .8 cm. What is the dia of each ball?

Question

A cylindrical vessel of diameter 4cm is partly filled with water and 300 lead balls are dropped into it which increases its level by .8 cm. What is the diameter of each ball?

Answer 1

Diameter of each ball is 8.6mm.
First find the volume of the water rise. V = 3.14 x r x r x h. V = 3.14 x 2 x 2 x 8 = 100.53. This is then equal to the volume of 300 lead balls.
Now Volume of a ball (sphere) is: V = 4/3 x 3.14 x r x r x r, so for 300 balls, 100.53 = 300*4/3*3.14*r*r*r. If you work this out, the answer comes to 4.3mm. As the question asks for diameter, then the final answer is 8.6mm.

Comments

its .8 cm not 8, do you like to correct your answer (use edit button)

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I can fix it.
Volume increase = pi * R^2 * h
Volume of all balls = N * 4/3 * pi * r^3
Both volumes equal:
pi * R^2 * h = N * 4/3 * pi * r^3
h = N * 4/3 * r^3/R^2
r^3 = 3/4 * h/N *R^2 = (2*d)^3
d = 0.5 * (3/4 * 0.8/300 * 4 * 4)^(-3)
d = 0.5 * (0.8 * 4/100)^(-3)
d = 0.5 * 0.032^(-3) (exact answer)
It gives a diameter of 1.59 mm per ball.

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Still this diameter issue... sorry, replace R by 2cm:
d = 0.5 * (3/4 * 0.8 / 300 * 2 * 2)^(-3)
d = 0.5 * 0.008^(-3)
Thus the exact answer should rather be a diameter of 1 mm per ball.

Answer 2

Volume of water rise
V1=Pi*R^2*h.......................(R is radius of cylinder=D/2=4/2=2cm=20mm)
V1=Pi*20*20*8...........................(all dimensions are in mm)
V1=3200Pi............................(i)
Volume of 300 sphere balls
V2=300*(4/3)*Pi*r^3
V2=400*Pi*r^3........................(ii)
as V1+V2 so
3200Pi=400Pir^3
3200=400r^3
r=2mm
d=4mm
Ans :Dia. of each ball is 4mm