Urgent halp in a math problem solving question for ninth grade(involves volume and etc) please?

Question

Ok, this question is taken from the IGCSE math textbook, i have no clue how to do it! please help me asap, thanks

A spherical ball is immersed in water contained in a vertical cylinder. The rise in water level is measured in order to calculate the raduis of the spherical ball. Calculate the radius of the ball in the following cases ..............................

a) cylinder of radius 10 cm, water level rises 4 cm
b) cylinder of radius 100 cm, water level rises 8 cm
please help! thankyou in advance..

Answer 1

Its simple.

The radius of the cylinder is 10cm, the rise in water level is 4cm, so the volume of the water displaced is
pi*10*10*4 = 1256.637 cu cm

The volume of water displaced by an object is the same as the volume of the object, so the volume of the sphere = volume of displaced water.

Volume of a sphere is 4*pi*r*r*r/3 = 1256.637

r^3 = 1256.637 * 3 / 4pi
r^3 = 300
r = 6.694cm

The second bit can be done the same way.

Answer 2

For this, you need to know the formulae of both a spehre and a cylinder.

For a sphere, the formula is 4/3πr³

For a cylinder, the formula is hπr².

In these problems, you need not know the actual height of the cylinders, just the change in height, or Δh, so the formula is Δhπr².

Since the radius of the cylinder is going to be greater than the radius of the sphere, we can denote that by using R to represent the radius of the cylinder and r to represent the radius of the sphere, giving us the formula ΔhπR²

Since we know that the variables of the cylindracal formula represent 4cm and 10cm, the volume of displacement of the water is 4cmπ10cm², or π400cm³.

So, that will also be the volume of the sphere.

4/3πr³=π400cm³.

So, now all it takes is to work it out.

The π's cancel each other out, leaving us with 4/3r³=400cm³

Multiplying both sides by 3 gives us 4r³=1200cm³.

Dividing both sides by 4 gives us r³=300cm³.

Cube rooting gives us r=*the cubed root of three hundred, cm.

Simplification of the cubed root of three hundred is not possible as the prime factors of three hundred are 3*5*5*2*2, none of which are triplicated, so that's the answer.

Or, we can apporoximate the answer to be 6.69cm.



Plugging the data for part b) into our formulae, we get

4/3πr³=8cm*π*(100cm)², or
4/3πr³=8*π*10000cm³, or
4/3πr³=π*80000cm³, or
4/3r³=80000cm³, or
4r³=240000cm³, or
r³=60000cm³, or
r=the cube root of 60000cm,
which can be simplified to be 2*5*the cube root of 4*3*5, cm, or

r=10*the cubed root of sixty.

Approximations would be 39.15cm.

Answer 3

When the ball is immersed the displacement of water is equal to the volume of the sphere.

For a) - first we need to work out the volume of water change (i.e. a 4 cm rise)

pi x radius x radius x rise of water = volume of sphere

3.142 x .10 x .10 x .04 (I work in SI units which are metres)
= 1.2568 x 10^-3 metres^3

Sphere volume = (4 x pi x radius^3) / 3

(3 x 1.2568 x 10^-3) / 4 x pi = radius of sphere^3

Radius of sphere = 0.06695 metres or 6.695 cm

b) Exactly the same principle....

pi x radius^2 x water rise = pi x 1^2 x .08 (again working in metres)
= 0.251

Vol of sphere = 0.251 metres^3

For the radius-:
Vol. of sphere = 4 x pi x radius^3 / 3
3 x 0.251 / 4 x pi = radius^3

radius^3 = 0.05992
radius = 0.39131 metres or 39.131 cm

Hope you understand how I got this, but it is a bit tough for 9th grade. I'm an MSc Physicist and certainly had to think about how to work this one out....

Anyway - ask your teacher how they would do it, probably similar method to mine. In physics we do this experiment with a 'Eureka Can' and measure an objects volume by how much water it displaces from the can.

Answer 4

Hint : Volume of water displaced is equal to volume of the sphere
ie Volume of water displaced = pi * r^2 * h , r - radius of cylinder, h - height of water raise
Volume of sphere = (4/3) * pi * r^3 r- radius of sphere
equating the two you can find out the radius of sphere

Answer 5

vol of cyl = pi x radius^2 x height
a.) vol cyl = pi x r^2 x h
vol of water that rises = pi x 10^2 x 4 = pi x 400 = vol sherical ball = 4/3 x pi x r^3 = pi x 400
r^3 = 300
r = cube root of 300 cm.
b.) same soln.
r^3 = 60000
r = cube root of 60000 cm.
hope u understand.

Answer 6

V(sphere) = 4/3(pi)(r^3)

Volume displaced = V(sphere)

For case a:
Volume displaced = (pi)(10)(10)(4)
= 400(pi)

4/3(pi)(r^3) = 400(pi)

r^3 = (400)(3/4)

therefore,

r = 6.69cm

Follow these steps and you’ll get the answer for other case.

Answer 7

the volume of the water that rises must be equal to the volume of the sphere...
volume of water = v=pi*r^2*h=3.14*10^2*4=1256.64 cm^3
volume of sphere is = v=(4/3)*pi*r^3
r=6.69cm

again
volume of water= v=251327.41cm^3
volume of sphere = v=4/3 pi* r^3
r=39.15cm

Answer 8

I think this question is way too hard for nineth grade! I dont have a clue and I'm in my final year of a degree!