Dripping Cones

The ratio of the volumes of similar cones is the ratio of one of their dimension, cubed (since there are 3 dimensions with the same ratio).

Total volume of water: $2^3 \times c = 8c$ (where c is a constant)

Volume of upper cone: $1^3 \times c = c$

Volume of lower cone: $8c - c = 7c$

The height of water in the lower cone is $\sqrt[3]{7}$ .

Its pretty easy. Most people will answer it wrongly (if at all) as they assume that the amount of water is same above and below 1 cm(and probably answer 1 ), whereas the cone has a larger volume towards its base (here the top of the cone).

Assume the cone has a radius r and height h . Now initially the amount of water is $pi \times r^2 \times h$ . Now since the new height is half of the initial height, the radius of the $\frac {h}{2}$ height cone is also half, using similarity. So final volume of water in top cone is $pi \times (\frac {r}{2})^2 \times (\frac {h}{2})$ .

So amount of water that drips down to bottom cone is $pi \times r^2 \times h - pi \times (\frac {r}{2})^2 \times (\frac {h}{2}) = \frac {7}{8}\times pi\times r^2 \times h$ . Now let the height of water in the bottom cone be x , then using similarity, its radius is $r \times \frac {x}{h}$ . So the amount of water in the bottom cone is $pi \times (r \times \frac {x}{h})^2 \times x = \frac {7}{8}\times pi\times r^2 \times h$ . So putting $h=2$ , we get $\boxed{x^3 = 7}$ . So the answer is $\boxed{\sqrt[3]{7}}$ .

Assume radius and height for both cube is 2 cm. ie base angle is 45 deg.

Now Full volume of upper cube = 8 pi/3

Volume with one cm height = pi/3

Net water flowed down = 8pi /3 - pi/3 = 7pi/3

Now since base angle is assumed 45 degree radius at all points = height.

Assume height at bottom cone is x => radius is also x

Therefore, (x)cube pi/3 = 7 pi/3

=> (x) cube = 7 => x = cube root 7

More generic solution may be worked out seperately.

assume some slope k. so r at any hight of cone is proportional to h thus volume is proportionla to h^3. now when height is reduced from 2 to 1 the volume of liquid is proportional to (2^3-1^3) =h2^3, where h2 is the height of the 2nd cone

Total volume of water is constant so $\frac { 1 }{ 3 } \pi { { R }_{ 1 } }^{ 2 }{ H }_{ 1 }+\frac { 1 }{ 3 } \pi { { R }_{ 2 } }^{ 2 }{ H }_{ 2 }=\frac { 1 }{ 3 } \pi { { R } }^{ 2 }{ H }$

Also, H and R are in the same proportion, so equation reduces to ${ H_{ 1 } }^{ 3 }+{ { H }_{ 2 } }^{ 3 }={ { H } }^{ 3 }$

So ${ H_{ 2 } }={ { \sqrt [ 3 ]{ { H }^{ 3 }-{ { H }_{ 1 } }^{ 3 } } } }$ = Cube Root of 7

For a right circular cone, the height $h$ and radius $r$ follow the following relationship:

$r=h\tan \theta$ ... Where $\theta$ is the semi-vertical angle

So as long as $\theta$ doesn't change, we can safely say that volume $V$ :

$V=\frac{1}{3}\pi r^{2}h=kh^{3}$ .... Where $k=\frac{1}{3}\pi\tan^{2}\theta$ , which does not change in this case.

Volume of water lost by first cone $V_{o}$ :

$V_{o}=k(h_{initial})^{3}-k(h_{final})^{3}$

$\therefore V_{o}=k(2^{3}-1^{3})$

$\therefore V_{o}=7k$ cu. cm.

This lost volume would be the volume of water in the second cone, now having water up to height $h^{'}$ .

$\therefore k(h^{'})^{3} = 7k$

$\therefore (h^{'})^{3}=7$

$\therefore h^{'}=\boxed{7^{\frac{1}{3}}}$ cm.

i did'nt done any solution,as volume of cone above half is more grater than below, 1is obviously not and sqrt(8) is more than 2 hence it not correct,cube root of 7 is more near to volume of cone than cube root of 6 hence i have marked it.

Let the original height of the cone be H and its original radius be R.For a right circular cone, Radius r=height h ---- (1)

r / R = 1/ H

Therefore change in volume = 1/3πR2H – 1/3π(R/H)2(1) =1/3πR2(H3-1)/H2 ---------(2)

Let the water in the lower cone be upto a height h. Again using (1),r/R=h/H, which gives r=Rh/H

Therefore volume of water in lower cone = 1/3π(Rh/H)2h = 1/3πR2h3/H2-------------------(3)

Now, change in volume=volume in lower cone i.e., (2)=(3) this gives,

1/3πR2(H3-1)/H2 = 1/3πR2h3/H2

After cancelling common terms i.e., (1/3), π, R2 in the numerator and H2 in the denominator

H3-1=h3

We need to find h and we are given that height of cone i.e., H=2cm, which gives

h3=(2)3-1= 8-1=7cm. The answer is cube root of 7.

Let r be he radius of each cone( equal, not necessarily =h) the total volume of the water in upper cone is pi r^2 h/3. When the height of the water is 1, we can find using similarity of triangles that the radius will be r/2 and the volume of water gone down is pi r^2 2/3 - pi (r/2)^2 1/3= 7 pi r^2/12. Now this volume of water is collected in lower cone. Let the height of the water be h, by similarity of triangles again we can find its radius r h/2 and so 7 pi r^2/12= pi (r h/2)^2 h/3 and solving this equation we get h^3=7 and so h= cube root of 7.

We know the volume of cone is (1/3)pi (r^2) h Considering angle of slant to be ᶿ. we get the formula for volume as (1/3)pi (h^3) tan ᶿ.

So, Amount of water in lower cone = amount water spilled from upper cone

let h be the height of water in lower cone.

then,

(1/3) pi (h^3) tan ᶿ ={ (1/3) pi (2^3) tan ᶿ } - { (1/3) pi (1^3) tan ᶿ}

Cancelling out (1/3) pi tan ᶿ from both sides.

h^3 = 8 -1

h^3 = 7

h = cube root of 7

total volume of a cone=1/3 * 22/7 * r^3. So when the upper cone is full of water the volume is (8 * 22/7)/3.When the height of water level in the upper cone is 1cm, then the volume is(22/7)/3.Therefore the volume in the lower cone should be the subtracted value of the full vol. of the upper cone - the vol. when the height is 1cm=(7 *22/7)/3.Deriving the height from here, comes out the value of cube root of seven

Volume of Top Cone when full = (1/3) pi R R H, R = radius, H = full height = 2cm Therefore, Volume of Top Cone when full = (2/3) pi R R. ----------------> (A) Volume of Top Cone when height is 1cm is (1/3) pi R R/4.---------------(B) The volume in Bottom cone = A-B = (1/3) pi R R (7/4). Now let h be the height if water in the second cone and r be the radius of the water filled part of the second cone. Then, the volume of water in the second cone is (1/3) pi r r h. This is equal to (1/3) pi R R (7/4). Hence, we have r r h = R R 7/4 ---------------->(C) Using symmetry, we know that h/r = H/R i.e. r = h*R/2, since H = 2cm Using this in C, we get h=7^(1/3).