Regular poster
Post Number: 36
| Posted on Sunday, 13 March, 2005 - 11:38 am: |
"An inverted cone represents a glass of depth 6cm when full. Find the percentage drop in depth when the glass is half full." Could anyone start me off in the right direction? Thanks
Frequent poster
Post Number: 101
| Posted on Sunday, 13 March, 2005 - 11:56 am: |
Let me know if that helps...
JG
Regular poster
Post Number: 42
| Posted on Sunday, 13 March, 2005 - 12:01 pm: |
Inverted cone means "A cone thats upside down".
I assume you know the volume of a cone. In this case the height is 6cm (unknown radius). This gives you an expression for the volume this particular cone, say V1 , which is full of liquid. If the glass is half full, the volume of liquid is halved, V1 /2. This occupies the bottom of the cone, in exactly the same shape, but a smaller cone. Giving you know V1 , the question asks for you to work out the height of this new cone of liquid, h1 , and hence the percentage drop, which would be ([h-h1 ]/h)*100 = 100(1-h1 /6).
ie if half the volume of water creates a depth of 4cm, then the drop would be 33% (6cm to 4cm).
If you still need pointers, just ask
Regular poster
Post Number: 37
| Posted on Sunday, 13 March, 2005 - 12:21 pm: |
I don't understand what you mean, I just have an expression for V1 and for V1/2
Regular poster
Post Number: 43
| Posted on Sunday, 13 March, 2005 - 12:39 pm: |
V=pr2 h/3 h=6cm in this case so V1=2pr2. If the volume is halved, V1 /2=pr2. If you look at John's picture you'll see the new situation. This volume, V1 /2, occupies the bottom of the glass, and you want to know how high the liquid is, h1. The problem is you've a new radius for the bottom of the cone of liquid, r1. The new cone of water has volume pr12/h1/3, but you can relate r1, r, h1 and h=6cm by considering similar triangles formed by the axis of the cone, the side of the cone and the surface of the water before and after. From this you can get an expression for h1 in terms of r and h=6. Then you can work out 1-(h1/6) (the fraction drop in water height).
Regular poster
Post Number: 38
| Posted on Sunday, 13 March, 2005 - 01:08 pm: |
Thanks JG and George for your help.