# Bell shrinking due to water pressure

1. Oct 31, 2009

### semc

A bell of diameter 3m and bulk modulus of 14x1010N/m2 is toss into the ocean. By how much did the diameter decrease as it sank to a depth of 10km? The bell is assumed to be a sphere of diameter 3m.

Well o what i did was to calculate the pressure at the depth of 10km so P=Po+$$\rho$$gh which gives me P=1.0135x105+1000x9.81x10000=9.82x107. Then Bulk modulus=$$\frac{P}{V/Vi}$$ where V is change in volume and Vi is the initial volume. This gives me change in diameter to be 0.267m however the answer provided is 0.722mm What is wrong ??

2. Nov 1, 2009

### rl.bhat

Re: Pressure!!

3. Nov 1, 2009

### semc

Re: Pressure!!

Hello rl! My working is in the post? Okie i guess my post is too messy haha sorry about that.

Okie here is what i did

First i calculate the pressure at the depth of 10 km including the atmospheric pressure Po which is taken to be 1.013x105 Pa.

P=Po + $$\rho$$gh
P=1.013x105 + 1000 x 9.81 x 10000 = 9.82 x 107 Pa

So using the formula of bulk modulas,

Bulk modulas = $$\frac{P}{V/Vi}$$
14x1010=$$\frac{P}{V/Vi}$$

Since formula for both volume are the same V/Vi is reduced to r3/ri3 where r is the change in radius and ri is the initial radius.

Using 14x1010=$$\frac{P}{V/Vi}$$ and the P i found previously, i got the change in radius which is r to be 0.267m

Yup basically that's how i did it

Last edited: Nov 1, 2009
4. Nov 1, 2009

### rl.bhat

Re: Pressure!!

Vi = 4/3*π*R^3
V' = 4/3*π*( R - ΔR )^3
V/Vi = (Vi - V')/Vi = [R^3 - (R-ΔR)^3]/R^3
Expand the brackets and simplify to find V/Vi.

5. Nov 1, 2009

### semc

Re: Pressure!!

I don't get it. If we are looking for the change in radius why i can't just find the change in volume and from there calculate the change in diameter?

6. Nov 1, 2009

### rl.bhat

Re: Pressure!!

Find the difference in original volume and compressed volume.
Change in volume = Bulk modulus*Pressure* Original volume.

7. Nov 1, 2009

### semc

Re: Pressure!!

I am assuming you meant Change in volume = Bulk modulus*Pressure*Original volume? How did you get this? From my working I get Change in volume =$$\frac{Pressure x Original volume}{Bulk Modulus}$$ ??

8. Nov 1, 2009

### rl.bhat

Re: Pressure!!

Yes. You are right. It is typo.

9. Nov 1, 2009

### semc

Re: Pressure!!

So you mean that's the way to do it? But my answer is different from the answer provided

10. Nov 1, 2009

### rl.bhat

Re: Pressure!!

11. Nov 2, 2009

### semc

Re: Pressure!!

I am afraid i don't get you, this is my calculation and i can't figure out what's wrong with this and why do you keep asking me to post this?

$$\Delta$$r3=$$\frac{9.82*10^7*1.5^3}{140*10^9}$$
$$\Delta$$r=0.1333m $$\Rightarrow$$ $$\Delta$$d=0.2666m

12. Nov 2, 2009

### rl.bhat

Re: Pressure!!

Sorry. Your calculation is not correct.
Vi = 4/3π*r^3
Vf = 4/3π*(r - Δr)^3
Chane in volume =ΔV = Vi - Vf = ...........?
Then ΔV = P*Vi/B gives you Δr.

13. Nov 2, 2009

### semc

Re: Pressure!!

ΔV = Vi - Vf
Δr = ri3-rf3
ΔV = P*Vi/B

Vi - Vf = (P/B)*Vi
$$\frac{4}{3}$$$$\Pi$$(ri3-rf3)=(P/B)$$\frac{4}{3}$$$$\Pi$$ri3

(ri3-rf3)=(P/B)[/tex]ri3
Δr3 = (P/B)ri3
Δr3 = $$\frac{9.82*10^7*1.5^3}{140*10^9}$$
Δr = 0.133 $$\Rightarrow$$ Δd=0.266

Whats wrong in my calculation?

14. Nov 2, 2009

### rl.bhat

Re: Pressure!!

Δr = (ri - rf)
Δr^3 is not equal to ri^3 - rf^3
In algebra
(a^3 - b^3) = (a - b)(a^2 + ab + b^2)