Bell shrinking due to water pressure

[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=P_o+$$\rho$$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 ??

 

[rl.bhat]

Show your calculations.

 

[semc]

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

P=P_o + $$\rho$$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

 

[rl.bhat]

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.

 

[semc]

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?

 

[rl.bhat]

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

 

[semc]

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 =$$\frac{Pressure x Original volume}{Bulk Modulus}$$ ??

 

[rl.bhat]

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 =$$\frac{Pressure x Original volume}{Bulk Modulus}$$ ??

Yes. You are right. It is typo.

 

[semc]

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

 

[rl.bhat]

Show your calculations instead of answer.

 

[semc]

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

 

[rl.bhat]

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.

 

[semc]

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?

 

[rl.bhat]

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)