A solid brass sphere is subjected to a pressure of 1.00 105 Pa due to the earth\

Question

A solid brass sphere is subjected to a pressure of 1.00 105 Pa due to the earth's atmosphere. On Venus the pressure due to the atmosphere is 9.00 106 Pa. By what fraction r/r0 (including the algebraic sign) does the radius of the sphere change when it is exposed to the Venusian atmosphere? Assume that the change in radius is very small relative to the initial radius. r r0 =

I was given the answer below, but it is not the correct answer for webassign. Can someone check this work?

Bulk modulus of brass is B = 6.1X1010N/m2

Now B = -VdP/dV

Volume of sphere is V = (4/3)r3

or dV/dr = 4r2

dV = (4r2)dr

B = -VdP/dV = -[(4/3)r3]dP/[(4r2)dr]

or B = -(1/3)rdP/dr

or dr/r = -(1/3)dP/B

so dr/r = -(1/3)( 8.9X106Pa)/( 6.1X1010N/m2)

or dr/r = -4.863X10-5

So, the fractional change in radius of sphere is = -4.863X10-5

sign is -ve as the radius decreases making the change -ve.

Explanation / Answer

bulk modulus of brass is 6.7*10^10.

B=-V*dP/dV

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

dV=(4/3)*pi*3*r^2*dr=4*pi*r^2*dr

B=-(4/3)*pi*r^3*dP/(4*pi*r^2*dr)

=(-1/3)*r*dP/dr

==>B*dr/r=(-1/3)*dP

integrating both sides,

B*ln(r)=(-1/3)*P

if radius on earth is r0 and radius on venus is r1

then B*ln(r1/r0)=(-1/3)*(pressure on venus-pressure on earth)

==>ln(r1/r0)=(-1/3)*(9*10^6-10^5)/(6.7*10^10)

==>r1/r0=0.999955722373318

==>(r1/r0)-1=0.999955722373318-1

==>delta_r/r0=-4.42776*10^(-5)

so fraction change is -4.42776*10^(-5)

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