Question

A solid brass (bulk modulus 6.7 x 1010 N/m2) sphere is subjected to a pressure of 1.0 x 105 Pa due to the earth's atmosphere. On Venus the pressure due to the atmosphere is 9.0 x 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.

Answer 1

Answer:

$\frac{r}{r_0}=-4.4x10^{-5}$

Explanation:

Using the equation to find the fraction does the radius of the sphere change when it is exposed to Venusian atmosphere.

$\frac{r}{r_0}=\frac{1}{3}*\frac{V}{V_0}$

Replacing to find the relation between the radius

$\frac{V}{V_0}=-\frac{P}{B}$

Replacing numeric to find the relation

$\frac{r}{r_0}=\frac{1}{3}*\frac{8.9x10^6pa}{6.7x10^{10}pa}$

$\frac{r}{r_0}=\frac{1}{3}*-1.33x10^{-4}$

So the relation is

$\frac{r}{r_0}=-4.4x10^{-5}$