Question

Normal forces are applied uniformly over the surface of a spherical volume of water whose radius is 20.0 cm. If the pressure on the surface is increased by 200 MPa, by how much does the radius of the sphere decrease?

Answer 1

r_i = initial radius = 20 cm = 0.20 m

r_f = final radius = ?

K = bulk modulus of water = 2.2 x 10⁹

initial volume is given as

V_i = 4r_i³/3

Change in volume is given as

V = (4/3) (r_f³ - r_i³)

P = Change in pressure = 200 MPa = 200 x 10⁶ Pa

Bulk modulus is given as

K = - P V_i /V

V = - P V_i /K

(4/3) (r_f³ - r_i³) = - P (4r_i³/3)/K

(r_f³ - r_i³) = - P r_i³ /K

r_f³ = r_i³ (1 - P /K)

r_f = r_i (1 - P /K)^1/3

Change in radius is given as

r = r_f - r_i

r = r_i (1 - P /K)^1/3 - r_i

inserting the values

r = (0.20)(1 - (200 x 10⁶) /(2.2 x 10⁹ ))^1/3 - (0.20)

r = - 0.006254

r = - 0.63 cm

hence the radius decrease by 0.63 cm