✔ 最佳答案
The centrifugal force F experienced by a small volume of liquid V in a test-tube of cross-sectional area A in a centrifuge is given by,
F = (pV)r.w^2
where p is the density of liquid and r is the radius of revolution of the volume of liquid.
Hence, pressure P produced = F/A = (pV)r.w^2/A = p(V/A)r.w^2
since V/A = L, the height of liquid in the tes-tube
P = pL.rw^2
Thus, pressure P increases with r
The presure gradient established in the liquid because of centrifugal force
= dP/dr = pLw^2
which is a constant
Hence, for particulates (with higher density p) in suspension in the liquid, they will experience a higher pressure gradient, and hence will be "thrown" to the end of the test tube by the centrifuge.