Physics · Mechanical Properties Of Fluids · NEET
Not always. Terminal velocity depends on radius squared AND the density gap (rho_body - rho_medium). A bigger radius pushes vt up strongly (r^2), but if the bigger ball is made of a lighter material, its smaller density gap can pull vt back down. You must compare both effects together using the ratio.
Because the ball is not just fighting its own weight. The surrounding fluid pushes up with a buoyant force (upthrust). The net downward pull is due to the effective density (rho_body - rho_medium). If the medium were as dense as the ball, the gap would be zero and terminal velocity would be zero, the ball would just float.
No. Equal mass with different radius means different volume, so different density. Since mass = density x volume and volume = (4/3) pi r^3, the smaller ball must be denser. In the NEET problem rho1 = 8 rho2 while r1 = 1 mm and r2 = 2 mm, and 8 = (2/1)^3, which confirms equal mass. Always check this link before comparing.
Not directly in the ratio formula. The formula vt = (2/9) r^2 (rho_body - rho_medium) g / eta uses radius and density, not mass. Mass is hidden inside density (rho = mass/volume). So do not plug mass in directly, convert to density and radius first.
Terminal velocity becomes 4 times larger, because vt is proportional to r^2 and 2^2 = 4. This is only true if the density gap and medium stay the same. This r^2 rule is the most common thing NEET tests.
Two small spherical metal balls of equal mass are made from materials of densities rho1 and rho2 (rho1 = 8 rho2) and have radii 1 mm and 2 mm respectively. They fall vertically from rest in a viscous medium of coefficient of viscosity eta and density 0.1 rho2. The ratio of their terminal velocities is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
vt1/vt2 = [r1^2 (rho1 - rho_m)] / [r2^2 (rho2 - rho_m)], where r is radius, rho is the ball density and rho_m is the medium density. The constants (2/9), g and eta cancel because both balls fall in the same medium.
Both balls fall in the same fluid under the same gravity, so g, eta and the constant 2/9 are identical for both. When you divide vt1 by vt2, these common factors cancel, leaving only radius and density terms.
Yes. If the ball density equals the medium density, the gap (rho_body - rho_medium) is zero, so terminal velocity is zero. The ball neither sinks nor rises, it stays suspended.
The formula vt = (2/9) r^2 (rho_body - rho_medium) g / eta and Stokes law apply to small smooth spheres in slow (laminar) flow. For other shapes or fast flow, this simple ratio does not hold.
Yes, the same density gap idea works. For a bubble, rho_medium is larger than rho_body, so the gap is negative and the terminal velocity points upward, the bubble rises steadily.
