Physics · Mechanical Properties Of Fluids · NEET
When the ball starts falling, it speeds up because of gravity. But the viscous drag from Stokes' law, F = 6 pi eta r v, grows as the speed v grows. At some speed the upward forces (drag + buoyancy) exactly cancel the downward weight. Now the net force is zero, so by Newton's first law there is no acceleration and the speed stays constant. This constant speed is the terminal velocity.
Three forces act. 1) Weight downward = (4/3) pi r^3 rho g, where rho is the density of the ball. 2) Buoyant force upward = (4/3) pi r^3 sigma g, where sigma is the density of the fluid (Archimedes' principle). 3) Viscous drag upward = 6 pi eta r v (Stokes' law). At terminal velocity: weight = buoyancy + drag.
Set net force to zero: (4/3) pi r^3 rho g = (4/3) pi r^3 sigma g + 6 pi eta r v. Move buoyancy over: 6 pi eta r v = (4/3) pi r^3 (rho - sigma) g. Divide both sides by 6 pi eta r: v = [(4/3) pi r^3 (rho - sigma) g] / (6 pi eta r). Simplify the numbers (4/3)/6 = 2/9 and cancel one r: v = (2/9) r^2 (rho - sigma) g / eta.
Yes, and strongly. From v = (2/9) r^2 (rho - sigma) g / eta, the terminal velocity is proportional to r^2. So a ball with double the radius has 4 times the terminal velocity (if densities and fluid are the same). This r-squared link is a very common NEET trap and appears in the two-ball comparison questions.
If sigma is greater than rho, then (rho - sigma) is negative, so v comes out negative. A negative terminal velocity means the ball does not sink at all; it rises up (like an air bubble in water). The formula still works, the sign just tells you the direction of motion.
The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be:
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 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.
It is the constant maximum speed a body reaches while falling through a fluid, reached when the net force on it becomes zero.
v = (2/9) r^2 (rho - sigma) g / eta, where r is radius, rho is the ball's density, sigma is the fluid density, g is gravity, and eta is the fluid's viscosity.
Because the ball is inside a fluid, the fluid pushes it up with a buoyant force equal to the weight of fluid displaced (Archimedes' principle). This upward force must be subtracted from the weight, otherwise the terminal velocity comes out too large.
Yes. Terminal velocity is proportional to (rho - sigma). A denser ball has a larger rho, so a larger (rho - sigma), and therefore a higher terminal velocity, if the radius and fluid stay the same.
It decreases. Since v is proportional to 1/eta, a thicker (more viscous) fluid like glycerine gives a smaller terminal velocity, so the ball settles more slowly.
