A sphere of lead of radius 1.5 mm falls at a constant velocity vertically through…

PHYSICS · Multi-part · CBSE · Physics · Mechanical Properties of Fluids · Mechanical Properties of Fluids

Problem

A sphere of lead of radius 1.5 mm falls at a constant velocity vertically through water of a viscosity 1.0×10−3 Nsm−21.0 \times 10^{-3} \text{ Nsm}^{-2} . (1) - a Draw a free-body force diagram of the sphere showing the forces acting on the sphere. (2) - b By equating the weight of the sphere with its buoyancy force and its drag force, show that the terminal velocity of the sphere is given by the expression:Vterminal=2r2g(ρlead−ρwater)9η,V_{\text{terminal}} = \frac{2r^2g(\rho_{\text{lead}} - \rho_{\text{water}})}{9\eta} ,where ρlead\rho_{\text{lead}} is the density of lead and ρwater\rho_{\text{water}} is the density of the fluid through which it is falling. (3) - c Given that the density of lead is 1.14×104 kgm−31.14 \times 10^4 \text{ kgm}^{-3} , calculate the speed at which the sphere is falling. ( g=9.81 Nkg−1g = 9.81 \text{ Nkg}^{-1} ) (4) - d In practice, such a sphere falling through water falls with a terminal velocity that is smaller than the value that Stokes' law predicts. Suggest a reason why this is. (5) - e If the same sphere were to fall through water of a higher temperature, state what would happen to the terminal velocity of the sphere and suggest a reason to support your statement.

Related theory

Browse the problem library