## Friday, September 22, 2006

### Two Questions on Properties of liquids

The weight of a floating body is equal to the weight of the displaced liquid. Question setters often find this law of floatation handy while setting Medical and Engineering test papers. Consider the following question:
A piece of wood having volume ‘V’ and density ‘d’ floats at the interface of two immiscible liquids of densities ‘ρ’ and ‘σ’ respectively. If ρ > d > σ, the ratio of the volumes of the parts of the wooden piece in the rarer and denser liquids is
(a) (ρ-d)/ (d-σ) (b) (d-σ)/(ρ-d) (c) (ρ-d)/(d-σ) (d) (ρ+d)/(ρ+σ) (e) (d-σ)/(ρ+σ)

Equating the weight of the wooden piece to the weight of the displaced liquids, we have,
Vdg = vσg + (V-v)ρg where ‘v’ is the volume of the part of the wooden piece in the rarer liquid (of density ‘ρ’). Rearranging this equation, we have,
v(ρ-σ) = V(ρ-d) so that v/V = (ρ-d)/ (ρ-σ).
Therefore, v/(V-v) = (ρ-d)/(d-σ)
The correct option is (a).
Now consider the following M.C.Q. which appeared in the Kerala Medical Entrance test paper of 2006:
Blood is flowing at the rate of 200cm3/s in a capillary of cross sectional area 0.5m2. The velocity of flow in mm/s is
(a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 (e) 0.5
This is a very simple question. From the equation of continuity (av = constant), we have,
0.5 v = 200×10-6 so that v = 4×10-4m/s = 0.4mm/s [option (d)].