Question

a cylindrical solid ,area of cross section 0.0002m² and length 0.40m is completely immersed in water calculate

(i)wt of solid in si system (ii)upthrust acting on the solid in si system (ii)apparent weight of the solid in (a)water (b)alchohol

take g=10m/s² density of water=1000kg/m³ density of alchohol=800kg/m3density of solid=1200kg/m³

Answer 1

volume of the cylinder, V = area of cross-section x length

or

V = 0.0002 m² X 0.40 m

so, V = 8 x 10^-5 m³

the density of solid, d = 1200 kg/m³

(1)

the true weight of the solid cylinder would be, W = mg = (Vd).g

or

W = (8 x 10^-5 X 1200) X 9.81

so,

W = 0.941 kg

(2)

the upthrust will be given as

F' = d'gV

d' = density of water = 1000 kg/m³

V = full volume of the cylinder = 8 x 10^-5 m³

so,

F' = 1000 X 8 x 10^-5 m³ X 9.81

thus, we get

F' = 0.784 N

and similarly, the buoyant force due to alcohol will be

F'' = d''gV

or

F'' = 800 X 8 x 10^-5 m³ X 9.81

so,

F'' = 0.627 N

(3)

The apparent weight of solid will be given as

W' = W - F

here

F is the buoyant force acting on it due to either liquid or water.

Now for water F = 0.784N

so, the apparent weight in water will be

W' = 0.941 - 0.784

or W' = 0.157N

and similarly the apparent weight in alcohol would be

W'' = 0.941 -0.627

thus,

W'' = 0.314N