A piece of wood of mass 40g and uniform cross-sectional area of 2cm² floats upright in water. The length of the wood immersed is

Acdas

1 year ago

Answer is C
Mass is 40g, so therefore , Volume is alos 40cm³.
Area is 2cm²

Recall, Volume = Area × Height
so,
40cm³ = 2cm² x Height.
Height = 40cm³ / 2cm² = 20cm.

Stock

10 years ago

C

Somysoph

5 years ago

When a body floats density of liquid= density of the body/object immersed since the mass is in g and area is in cm2 the density of the body will be in gcm-3. The density of water (the liquid the body was immersed )=1gcm-3 meaning density of the body =1. 1=mass/area ×l(I.e density =mass/volume.)

Refhord

3 years ago

Just use your number 6

SodiumM

7 years ago

correct option is B.
U = weight of equal volume of water displaced = weight of the body
U = density x Volume x g = mg
density x area x height x g = mg
h = m / density x area = 0.04 / 1000 x 10-4 = 0.4m = 40cm.

Just understand the basic principles. Thanks