Question

Calculate the buoyant force acting on a body of volume 200 cm³ when it is dipped in water with :

a) Half the volume in water.

b) Completely submerged in water.

[Given : Density of water = 1000 kg/m³, g = 9.8 m/s²]

Explain with complete calculations & justifications.

Points : 20 ☺

Answer 1

According to Archimedes' principle ; The buoyant force (say B) acting on a body kept in a liquid(say of density 'd') is equal to weight of liquid displaced by it.
Mathematically ,
B = Vdg
where 'V' is volume of body dipped inside a liquid and 'g' acceleration due to gravity .

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NOTE : weight of liquid displaced (W);
W= mg
where 'm' is the mass of the liquid displaced .
and using ;
density = mass/Volume .
=> mass = density ×Volume .
Hence :
m = d × V
=> W = Vdg = B
______________________
a) When body is dipped with half the volume in water ;
V = ½×200/1000000 m³
= 1/10000 m³
B= (1/10000)×(1000)×(9.8)
= 0.98 N
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b) when completely submerged in water then ,
V = 2/10000 m³
B = (2/10000)×(1000)×(9.8)
= 1.96 N
_______________________
hope it helps!

Answer 2

b) Given,

Volume of the water = 200cm^3 or 200/10^6 m^3

Now according to Archimedes law,

weight of the water which is displaced = the loss of weight = buoyant force

The weight of the displaced water= Vdg [V = Volume ; d = density of water ; g = gravitational force]

W = 200/10^6 × 1000 × 9.8

= 2/10000 × 1000 × 98/10

=1.96N

a)Here the volume is half,

Weight of the water displaced = V/2dg = buoyant force

W = V/2 × d × g

= 2/20000 × 1000 × 98/10

= .98N

Hope it helps.

P.S : I answered the second question first.