Prove that the loss in weight of a body when immersed wholly or partially in a liquid is equal to the buoyant force (or upthrust) and this loss is because of the difference in pressure exerted by liquid on the upper and lower surfaces of the submerged part of the body.
Prove that the loss in weight of a body when immersed wholly or partially in a liquid is equal to the buoyant force (or upthrust) and this loss is because of the difference in pressure exerted by liquid on the upper and lower surfaces of the submerged part of the body.
Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density ρ as shown in the figure above. Let the upper surface PQ of the body is at a depth h1 while its lower surface RS is at depth h2 below the free surface of liquid.
At depth h1, the pressure on the upper surface PQ,
P1 = h1 ρg.
Therefore, the downward thrust on the upper surface PQ,
F1 = Pressure x Area = h1 ρgA ……………….(i)
At depth h2, pressure on the lower surface RS,
P2 = h2 ρg
Therefore, the upward thrust on the lower surface RS,
F2 = Pressure x Area = h2 ρgA …………………(ii)
The horizontal thrust at various points on the vertical sides of body get balanced because the liquid pressure is the same at all points at the same depth.
From the above equations (i) and (ii), it is clear that F2 > F1 because h2 > h1 and therefore, body will experience a net upward force.
Resultant upward thrust or buoyant force on the body,
FB = F2 – F1
= h2 ρgA – h1 ρgA
= A (h2 – h1) ρg
However, A (h2 – h1) = V, the volume of the body is submerged in a liquid.
Therefore, upthrust FB = V ρg.
Now, V g = Volume of solid immersed x Density of liquid x Acceleration due to gravity
= Volume of liquid displaced x Density of liquid x Acceleration due to gravity
= Mass of liquid displaced x Acceleration due to gravity
= Weight of the liquid displaced by the submerged part of the body
Thus, Upthrust FB = weight of the liquid displaced by the submerged part of the body…..(iii)
Now, let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight.
Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid is collected in the measuring cylinder.
When the water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.
From the diagram, it is clear that
Loss in weight (Weight in air – Weight in water) = Volume of water displaced.
Or, Loss in weight = Volume of water displaced x 1 gcm-3 [Because the density of water = 1 gcm-3]
Or, Loss in weight = Weight of water displaced ……………(iv)
From equations (iii) and (iv),
Loss in weight = Upthrust or buoyant force
Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density ρ as shown in the figure above. Let the upper surface PQ of the body is at a depth h1 while its lower surface RS is at depth h2 below the free surface of liquid.
At depth h1, the pressure on the upper surface PQ,
P1 = h1 ρg.
Therefore, the downward thrust on the upper surface PQ,
F1 = Pressure x Area = h1 ρgA ……………….(i)
At depth h2, pressure on the lower surface RS,
P2 = h2 ρg
Therefore, the upward thrust on the lower surface RS,
F2 = Pressure x Area = h2 ρgA …………………(ii)
The horizontal thrust at various points on the vertical sides of body get balanced because the liquid pressure is the same at all points at the same depth.
From the above equations (i) and (ii), it is clear that F2 > F1 because h2 > h1 and therefore, body will experience a net upward force.
Resultant upward thrust or buoyant force on the body,
FB = F2 – F1
= h2 ρgA – h1 ρgA
= A (h2 – h1) ρg
However, A (h2 – h1) = V, the volume of the body is submerged in a liquid.
Therefore, upthrust FB = V ρg.
Now, V g = Volume of solid immersed x Density of liquid x Acceleration due to gravity
= Volume of liquid displaced x Density of liquid x Acceleration due to gravity
= Mass of liquid displaced x Acceleration due to gravity
= Weight of the liquid displaced by the submerged part of the body
Thus, Upthrust FB = weight of the liquid displaced by the submerged part of the body…..(iii)
Now, let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight.
Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid is collected in the measuring cylinder.
When the water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.
From the diagram, it is clear that
Loss in weight (Weight in air – Weight in water) = Volume of water displaced.
Or, Loss in weight = Volume of water displaced x 1 gcm-3 [Because the density of water = 1 gcm-3]
Or, Loss in weight = Weight of water displaced ……………(iv)
From equations (iii) and (iv),
Loss in weight = Upthrust or buoyant force
