Archimedes' principle - class-XI

According to Archimedes' principle, a floating object displaces a weight of fluid equal to its own weight. Since the weight of the block remains constant regardless of its orientation, the weight (and thus the volume) of the water displaced remains the same, so the water level does not change.

When the iron pieces are in the boat, they displace a volume of water equal to their weight divided by the density of water. When thrown into the water, they sink and displace only a volume of water equal to their own volume. Since the density of iron is much greater than that of water, the volume displaced when submerged is less than the volume displaced when floating, causing the water level to decrease.

The loss in weight in water equals the buoyant force, which is the weight of the displaced water. Using density of water (1 g/cm^3), the volume of the coil is 10 cm^3. Given the cross-section of 0.5 mm^2 (0.005 cm^2), the length is Volume/Area = 10 / 0.005 = 2000 cm, which matches 10^5 / 50.

FALSE
If hollow metal container  displaces more weight of water than its own weight it will sink.
Because, we know that by Archimedes principle, the weight of water displaced should be equal to the weight of the part submerged in water. If the weight is greater than the weight of the object, it will sink.

Density of ice is less than that of water. If it melts it becomes water. But still, the density of liquid is greater than water therefore, water remains on top and thus it floats and then overflows.

Objects weigh more in air than in water. When submerged, the 3 N block appears to weigh only 1 N. The missing weight is equal to the weight of the water displaced. The 2 N weight is equal to the buoyant force.
That is, Loss in weight of body = Weight of water (liquid) displaced by the body = Buoyant force or upthrust exerted by water (any liquid) on the body.

Any solid floats on water if its density is less than the density of water(i.e. 1).So, it will sink.

Since condition for sinking of a body of density $\sigma $ in a liquid of density S is 

Conditions for sinking of a body of density $\sigma $ in a liquid of density S is 

a body of higher density would sink in a liquid of lower density 

Conditions of a body of density $\sigma $ to float in liquid of density S is 

The buoyant force depends on the density of liquid. When the density of the object is more than the density of the liquid, it sinks because the gravitational force acting on the object is more than the buoyant force of water.

Objects that sink will always displace the same volume of water as that of body which gives the net buoyant force applied on the body.

In a lifting balloon, force of gravitation acts downwards as on all other materials present on the earth, given by $mg$.

Whenever a body is wholly or partially immersed in a liquid it experience an upthrust or buoyant force.

Hare`s apparatus is used  to compare the densities of two liquids and is base on Archimedes principle .

when a body is immersed in a fluid an downward force mg acts on it and to balance this force an upward force kown by the name buoyant force is exerted by the fluid .

Before pushing body was in the position of equilibrium so if we disturbs the eqm than with the passage of time it would come back to its eqm position .

Archimedes principle states that  loss in weight of a partially or fully immersed body in a fluid is equal to the weight of the fluid displaced .

The ball floats due to the buoyant force, which depends on the pressure gradient in the fluid. When the air is removed, the pressure at the surface becomes zero, but the pressure gradient (rhogh) remains the same as long as the fluid is incompressible; however, if the ball is held up by air pressure or if the surface tension/meniscus changes, it might sink. In standard physics problems of this type, the removal of atmospheric pressure does not change the buoyancy of a submerged object, but if the ball is floating, the lack of air pressure on the top surface while the water pressure remains below causes it to sink deeper.

As the wooden block rises, the buoyant force remains constant (rhoVg) while the weight is constant. Since the density of wood is less than water, the net upward force is constant, resulting in constant acceleration.

When the 200 gm mass is removed, the buoyant force decreases by an amount equal to the weight of 200 gm, which corresponds to the volume of the cube of side L that rises by 2 cm (h = 2 cm). Thus, buoyant force difference = A * h * rho_w * g = 200 gm * g. Solving L * L * 2 cm * (1 g/cm^3) = 200 g gives L^2 = 100 cm^2, so L = 10 cm.

Let $\rho $=density of sugar solution $> \rho _{w}$
m=mass of ice.
In floating condition weight=upthrust
$\therefore $   $mg=V _{I}\rho g$
or $\displaystyle V _{I}=\frac{m}{\rho }$          (i)
When ice melts, m mass of ice becomes m mass of water. Volume of this water formed.
$\displaystyle V _{F}=\frac{m}{\rho _{w}}$          (ii)
Since $\rho _{w}< \rho $, $V _{F}> V _{I}$. Hence level will increase.

At room temperature and standard pressure, the first family of alcohol (methanol, ETHANOL and iso-propyl) all have densities of about 0.79g/cc compare to water at the same conditions with a density of 1.0g/cc.

Graduated cylinder  is used to measure the volume of a solid by water displacement. The volume of water displaced will be equal to the volume of the solid.

When the balls are in the boat, they displace water equal to their weight. When thrown into the tank, they displace water equal to their volume. Since the density of steel is much greater than water, the volume of water displaced when submerged is less than the volume of water displaced when floating, causing the water level to fall.

There Will be no change in the water level of the pond, as the water which was drawn from the pond went to the man's body, thereby increasing his weight. So the decrease in level of water will be compensated by volume displaced by increase in weight of the man.

False because the liquid displaced has the same weight as the wooden block has The weight of the displaced liquid from the vessel is same as the weight of the wooden block.
Hence,  the weight will remain the same.

volume of liquid inside the liquid1=2/3rd of total volume

volume of liquid inside the liquid2 = 1/4th of total volume

density of liquid= (2/3)/(1/4) = 8/3 g/cc

Since the weight remains constant & so does buoyant force on a floating body. Hence there will be no change in level of water.

This is because density of mercury is more than that of water.
An object floats on a liquid, if its density is lower than that of the liquid. 
The density of iron nail is more than that of the water but less than that of mercury.