Applications of floatation - class-IX

Let $x$ be the initial depth upto which the cube is sinked in water.

As there is no atmosphere on the surface of the moon  so no bouyancy will act Hence it will fall with acceleration $9.8/6 m/s^2$

In a freely falling frame, the effective gravity is zero. Since buoyant force depends on the pressure gradient caused by gravity, the pressure gradient vanishes, making the buoyant force zero.

TRUE
Ice floats in water because it is less dense than water. So any substance that has a lower density in its solid state than in its liquid state will float.

For a floating object, the weight of the object equals the weight of the displaced fluid. Thus, density_wood * V_total = density_water * V_submerged. Given V_submerged = (2/3)V_total, the density of wood is (2/3) * 1000 kg/m^3 = 666.7 kg/m^3, which is 0.667 g/cm^3.

According to archimedes principle a body floats in a fluid if the density of the body is less than that of the fluid.
If the density of body is greater than that of fluid, it will sink.
In the above case, Since the body has  a higher density, therefore it will sink.

Hot air balloon has hot air inside.  Also hot air is less dense than cooler air which surrounds the hot air balloon. This means hot air is lighter than equal volume of cold air.  Hence due to buoyant force of cold air, the balloon rises.

When a piece of ice is placed in a beaker containing water, some of its portion remains out of the water level. Since the volume of ice is greater than the water hence after melting the volume of the piece will decrease and the water level will remain the same.

Answer is A.

As water is cooled down, however, the molecules have less energy and hydrogen bonding takes over. The molecules form a ordered crystal through hydrogen bonding that spaces the molecules farther apart than when they were in a liquid. This makes ice less dense than water allowing it to float.
Hence, the statement is true.

Average density of ship is less than the density of water.

Sea water contains dissolved salts which makes its density more than river water Hence upthrust is more in sea water than river water So it is easier to swim in sea water than in river water 

Density of cork or wood are less than that of water hence they float in water

Bodies like stones ans metals sink in water as these substances have density more than that of water

More denser objects stays at the bottom.

the buoyant force depends on the density of liquid. When the density of the object is less than that of the liquid, it floats on the liquid because the upthrust or buoyant force of water is more than the gravity acting on the object.

If the weight of the body immersed in the liquid is equal to the force of buoyancy acting on it, then the body will float just inside the liquid surface.

Remains unchanged because there is no change in mass.

Since there is no external horizontal force, the center of mass of the system remains stationary. mx + MX = 0, where x and X are displacements. The person moves distance L relative to the plank, so x_person = L - X_plank. Solving m(L - X) + MX = 0 gives X = -mL / (m+M). The magnitude is mL / (m+M).

For a body rising with constant velocity, the forces are balanced: Buoyancy = Weight + Viscous Force. F_b = rho_liquid * V * g = 4 * rho_ball * V * g = 4 * Weight. Therefore, Viscous Force = F_b - Weight = 4W - W = 3W. The ratio of viscous force to weight is 3.

Archimedes' principle states that a floating object displaces its own weight in fluid. When ice melts, the mass of the resulting water is equal to the mass of the ice, so the volume of water produced is exactly equal to the volume of water displaced by the ice. Thus, the level remains the same.

For a floating body, the fraction submerged is equal to the ratio of densities. 0.80 = rho_body / rho_water. rho_body = 0.80 * 1000 kg/m^3 = 800 kg/m^3.

Apparent weight = Actual weight - Buoyant force. Actual weight = 6 kg * 9.8 m/s^2 = 58.8 N. Buoyant force = weight of displaced water = 0.1 kg * 9.8 m/s^2 = 0.98 N. Apparent weight = 58.8 - 0.98 = 57.82 N, which is approximately 58 N.

The boat sinks when the volume of water displaced equals the volume of the boat. Using the thermal expansion formula V = V0(1 + gamma*deltaT), we set the submerged volume to 100% (1.0) and solve for deltaT given the initial 97% (0.97) submerged volume at 0 degrees Celsius.

Since there is no external horizontal force, the center of mass of the dog-boat system remains stationary. Using the conservation of center of mass: m_dog * delta_x_dog + m_boat * delta_x_boat = 0, we can find the displacement of the boat relative to the shore.

When an object floats with half of its volume immersed, the buoyant force equals its total weight. According to Archimedes' principle, the ratio of the immersed volume to total volume equals the ratio of the object's density to the fluid's density, making the wood's density half that of water.

The additional weight of 200g causes an additional displacement of water equal to the volume of the cube submerged by 2cm. Thus, 200g = (Area * 2cm) * density_water. Solving for the side length L where Area = L^2 gives 10cm.

Buoyant force is equal to the weight of the displaced liquid. Volume of the ball = mass / density = 4kg / 4000kg/m^3 = 0.001 m^3. Buoyant force = volume * density_water * g = 0.001 * 1000 * 10 = 10 N.

The net upward force (lift) is equal to the difference in weight between the displaced air and the hydrogen gas inside the balloon. Lift = V * (rho_air - rho_hydrogen) = 1500 * (1.29 - 0.09) g/L = 1500 * 1.20 g/L = 1800 kg.

The volume of the object is 2 * 2 * 5 = 20 cm^3, and its mass is 16 g, giving a density of 16/20 = 0.8 g/cc. The fraction of volume submerged is equal to the relative density (0.8/1.0 = 0.8 or 80%), which means the percentage of volume outside water is 100% - 80% = 20%.

When ice melts in water, the level remains unchanged because the weight of the displaced water equals the weight of the ice. In a liquid with specific gravity > 1, the ice displaces a smaller volume of liquid than its own melted volume, so the level decreases.

When ice with embedded stones melts, the stones sink. While floating, the stones displace liquid equal to their weight (due to the ice). Once they sink, they only displace their own volume, which is less than the volume of liquid required to support their weight.

The buoyant force depends on the weight of the displaced fluid. Since the block is floating, the buoyant force equals the weight of the block. Increasing air pressure does not change the density of the water or the weight of the block, so the submerged volume remains the same.

Since 2/5th of the volume is above the surface, 3/5th of the volume must be immersed in water. The density of the wood is therefore 3/5 (or 60%) of the density of water, which equals 0.6 g/cm^3.

If solid object has higher density than liquid then it will sink in that liquid otherwise it will float.
R.D. of liquid A = 0.75
R.D. of liquid B = 1.0
R.D. of object P = 0.6
R.D. of object Q = 0.9

• P has R.D. less then both the liquids. so it will float in both the liquids.
  
• Q has R.D. more than liquid A, so it will sink in A.
• Q has R.D. less than liquid B, so it will float in B.

For a wire of length l and radius r to stay afloat on a liquid surface due to surface tension, the maximum downward gravitational force (weight) must equal the upward force due to surface tension acting on both sides: mg = 2Tl. Substituting mass as volume times density (pi * r^2 * l * rho) and solving for radius r gives sqrt(2T / (pi * rho * g)).

The weight of the 200g mass equals the weight of the water displaced by the additional 2cm immersion. 200g = (Area * 2cm) * density_water. Area = 100 cm^2. Side length = 10 cm.

For a floating body, the weight of the floating object equals the buoyant force, which is the weight of the displaced water. Using the dimensions of the cubical box, the mass divided by the area of the base gives the volume of water displaced per unit area, yielding the depth of immersion. Calculation: (21.6 kg) / (0.3 m * 0.3 m * 1000 kg/m^3) = 0.24 m or 24 cm.

The weight of the wire (pi * r^2 * L * rho_wire * g) is balanced by surface tension (2 * T * L). Solving for diameter d = 2r: pi * (d/2)^2 * rho_wire * g = 2 * T. Plugging in values gives d approximately 1.5 mm.

Weight of the cylinder = buoyant force = (3/5 * 25 cm * 15 cm^2) * 1 g/cm^3 = 225 gf. This confirms option B.

For a floating object, density_object / density_fluid = fraction_submerged. Density_A / Density_water = 1/2. Density_B / Density_water = 2/3. Ratio A:B = (1/2) / (2/3) = 3/4.

According to Archimedes principle, when a body is immersed in a liquid, the upward buoyant force acting on it is equal to the weight of the liquid displaced by the body. This upward force is termed upthrust.

According to Archimedes principle, A body immersed in water experiences an upward force equal to the mass of the fluid displaced by the body. If the weight of an object is greater than the weight of displaced fluid, it will float. If the two are equal, it is suspended, neither floating nor sinking. For example, when an object is placed in water, it will displace its own volume of water, and that water will push back against it proportionally, producing an upthrust.
Water has a weight density of $62$ pounds per cubic foot. It an object weighing $62$ pounds has a volume that displaces $2$ cubic feet of water, it will float. The displaced water will weigh $124$ pounds and the pressure of that water would be enough to keep the object floating.
Hence, Ice floats in water, because the weight of water is displaced by the immersed part of the ice is more than its own weight and the statement is true.

Let $v$ be the volume of the ice-berg outside the sea water and $V$ be the total volume of ice-berg. Then as per question
$0.92V = 1.03(V-v)$

C. less than 4cm  
Because more volume of alcohol needs to be displaced to displace the equal weight of water. since density of alcohol is lesser.

We know that the volume of part submerged in a liquid is given by, $V= \dfrac{Density\ of\ body}{Density\ of\ liquid}$
Density of cube $= \dfrac{Mass}{Volume}$$ =\dfrac{700 g }{ 10^{3}} = 0.7g/cm^{3}$
So, density of water $= 1 g/cm^{3}$
So, part of cube submerged in water $= \dfrac{Density\ of\ body}{Density\ of\ water} = 0.7/1 = 7/10$
$\therefore$ Part of cube above water $= 1 - 7/10 = 3/10$
i.e. $3cm$ of cube is above water.

Sinks.
On heating, the density of water decreases. So, upthrust on block decreases and weight of block exceeds upthrust due to which it sinks.

Floats with some part outside water. 
On adding some salt to water, the density of water increases, so upthrust on block of wood increases and hence the block rises up till the weight of salty water displaced by the submerged part of block becomes equal to the weight of block.

Float
We know according to Archimedes Principle, Object having density less than that of the liquid in which they are immersed, float on the surface of the liquid.
And Object having density greater than that of the liquid in which they are immersed, sink in the liquid.

Let say $m _1=$ mass of the man = 100kg
and $m _2=$ mass of the ice $= 0.9 \times 1000V=900V$, where $V$ is the volume of the ice block.
For equilibrium,
Total downward weight = total upthrust
$100g +900 Vg=1000Vg \\Rightarrow V=1m^3$
Volume = Area $\times $ height
$\Rightarrow A=\frac{Volume}{Height}=\frac{1}{0.5}=2m^2$

Let $V _L$ and $V _S$ be the volume of water displaced and volume of the ice respectively; $\rho _L$ and $\rho _S$ be the density of water and density of ice respectively. Since ice is floating on water, $F _B=W$
or $\rho _LV _Lg=\rho _SV _Sg$ or $\rho _LV _L=\rho _SV _S$
or $\frac{V _L}{V _S}=\frac{\rho _S}{\rho _L}$
This is the fraction of volume of the iceberg that is inside the water. Therefore, the fraction of volume of the iceberg that is outside the water is given by, 1 - 0.893 = 0.107
Hence, the fraction of the total volume of an iceberg is outside the water is 0.107.

Answer is C.

An egg will sink in fresh water but it will float in very salty water; the density of the egg is greater than the density of fresh water but less than the density of the salty water.
The density of salty water can be a much as $10%$ greater than that of fresh water i.e. up to $1.1 g/cm^{ 3 }$.
Hence, on dissolving a lot of salt in the water, the egg will rise and then float.

When an object is floating, the net force on it will be zero. This happens when the volume of the object submerged displaces an amount of liquid whose weight is equal to the weight of the object.

Given that $m _1=V _1/2 \times d \implies d _1=d/2$
and $m _2=2V _2/3 \times d \implies d _2=2d/3$
$\therefore d _1:d _2=3:4$

$\rho _{salt water0}>\rho _{ice}$

according to the Archimedes principle , the floating substance displaces some liquid, so when the ice melts ,there will be no change in the water level as the melted ice will occupy the same volume as it was occupying earlier.

Let $F _f = $ Force of friction  

When the cube melts, the water level will remain the same.An floating object displaces an amount of water equal to its own weight. Since, water expands when it freezes, one ounce of frozen water has a larger volume than one ounce of liquid water.

Let $V$ and $\rho$ be the volume and density of the cylindrical block.

Given that solid floats in a liquid, in the partially submerged position.

Given, That the ice cube is floating in the liquid.

Since density of a hollow ice cube is less than water. Hence it will float and when ice melts, then level of water decreases due to loss in volume.