Tag: applications of floatation

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Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

An iceberg of density $900 kg/m3$ is floating in water of density $1000 kg/m3$. the percentage of volume of ice-cube outside the water is

  1. $10$ percent
  2. $20$ percent
  3. $31$ percent

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let  V is the total volume of iceberg
       ${ V } _{ sub }$ = volume of iceberg submerged
        ${ \rho  } _{ b }$= density of iceberg = 900 Kg/m3
         ${ \rho  } _{ w }$= density of water = 1000 Kg/m3
 So, for the flotation of body,
             weight of body= weight of water displaced
           $\Rightarrow{ \rho  } _{ b }V={ \rho  } _{ w }{ V } _{ sub }$
           $\Rightarrow \dfrac { { V } _{ sub } }{ V } =\dfrac { { \rho  } _{ b } }{ { \rho  } _{ w } } $
substracting both side from 1, we get
             $\Rightarrow 1-\dfrac { { V } _{ sub } }{ V } =1-\dfrac { { \rho  } _{ b } }{ { \rho  } _{ w } } $
            $\Rightarrow \dfrac { V-{ V } _{ sub } }{ V } =\dfrac { { \rho  } _{ w }-{ \rho  } _{ b } }{ { \rho  } _{ w } } $
             Converting it in percentage,
              $\Rightarrow \dfrac { V-{ V } _{ sub } }{ V } \times 100=\dfrac { { \rho  } _{ w }-{ \rho  } _{ b } }{ { \rho  } _{ w } } \times 100$
by  substituting values, we get
            percentage volume outside the water= $\dfrac { 1000-900 }{ 1000 } \times 100=10$%

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A wooden cylinder floats vertically in water with half of its length immersed, Density of wood is

  1. Equal to that of water

  2. Half the density of water

  3. Double the density of water

  4. One fourth the density of water
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A wooden cube floats in water partially immersed. When 200 g weight is put on the cube, it further immersed by 2 cm. The length of the side of the cube is

  1. $1.0 cm$
  2. $\sqrt{10}cm$
  3. $10 cm$
  4. $20 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A ball weighing  $4 kg$  of density  $4000{ kg }\quad { m }^{ { -3 } }$  is completely immersed in water of density  $1000{ kg }\quad { m }^{ { -3 } }.$  Find the force of buoyancy on it. (Given $g = 10 { ms } ^ { - 2 }$)

  1. $100{ N }$
  2. $1{ N }$
  3. $20{ N }$
  4. $10{ N }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A balloon has volume of $1500\, m^3$. It is filled with hydrogen $(\rho = 0.09\, gL^{-1})$. If the density of air is $1.29\, gL^{-1}$, it can lift a total weight of 

  1. $2400\, kg$
  2. $1600\, kg$
  3. $2700\, kg$
  4. $1800\, kg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

An object measuring $2\times 2\times 5{cm}^{3}$ has a mass of $16g$. It is put in water of density $1g/cc$. Percentage of its volume outside water while floating is

  1. $10$%
  2. $20$%
  3. $30$%
  4. $40$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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%.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

lce pieces are floating in a beaker  $A$  containing water and also in a beaker  $B$  containing miscible liquid of specific gravity  $1.2 .$  When ice melts, the level of

  1. Water increases in $A$
  2. Water decreases in $A$
  3. Liquid in $B$ decreases
  4. Liquid in $B$ increases
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

Statement I:- An ice ball is floating in water. Some stone pieces are embedded inside the ice. When ice will melt level of water will fall.
Statement I:- In floating condition,stone pieces will displace more liquid compared to the condition when they sink.

  1. Statement I is true,statement II is true and statement II is a correct explanation for statement I.

  2. Statement I is true,statement II is true and statement II is NOT the correct explanation for statement I.

  3. Statement I is true,Statement II is false.

  4. Statement I is false,Statement II is true.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A block of wood is floating in water in a closed vessel as shown in the figure. The vessel is connected to an air pump. When more air is pushed into the vessel, the block of wood floats with : (neglect compressibility of water)

  1. Larger part in the water

  2. Smaller part in the water

  3. Same part in the water

  4. At some instant it will sink
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.