Tag: upthrust in fluids, archimedes' principle and floatation

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Questions Related to upthrust in fluids, archimedes' principle and floatation

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?

Consider a small balloon filled with an ideal gas which is submerged in water. Assuming that the temperature is the same everywhere in the water, the buoyant force on the balloon when it is at a depth d below the surface, in terms of its volume at the surface $V _ { 0 }$ , the atmospheric pressure $P _ { 0 }$ , the density of water $\rho _ { 0 }$ , and the acceleration due to gravity g.

  1. $F _ { B } = \frac { P _ { 0 } V _ { 0 } } { d + \frac { P _ { 0 } } { \rho g } }$
  2. $F _ { B } = \frac { P _ { 0 } V _ { 0 } } { d \rho g + P _ { 0 } }$
  3. $F _ { B } = \frac { d \rho g + P _ { 0 } } { P _ { 0 } V _ { 0 } }$
  4. $F _ { B } = \frac { P _ { 0 } V _ { 0 } } { d + \frac { \rho g } { P _ { 0 } } }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 body is floating in water with $80$% of its volume below the surface of water. What is the density of body?

  1. $666.7kg/{ m }^{ 3 }$
  2. $777.6kg/{ m }^{ 3 }$
  3. $800kg/{ m }^{ 3 }$
  4. $876.6kg/{ m }^{ 3 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

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 body of mass $6kg$ immerses in water partially. If the body displaces $100$ g of water, then the apparent weight of the body is

  1. $59$ N
  2. $40$ N
  3. $49$ N
  4. $60$ N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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 boat is floating in water at $0^{ \circ  }C$ such that 97% of the volume of the boat is submerged in water . The temperature at which the boat will just completely sink in water is $(\gamma _{ R }=3\times { 10 }^{ -4 }/{ ^{ 0 }C })(nearly)$ 

  1. ${ 100 }{ ^{ 0 }C }$
  2. ${ 103 }{ ^{ 0 }C }$
  3. ${ 60 }{ ^{ 0 }C }$
  4. ${ 50 }{ ^{ 0 }C }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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 dog weighing 5n kg is standing on a flat boat so that it is 10m from the shore . The dog walks 4 m on the boat towards the shore and then halts. The boat weighs 20kg and one can assume that there is no friction between it and the water .How far is the dog from the shore at the end of this time ?

  1. 3.2 m

  2. 0.8 m

  3. 10 m

  4. 6.8 m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

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.