Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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 float inside liquid. If we increase temperature then what charges occur in buoyancy force?(Assume body is always in floating condition )

  1. Buoyancy force will condition

  2. Buoyancy force will increase

  3. Buoyancy force remains constant

  4. Cannot be calculated from given statement
Reveal answer Fill a bubble to check yourself
A 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 wooden cube floats just inside the water, when a mass of $x$ (in grams) is placed on it. If the mass is removed, the cube floats with a height $\dfrac{x}{100}\ cm$ above the water surface. The length of the side of cube is (density of water is $1000\ kg/m^{3}$)

  1. $10\ cm$

  2. $15\ cm$

  3. $20\ cm$

  4. $30\ cm$
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?

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= $\dfrac { 1000-900 }{ 1000 } \times 100=10$%