Tag: upthrust in fluids, archimedes' principle and floatation

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

When a ship floats on water :

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. it displaces no water

  2. the mass of water displaced is equal to the mass of the ship

  3. the mass of water displaced is lesser than the mass of the ship

  4. the mass of water displaced is greater than the mass of the ship

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Correct Option: B

A boat 3 m long and 2 m wide is floating in a lake. When a man climbs over it, it sinks 1 cm further into water. The mass of the man is:

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. 60 kg

  2. 64 kg

  3. 70 kg

  4. 72 kg

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Correct Option: A
Explanation:

$\delta Vg$=Weight of mass.

$\begin{array}{l} 1000\times \left( { 3\times 2\times \frac { 1 }{ { 100 } }  } \right) \times 10=mg \ m=60kg \end{array}$
$ \therefore$ Option $A$ is correct.

Calculate the height $h$ of the portion of the slab the is above the water surface.

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $0.15\ m$

  2. $0.16\ m$

  3. $0.17\ m$

  4. $0.19\ m$

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Correct Option: A

A ball floats on the surface of water in a container exposed to the atmosphere. Volume $V _{1}$ of its volume in inside the water. If the container is now covered and the air is pumped out. Now let $V _{2}$ be the volume now immersed in water. Then

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $V _{1}=V _{2}$

  2. $V _{1}>V _{2}$

  3. $V _{2}>V _{1}$

  4. $V _{2}=0$

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Correct Option: A

A vessel in the shape of a hollow hemisphere surmounted by a cone is held with the axis vertical and vertex uppermost. If it be filled with a liquid so as to submerge half the axis of the cone in the liquid, and the height of the cone be double the radius of its base, find the value of $x$, where the resultant downward thrust of the liquid on the vessel is $x$ times the weight of the liquid that the hemisphere can hold.

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $15/8$

  2. $1/8$

  3. $5/8$

  4. $15/2$

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Correct Option: C

What is the area of the smallest block of ice $0.5\ m$ thick that will just support a man of mass $100\ kg$? The block of ice is floating in fresh water.$\left(SG\ of\ ice\ =0.9\right)$.

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $2m^{2}$

  2. $4m^{2}$

  3. $10m^{2}$

  4. $0.25m^{2}$

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Correct Option: B

A block of ice of density $0.9\ gm\ cc^{-1}$ and of volume $100\ cc$ floats in water. Then volume of ice inside the water 

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $10\ cc$

  2. $90\ cc$

  3. $50\ cc$

  4. $9\ cc$

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Correct Option: B

A wooden cube floating in water supports a mass m = 0.2 kg on its stop. When the mass is removed the cube rises by 2 cm. The side of the cube is - (density of water $10^3 kg/m^3$)

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. 6 cm

  2. 12 cm

  3. 8 cm

  4. 10 cm

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Correct Option: D
Explanation:
Cube is floating, therefore
Weight of liquid displaced $=$ Weight of cube
let  the volume of liquid displaced $= V$
Density of liquid $= \rho$
Mass of cube $= M$

Then,

$(M+m)g=\rho v g$

$\implies M+m=\rho v$       ............(1)

When mass is removed, cube reise by $l=2\,cm$

Therefore,

$Mg=(V-lA)\rho g$

where, $AS$ is the are of face of cube

$\implies M=\rho(V-lA)$          .................(2)

Eliminating $M$ by subtracting (2) from (1)

$m=\rho l A$

$\implies A=\dfrac{m}{\rho l}$

Let side of cube $=a$

Then,

$a^2=A=\dfrac{m}{\rho l}$

$\implies a=\sqrt{\dfrac{m}{\rho l}}$

If liquid is water, then $\rho=1000\,kg/m^3$

$I=0.02\,m$ (given)

$m=0.2\,kg$ (given)

Putting these value, we get $a=0.1\,m=10\,cm$

Three cubes of volume $V , V / 4$ and $V / 8$ respectively are immersed in thesame liquid. If the buoyant force exerted on the $3 ^ { rd }$ cube is 40 units. Then 

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. Buoyant force on $1 ^ { st }$ cube is 80 units.

  2. Buoyant force on $1 ^ { st }$ cube is 320 units.

  3. Buoyant force on $2 ^ { nd }$ cube is 80 units.

  4. Buoyant force on $2 ^ { nd }$ cube is 320 units.

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Correct Option: C

A body of mass  $40 kg$  floats on a lake with  $5 { cm }$  in water, when another block placed on the block,  $7 { cm }$  are submerged, the mass of second block is :

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $20{ kg }$

  2. $10{ kg }$

  3. $16{ kg }$

  4. $13{ kg }$

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Correct Option: A