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Upthrust is equal to the weight of displaced liquid - class-VII

Description: upthrust is equal to the weight of displaced liquid
Number of Questions: 43
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Tags: fluids forces in fluids upthrust in fluids, archimedes' principle and floatation pressure physics
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A bob's weight is measured by a spring balance. Its volume is $10 cm^3$. In vacuum it measures $80 g$. What does it weigh in water?

  1. $70 g$

  2. $80 g$

  3. $90 g$

  4. $76 g$


Correct Option: A
Explanation:

Weight of bob $= 80g/1000 N$ where $g$ is acceleration due to gravity

Weight of water displaced $= 10\times1\times g/1000 = 10g/1000 N$
Apparent weight$ = 80g/1000 - 10g/1000 = 70g/1000 N$
Weight measured $= 70/1000kg = 70 g$

Upthrust due to fluid equals

  1. Volume of liquid displaced

  2. Mass of liquid displaced

  3. Weight of liquid displaced

  4. None of the above


Correct Option: C
Explanation:

Upthrust equals weight of fluid displaced by the object.

A ship has volume of $1,50,000 m^3$. It submerges $10\%$ in water. Upthrust felt by it equals

  1. $15 \times 10^6 N$

  2. $15 \times 10^7 N$

  3. $15 \times 10^8 N$

  4. $15 \times 10^5 N$


Correct Option: B
Explanation:

Volume submerged$ = 15,000 m^3 $

Upthrust $= 15000m^3 \times 1000 kg/m^3 \times 10 m/s^2$
$=15 \times 10^7 N$

Consider a cylinder of height $h$ and Cross sectional area $A$ completely submerged in a fluid of density $\rho$. What is upthrust?

  1. $Ah\rho g$

  2. $Ah\rho /g$

  3. $A\rho g$

  4. $Ah^2 \rho g$


Correct Option: A
Explanation:

Upthrust = weight of fluid displaced

= Vol of fluid displaced x density of fluid x acc. due to gravity
$=Ah\rho g$

An object having relative density 0.56 is dropped in water. Will it float or sink in water?

  1. Sink

  2. Float

  3. Depends on the density of water

  4. Unsure


Correct Option: B
Explanation:

Explanation: the relative density of water is 1 and that of the object is less than 1. Thus the upthrust experienced by the object will be more than the weight of the object making it float on the surface of water.

A body of density equal to density of fluid is sinking at speed of $1 m/s$ at $t=0$. What is distance covered by object in $1s$? 

  1. $0.5m$

  2. $1m$

  3. $0.75m$

  4. $0.875m$


Correct Option: B
Explanation:

Upthrust = weight of the body since density of body equals density of fluid

Hence, Net force $= 0$ and acceleration$ = 0$
Distance = speed x time$ = 1m$

Analyse the given statements and choose the correct option.
Statement I: The buoyant force of water on a submerged wooden cube is greater than the buoyant force of water on a steel cube of equal volume.
Statement II: The buoyant force on a body is equal to the weight of the liquid displaced by the body.

  1. Both statement I and statement II are correct and statement II is the correct explanation of statement I

  2. Both statement I and statement II are correct, but statement II is not the correct explanation of statement I

  3. Statement I is correct, but statement II is incorrect

  4. Statement I is incorrect, but statement II is correct


Correct Option: D
Explanation:

Since buoyant force experienced by a body is equal to weight of liquid displaced by the body. Hence it remains same for all the bodies of equal volume.

The reading of a spring balance when a block is suspended from it in air is 60 newton. This reading is changed to 40 newton when the block is fully submerged in water. The specific gravity of the block must be therefore " 

  1. 3

  2. 2

  3. 6

  4. 3/2


Correct Option: A
Explanation:

Specific gravity or relative density of block is given as

$R.D.=\dfrac { mass\quad of\quad block }{ loss\quad in\quad mass\quad in\quad liquid } =\dfrac { 60 }{ 60-40 } $
$\boxed { R.D.=3 } $

The body when immersed fully or partially in a fluid, experiences an upward force, equal to ............. of displaced fluid.

  1. Mass

  2. Weight

  3. Volume

  4. Area


Correct Option: C

A solid sphere weighs $40\ N$ in water and $45\ N$ in water and $45N$ in a liquid of density $0.75$. The weigh of sphere in air will be

  1. $60\ N$

  2. $53\ N$

  3. $50\ N$

  4. $80\ N$


Correct Option: C

A piece of ice having a stone frozen in its melts in a glass vessel filled with water. How will the level of water in the vessel change when the ice melts?

  1. The level will rise

  2. The level will not change

  3. The level will drop

  4. Some water will flow out


Correct Option: C

A metal wire of density $d$ floats, in a liquid of surface tension $\sigma$, in the horizontal position. The maximum radius of the wire so that it may not sink is

  1. $\sqrt{\dfrac{2\sigma}{\pi d g}}$

  2. $\sqrt{\dfrac{3\sigma}{\pi d g}}$

  3. $\sqrt{\dfrac{5\sigma}{\pi d g}}$

  4. $\sqrt{\dfrac{7\sigma}{\pi d g}}$


Correct Option: C

A cubical block floating in a liquid with half of its volume immersed in the liquid.when the hole system accelerates downward with an accelerations $g/3$. The fraction of volume immersed in liquid will be 

  1. $1/3$

  2. $3/8$

  3. $2/3$

  4. $3/4$


Correct Option: C

Two solids $A$ and $B$ floats in a liquid. it is observed that $A$ floats with half its volume immersed an d$B$ floats with $(2/3$ of its volume immersed. Compare the densities of $A$ and $B$ :

  1. $4:3$

  2. $2:3$

  3. $3:4$

  4. $1:3$


Correct Option: B

A block wood side 40 cm flods in wateer ina way that it lower fece 5 cm below the free surface or water.

  1. ${1600 cm{3}}$

  2. ${64000 cm{3}}$

  3. ${3000cm{3}}$

  4. ${40000cm{3}}$


Correct Option: A

A frictionless sloping plank 6m long. An oil drum of 300 N is raised up the sloping plank by an effort of 15 N. Calculate height to which oil drum rises is _____

  1. $2m$

  2. $7m$

  3. $3m$

  4. $0.3m$


Correct Option: C

When a ship floats on water :

  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


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:

  1. 60 kg

  2. 64 kg

  3. 70 kg

  4. 72 kg


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.

  1. $0.15\ m$

  2. $0.16\ m$

  3. $0.17\ m$

  4. $0.19\ m$


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

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

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

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

  4. $V _{2}=0$


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.

  1. $15/8$

  2. $1/8$

  3. $5/8$

  4. $15/2$


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.$\left(SG\ of\ ice\ =0.9\right)$.

  1. $2m^{2}$

  2. $4m^{2}$

  3. $10m^{2}$

  4. $0.25m^{2}$


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 

  1. $10\ cc$

  2. $90\ cc$

  3. $50\ cc$

  4. $9\ cc$


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 $10^3 kg/m^3$)

  1. 6 cm

  2. 12 cm

  3. 8 cm

  4. 10 cm


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 

  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.


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 :

  1. $20{ kg }$

  2. $10{ kg }$

  3. $16{ kg }$

  4. $13{ kg }$


Correct Option: A

A solid sphere weighs $40 N$ in water and $45 N$ in a liquid of relative density $0.7$5 when completely immersed. The weight of sphere in air will be 

  1. $50N$

  2. $53N$

  3. $80N$

  4. $60N$


Correct Option: A

An ice block contains a glass ball when the ice melts within the water containing vessel, the level of water

  1. Rises

  2. Falls

  3. Unchanged

  4. First rises and then falls


Correct Option: B

A body is floating in liquid with  $50\%$  of its volume out side the liquid. When the entire system accelerates upwards with an acceleration  $g / 3 ,$  the percentage of its volume outside the liquid is

  1. $33 \%$

  2. $50 \%$

  3. $25 \%$

  4. $66 \%$


Correct Option: C

Ice 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


Correct Option: C

A body floats in a liquid if the buoyant force is

  1. zero

  2. less than its weight

  3. equal to its weight

  4. none of these


Correct Option: B

A block of wood floats in water with (2/3)rd its volume submerged. The density of wood is 

  1. ${ 67gm/cm^{ 3 } }$

  2. ${ 6.7gm/cm^{ 3 } }$

  3. ${0. 67gm/cm^{ 3 } }$

  4. $7\times { 10 }^{ 3 }kg/{ m }^{ 3 }$


Correct Option: C

Initially A cubical block of density  $\rho$  is floating in water. in a vessel Out of its height  $L ,$  length  $x$  is submerged in water. Suddenly vessel is kept in an elevator accelerating upwards with acceleration  $a.$   What is the length immersed in water?

  1. $L - X$

  2. $2 x$

  3. $x$

  4. $L$


Correct Option: C

A ball falls through a liquid at a constant speed. It is acted upon by three forces: an upthrust, a drag-force and its weight.
Which statement is correct?

  1. The drag-force increases with increasing depth

  2. The drag-force is equal to the sum of the upthrust and weight

  3. The upthrust is constant with increasing depth

  4. The weight is greater than the sum of the drag-force and the upthrust


Correct Option: C
Explanation:

A ball falls through a liquid at a constant speed .

It acted upon by the three force ,
1. up thrust : it is constant
2. drag force
3, weight ; it is downward
Hence, C is correct option.

A cube of wood supporting $200g$ mass just floats in water. When the mass is removed, the cube rises by $1cm$, the linear dimension of cube is:

  1. $10cm$

  2. $20cm$

  3. $10\sqrt{2}cm$

  4. $5\sqrt{2}cm$


Correct Option: C
Explanation:

cube of wood supports$=200gm$
As, the mass is removed it rises by $1cm$
So, this implies that the weight of the $200g$ mass is equal to the weight of the water displaced by immersing $1cm$ of the cube,

Mathematically,
$a _{2}\times 1\times 1gm/cm^{3}\times g=g\times 200$
$a^{2}=10\sqrt{2}200cm^{2}$
$a=10\sqrt{2}cm$

A boat having a length of 3 metre and breadth 2 metre is floating on a lake. The boat sinks by one cm when a man gets on it. The mass of the man is:

  1. 60 kg

  2. 62 kg

  3. 72 kg

  4. 128 kg


Correct Option: A
Explanation:

Amount of water displaced is $V=3\times 2\times 0.01=0.06m^3$
$\therefore F=1000\times 0.06 \times g=60g$
hence the man weighs 60kgs

A piece of iron of density $7800kg/m^3$ and volume $100cm^3$ is completely immersed in water. Calculate the upthrust on the iron piece.

[Take, $g=10m/s^2$.]

  1. $1N$

  2. $2.8N$

  3. $7.8N$

  4. $6.8N$


Correct Option: A

if a block of wood is floating in a river water, then the apparent weight of the floating block is

  1. equal to the weight of the displaced water

  2. zero.

  3. greater than the weight of the displaced water

  4. equal to the actual weight of the block


Correct Option: B
Explanation:
Apparent weight in calculated as
Apparent weight$=$(weight of body)$-$(Buoyancy force)
Where
weight of body$=$(mass of body)$\times$(Gravitational force)
Buoyancy force$=$weight of fluid displaced
for the floating body:-
weight of body$=$Buoyancy force
$\therefore$ The body floating and
apparent weight equals to zero.
$\therefore$ Answer is B.

A body of mass 50 kg has a volume 0.049 $m^3$ and is kept on a table. The buoyant force on it due to air is.

  1. 0.588 N

  2. 0.49 N

  3. 58.8 N

  4. 49 kgf


Correct Option: D
Explanation:

since the mass is in equilibrium. So, the net downward force =buoyant force

buoyant force=50x9.8=49kgf

Calculate the upthrust acting on a ball of volume $\displaystyle 0.2{ m }^{ 3 }$ immersed in a liquid having density $\displaystyle { 1kg }/{ { m }^{ 3 } }$.

  1. 0.2f

  2. 2f

  3. 0.2g f

  4. 2g f


Correct Option: C
Explanation:

Upthrust = weight of the liquid displaced by the submerged part of the body Upthrust = mass of liquid displaced x Acceleration due to gravity Upthrust = volume of liquid displaced x density of liquid displaced x Acceleration due to gravity,Since volume of solid immersed is equal to the volume of the liquid displaced, Upthrust = volume of solid immersed x density of liquid displaced x Acceleration due to gravity Upthrust = 0.2 x 1 x g = 0.2g f.

A cube of wood floats in water, with $42$% of its volume is submerged, then the density of the wood is

  1. $42\ g\ cm^{-3}$

  2. $0.42\ g\ cm^{-3}$

  3. $0.58\ kg\ cm^{-3}$

  4. $600\ g\ cm^{-3}$


Correct Option: B
Explanation:

By Archimedes Principle, the buoyant force on a body partially or fully immersed in a fluid is given by the weight of the fluid displaced.


Let the volume of wood be $V$
Thus, volume of wood submerged is $0.42V$

Thus, the buoyant force acting on the wood is $B = 0.42\rho gV$
Weight of wood is $\rho _\textrm{wood}gV$

Thus, in equilibrium, $0.42\rho gV = \rho _\textrm{wood}gV \Rightarrow \rho _\textrm{wood} = 0.42\rho$

As Density of water is $\rho = 1\textrm{ g cm}^{-3}$, we have $\rho _\textrm{wood} = 0.42 \textrm{ g cm}^{-3}$

A: Diver dives in and reaches a depth of 100m.
B: Diver swims up from the depth of 100m to the surface.
Choose the correct alternative:

  1. A is easier than B

  2. B is easier than A

  3. A and B are equally hard

  4. A is easier than B if speed of descent and ascent is same and more.


Correct Option: A
Explanation:
Assume:
$F _d: \text{Force applied by diver during descent in downward direction}$
$F _u: \text{Force applied by diver during ascent in upward direction}$
$U: \text{Upthrust}$
$m: \text{Mass of the diver}$
$g: \text{Acceleration due to gravity}$
Uniform ascent and descent.

Diver has more density than water. Hence, weight of diver is more than the upthrust and without any effort, the diver sinks.
i.e. $mg>U..................(1)$

During upward motion, $F _u+U-mg=0............(2)$
During downward motion, $F _d-U+mg=0...............(3)$

From (1),(2) and (3), 
$F _u-F _d=2mg-2U$
$F _u-F _d>0$
$F _u>F _d$

Mathematical proof of upthrust is based on 

  1. Definition of pressure

  2. Weight of object

  3. Pressure exerted by a column of fluid

  4. Viscosity


Correct Option: C
Explanation:
Mathematical proof:
Consider a cylinder of cross section area $A$ and height $L$ completely submerged in water. Let depth of upper surface be $h$.
Using, pressure exerted by fluid column:
Force on the upper face of the cylinder = $hρgA$
Force on the lower face of the cylinder = $[h + L]ρgA$
Difference in force = $LAρg$

But $LA$ is the volume of liquid displaced by the cylinder, and $LrgA$ is the weight of the liquid displaced by the cylinder.

Therefore there is a net upward force on the cylinder equal to the weight of the fluid displaced by it.
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