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?

A body floats in water because of:

  1. No force is acting on it

  2. The buoyant force acting on it

  3. Gravitational pull

  4. Friction between body and the water
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When an object is floating, the net force on it will be zero. This happens when the volume of the object submerged displaces an amount of liquid whose weight is equal to the weight of the object.

Answer (B) the net force acting on this body is zero

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?

Two solids A and B float in water. It is observed that A floats with half its volume immersed and 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. $1:1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that $m _1=V _1/2 \times d \implies d _1=d/2$
and $m _2=2V _2/3 \times d \implies d _2=2d/3$
$\therefore d _1:d _2=3:4$

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 piece of ice is floating in a concentrated solution of common salt (in water) in a pot. When ice melts completely, the level of solution will 

  1. Go up

  2. Remain the same

  3. Go down

  4. First go up then go down
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\rho _{salt water0}>\rho _{ice}$

When ice floats, volume of ice in the salt water is given by
$V _{imm}\rho _{salt water}g=V _{ice}\rho _{ice}g$
$\implies V _{imm}=V _{ice}\dfrac{\rho _{water}}{\rho _{salt water}}<V _{ice}$

Hence when ice melts, whole of the volume would add to water and level of solution would rise.

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 ice cube is floating in a glass of water. What happens to the water level when the ice melts?

  1. Rises

  2. Falls

  3. Remains the same

  4. First rises and then falls
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

according to the Archimedes principle , the floating substance displaces some liquid, so when the ice melts ,there will be no change in the water level as the melted ice will occupy the same volume as it was occupying earlier.

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 rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The frictional force of the liquid on the rising ball is greater than the weight of the ball by a factor of

  1. $2$
  2. $3$
  3. $4$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $F _f = $ Force of friction  

$F _b = $ Force of bouyancy
$F _w = $ Weight of ball

Archimedes' principle:

$F _b = F _w + F _f$

Given: $F _b = (4P _B)gV$

$F _w = VP _Bg$

$F _f = F _b - F _w = 3P _B gV$

$\Rightarrow \dfrac{F _f}{F _w} = \cfrac{3P _B gV}{P _BgV} = 3 : 1 $

$\Rightarrow F _f = 3F _w$

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 ice cube contains a glass ball. The cube is floating on the surface of water contained in a trough on the surface of water contained in a trough. What will happen to the water level, when the cube melts?

  1. It will remain unchanged

  2. It will fall

  3. It will rise

  4. First it will fall and then rise
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the cube melts, the water level will remain the same.An floating object displaces an amount of water equal to its own weight. Since, water expands when it freezes, one ounce of frozen water has a larger volume than one ounce of liquid 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 cylindrical block floats vertically in a liquid of density ${\rho} _1$ kept in a container such that the fraction of volume of the cylinder inside the liquid is $x _1$. then some amount of another immiscible liquid of density ${\rho} _2 ({\rho} _2 < {\rho} _1)$ is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with $x _2$ fraction of volume of the cylinder inside the liquid of density ${\rho} _1$. The ratio ${\rho} _1 / {\rho} _2$ will be

  1. $\dfrac{1 - x _2}{x _1 - x _2}$
  2. $\dfrac{1 - x _1}{x _1 + x _2}$
  3. $\dfrac{x _1 - x _2}{x _1 + x _2}$
  4. $\dfrac{x _2}{x _1}-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $V$ and $\rho$ be the volume and density of the cylindrical block.

Case 1 : When $x _1$ fraction of block's volume is immersed in liquid of density $\rho _1$
Using Archimede's principle :    Weight of cylindrical block = Weight of liquid displaced
$\therefore$  $\rho V g = \rho _1 x _1 V g$             ........(1)

Case 2 : When $x _2$ fraction of block's volume is immersed in liquid of density $\rho _1$ and $1-x _2$ fraction of block's volume is immersed in liquid of density $\rho _2$
Using Archimede's principle :    Weight of cylindrical block = Weight of liquid displaced
$\therefore$  $\rho V g = \rho _1 x _2 V g + \rho _2 (1-x _2) V g$             ........(2)

Equating (1)  and (2) we get    $\rho _1 x _1V g = \rho _1 x _2 V g + \rho _2 (1-x _2) V g$
OR    $\rho _1 x _1 = \rho _1 x _2 + \rho _2 (1-x _2)$
OR   $\rho _1 (x _1 - x _2) = (1-x _2)\rho _2$
$\implies$  $\dfrac{\rho _1}{\rho _2} = \dfrac{1-x _2}{x _1-x _2}$

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 solid floats in a liquid in the partially submerged position:

  1. the solid exerts a force equal to its weight on the liquid

  2. the liquid exerts a force of buoyancy on the solid which is equal to the weight of the solid

  3. the weight of the displaced liquid equals the weight of the solid

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that solid floats in a liquid, in the partially submerged position.

The weight of the displaced liquid will be equal to the weight of the solid, and the solid exerts a force equal to the weight of the liquid.
The weight of solid is equal to the buoyancy force exerted by the liquid on the solid.
Therefore option $D$ is correct.

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?

In a beaker containing liquid, an ice cube is floating. When ice melts completely, the level of liquid rises. Then the density of the liquid is:

  1. more than the density of ice

  2. less than the density of ice

  3. same as the density of ice

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, That the ice cube is floating in the liquid.

Let the height of ice cube be $h,$
Given, If ice cube completely melts, the level of liquid raises. So initially the length of ice cube submerged in liquid be $l <h,$
Let the density of liquid be $d _{l}$ and density of ice cube be $d _{i}$
In equilibrium , $Mg=M _{l}g$
$\Rightarrow d _{i}Ahg=d _{l}Alg$
$\Rightarrow \frac{d _{l}}{d _{i}}=\frac{h}{l}>1$
$\Rightarrow d _{l} > d _{i}$
Therefore the density of liquid is more than the density of ice.
So option $A$ is correct.

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 ice cube contains a large air bubble. The cube is floating on the surface of water contained on a trough. What will happen to the water level, when the cube melts?

  1. $It\ will\ remain\ unchanged$
  2. $It\ will\ fall$
  3. $It\ will\ rise$
  4. $First\ it\ will\ and\ then\ rise$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since density of a hollow ice cube is less than water. Hence it will float and when ice melts, then level of water decreases due to loss in volume.