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PHYSICS [5 Mark Question Answer]

Upthrust in Fluids, Archimedes' Principle and Floatation · ICSE STD 9 (ICSE - English Medium). 19 questions.

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Question 15 Marks
A weather forecasting plastic balloon of volume $15 m^3$ contains hydrogen of density $0.09 kg m ^{-3}$. The volume of equipment carried by the balloon is negligible compared to its own volume. The mass of an empty balloon alone is 7.15 kg . The balloon is floating in air of density of $1.3 kg m ^{-3}$. Calculate:
(i) The mass of hydrogen in the balloon,
(ii) The mass of hydrogen and balloon,
(iii) The total mass of hydrogen, balloon and equipment if the mass of equipment is $x kg$,
(iv) The mass of air displaced by balloon and
(v) The mass of equipment using the law of floatation.
Answer
Volume of plastic balloon $=15 m^3$ Mass of empty balloon $=7.15 kg$ Density of hydrogen $=0.09 kgm ^{-3}$ Density of air $=$ $1.3 kgm ^{-3}$ (i) Mass of hydrogen in the balloon= Volume of balloon $\times$ Density of hydrogen
Mass of hydrogen in the balloon $=(15 \times 0.09) kg =1.35 kg$ (ii) Mass of hydrogen and balloon $=$ Mass of empty balloon
+ Mass of hydrogen in the balloon
Mass of hydrogen balloon $=[7.15+1.35) kg =8.5 kg$ (iii) Given mass of equipment $= x$
Total mass of hydrogen, balloon and equipmemt $=(8.5+x) kg$ (iv) Weight of air displaced by the balloon $=$ upthrust
$=$ Volume of balloon x density of air x g
Mass of air displaced = Volume of balloon $x$ density of air
$=15 \times 1.3=19.5 kg( v )$ Using the law of floatation,
Mass of air displaced= Total mass of hydrogen, balloon and equipmemt
or, $19.5=8.5+x$
or, $x=11 kg$
Thus, mass of the equipment 11 kg
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Question 25 Marks
QUESTION The density of ice is $0.92 g cm ^{-3}$ and that of sea water is $1.025 g cm ^{-3}$. Find the total volume of an iceberg which floats with its volume $800 cm^3$ above water.
Answer
Let $V$ be the volume of the iceberg.
Volume of iceberg above water $=800 cm ^3$
Volume of iceberg submerged in water $=v$
Density of ice $\left(\rho_{\text {ice }}\right)=0.92 g cm ^{-3}$
Density of sea water $\left(\rho_{\text {sea water }}\right)=1.025 gcm ^{-3}$
According to the law of floatation,
$ \frac{\text { Volume of immersed part of body }}{\text { Total volume of body }}=\frac{\text { Density of body }}{\text { Density of liquid }}$
$\frac{v}{ V }=\frac{\rho_{\text {ice }}}{\rho_{\text {seawater }}}$
$\Rightarrow \frac{v}{ V }=\frac{0.92}{1.025}$
$\Rightarrow \frac{v}{ V }=\frac{0.92}{1.025}$
$\Rightarrow \frac{v}{ V }=0.8976$
$\Rightarrow v=0.89756 V $
Now,
$\therefore$ Floating Volume of iceberg $= V -v$
$\Rightarrow 800=V-0.89756 V$
$\Rightarrow 800= V (1-0.89756)$
$\Rightarrow \frac{800}{0.10244}= V$
$\Rightarrow V =7809.45 cm ^3 \approx 7809.5 cm ^3$
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Question 35 Marks
A wooden block floats in water with two-third of its volume submerged.
(a) Calculate the density of wood.
(b) When the same block is placed in oil, three-quarters of its volume is immersed in oil. Calculate the density of oil.
Answer
Volume of wooden block submerged in water $( v )=\frac{2}{3} \times$ total volume $( V )$
Volume of wooden block submerged in oil $\left( v ^{\prime}\right)=\frac{3}{4} \times$ total volume (V)
Say density of wood $=\rho_{\text {wood }} gcm ^{-3}$
Say density of oil $=\rho_{\text {oil }} gcm ^{-3}$
According to the law of floatation,
$
\frac{ v }{ V }=\frac{\rho_{\text {wood }}}{\rho_{\text {water }}}
$
or,$\frac{2}{3}=\frac{\rho_{\text {wood }}}{\rho_{\text {water }}}=\frac{\rho_{\text {wood }}}{1000}$
or , $\rho_{\text {wood }}=1000 \times \frac{2}{3}=667 kgm ^{-3}$
Again, according to the law of floatation,
$
\frac{ v ^{\prime}}{ V }=\frac{\rho_{\text {wood }}}{\rho_{\text {oil }}}
$
or,$\frac{3}{4}=\frac{\rho_{\text {wood }}}{\rho_{\text {oil }}}$
or , $\frac{3}{4}=\frac{667}{\rho_{\text {oil }}}$
or , $\rho_{\text {oil }}=\frac{4}{3} \times 667=889 kgm ^{-3}$
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Question 45 Marks
A wooden cube of side 10 cm has mass 700 g. What part of it remains above the water surface while floating vertically on water surface?
Answer
Given, Side of wooden cube $=10 cm$
Hence, Volume of wooden cube $=10 cm \times 10 cm \times 10 cm =1000 cm ^3$
Mass $=700 g$
Density $=\frac{\text { mass }}{\text { volume }}$
$\therefore$ Density of wooden cube $=\frac{700}{1000}=0.7 g cm ^{-3}$
By the principle of floatation,
$\frac{\text { Volume of immersed part }}{\text { Total volume }}=\frac{\text { Density of wood }}{\text { Density of water }}$
Density of water $=1 g cm ^{-3}$
Density of wooden cube $=0.7 g cm ^{-3}$
$\therefore \frac{\text { Volume of immersed part }}{\text { Total volume }}=\frac{0.7}{1}$
Hence, fraction submerged $=0.7$
Height of wooden cube $=10 cm$
Part of wooden cube which is submerged $=10 \times 0.7=7 cm$
Therefore, part above water $=10-7=3 cm$
Hence, $3 cm$ of height of wooden cube remains above water while floating.
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Question 55 Marks
A block of wood of mass 24 kg floats on water. The volume of wood is $0.032 m^3$. Find : (a) the volume of block below the surface of water,
(b) the density of wood.
(Density of water $=1000 kg m ^{-3}$ )
Answer
Mass of block of wood $=24 kg$
Volume of wood $=0.032 m ^3$
(a) Upthrust $=$ Volume of block below the surface of water $(v) \times$ density of liquid $\times g$
Now for floatation, Upthrust $=$ weight of the body $=24 kgf$
or , $24 kgf = v \times 1000 \times g$
or,$v=\frac{24}{1000}=0.024 m ^3$
(b) According to the law of floatation,
$\frac{\text { Volume of the submerged block }}{\text { Total volume of block }}=\frac{\text { Density of wood }}{\text { Density of water }}$
or,$\frac{0.024}{0.032}=\frac{\text { Density of wood }}{1000}$
or , Density of wood $=1000 \times \frac{0.024}{0.032}=750 kgm ^3$
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Question 65 Marks
A body is held immersed in a liquid. (i) Name the two forces acting on the body and draw a diagram to show these forces. (ii) State how the magnitudes of two forces mentioned in part (i) determine whether the body will float or sink in liquid when it is released. (iii) What is the net force on the body if it (a) sinks and (b) floats?
Answer
(i) Two forces acting on the body are as listed below:

(a) Weight of the body (downwards)

(b) Upthrust of the liquid (upwards)

(ii) If the weight of the body is greater than the upthrust acting on it, the body will sink

If the weight of the body is equal to or less than the upthrust acting on it, the body will float.

(iii) (a) The net force acting on the body when it sinks is body's own weight.

(b) The net force acting on the body when it floats is the upthrust due to the liquid.
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Question 75 Marks
A body of volume V and density $\rho_{ s }$, floats with volume v inside a liquid of density $\rho_{ L }$. Show that $\frac{v}{V}=\frac{\rho_s}{\rho_L}$
Answer
Let $V$ be the volume of a body of density $\rho_5$
Let the body be floating with its volume $v$ immersed inside a liquid of density $\rho_L$

Then, weight of the body,
$W =$ Volume of body $\times$ density of body $\times g$
or, $W=V \rho_s g$

Weight of liquid displaced by body or upthrust,
$F_B=$ Volume of displaced liquid $\times$ density of liquid $\times g$
or, $F _{ B }= v \rho_{ L } g$\

For floatation, $W = F _{ B }$
i.e. , $V \rho_{ s } g=v \rho_L g$
or,$\frac{ v }{V}=\frac{\rho_s}{\rho_L}$
Thus, $\frac{\text { Volume of immersed part of body }}{\text { Total volume of body }}=\frac{\text { Density of body }}{\text { Density of liquid }}$
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Question 85 Marks
 A body of volume $100 cm^3$ weighs 1 kgf in air. Find:
(i) Its weight in water and
(ii) Its relative density.
Answer
Volume of body $=100 cm ^3$

Weight in air , $W_1=1 kgf =1000 gf$

Mass of body $=1 kg =1000 g$
R.D. of solid $=10$
R.D. of water $=1$
(i) Let $W_2$ be the weight of the body in water.
R.D. of body $=\frac{W_1}{W_1-W_2} \times$ R.D. of water
or, $10=\frac{1000}{\left(1000-W_2\right)} \times 1$
or, $10\left(1000-W_2\right)=1000$
or, $1000- W _2=100$
or, $W_2=900 gf$
(ii) R.D. of body = Density in C.G.S. (without unit)
or , R.D. $=\frac{\text { Mass }}{\text { Volume }}=\frac{1000}{100}=10$
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Question 95 Marks
A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate:
  1. The weight of the piece of stone in air,
  2. The volume of the piece of stone,
  3. The relative density of stone,
  4. The relative density of the liquid.
Answer
1. The mass of stone is $15.1 g$. Hence, its weight in air will be $W_a=15.1 gf$
2. When stone is immersed in water its weight becomes $9.7 gf$. So, the upthrust on the stone is $15.1-9.7=5.4 gf$,
Since the density of water is $1 g cm ^{-3}$, the volume of stone is $5.4 cm ^3$.
3. Weight of stone in liquid is $W_1=10.9 gf$
Weight of stone in water is Ww $=9.7 gf$
Therefore, the relative density of stone is
$\text { R.D } \text { D }_{\text {stone }}=\frac{W_a}{W_a-W_w}=\frac{15.1 gf }{15.1-9.7 gf }$
$\text { R. } D _{\text {stone }}=\frac{15.1}{5.4}=2.8$
Relative density of liquid is
$ \text { R.D }_{\text {liquid }}=\frac{W_a-W_l}{W_a-W_w}=\frac{15.1-10.9}{15.1-9.7}=\frac{4.2}{5.4}$
$\therefore \text { R.D }_{\text {Stone }}=0.7777 \approx 0.78 $
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Question 105 Marks
Describe an experiment, using Archimedes principle, to find relative density of a liquid.
Answer
Relative density is the ratio of weight of a given volume of liquid to the weight of the same volume of water. Using Archimedes principle, we can perform an experiment which measures the weight of a liquid displaced by a body and weight of water displaced by the same body. Weight of liquid displaced by a body is given by the difference of weight of a body in air and weight of a body in liquid. Weight of the water displaced by the body can be found by knowing the difference of the weight of the body in air and the weight of the body in water. Therefore, using Archimedes principle, the relative density can be calculated using the formula : RD of Liquid = $\frac{W_1-W_2}{W_1-W_3}$
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Question 115 Marks
With the use of Archimedes' principle, state how you will find relative density of a solid denser than water and insoluble in it. How will you modify your experiments if the solid is soluble in water?
Answer

Steps:

(i) With the help of a physical balance, find the weight, $W_1$ of the given solid.
(ii) Immerse the solid completely in a beaker filled with water such that it does not touch the walls and bottom of beaker, and find the weight $W_2$ of solid in water.
Observations:
Loss in weight of solid when immersed in water $= (W_1 - W_2) gf$
R.D. = Weight of solid in air/Loss of weight of solid in water
$R.D. = W_1/(W_1 - W_2).$
If the solid is soluble in water, then instead of water, take a liquid in which the solid is insoluble and it sinks in the liquid
Then, R.D. = (Weight of solid in air/Loss of weight of solid in liquid) x R.D. of the liquid
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Question 125 Marks
 A solid weighs 32 gf in air and 28.8 gf in water. Find: (i) The volume of solid, (ii) R.D. of solid and (iii) The weight of solid in a liquid of density $0.9 g cm ^{-3}$.
Answer
Weight of solid in air, $W_1=32 gf$
Weight of solid when completely immersed in water $W_2=28.8 gf$
(i) Volume of solid = Mass / density of solid
$
=\frac{32}{10}=3.2 m ^3
$
(ii) R.D. of solid $=\frac{W_1}{W_1-W_2} \times$ R.D. of water
R.D. of solid $=\frac{32}{32-28.8} \times 1$
R.D. of solid $=10$
(iii) Weight of solid in liquid fo density $0.9 gcm ^{-3}=W_3$
R.D. of solid $==\frac{W_1}{W_1-W_3} \times$ R.D. of liquid
or , $10=\frac{32}{32-W_3} \times 0.9$
or, $W_3=29.12 gf$
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Question 135 Marks
QUESIION A metal cube of edge 5 cm and density $9 g cm ^{-3}$ is suspended by a thread so as to be completely immersed in a liquid of density $1.2 g cm ^{-3}$. Find the tension in thread. (Take $g =10 m s ^{-2}$ )
Answer
Given , side of the cube = 5 cm ∴ volume of the cube $= 5 \times 5 \times 5 = 125 cm^3$​​​​​​​
Mass of the cube = volume × density $= 125 × 9 = 1125 g$
$\therefore $ weight of the cube = 1125 gf (downwards)
Upthrust on cube = weight of the liquid displaced
= volume of the cube × density of liquid × g = 125 × 1.2 × g = 150 gf (upwards)
Tension in thread = Net downward force = Weight of cube - Upthrust on cube
$= 1125 - 150 = 975 gf = 9.75 N$
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Question 145 Marks
Describe an experiment to show that a body immersed in a liquid appears lighter than it really is.
Answer
Experiment to show that a body immersed in a liquid appears lighter:

Take a solid body and suspend it by a thin thread from the hook of a spring balance as shown in the above figure (a). Note its weight. Above figure (a) shows the weight as 0.67 N.

Then, take a can filled with water. Immerse the solid gently into the water while hanging from the hook of the spring balance as shown in figure (b). Note its weight. Above figure (b) shows the weight as 0.40 N.

The reading in this case (b) shall be less than the reading in the case (a), which proves that a body immersed in a liquid appears to be lighter.
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Question 155 Marks
Two spheres $A$ and $B$, each of volume $100 cm^3$ is placed on water (density $=1.0 g cm ^{-3}$ ). The sphere A is made of wood of density $0.3 g cm ^{-3}$ and sphere B is made of iron of density $8.9 g cm ^{-3}$.
  1. Find:
    1. The weight of each sphere, and
    2. The upthrust on each sphere.
  2. Which sphere will float? Give reason.
Answer
Volume of sphere $A \& B=100 cm^3$
Density of water $=1 gcm ^{-3}$
Density of sphere $A=0.3 gcm ^{-3}$
Density of sphere $B=8.9 gcm ^{-3}$
(a)
(i) Weight of sphere $A=($ density of sphere $A \times$ volume $) \times g$
$=0.3 \times 100 \times g=30 gf$
Weight of sphere $B=($ density of sphere $B \times$ volume $\times g$ )
$=8.9 \times 100 \times g=890 gf$
(ii) Upthrust on sphere $A =$ Volume of sphere $A \times$ density of water $\times g$
$=100 \times 1 \times g=100 gf$
Upthrust on sphere $B=$ Volume of sphere $B \times$ density of water $\times g$
$=100 \times 1 \times g=100 gf$
Since the volume of both spheres is the same inside water, the upthrust acting on them will also be same.
(b) The sphere A will float because the density of wood is less than the density of water.
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Question 165 Marks
What is the cause of upthrust? At which point can it be considered to act?
Answer
A liquid contained in a vessel exerts pressure at all points and in all directions. The pressure at a point in a liquid is the same in all directions - upwards, downwards and sideways. It increases with the depth inside the liquid.
When a body is immersed in a liquid, the thrusts acting on the side walls of the body are neutralized as they are equal in magnitude and opposite in direction. However, the magnitudes of pressure on the upper and lower faces are not equal. The difference in pressure on the upper and lower faces cause a net upward force (= pressure x area) or upthrust on the body. It acts at the centre of buoyancy.
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Question 175 Marks
Describe an experiment to verify the Archimedes' principle.
Answer
Let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight .
Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid
gets collected in the measuring cylinder.


When water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.
From diagram, it is clear that
Loss in weight (Weight in air – weight in water) = 300 gf – 200 gf = 100 gf
Volume of water displaced = Volume of solid $= 100 cm^3$
Because density of water $= 1 gcm^{-3}$​​​​​​​
Weight of water displaced = 100 gf = Upthrust or loss in weight
This verifies Archimedes' principle.
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Question 185 Marks
What do you understand by the term upthrust of a fluid? Describe an experiment to show its existence.
Answer
When a body is partially or wholly immersed in a liquid, an upward force acts on it. This upward force is known as an upthrust. Upthrust can be demonstrated by the following experiment: Take an empty can and close its mouth with an airtight stopper. Put it in a tub filled with water. It floats with a large part of it above the surface of water and only a small part of it below the surface of water. Push the can into the water. You can feel an upward force and you find it difficult to push the can further into water. It is noticed that as the can is pushed more and more into the water, more and more force is needed to push the can further into water, until it is completely immersed. When the can is fully inside the water, a definite force is still needed to keep it at rest in that position. Again, if the can is released in this position, it is noticed that the can bounces back to the surface and starts floating again.
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Question 195 Marks
Prove that the loss in weight of a body when immersed wholly or partially in a liquid is equal to the buoyant force (or upthrust) and this loss is because of the difference in pressure exerted by liquid on the upper and lower surfaces of the submerged part of body.
Answer
Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density as shown in the figure above. Let the upper surface PQ of the body is at a depth $h_1$ while its lower surface RS is at depth $h_2$ below the free surface of liquid.
At depth $h _1$, the pressure on the upper surface $P Q$,
$P_1=h_1 \rho g .$
Therefore, the downward thrust on the upper surface $P Q$,
$F_1=\text { Pressure } \times \text { Area }=h_1 \rho gA$
At depth $h_2$, pressure on the lower surface RS,
$P_2=h_2 \rho g$ Therefore, the upward thrust on the lower surface RS,
$F _2=$ Pressure $\times$ Area $=h_2 \rho gA$.
The horizontal thrust at various points on the vertical sides of body get balanced because the liquid pressure is the same at all points at the same depth.
From the above equations (i) and (ii), it is clear that $F_2>F_1$ because $h_2>h_1$ and therefore, body will experience a net upward force.
Resultant upward thrust or buoyant force on the body,
$F_B=F_2-F_1$
$=h_2 \rho g A-h_1 \rho g A$
$=A\left(h_2-h_1\right) \rho g$
However, $A\left(h_2-h_1\right)=V$, the volume of the body is submerged in a liquid.
Therefore, upthrust $F _{ B }= V \rho g$. Now, $V \rho g =$ Volume of solid immersed x Density of liquid x Acceleration due to gravity
$=$ Volume of liquid displaced $x$ Density of liquid $x$ Acceleration due to gravity
$=$ Mass of liquid displaced $x$ Acceleration due to gravity
$=$ Weight of the liquid displaced by the submerged part of the body
Thus, Upthrust $F_B=$ weight of the liquid displaced by the submerged part of the body
Now, let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight.
Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid is collected in the measuring cylinder.
When the water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.
From the diagram, it is clear that
Loss in weight (Weight in air - Weight in water) = Volume of water displaced.
Or, Loss in weight = Volume of water displaced x 1 $gcm^{-3}$ [Because the density of water = $1 gcm^{-3}$] Or, Loss in weight = Weight of water displaced ……………(iv)
From equations (iii) and (iv),
Loss in weight = Upthrust or buoyant force
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