# FRANK Solutions for Class 9 Physics Chapter 4 - Fluids

Revise Science syllabus concepts with TopperLearning’s Frank Solutions for ICSE Class 9 Physics Chapter 4 Fluids. Revisit the definition of Pascal, Fluid, Fluid Pressure, Thrust and other important terms. Also, practise the laws of liquid pressure and learn to apply it in real-life scenarios with our expert solutions.

With the support of Frank Solutions, you can also run through topics like Buoyancy, Archimedes principle, etc. To further revise your Physics lessons, use our ICSE Class 9 video lessons, sample papers, practice tests and other e-learning resources.

## Chapter 4 - Fluids Exercise 157

^{-2}.

P = h Xp Xg.

Where h is height of liquid column, p is density of liquid, g is acceleration due to gravity.

Density of mercury is = 1.36 X10

^{4}kg/m

^{3}.

h= height of mercury column which is given = 75 cm = 0.75 m.

So pressure = 0.75 X 1.36 X10

^{4}X9.8 = 9.996X104Nm

^{-2}.

^{2}by a force of 1 Newton acting normally on the surface.

SI unit of thrust is N.

(i) It depends on the point below the free surface (h).

(ii) It depends on density of liquid (p).

(iii) It depends upon acceleration due to gravity (g) of the place.

A fluid exerts pressure on the bottom due to its weight and on the walls of the container in which it is enclosed by virtue of its ability to flow. This is called fluid pressure.

(i) Pressure inside the liquid increases with the depth from the free surface of the liquid.

(ii) Pressure is same at all points on a horizontal plane, in case of a stationary liquid.

(iii) Pressure is same in all directions about a point inside the liquid.

(iv) Pressure at same depth is different in different liquids. It increases with the increase in the density of the liquid.

(v) A liquid will always seek its own level.

Manometer is a U shaped tube containing water whose one limb is dipped in vessel and vessel is tightly covered with plastic sheet. U shaped tube has two limbs one towards the vessel and other is opened to atmosphere.

Now if level of water toward atmospheric open limb is more than level of water in limb towards apparatus end then liquid is said to be at higher pressure than atmosphere. And if level of water toward atmospheric open limb is less than level of water in limb towards apparatus end then liquid is said to be at lower pressure than atmospheric pressure.

## Chapter 4 - Fluids Exercise 158

Hydraulic machines such as hydraulic press, hydraulic brakes and hydraulic jack are application of pascal's law.

(i) It can be obtained in pure form.

(ii) It does not vaporizeat ordinary temperatures.

(iii) Its density is high and hence the length of the mercury column supported by atmospheric pressure is 76 cm which is practically possible.

## Chapter 4 - Fluids Exercise 173

(i) Volume of body submerged in the liquid.

(ii) Density of the liquid.

(iii) Acceleration due to gravity.

Apparent Weight of the completely immersed body in water = 280 gf.

(i) Loss in weight of the body = Weight of body in air - apparent weight of immersed body.

Loss in weight = 300 gf - -280 gf = 20 gf.

(ii) As upthrust on the body = loss in weight

(iii) So uptrust = 20 gf.

Density of the metal cube = 9 gcm

^{-3}= 9 X 10

^{3}kgm

^{-3}.

Volume of the metal cube = 125 cm

^{3}= 125X10

^{-6}m

^{3}.

Mass of the metal cube =9 X 10

^{3}X125X10

^{-6}= 1125 X10

^{-3}=1.125 kg.

Weight of the liquid = mass X gravity = 1.125 X10 = 11.25 N.

Density of liquid = 1.2 gcm

^{-3}= 1.2 X10

^{3}kgm

^{-3}.

Upthrust of the liquid = V X p Xg.

Upthrust = 125X10-6 X1.2 X103 X10 = 1.5 N.

Apparent weight of the body = weight of liquid - upthrust

Apparent weight = 11.25 N - 1.5 N = 9.75 N

Tension in the string is equal to the apparent weight of the body

So, tension in string would be 9.75 N.

## Chapter 4 - Fluids Exercise 174

^{o}C.

(ii) SI unit of density is Kgm

^{-3}.

(iii) SI unit of weight of body is N.

(iv) Relative density is a pure ratio it has no dimension.

(Density of floating body / Density of liquid) = fraction submerged.

The Fraction of ice submerged in water remain same as density of ice and water remain same during melting. As ice melts some volume of ice decrease and convert into water and volume of water increase by same amount. So, level of water remains same during melting.

(Density of floating body / Density of liquid) = fraction submerged.

Height of wooden piece = 15 cm.

Height of wooden piece sinks in water = 10 cm.

Fraction of wooden piece submerged in water = 10/15 = 0.67.

As liquid is water so ratio of Density of wooden by density of water gives relative density of floating wooden piece. So, relative density of wooden block is 0.67.

Height of wooden piece = 15 cm.

Height of wooden piece sinks in spirit = 12 cm.

Fraction of wooden piece submerged in water = 12/15 = 0.8.

We know density of wooden piece = 0.67

(Density of floating body / Density of liquid) = fraction submerged.

Density of liquid/spirit = (Density of floating body /fraction submerged)

Density of liquid/spirit = 0.67/0.8 = 0.83.

Relative density of spirit is 0.83.

(ii) Sea water contains mineral salts and density of sea water increase due to presence of these. As density of sea water is more than the normal water so it apply more buoyant force than usual one and a person find it easy to swim in sea water.

Relative density of sea water = 1.025

Let total volume of iceberg = X cm

^{3}.

Volume of iceberg above water = 800 cm

^{3}.

Volume of iceberg in submerged in the water = (X - 800) cm

^{3}.

Fraction of iceberg submerged = (X- 800)/X

Now we know that fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.

(Density of ice / Density of sea water) = fraction submerged

0.92/1.025 = (X-800)/X

0.8975 X = X - 800

X - 0.8975 X = 800

0.1025 X = 800

X = 800/0.1025 = 7804.8 cm

^{3}.

Total volume of iceberg = 7804.8 cm

^{3}.

Relative density of brine = 1.1

(Density of wax/ Density of brine) = fraction submerged

0.95/1.1 = fraction of volume submerged

Fraction of volume submerged = 0.86

Relative density of sea water = 1.1 cm

^{}

(Density of ice / Density of sea water) = fraction submerged of iceberg

0.9/1.1 = fraction of iceberg submerged

Fraction of iceberg submerged = 9/11.

^{o}c in SI system is = 1000 Kgm

^{-3}.

Volume of wooden cube = 10 X 10 X 10 = 1000 cm

^{3}.

Mass of wooden cube = 700 g.

Density of wooden cube = mass/volume = 700/1000 = 0.7 gcm

^{-3}.

Density of water = 1 gcm

^{-3}.

(Density of floating body / Density of liquid) = fraction submerged

0.7/1 =fraction submerged

Fraction of wooden cube submerged in water = 0.7

Height of wooden cube = 10 cm

Part of wooden cube which is submerged = 10 X 0.7 = 7 cm

So, wooden cube will float in water with 3 cm height above the water surface.

^{3}.

Mass of wooden block = 24 Kg.

Density of wooden block = mass/volume = 24/0.032 = 750 Kgm

^{-3}.

Density of water = 1000 Kgm

^{-3}.

(Density of floating body / Density of liquid) = fraction submerged

750/1000 =fraction submerged

Fraction of wooden block submerged in water = 0.75

Total volume of wooden block = 0.032 m

^{3}.

Part of volume of wooden block which is submerged = 0.032 X 0.75 = 0.024 m

^{3}.

^{0}C.

As relative density of platinum is 21.50, this means platinum is 21.5 times denser than water at 4

^{o}C.

## Chapter - Exercise

## Chapter 4 - Fluids Exercise 175

^{-3}.

Density of water at 4

^{o}C = 1000 kg m

^{-3}.

Relative density = density of substance /density of water at 4

^{0}C.

Relative density of mercury = 13600 Kgm

^{-3}/1000 kg m

^{-3}= 13.6.

^{3}.

Weight of body = 1 kgf = 1000 gf

Mass of body= 1000 gm.

Density of liquid = 1000 gm/100cm

^{3}= 10 gcm

^{3}.

Density of water at 4

^{o}= 1gcm

^{-3}.

Relative density = density of substance /density of water at 4

^{0}C

Relative density = 10 gcm

^{3}/1 gcm

^{3}= 10

Mass of body= 1000 gm.

Density of water = 1 gcm

^{-3}

Acceleration due to gravity = 10 ms

^{-2}.

Upthrust = V X p Xg.

Upthrust = 100 X 1 Xf = 100 gf.

Resultant weight of the body = weight - upthrust = 1000 gf - 100 gf = 900 gf.

So, volume of body = 20000 cm

^{3}.

Mass of body = 70 kg = 70000 gm

Density of body = mass /volume = 70000/20000= 3.5 gm cm

^{-3}.

Density of water in C.G.S system = 1g cm

^{-3}.

Relative density of body = density of body /density of water =3.5 gm cm

^{-3}/1g cm

^{-3}.

Relative density = 3.5.

Density of mercury = relative density X density of water.

Relative density = 13.6.

Density of water in C.G.S system = 1g cm

^{-3}.

So, density of mercury in C.G.S system = 13.6 X1 = 13.6 gcm

^{-3}.

Density of water in SI system = 1000 Kg m

^{-3}.

So, density of mercury in SI system = 13.6 X1000 = 13.6 X10

^{3}Kgcm

^{-3}.

^{3}Kg m

^{-3}.

Density of water at 4

^{o}C = 10

^{3}Kg m

^{-3}.

Relative density of a substance is the ratio of the density of the substance to the density of water at 4

^{o}C.

So, relative density of iron is = 7.8 X 10

^{3}Kg m

^{-3}/10

^{3}Kg m

^{-3}= 7.8

(ii) Volume of metallic piece increases with increase in temperature.

(iii) Density of metallic piece decreases with increases in temperature.

## Chapter 4 - Fluids Exercise 177

^{-3}.

^{o}C.

(ii) A hydraulic press can be used for pressing cotton bales, quilts, books etc.

^{5}N/m

^{2}.

## Chapter 4 - Fluids Exercise 178

Means when pressure is applied at a point in a confined fluid, it is transmitted undiminished and equally in all directions throughout the liquid.

(i) Weight of gaseous column.

(ii) Density of gaseous column.

Simple barometer has two main defects

(i) It is not suitable for making accurate measurement of atmospheric pressure as any change in the level of mercury in the tube changes the level of the free surface of mercury is trough and fixed scale cannot be used with it.

(ii) Simple barometer is not portable. So, it cannot be used by airmen, navigators, mountaineers, who need a portable barometer.

Reading of a barometer would fall if it is taken to a hill as pressure decreases with increase in height.

Relative density of solid = 8.4

Now, Relative density = weight of solid in air/ loss of weight of solid in water.

Loss of weight of solid in water = weight of solid in air/ Relative density.

Loss of weight of solid in water = 2.10/8.4 = 0.25 N.

Weight of solid in water = weight in air - loss of weight in water

Weight of solid in water = 2.10 - 0.25 =1.85 N.

Relative density of liquid =1.2

We know

Relative density of liquid = Loss of weight of solid in liquid/loss of weight of solid in water.

Loss of weight of solid in liquid = Relative density X loss of weight of solid in water.

Loss of weight of solid in liquid = 1.2 X 0.25 = 0.3 N.

Weight of solid in liquid = weight of solid in air - loss of weight of solid in liquid.

Weight of solid in liquid = 2.10 - 0.3 = 1.8 N.

^{-3}.

This means a cube of iron having side 1m would weigh 7800 Kg.

Density of water at 4

^{o}C is 1000 Kgm

^{-3}.

Density of water at 4

^{o}C = 1000 Kgm

^{-3}.

Density of body = 0.52 X 1000 Kgm

^{-3}= 520 Kgm

^{-3}

We know density = mass X volume.

Mass = density X volume

Mass = 520 X 2 =1040 Kg.

Mass of given body is 1040 Kg.

Piece of metal weighs in liquid = 39.5 f.

Loss of weight of metal in liquid = 44.5 - 39.5 = 5f.

Relative density = weight of solid in air/ loss of weight of solid in water.

Relative density of liquid =44.5f/5f =8.9

Relative density of liquid = 8.9

Ship is designed in such a manner that it encloses large quantity of air in air tight bags and in rooms and corridors which makes the average density of ship less than that of water.

(ii) Density of egg is greater than fresh water so it sinks in fresh water but due to addition of salt density of water increases which makes the density of salt water greater than egg and hence floats in a strong solution of salt.

(iii) The bottom of the hydrometer is made heavier by loading it with lead shots so that it floats vertically with some of its portion outside the surface of water in the jar.

(iv) Relative density of Ice is = 0.9 cm

^{-3}

Relative density of sea water is = 1 cm

^{-3}

(Density of ice / Density of sea water) = fraction submerged of iceberg

0.9/1 = fraction of iceberg submerged

Fraction of iceberg submerged = 9/10.

Thus in colder countries where there are icebergs in oceans, only about 1/10 is seen above water and the remaining water 9/10 remain submerged. Hence, there is danger of these icebergs to the ships sailing in these oceans.

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