# Frank Solutions for Chapter 4 Fluids Class 9 Physics ICSE

### Exercise 4.1

1. What do you mean by pressure? Write its SI unit.

The thrust on the unit surface is known as pressure. The SI unit of pressure is Nm-2

2. The atmospheric pressure is 75 cm of mercury. Express it in Nm-2.

Pressure is given by

P = h × p × g

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

Density of mercury is = 1.36 × 104 kg/m3

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

So, pressure = 0.75 × 1.36 × 104 × 9.8

= 9.996 × 104 Nm-2

3. Is pressure a scalar or vector physical quantity?

Pressure is a scalar physical quantity.

4. Define one Pascal.

One pascal is defined as the pressure exerted on a surface of area 1 m2 by a force of 1 Newton acting normally on the surface.

5. What do you mean by thrust? Write its SI unit.

The force acting normally on a surface is known as thrust.

SI unit of thrust is N.

6. What length of water column is equivalent to 0.76 m mercury column?

we know pressure exerted by a liquid column of height h,

density p is

P = h × p×g.

Pressure exerted by mercury column of height 76 cm.

Density of mercury = 13.6 g/cc = 1.36 × 104 kg/m3

Pmercury = 0.76× 1.36 × 104 ×9.8 = 10.12× 104 Nm-2

Let height of water column = h m.

Density of water = 1g/cc = 103 kg/m3

P water = h × 10×9.8 = 9.8h× 103 Nm-2

Now put P mercury = Pwater

9.8h× 103 = 10.12× 104

⇒ h = 10.12/9.8 = 10.34 m.

So, 10.34 m height of water column would exert same pressure on its base as 76 cm column of mercury.

7. Why can't water be used in place of mercury in a barometer?

Water can't be used in place of mercury in a barometer because of its low density. It would require 10.34 m long tube to measure 1 atmospheric pressure which is not practically possible while mercury having high density (13.6 g/cc) would require only 0.76 m long pipe which is practically possible.

8. What physical quantity is measured in bar?

Pressure is the physical quantity which is measured in bar.

9. State whether thrust is scalar or vector.

Thrust is a vector quantity.

10. What is the difference between thrust and pressure?

Thrust on a surface is the force acting normally on a surface while pressure on a surface is thrust acting on the unit area of a surface.

11. (i) Calculate the height of a water column which will exert on its base the same Pressure as the 70 cm column of mercury.

(ii) Will the height of the water column change if the cross section of the water column is made wider?

(i) we know pressure exerted by a liquid column of height h, density p is

P = h × p×g.

Pressure exerted by mercury column of height 70 cm.

Density of mercury = 13.6 g/cc = 1.36×104 kg/m3,

P mercury = 0.7×1.36× 104 ×9.8 = 9.32× 104 Nm-2

Let height of water column = h m.

Density of water = 1g/cc = 103 kg/m3

Pwater = h× 103 ×9.8 = 9.8h× 103 Nm-2

Now put Pmercury = P water

9.8 h × 103 = 9.32 x 104

⇒ h = 93.2/9.8 = 9.52 m.

So, 9.52 m height of water column would exert same pressure on its base as 70 cm column of mercury.

(ii) Height of the water column would not change if the cross section of the water column is made wider.

12. Explain, why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of the lake.

Lake has greater pressure at the bottom than the surface as pressure increases with depth. So when gas bubble is released at the bottom of the lake it experiences more pressure and is small in size but as it rises upwards the pressure experienced by it decreases. So it grows in size as it moves towards the surface from bottom.

13. A dam has broader walls at the bottom than at the top. Explain.

A dam has broader walls at the bottom than at the top because the pressure exerted by a liquid increases with its depth, and at any point at a particular depth liquid pressure is same in all directions. Now as more pressure is exerted by water on the wall of the dam as depth increases. Hence a thick wall is constructed at the bottom of dam to withstand greater pressure.

14. The base of a cylindrical vessel measures 300 cm2. Water (density = 1000 kg m-3) is poured into it upto a depth of 6 cm. Calculate the pressure and thrust of water on the base. (g = 10m s-2

Depth of water = 6 cm = 0.06 m.

Density of water = 1000 kgm-3.

Acceleration due to gravity = 10 ms-2.

We know pressure

P = h× p×g.

⇒ P = 0.06×1000× 10 = 600 Nm-2 = 600 Pa.

Area of base of cylindrical vessel = 300 cm2 = 300× 10-4 m2 = 0.03 m2

We know

Thrust = pressure x area.

⇒ Thrust = 600× 0.03 = 18N.

So thrust acting on base of vessel is 18 N.

15. State three factors on which the pressure at a point in a liquid depends.

The pressure at a point in a liquid depends upon on the following three factors:

1. It depends on the point below the free surface (h).
2. It depends on density of liquid (p).
3. It depends upon acceleration due to gravity (g) of the place.

16. Deduce an expression for the pressure at a depth inside the liquid.

For calculating the total pressure in a liquid at a depth, we have to add the atmospheric pressure that acts on the free surface of the liquid.

Total pressure inside a liquid at depth ‘h’ = atmospheric pressure + pressure due to liquid column.

So Total pressure inside a liquid at depth = Po + hpg.

Here Po is atmospheric pressure acting on the free surface of liquid.

17. What is meant by a fluid? What is a fluid pressure?

A substance having a tendency to flow is called fluid.

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.

18. State the laws of liquid pressure.

The laws of liquid pressure are

1. Pressure inside the liquid increases with the depth from the free surface of the liquid.
2. Pressure is same at all points on a horizontal plane, in case of a stationary liquid.
3. Pressure is same in all directions about a point inside the liquid.
4. Pressure at same depth is different in different liquids. It increases with the increase in the density of the liquid.
5. A liquid will always seek its own level.

19. How does the liquid pressure on a diver change according to the following conditions:

(i) When the driver moves horizontally, and

(ii) When the driver moves to the greater depth?

1. When the diver moves horizontally the depth from the surface remains same and as we know that pressure at depth h is given by
2. P = h× p× g so as depth remains same pressure would remains same while moving horizontally. (Density of liquid, g also remains same).
3. When diver moves to depth, depth h increases and thus pressure increases as diver moves to depth.

20. What do you mean by diver's suit?

A diving suit is a garment or device designed to protect a diver from the underwater environment.

21. Name some ambient pressure diving suits.

There are five main types of ambient pressure suits. These are wetsuits, dry suits, semidry suits, dive skins etc.

22. What fact about liquid pressure does the following diagram in fig. 31 illustrate?

This diagram illustrates that pressure of liquid increases with depth. Hole at greater depth has more pressure and hence flow of liquid from there is more.

Other hole is at less depth so the pressure of liquid there is less and hence flow of liquid from there is less.

So the diagram shows that pressure increases with increase in depth.

23. What is a manometer? How does it show whether the pressure inside a vessel connected to one arm of it, is lower or above the atmospheric pressure?

Manometer is a simple pressure gauge that measures differences in pressure at the two ends of the apparatus.

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.

24. The fig. 32 shows a manometer containing a liquid of density p. The limb P of the manometer is connected to a vessel V and the limb Q is open to atmosphere. The difference in the levels of liquid in the two limbs of manometer is h as shown in the diagram. The atmospheric pressure is Pp.

(i) What is the pressure on the liquid surface in the limb Q?

(ii) What is the pressure on the liquid surface in the limb P?

(i) The pressure on the liquid surface in the limb Q is equal to atmospheric pressure i.e Po.

(ii) According to manometer principle, difference in atmospheric pressure in two limbs is equal to difference in height of liquid in two limbs.

So, pressure at P = pressure at Q + h×p×g.

⇒ Pp = PQ + h×p×g.

25. State the principle on which a hydraulic press works. Write one use of hydraulic press.

A hydraulic press works on the principle of pascal's law. A hydraulic press can be used for extracting juice of sugarcane, sugar beet etc.

28. State Pascal's law of transmission of pressure. Name three applications of Pascal's law.

Pascal's law states that pressure applied to an enclosed liquid, is transmitted equally to every part of the liquid or in other words when pressure is applied at a point in a confined fluid, it is transmitted undiminished and equally in all directions throughout the liquid.

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

29. What is the use of altimeter?

Altimeter is a device which is used in an aircraft to measure its altitude.

30. How does atmosphere pressure very with height?

Atmospheric pressure decreases with increase in height. Our atmosphere comprises of a large number of parallel layers. The pressure on a layer is equal to the thrust or weight of the gaseous column on the unit area of that layer. Hence, as we go up, the weight of the gaseous column decreases, which decrease the pressure of the gaseous column.

31. Why is aneroid barometer called so?

Aneroid means containing no liquid and aneroid barometer is evacuated so it tends to collapse under the pressure of air. The stout spring balances the thrust on the metal box due to normal air pressure and prevents the box from collapsing. As this type of barometer doesn't contain any liquid so it got its name aneroid barometer.

32. What is a barometer write two uses of a barometer?

Barometer is a device which is used for measuring atmospheric pressure. Barometers are used in weather forecasting and in measuring altitudes.

33. Why is mercury used in a barometer?

Mercury is used in barometer because

1. It can be obtained in pure form.
2. It does not vaporize at ordinary temperatures.
3. Its density is high and hence the length of the mercury column supported by atmospheric pressure is 76 cm which is practically possible.

### Exercise 4.2

1. What is meant by the term buoyancy?

All liquid exerts an upward force on the body placed in it. This Phenomenon is called buoyancy.

2. Define up thrust and state its SI unit.

The upward force which any liquid exerts upon a body placed in it is called the upthrust. The SI unit of up thrust is N.

3. In what direction does the buoyant force act on a body due to a liquid?

Buoyant force act on a body in upward direction.

4. What do you mean by the term upthrust of a liquid?

Upthrust is defined as the upward force on the object provided by the liquid because the object has displaced some of the fluid.

5. Why is a force needed to keep a block of cork inside water?

When block of cork is immersed in water buoyant force acts on it in upward direction so to overcome this force we have to apply an equal force in downward direction to keep block of cork inside water.

6. A piece of wood when left under water again comes to the surface. Explain with reason.

Wood has density less than water so volume of water displaced by it is more than the volume of wooden block submerged so force of upthrust is greater than the weight of wood which pushes wooden block on the surface. Hence, a piece of wood when left under water again comes to the surface.

7. Will a body weigh more in air or in water when weighed with a spring balance? Why?

A body will weigh more in air as weight of body acts in downward direction and there is no force in upward direction while body submerged in water weigh less because an upthrust act on the body in upward direction so the resultant weight of the body decreases.

8. A body dipped into a liquid experiences an upthrust. State two factors on which upthrust on the body depend.

Upthrust or buoyant force depends on the following factors:

1. Volume of body submerged in the liquid. -
2. Density of the liquid.
3. Acceleration due to gravity.

9. A small stone of mass m ( = 200 g) is held under water in a tall jar and allowed to fall as shown in the fig. 20. The free body diagram of the stone is also shown.

(i) What does F2 represent?

(ii) What does m2 represent?

(iii) What is the net force acting on the stone?

(iv) What is the acceleration of the stone as it falls through water? Neglect the force due to viscosity. Assume that volume of stone = 80 cm3, density of water = 1.0 g cm-3 and acceleration due to gravity g = 10 ms-2.

(i) F2 represent the buoyant force acting on the stone in upward direction.

(ii) m1 represent the apparent mass of the stone during motion through water.

(iii) Net force acting on the stone = F1 - F2 = (m – m1)g.

(iv) Mass of stone = 200 gm = 0.2 kg.

Volume of stone = 80 cm3 = 80× 10-6 m3

Density of water = 1 gcm-3 = 1000 kgm3

Acceleration due to gravity = 10 ms-2

Upthrust = V×p× g.

Upthrust = 80× 10-6 ×1000× 10 = 0.8 N.

Weight of the stone = mass× gravity = 0.2 x 10 = 2N.

Resultant weight of the stone = weight - upthrust = 2 N - 0.8N = 1.2 N

Resultant acceleration of the gravity = m x a’

m× a’= 1.2 N

⇒ a’ = 1.2/0.2 = 6ms-2

Resultant acceleration of the stone as it falls through the water is 6 ms-2

10. A body weighs 300 gf in air and 280 gf when completely immersed in water. Calculate:

(i) The loss in weight of the body,

(ii) The upthrust on the body.

Weight of the body in air = 300 gf.

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.

11. A metal cube of 5 cm edge 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 the thread. (Take g = 10 ms-2).

Edge of metal cube = 5 cm.

Density of the metal cube = 9 gcm-3 = 9× 103 kgm-3

Volume of the metal cube = 125 cm3 = 125×10-6 m3

Mass of the metal cube = 9×103 ×125×10-6 = 1125×10-3 = 1.125 kg.

Weight of the liquid = mass x gravity = 1.125×10 = 11.25 N.

Density of liquid = 1.2 gcm-3 = 1.2×103 kgm-3

Upthrust of the liquid = V× p×q.

Upthrust = 125×10-6×1.2 ×103 x 10 = 1.5. N.

Apparent weight of the body = weight of liquid - upthrust

Apparent weight = 11.25N - 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.

12. It is easier to lift a heavy stone under water than in air. Explain.

It is easier to lift a heavy stone under water because in water an upthrust acts on the upward direction which reduces the apparent weight of the stone and makes it easy to lift.

13. State Archimedes’ principle.

Principle of Archimedes’ states that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.

14. Describe an experiment to verify the Archimedes’ principle.

We can verify Archimedis’ principle experimentally by doing this experiment.

The stone weighed 0.67 N in air and 40 N when immersed in water. The displaced water weighed is 0.27 N (= 0.67 – 0.40)

Pour water into eureka can till the water starts overflowing through the spout.

When the water stops dripping replace the beaker by another one of known weight.

Suspend a stone with the help of a string from the hook of a spring balance and record the weight of the stone.

Now, gradually lower the body into eureka can containing water and record its new weight in water when it is fully submerged in water. When no more water drips from the spout, weigh the beaker containing water.

After observation, we can conclude that apparent loss of weight of stone (calculated by differences in weight measured by spring balance) = weight of water displaced (weight of water in beaker).

This proves the Archimedis’ principle that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced.

15. Fig. 21 shows the same block of wood floating in three different liquids P, Q and R of densities d1, d2 and d3 respectively. Which of the liquid has the highest density? Give a reason to your answer.

The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.

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

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

Now in this case as density of floating body is same in all three cases. So, density of liquid is maximum when fraction submerged of wooden block is minimum.

As liquid R has least fraction submerged of wooden block in it So, it has maximum density.

16. When a piece of wood is suspended from the hook of a spring balance, it reads 90 gf. The wood is now lowered into water. What reading do you expect on the scale of the spring balance?

Wood has density less than water so volume of water displaced by it is more than the volume of wooden block submerged so force of upthrust is greater than the weight of wood which makes it float on the water surface. And the apparent weight of the piece of the wood would be zero.

17. An iron ball of mass 600 g is dropped in mercury contained in a beaker. Will the ball float or sink? What will be its apparent weight?

Density of iron is less than the density of mercury so it will float on the surface of the mercury. Apparent weight of the floating iron ball is zero.

18. Explain, why an iron nail floats on mercury, but it sinks in water.

Iron nail has density less than that of mercury so it will float on the surface of mercury but in the case of water it will sink because the density of iron nail is more than that of water.

19. Is density of a body same as its relative density?

No, the relative density of a substance is the ratio of the density of the substance to the density of water at 4°C.

20. Write the SI units of

(i) Buoyant force

(ii) Density

(iii) Weight of a body

(iv) Relative density

(i) SI unit of buoyant force is N.

(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.

21. An iron nail sinks in water while an iron ship floats on water. Explain the reason.

What can you say about the average density of a ship floating on water?

Density of iron is more than the density of water so it sinks down in the water but in case of ship, it is design 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 and ship floats on the surface of the water.

22. A piece of ice floating in a glass of water melts but level of water in the glass does not change. Give reason.

The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.

(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.

23. A piece of wood of uniform cross-section and 15 cm height sinks 10 cm in water and 12 cm in spirit. Find the R.D. of wood and spirit.

The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.

(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.

24. A man first swims in sea water and then in river water.

(i) Compare the weight of sea water and the river water displaced by him.

(ii) Where does he find it easier to swim and why?

(i) When a body is completely immersed in water then it displaces equal volume of water to its own weight. Volume of body of man is same in both river and sea so weight of water of sea displaced by him is equal to the weight of water of river displaced by him. And ratio of weights would be 1:1.

(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.

25. The R.D. of ice is 0.92 and that of sea water is 1.025. Find the total volume of an iceberg which floats with its volume 800 cm3 above water.

Relative density of Ice = 0.92

Relative density of sea water = 1.025

Let total volume of iceberg = X cm3

Volume of iceberg above water = 800 cm3

Volume of iceberg in submerged in the water = (X - 800) cm3

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 cm3

Total volume of iceberg = 7804. 8 cm3

26. A piece of wax floats in brine. What fraction of its volume will be immersed?

R.D. of wax = 0.95, R.D. of brine = 1.1.

Relative density of wax = 0.95

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

27. If the density of ice is 0.9 g cm, what portion of an iceberg will remain below the surface of water in the sea? (Density of sea water = 1.1 g cm-3)

Relative density of Ice = 0.9 cm

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.

28. What is the common use of a lactometer?

Lactometer is commonly used for testing the purity of milk.

29. What is the density of water at 4°C in SI system?

Density of water at 4°c in SI system is = 1000 Kgm-3

30. A wooden cube of side 10 cm has mass 700 g. It will float in water with:

(a) Half of its volume inside water

(b) 3 cm height above the water surface

(c) 7 cm height above the water surface

(d) Just inside the water surface.

Side of wooden cube = 10 cm.

Volume of wooden cube = 10 ×10×10 = 1000 cm3

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 = 10×0.7 = 7.cm

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

31. A block of wood of mass 24 kg floats in water. The volume of the wood is 0.032 m3. Find:

(a) The volume of the block below the surface of water.

(b) The density of the wood.

(Density of water = 1000 kg m-3)

Volume of wooden block = 0.032 m3

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 m3

Part of volume of wooden block which is submerged = 0.032×0.75 = 0.024 m3

32. If R.D. of platinum is 21.50, what does it signify?

Relative density = density of substance/density of water at 4°c.

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

33. Density of mercury is 13600 kg m’. What is its relative density?

Density of mercury = 13600 Kgm-3

Density of water at 4°C = 1000 kg m-3

Relative density = density of substance /density of water at 4°c.

Relative density of mercury = 13600 Kgm-3/1000 kg m-3 = 13.6

34. A body of volume 100 cm3 weighs 1 kgf in air. Find its weight in water and its relative density.

volume of body = 100 cm3

Weight of body = 1 kgf = 1000 gf

Mass of body = 1000 gm.

Density of liquid = 1000 gm/100cm3 = 10 gcm3

Density of water at 4° = 1 gcm-3

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

Relative density = 10 gcm3/1gcm3 = 10

Mass of body = 1000 gm.

Density of water = 1 gcm-3

Acceleration due to gravity = 10 ms-2

Upthrust = V× p× g.

Upthrust = 100×1×f = 100 gf.

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

35. A body of mass 70 kg, when completely immersed in water, displaces 20,000 cm3 of water. Find the relative density of material of the body.

When a body is completely immersed in water then it displaces equal volume of water to its own weight.

So, volume of body = 20000 cm3

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

36. The relative density of mercury is 13.6. State its density in

(i) CGS. unit,

(ii) S.I. unit.

Relative density = density of mercury /density of water.

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 ×1 = 13.6 gcm-3

Density of water in SI system = 1000 Kg m-3

So, density of mercury in SI system = 13.6×1000 = 13.6× 103 Kgcm-3

37. The density of iron is 7.8 x 103 kg m-3. What is its relative density?

Density of iron is = 7.8 x 10° Kg m-3

Density of water at 4°C = 103 Kg m-3

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

So, relative density of iron is = 7.8×103 Kg m-3/103 Kg m-3 = 7.8

38. How is

(i) mass,

(ii) volume, and

(iii) Density of a metallic piece affected, if at all, with increase in temperature?

(i) Mass of a metallic piece remains unchanged with increase in temperature.

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

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

39. How does the density of water change with temperature?

Density of water decreases with the increase in temperature and increases with decreases in temperature.

40. Complete the following results:

(i) Mass = ....... × density

(ii) S.I. unit of density is ............

(iii) Density of water is .......... Kg m-3

(iv) Density in kg m-3 = …………. × density in g cm-3

(i) Mass = VOLUME × density.

(ii) SI unit of density is Kgm-3.

(iii) Density of water is 1000 Kgm-3,

(iv) Density in Kgm-3 = 1000 × density in gcm-3.

### Exercise 4.3

1. What do you mean by buoyancy?

All liquid exerts a upward force on the body placed in it. This Phenomenon is called buoyancy.

2. What is meant by up thrust?

The upward force which any liquid exerts upon a body placed in it is called the upthrust.

3. Is pressure a vector quantity?

Pressure is a scalar quantity.

4. Is thrust a scalar or vector quantity?

Thrust is a vector quantity.

5. What is the SI unit of density?

SI unit of density is Kgm-3

6. What do you mean by relative density?

The relative density of a substance is the ratio of the density of the substance to the density of water at 4°C.

7. What are the factors on which the pressure of a liquid depends?

The pressure at a point in a liquid depends upon on the following three factors:

1. It depends on the point below the free surface (h).
2. It depends on density of liquid (p).
3. It depends upon acceleration due to gravity (g) of the place.

8. State Archimedes’ principle. Does it apply to gases?

Principle of Archimedes’ states that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced. Yes, it applies to gases also.

9. How is thrust different from pressure?

Thrust on a surface is the force acting normally on a surface while pressure on a surface is thrust acting on the unit area of a surface.

10. State Pascal's law.

Pascal's law states that pressure applied to an enclosed liquid, is transmitted equally to every part of the liquid or in other words when pressure is applied at a point in a confined fluid, it is transmitted undiminished and equally in all directions throughout the liquid.

11. Do all liquids exert pressure?

Yes, all liquid exert pressure.

12. State two applications of Pascal's law.

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

13. What is the principle of a hydraulic machine?

Pascal's law is principle of hydraulic machines.

14. State the principle on which Brahma press depends.

Brahma press depends upon Pascal's law.

15. State two uses of a hydraulic press.

1. A hydraulic press can be used for extracting juice of sugarcane, sugar beet etc.
2. A hydraulic press can be used for pressing cotton bales, quilts, books etc.

16. What is meant by atmospheric pressure?

Atmospheric point at any point in air at rest is equal to the weight of a vertical column of air on a unit area surrounding the point, the column extending to the top of atmosphere.

17. What is the value of atmospheric pressure?

Atmospheric pressure at sea level is about 105 N/m2.

18. Name the instrument used to measure atmospheric pressure.

Barometer is used for measuring the atmospheric pressure.

19. What is the use of altimeter?

Altimeter is a device which is used in an aircraft to measure its altitude.

20. What does a falling barometer indicate?

A falling barometer indicates the approach of rain or storm or both.

21. Define atmospheric pressure diving suits. Does a diving suit create buoyancy?

A atmospheric pressure diving suit is a garment or device designed to protect a diver from the underwater environment. Yes, diving suits create buoyancy.

22. What do you mean by the term fluid pressure?

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.

23. A dam has broader walls at the bottom than at the top. Explain.

A dam has broader walls at the bottom than at the top because the pressure exerted by a liquid increases with its depth, and at any point at a particular depth liquid pressure is same in all directions. Now as more pressure is exerted by water on the wall of the dam as depth increases. Hence a thick wall is constructed at the bottom of dam to withstand greater pressure.

24. State and explain Pascal's law of transmission of pressure.

Pascal's law states that pressure applied to an enclosed liquid, is transmitted equally to every part of the liquid.

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

25. State two factors which affect the atmospheric pressure as we go up.

Factors which affect the atmospheric pressure as we go up are

1. Weight of gaseous column.
2. Density of gaseous column.

26. How does the atmospheric pressure change with altitude?

Atmospheric pressure decreases with increase in height. Our atmosphere comprises of a large number of parallel layers. The pressure on a layer is equal to the thrust or weight of the gaseous column on the unit area of that layer. Hence, as we go up, the weight of the gaseous column decreases, which decrease the pressure of the gaseous column.

27. What is the purpose of a barometer? State two major defects of a simple barometer.

Barometer is a device used for measuring atmospheric pressure.

Simple barometer has two main defects

1. 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.
2. Simple barometer is not portable. So, it cannot be used by airmen, navigators, mountaineers, who need a portable barometer.

28. Why do not we feel uneasy even under enormous pressure of the atmosphere above as?

We don't feel uneasy even under enormous pressure of the atmosphere above us because our blood also exerts a pressure called blood pressure, which is greater than atmospheric pressure. So, there is balance between our blood pressure and atmospheric pressure.

29. How is the reading of a barometer affected when it is taken to (a) A mine, and (b) A hill?

Reading of a barometer would rise if it is taken to the mine as pressure increases with depth.

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

30. A solid weighs 2.10 N in air. It has a relative density of 8.4. How much will the body weigh if placed (i) In water (ii) In a liquid of relative density 1.2?

(i) Weight of solid in air = 2.10 N

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 =1.85N.

(ii) 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 × loss of weight of solid in water.

Loss of weight of solid in liquid = 1.2 × 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.

31. The density of iron is 7800 kg m-3. What do you mean by this statement? What is density of water at 4°C?

Density of iron is 7800Kgm-3

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

Density of water at 4°C is 1000 Kgm-3.

32. Calculate the mass of a body whose volume is 2m3 and relative density is 0.52.

Relative density of body = 0.52

Density of water at 4°c = 1000 Kgm-3

Density of body = 0.52 × 1000 Kgm-3 = 520 Kgm-3

We know,

density = mass × volume.

⇒ Mass = density × volume

⇒ Mass = 520 × 2 = 1040 Kg.

Mass of given body is 1040 Kg.

33. A piece of metal weighs 44.5 gf in air, 39.5 gf in water. What is the R.D. of the metal?

Piece of Metal weighs in air = 44.5 f

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

34. A body of volume 100 cm3 weighs 1 kgf in air. Calculate its weight in water. What is its relative density?

Volume of body = 100 cm3

Weight of body = 1 kgf = 1000 gf

Mass of body = 1000 gm.

Density of liquid = 1000 gm/100 cm3 = 10 gcm3

Density of water at 4° = 1gcm-3

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

Relative density = 10 gcm3/1 gcm3 = 10

Mass of body = 1000 gm.

Density of water = 1 gcm-3

Acceleration due to gravity = 10 ms-2

Upthrust = V× p× g.

⇒ Upthrust = 100× 1× f = 100 gf.

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

35. State the principle of floatation. What can you say about the average density of a ship floating on water?

Principle of floatation states that a floating body displaces an amount of fluid equal to its own weight.

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.

36. For what purpose is the acid battery hydrometer used?

Acid battery hydrometer is used to check the concentration of sulphuric acid in an acid battery.

37. Explain why an iron nail floats on mercury, but it sinks in water.

Iron nail has density less than that of mercury so it will float on the surface of mercury but in the case of water it will sink because the density of iron nail is more than that of water.

38. State the principle of floatation. Name an instrument based on this principle. State two uses of the instrument that you describe.

Principle of floatation states that a floatation body displaces an amount of fluid equal to its own weight.

Hydrometer is based on principle of floatation.

A hydrometer is a device used for measuring the relative density of a liquid directly.

It usually consists of a glass float with a long thin stem which is graduated. The glass float is a large hollow bulb which increases with buoyancy. The narrow stem increases the sensitivity of hydrometer. The bottom of hydrometer is made heavier by loading it with lead shots so that it floats vertically.

Hydrometer works on the principle of floatation. Consider a thin walled and flat bottomed test tube. Add some lead shots in the test tube and place it in jar containing water. Adjust the number of lead shots such that it floats vertically with some of its portion outside the surface of water in jar.

If l is the length of the test tube inside water, a is area of cross section of the test tube, d is density of water in which it floats, then the weight of water displaced = aldg and it is equal to the weight of loaded test tube.

Now allow the test tube to float in the other jar filled with a liquid of density d1 and note the level l2 at which it floats in that liquid.

Weight of the liquid displaced = al1d1g

Using law of floatation

aldg = al1d1g

⇒ ld = l1d1

⇒ l/l1 = d1/d

or l is inversely proportional to the density.

It sinks more in a lighter liquid so as to displace more volume of the lighter liquid whose weight is equal to the weight of Hydrometer. Hence, it will sink less in a denser liquid so that it has displace less volume of the denser liquid whose weight will be equal to the hydrometer. In this way this measures relative density of a liquid.

Barometer can be used as lactometer which can be used to check the purity of milk.

Barometer can be used as acid battery hydrometer which can be used to check the concentration of sulphuric acid in an acid battery.

39. Explain the following:

(i) A balloon filled with hydrogen rises to a certain height and then stops rising further.

(ii) An egg sinks in fresh water, but floats in a strong solution of salt.

(iii) A hydrometer is made heavy near the bottom.

(iv) Icebergs floating in sea are very dangerous for ships.

(i) A balloon filled with hydrogen has low density than air go it rises over the air but as height increases density of air decreases and at a certain height the density of hydrogen in balloon and density of air become equal. And as there is no density difference there is no pressure difference also and hence balloon stops rising further.

(ii) Density of egg is greater than fresh water so it sinks in fresh water but due to addition of sat density of water increases which makes the density of salt water greater than egg andhence 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.

40. Using Archimedes’ principle, describe an experiment to find the relative density of a solid which floats on water.

We can find the relative density of a solid which floats on water by following experiment.

Choose a sinker and find its weight in water by suspending tin water.

Tie the solid to the string attached to the sinker and find its weight in air but sinker in water,

Remove the solid and tie it together with the sinker and suspend it in water and find the find the weight of the solid together with the sinker in water.

Record your observation as shown below:

Weight of sinker in water = x gf

Weight of sinker in water + solid in air = y gf

Weight of solid in air = (y - x) gf

Weight of solid + sinker in water = z gf

upthrust on solid in water = (y - z) gf.

The upthrust on in water also represents the weight of the water displaced by the solid.

Relative density of solid = (weight of cork in air)/(weight of equal volume of water}

Relative density = (y - x)/(y - z)

41. Using Archimedes’ principle, describe an experiment to find the relative density of a solid denser than water.

We can find the relative density of a solid denser than water by following experiment.

Weighing a solid in air and water

Find the weight (W1 gf) of a solid in air using a hydrostatic balance.

Tie the solid firmly with a thread and suspend it from the hook. Lower the solid in water and find its weight.

Record the result as shown below:

Weight of solid in air = W1 gf.

Weight of solid in water = W2 gf.

Apparent loss of weight of solid = (W1 - W2) gf.

Relative density = W1/(W1 – W2)

Relative density of the solid = (weight of solid in air)/(Apparent loss of weight of solid in water).

42. State Archimedes’ principle, describe an experiment to verify Archimedes’ principle.

Principle of Archimedes' states that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced.

We can verify Archimedes' principle experimentally by doing this experiment.

Pour water into eureka can till the water starts overflowing through the spout.

When the water stops dripping replace the beaker by another one of known weight.

Suspend a stone with the help of a string from the hook of a spring balance and record the weight of the stone.

Now, gradually lower the body into eureka can containing water and record its new weight in water when it is fully submerged in water. When no more water drips from the spout, weigh the beaker containing water.

After observation, we can conclude that apparent loss of weight of stone (calculated by differences in weight measured by spring balance) = weight of water displaced (weight of water in beaker).

This proves the Archimedes'’ principle that when a body !s totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid

43. What is a barometer? How will you construct a simple barometer?

Barometer is a device for measuring atmospheric pressure.

It consists of a long glass tube and of bore 1 cm2 having thick walls and closed at one end. The tube is made wide to minimize the depression of mercury in the tube due to surface tension. The tube contains pure and dry mercury and stands inverted with its open end Immersed deep in mercury contained in trough called reservoir. The lower end of the tube is drawn out to avoid oscillations of mercury in the tube while making adjustment.

The barometer tube is enclosed for most of its length in a tube of brass. Near the upper end of the brass tube, there are two vertical rectangular slits cut diametrically opposite to each other. They enable us see the upper level of mercury through them. Two scales are engraved on the brass tube along the edges of front slit. The scale along one edge is graduated in centimeters and along the other in inches. The scales have graduations from 68 cm to 80 cm and 27 inch to 32 inch. It is so done as ordinarily the pressure varies only between these limits. The zero of both the scale is at the tip of an ivory pointer which projects downward from the ceiling of the reservoir. The pressure is read with a vernier scale.

44. What is an aneroid barometer? Draw a neat and labelled diagram to explain its construction and working.

Aneroid barometer is a portable type of barometer.

In this no liquid is used. It consists of a metal box corrugated to make it flexible.

The lower side of the box is fixed to the base of the instrument and the upper side is supported by a stout spring S. The box is partially evacuated. So it tends to collapse under the pressure of air. The stout spring balances the thrust on the metal box due to normal air pressure and prevents the box from collapsing.

When atmospheric pressure changes, the surface supported by the spring S moves in if pressure increases or out if pressure decreases slightly. This small movement is increased considerably by a system of levers one end of which is connected to the spring S. The other end of the lever system is connected to a chain which is wrapped round a spindle carrying a pointer. The pointer moves over a scale which is graduated to read the pressure in centimeters or inches.

45. Draw a simple diagram of a Fortin's barometer and state how it is used to measure the atmospheric pressure.

Fortin barometer is used to measure the atmospheric pressure.

Before reading the Fortin's barometer, the free surface of mercury in the reservoir is made to touch the tip of ivory pointer by raising or lowering the level with the help of the screw S2. The Vernier scale is then adjusted with the screw S, till its lower edge touches the upper meniscus of mercury in the tube. While making this adjustment the eye is kept in level with the mercury meniscus. For this adjustment, the Vernier is moved till the wall of glass plate just ceases to be visible through the slits. The Vernier can read to 1/20 mm or 1/500 inch. The reading of the barometer then gives the true atmospheric pressure.

A thermometer is usually provided with the barometer which is also read and correction is made for the expansion of scale with rise in temperature and also for the change in density of mercury with temperature. The scale graduations are correct only for the temperature at which the graduations are made.

46. Describe an experiment to demonstrate that air exerts pressure.