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## Chapter: 9th Science : Physics : Fluids

Science : Physics : Fluids : Book Back Important Questions, Answers, Solutions : Answer very briefly, Answer briefly, Answer in detail, Exercise Numerical Problems, Intext Activities

### TEXT BOOK EXERCISES

1. On what factors the pressure exerted by the liquid depends on?

The pressure exerted by the liquid depends on the

(i) Depth

(ii) Density of the liquid

(iii) Acceleration due to gravity.

2. Why does a helium balloon float in air?

Helium balloon floats in air because helium gas is less dense than air.

3. Why it is easy to swim in river water than in sea water?

The question itself is wrong. It is easier to swim in sea water than in the river water. It is because sea water has (i) greater density and (ii) larger buoyant force than river water.

4. What is meant by atmospheric pressure?

The pressure exerted by the weight of the atmosphere is called atmospheric pressure.

5. State Pascal’s law.

Pascal’s law: The external pressure applied on an incompressible liquid is transmitted uniformly throughout the liquid.

1. With an appropriate illustration prove that the force acting on a smaller area exerts a greater pressure.

1. Take a nail. It has two ends. One end is sharp and other end is a bulged head.

2. We usually keep the pointed end on the wall or wood and hammer on the bulged head.

3. So very small area creates a large pressure.

4. Thus the nail penetrates into the wall or wood.

2. Describe the construction and working of mercury barometer.

Mercury Barometer

1. It is designed by Torricelli.

Construction:

1. Mercury Barometer consists of long glass tube closed at one end and opened at other.

2. Mercury filled through open end and close that end by thumb and open it after immersing it into a trough of mercury.

Working:

1. The Barometer works by balancing the mercury in the glass tube against the outside air pressure.

2. If air pressure increases, it pushes more of the mercury up into the tub.

3. If air pressure decreases, more mercury drains from the tub.

4. As vacuum cannot exert pressure, Mercury in the tube provides a precise measure of air pressure which is called atmospheric pressure.

5. It is used in a laboratory or weather station.

3. How does an object’s density determine whether the object will sink or float in water?

l. Whether an object sinks or floats is determined by density of the object compared with density of liquid.

2. If density of object is less than the density of the liquid, the object will float, (e.g) less density object, wood will float on water.

3. If density of object is more than the density of liquid, the object will sink.

(e.g) more dense object, stone sinks into water.

4. Explain the construction and working of a hydrometer with diagram.

1. Hydrometer consists of a cylindrical stem having a spherical bulb at its lower end and a narrow tube at its upper end.

2. The lower spherical bulb is partially filled with lead shots or mercury.

3. This helps hydrometer to float or stand vertically in liquids.

4. The narrow tube has markings so that relative density of a liquid can be read directly.

Lower end of hydrometer:

A cylindrical stem having a spherical bulb which partially filled with lead shots or mercury which helps to float or stand vertical in liquids

Upper end of hydrometer:

A narrow tube has markings so that relative density of liquids can be read off directly.

Working:

1. Liquid to be tested is poured into the glass jar.

2. The hydrometer is gently lowered into the liquid until it floats freely.

3. The reading against the level touching the tube gives the relative density of the liquid.

5. State the laws of flotation.

Laws of flotation

1. The weight of a floating body in a fluid is equal to the weight of the fluid displaced by the body.

2. The centre of gravity of the floating body and the centre of buoyancy are in the same vertical line.

### VII. Assertion and reason type questions:

(a) If both assertion and reason are true and reason is the correct explanation of assertion.

(b) If both assertion and reason are true but reason is not the correct explanation of assertion.

(c) If assertion is true but reason is false.

(d) If assertion is false but reason is true.

1. Assertion: To float, body must displace liquid whose weight is equal to the actual weight.

Reason: The body will experience no net downward force in that case.

[Ans: (a) Both assertion and reason are true and reason is the correct explanation of assertion]

2. Assertion: Pascal’s law is the working principle of a hydraulic lift.

Reason: Pressure is thrust per unit area.

[Ans: (b) Both assertion and reason are true but reason is not the correct explanation of assertion.]

Reason: Pascal's law is the working principle of Hydraulic lift. In Hydraulic lift, applied pressure is transmitted uniformly and multiplied through out the system.

### VIII. Numerical Problems:

1. A block of wood of weight 200 g floats on the surface of water. If the volume of block is 300 cm3, calculate the upthrust due to water.

Weight of wood block, m = 200 g

Volume of the wood block, V = 300 cm3.

Upthrust = Weight of the fluid displaced

= Volume of the wood block

Upthrust = 300 cm3

2. Density of mercury is 13600 kg m-3. Calculate the relative density.

Density of Mercury = 13600 kg m-3

Density of water at 4°C = 1000 kg m-3

Relative density = Density of mercury / Density of water at 4° C

= 13600 kg m−3 / 1000 kg m−3

Relative Density = 13.6

3. The density of water is 1 g cm−3. What is its density in S.I. units?

Answer: Density of water in SI units = 1000 kg / m3.

4. Calculate the apparent weight of wood floating on water if it weighs l00g in air.

Mass of wood = 100 g.

As the wood floats on the water, water will not be displaced.

So, actual weight of wood is equal to Apparent weight of wood.

### IX. Higher Order Thinking Skills :

1. How high does the mercury barometer stand on a day when atmospheric pressure is 98.6 kPa?

Pressure of Atmosphere Patm = 98.6 kPa.

Density of Mercury , ρHg = 13.6 × 103 kg/cm3

Acceleration due to gravity, g = 9.8 m/s2

Pressure, Patm = h × ρHg × g

h = Patm / (ρHg×g)

= 98.6 kPa / (13.6 × 103 kg/cm3 × 9.8 m/s2)

= 98.6 × 103 Pa / (13.6 × 103 kg/cm3 × 9.8 m/s2)

Height of Barometer, h = 0.7397 m = 739.7 mm

2. How does a fish manage to rise up and move down in water?

(i) Fish manages to rise up in water by reducing its density by filling oxygen in bladder via the gills. Thus volume will be increased to support its ascending motion. ‘

(ii) Fish moves down by decreasing its volume by releasing oxygen from bladder. Thus volume will be decreased so it will sink in the water.

3. If you put one ice cube in a glass of water and another in a glass of alcohol, what would you observe? Explain your observations.

Ice cube in water: As the density of ice cube is less than water, the ice cube floats in water.

Ice cube in alcohol: As the density of ice cube is greater than alcohol, the ice cube will sink in alcohol.

[Note: Density: Water = 1.00, Ice cube = 0.917, Alcohol = 0.78]

4. Why does a boat with a hole in the bottom would eventually sink?

A boat with a hole in the bottom eventually sinks due to:

(1)The water entered through a hole will increase the weight of boat.

(2) The boat becomes heavier so it cannot displace more water. So the boat sinks.

### Intext Activity

ACTIVITY - 1

Stand on loose stand. Your feet go deep into the sand. Now, lie down on the sand. What happens? You will find that your body will not go that deep into the sand. Why?

Aim:

To demonstrate the effect of thrust

Materials Required:

Sand

Procedure:

1. First, you stand on the sand on your feet.

2. Lie down on the sand with your whole body.

Observation:

1. While standing on your feet on sand, your feet go deep into the sand.

2. While lying down with your body on sand, your body will not go deep into the sand.

Conclusion:

1. Pressure depends upon the area on which it acts.

2. The effect of thrust on sand is larger while standing than lying.

ACTIVITY-2

Take a transparent plastic pipe. Also take a balloon and tie it tightly over one end of the plastic pipe. Pour some water in the pipe from the top. What happens? The balloon tied at the bottom stretches and bulges out. It shows that the water poured in the pipe exerts a pressure on the bottom of its container.

Aim: To demonstrate that water exerts pressure on the bottom of the container.

Materials Required: Plastic pipe, Balloon, Water.

Procedure:

1. Take a transparent plastic pipe and a balloon.

2. Tie the balloon tightly over one end of plastic pipe.

3. Keep the pipe with the closed end at the bottom.

4. Pour some water in the pipe from the top.

Observation: The balloon tied at the bottom stretches and bulges out.

Conclusion: Water poured in the pipe exerts pressure on the bottom of its container.

ACTIVITY - 3

Take a large plastic can. Punch holes with a nail in a vertical line on the side of the can as shown in figure. Then fill the can with water. The water may just dribble out from the top hole, but with increased speed at the bottom holes as depth causes the water to squirt out with more pressure.

Aim:

To demonstrate that pressure increases as depth increases.

Materials Required:

1. Large plastic can.

2. A sharp nail.

Procedure:

1. Take a large plastic can.

2. Punch holes with a nail in a vertical line up on the side of can every inch or several centimetres.

Observation:

1. Water dribbles out from top hole.

2. Water from bottom hole flows with increased speed.

Conclusion:

Depth causes water to squirt out with more pressure.

ACTIVITY - 4

Take two liquids of different densities say water and oil to a same level in two plastic containers. Make holes in the two containers at the same level. What do you see? It can be seen that water is squirting out with more pressure than oil. This indicates that pressure depends on density of the liquid.

Aim:

To demonstrate pressure depends on density of the liquid.

Materials Required:

1. Two plastic containers, 2.Water, 3. Oil (Both same volume), 4. Sharp nail

Procedure:

1. Take a water and oil to a same level in two plastic containers.

2. Make a hole at same level in two containers.

Observation:

Water squirts out with more pressure than that of oil.

Conclusion:

Pressure depends on density of the liquid.

ACTIVITY - 5

Take two identical flasks and fill one flask with water to 250 cm3 mark and the other with kerosene to the same 250 cm3 mark. Measure them in a balance. The flask filled with water will be heavier than the one filled with kerosene. Why? The answer is in finding the mass per unit volume of kerosene and water in respective flasks.

Aim:

To prove that density of a substance is the mass per unit volume of given substance.

Materials Required:

2. Water

3. Kerosene (same volume as water)

Procedure:

2. Fill one flask with water to 250 cm3 mark.

3. Fill the other flask with kerosene to same 250 cm3 mark.

4. Measure both flasks in balance separately.

Observation:

The flask filled with water will be heavier than that of the flask filled with kerosene.

Conclusion:

In the above activity, we know that

1. Both water and kerosene have same volume (i.e.) 250 cm3.

2. The density of the water lg / cm3 and density of kerosene is 0.8g / cm3

3. Density = (mass/ volume) , therefore mass = Density × volume.

Hence mass of water = lg / cm3 × 250 cm3 = 250g

mass of kerosene = 0.8 g / cm3 × 250 cm3 = 200g

4. Even though, water and kerosene have same volume, they have different densities. So water and kerosene have different masses.

5. Water has more mass than kerosene.

Hence, we proved that density of the substance is the mass per unit volume of the substance.

### VIII. Numerical Problems:

1. A vessel with water is placed on a weighing pan and it reads 600 g. Now a ball of mass 40 g and density is 0.80g / cm3 is sunk into the water with a pin of negligible volume as shown in figure. The weighing pan will show the reading of….?

Solution:

Weight of vessel with water = 600g

Mass of ball = 40g

Density of ball = 0.80 g /cm3

Volume of the ball = ( mass / density ) = (40/0.80 )= 50g

So, weight of vessel + volume of ball = 600 + 50 g

The weighing pan will show = 650 g

The weighing pan will show = 650 g

2. 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 submerged in water. Calculate the specific gravity of block.

Solution:

Weight of block in air = 60 newton

Loss of weight of block in water = 60 - 40 = 20 newton

Relative density (or) specific gravity = (Weight of block in air / Loss of weight in water)

= (60 newton / 20 newton)

Specific gravity of block = 3

3. The mass of a body is 4 kg and its volume is 500 cm3. Find its relative density.

Solution:

Mass of the body m = 4 kg = 4000 g

Volume of the body v =500 cm3

∴ Density of the body = Mass (m) / Volume (v)

= 4000 / 500

The relative density of the body = 8 g cm-3

= (Density of substance / Density of water)

= (8 g/cm3 ) / (1 g/cm3) =8

Relative density of the body = 8

4. Calculate the pressure produced by a force of 800 N acting on an area of 2.0m2.

Solution:

Force = 800 N

Area = 2.0m2

Pressure, P = (Force / Area) = (800 / 2.0) = 400Nm-2

Pressure, P = 400Nm-2 (or) 400 Pa

5. A swimming pool of width 9.0 m and length 24.0 m is filled with water of depth 3.0 m. Calculate the pressure on the bottom of the pool due to the water.

Solution:

Width of the pool, b = 9.0m

Length of the pool, l = 24.0 m

Depth of the pool, h = 3.0m

Density of water, ρ = 1000 kg/m3

Pressure due to column of Fluid,p = ρhg

Acceleration due to gravity, g = 9.8 m/s2

Substituting the values, p = ρhg

P = (1000kgm−3) × (3.0m) × (9.8ms-2)

Pressure P = 2940 kgm-1s-2   1pa = 1kgm-1 s-2

P = 29400Nm-2 (or) 29400pa

6. A body of volume 100 cc is immersed completely in water contained in a jar. The weight of water and the jar before immersion of the body was 700 g. Calculate the weight of water and jar after immersion.

Solution:

Volume of body completely immersed in water V = 100cc.

Weight of water and jar before Immersion = 700g.

Volume of jar immersed in water = Volume of water displace

= 100cc.

Density of water = 1g/cm3

Mass of water displaced = Apparent weight loss

Mass of water displaced = Volume × Density.

= 100cc × 1 g/cm3.

Apparent weight loss of body = 100g

Weight of jar and water after immersion = Weight of water and jar before immersion − Apparent weight loss

= 700g − 100g

= 600g

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