Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element – hydrogen

Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element – hydrogen

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Selina ICSE Solutions for Class 9 Chemistry Chapter 9 Study of the First Element – hydrogen

Exercise 9

Solution 1.

(a) The position of hydrogen in the periodic table
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 1
Hydrogen is first element in the periodic table. It has an atomic number l and an atomic mass of 1.00794 amu, occupying group – IA. Its position is peculiar because it is grouped with metals although it is a non-metal properties. Hydrogen relate to Group IA as well as Group VII A.

(b) The properties of hydrogen resemble the properties of Group IA elements (Alkali metals), and some of it resembles the properties of Halogens (VIIA), so Hydrogen was put at the top of the periodic table so that the symmetry of the modern periodic table is not disturbed.

  1. All elements in Group- IA have one electron in outermost shell, so they havevalency one.
  2. These elements in Group-IA are good reducing agents.
  3. All elements of this group formsoxide which are highly basic and dissolves in water to form strong alkalis.
  4. They impartcolour to a flame.

Solution 2.

Similarity of hydrogen with alkali metals and halogens

Similarity of hydrogen with alkali metals [Group 1 (IA)]Similarity of hydrogen with halogens [Group 17 (VIIA)]
Electronic configurationElectronic configuration = 1. Thus, 1electro in the outermost valence shell.

Example:
H=1; Li=2, 1; Na=2,8,1; K=2,8,8,1

One electron less than the nearest noble gas.

Example:
H= 1 (He=2)
F= 2,7 (Ne=2,8)
Cl= 2,8,7 (Ar=2,8,8)

Ion formationElectropositive character exhibited.

H 1e → H1+
Li 1e → Li1+
Na 1e → Na1+

Electronegative character exhibited.

H + 1e → H1-
F + 1e → F1-
Cl + 1e → Cl1-

ValencyElectrovalency of one exhibited.

H1+ , Li1+ , Na1+

Electrovalency and covalencyexhibited.

Hydrogen:

forms NaH (electrovalent)forms CH4(covalent)

Chlorine:

forms NaCl (electrovalent)
forms CCl4 (covalent)

ReactionsStrong affinity for non-metals (example: O, S, Cl)

Hydrogen: forms H2O; H2S; HCl

Sodium: forms Na2O; Na2S; NaCl

__
Reducing agentActs as a reducing agent.

Hydrogen:

CuO + H2 → Cu + H2O

Sodium:

CuO + 2Na → Cu + Na2O

__
Atomicity__

Diatomic molecules are formed. (Two atoms linked by a single bond)

Hydrogen
H:H or H-H → H2 

Chlorine
Cl:Cl or Cl-Cl → Cl2

Resemblance with Halogens:

  1. Both exist in the form of diatomic molecules.
  2. Both show gaseous nature.
  3. Both have a valency of 1.
  4. Both are non-metals.
  5. Both lose electron to term anions.

Solution 3.

(a) Hydrogen is found in minute traces in the Earth’s crust and the Earth’s atmosphere. The atmosphere around the sun and stars is found to contain 1.1 % hydrogen.

(b) Henry Cavendish when prepared this gas from iron and dil. acids, he established its elementary nature and showed that when the gas burns in air, water is formed. It was on account of this property that Lavoisier in 1783 named it hydrogen (Greek word meaning water-former).

Solution 4.

(a) A monovalent metal
2Na + H2 → 2NaH
(Sodium hydride)

(b) A divalent metal
Ca + H2 → CaH2
(Calcium hydride)

Solution 5.

(a) Calcium: is not used in lab preparation of hydrogen because:

  1. The reaction and very violent and exothermic hence dangerous.
  2. The heat liberated ignites the hydrogen.
  3. Calcium is expensive.

(b) Iron: Iron reacts slowly at ordinary temperatures, hence requires heating. The hydrogen produced also contain impurities like sulphur dioxide and hydrogen sulphide. Hence, it is not used in lab preparation of hydrogen.

(c) Aluminium: It is not used in the lab preparation of hydrogen because oxides of this metal keep sticking to the surface of the metal. Thus the steam does not come in contact with metal and hence reaction stops. .

(d) Sodium: It is riot used in the lab preparation of hydrogen because the reaction is violent. The sodium melts into a globule and darts about freely on the surface of water hence the collection of hydrogen is difficult.

Solution 6.

Depending upon the nature of reaction taking place between metals and substances like air, water and acids, metals are arranged in a vertical series in order of their activity. Such a series is called activity series of metals.

The metals places near the top of the series are the most reactive, while those placed near the bottom are the least reactive.

When dilute hydrochloric acid or dilute sulphuric acid react with the metals above hydrogen in the activity series, they produce hydrogen. But the metals below hydrogen in the activity series do not.

Solution 7.

(a) Reactants: Nitrogen and hydrogen (Haber process)

Chemical equation:
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 2

Observation and conditions
Three volumes of hydrogen and one volume of nitrogen react at temperature 450oC-500oC at the pressure of 200-900 atm, in presence of a finely divided iron which acts as a catalyst, and promoter molybdenum.

(b) Reactants: Chlorine and hydrogen

Chemical equation:
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 3

Observation and conditions
Hydrogen and chlorine (in their equal volumes) react slowly in diffused sunlight but reacts explosively in direct sunlight. A spontaneous reaction takes place with the release of a large amount of energy.

(c) Reactants: Sulphur and hydrogen

Chemical equation:
H2 + S → H2S

Observation and conditions
Hydrogen gas when passed through molten sulphur, it reacts to give another gas, hydrogen sulphide.

(d) Reactants: Oxygen and hydrogen

Chemical equation:
2H2 + O2 → 2H2O

Observation and conditions
Hydrogen burns with a pop sound in oxygen. It burns with a pale blue flame forming water.

Solution 8.

(a) Among the given metals Zinc is most suitable.
(i) Copper: In case of copper, It is placed below hydrogen in the activity series. So it does not displace hydrogen from acid.
Cu + HCl → No reaction
(ii) In case of Mg; it is a very expensive metal.
(iii) In case of sodium, it reacts with explosively and violently.

(b) Among the given acids we prefer dilute sulphuric acid.
We reject concentrated sulphuric, dilute nitric and concentrated nitric acid because these are powerful oxidising agents and oxygen formed due to its decomposition oxidises the hydrogen to water.

(c) Modification: Collect the gas by downward displacement of water when all the air from the apparatus has been expelled. Drying Agent used is Calcium Chloride.

Solution 9.

(a) Iron reacts reversibly with steam. Hence the hydrogen formed is removed as it is released to prevent reduction of triferric tetraoxide.
Fe + 4H2O → Fe3O4 + 4H2 ?
(Steam)

(b)
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 4

Solution 10.

(a) The metal is magnesium
(b) Mg + H2O → MgO + H2

Solution 11.

(a) Substance A is CuO and substance B is Cu.

(b) Test for water
(i) It is neutral to litmus
(ii) It changes anhydrous copper sulphate into blue salt.

(c) When substance A i.e, CuO reacts with hydrogen, it removes oxygen and we get free metal i.e. Cu.

(d) CuO + H2 → Cu + H2O

(e) No, there is no reaction between substance B and dilute hydrochloric acid because copper does not displace hydrogen from acids.

(f) Cu + HCl → No reaction

Solution 12.

Magnesium lies above Hydrogen in reactivity series and can displace hydrogen from acid whereas, Mercury and silver lie below hydrogen in reactivity series and cannot displace hydrogen from acid and hence nothing happens.

Solution 13.

Soap bubbles containing hydrogen rapidly rise up in air as hydrogen is lighter than air.

Solution 14.

Bosch Process
Bosch process consists of following steps.

Step 1 :

Steam is passed over a hot coke (at 1000oC) in a special type of a furnaces called converters.
In this step carbon reacts with water to form carbon monoxide and hydrogen gas. This mixture is called water gas.Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 5

Step 2 :

In this step excess of steam is mixed with water gas and entire mixture is passed over heated ferric oxide and chromic oxide. Ferric oxide acts as catalyst and chromic oxide as promoter.
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 6

Step 3 :

In this step removal of carbon dioxide from reaction mixture takes place. The mixture of carbon dioxide and hydrogen is forced through cold water under pressure at 30 atmospheric pressure or through caustic potash solution which dissolve carbon dioxide leaving behind hydrogen gas.
2KOH + CO2 → K2CO3 + H2O

Step 4 :

In this last step, mixture is passed through ammonical solution of cuprous chloride solution so as to dissolve carbon monoxide. CO is removed by bubbling the gas through ammoniacal cuprous chloride solution. The moisture is removed by cooling the gas about 20°C when water vapoursfreeze. The pure and dry hydrogen gas is collected in steel cylinders.
CuCl + CO + 2H2O → CuCl.CO.2H2O

Solution 15.

(a) Cold water:

Sodium metal wrapped in small piece of wire gauze or Sodium amalgamated with mercury is used. This prevents sodium from darting about.
2Na + 2H2O → 2NaOH + H2 ↑

(b) Hot water:

Zn or Mg can be used.
Mg + H2O → MgO + H2 ↑
(boiling water)

(C) Steam:

Iron reacts with the steam and the reaction is reversible. Iron reacts with steam when it is red hot as hydrogen is blown out of contact with iron by the force of current of the steam.
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 7

Solution 16.

Metals

Mg, Al, Zn, Fe do not react with cold water. It reacts with boiling water liberating hydrogen gas but the reaction is very slow.
Mg, Al, Zn, Fe react with the hot steam in the heated state and form the corresponding oxide and hydrogen gas.
Iron reacts with the steam and the reaction is reversible.

Magnesium
Reaction of boiling water with steam is slow but Mg liberates Hydrogen rapidly with steam.

Mg + H2O → MgO + H2 ↑
(boiling water)

Aluminium

Gets coated with Al2O3 on rubbing with sand paper its oxide coating is removed and then it reacts with steam to produce hydrogen.

2Al + H2O → Al2O3 + 3H2 ↑
(steam)

Zinc

Zinc reacts with steam and produce zinc oxide and H2

Zn + H2O → ZnO + H2 ↑
(steam)

Iron

Iron produces H2 when red hot the reaction is reversible but as soon as hydrogen is produced it is blown out of contact with iron by the force of the current of steam.

Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 8
(steam)

Non-metals

Bosch Process (From Coke)
Bosch process consists of following steps.

Step 1 :

Steam is passed over a hot coke (at 1000oC) in a special type of a furnaces called converters.
In this step carbon reacts with water to form carbon monoxide and hydrogen gas. This mixture is called water gas.Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 9

Step 2 :

In this step excess of steam is mixed with water gas and entire mixture is passed over heated ferric oxide and chromic oxide. Ferric oxide acts as catalyst and chromic oxide as promoter.
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 10

Step 3 :

In this step removal of carbon dioxide from reaction mixture takes place. The mixture of carbon dioxide and hydrogen is forced through cold water under pressure at 30 atmospheric pressure or through caustic potash solution which dissolve carbon dioxide leaving behind hydrogen gas.

2KOH + CO2 → K2CO3 + H2O

Step 4 :

In this last step, mixture is passed through ammonical solution of cuprous chloride solution so as to dissolve carbon monoxide. CO is removed by bubbling the gas through ammoniacal cuprous chloride solution. The moisture is removed by cooling the gas about 20C when water vapoursfreeze. The pure and dry hydrogen gas is collected in steel cylinders.

CuCl + CO + 2H2O → CuCl.CO.2H2O

Solution 17.

(a) Zinc and iron lie above hydrogen in reactivity series and can displace hydrogen from acid.

Zn + H2SO4 → ZnSO4 + H2 ↑
(dilute)

Fe + H2SO4 → FeSO4 + H2 ↑
(dilute)

(b) Zn + 2HCl → ZnCl2 + H2 ↑
(dilute)

Fe + 2HCl → FeCl2 + H2 ↑
(dilute)

Copper lies below hydrogen. Thus, it cannot displace hydrogen from acids.

Solution 18.

Two alkalis which can displace hydrogen are NaOH and KOH.

Aluminium

2Al + 6NaOH → 2Na3AlO3 + 3 H2 ↑
(Sodium aluminate)

2Al + 2KOH + 2H2O → 2KAlO2 + 3 H2 ↑
(Potassium meta aluminate)

Zinc

Zn + 2NaOH → Na2ZnO2 + H2 ↑
(Sodium zincate)

Zn + 2KOH → K2ZnO2 + H2 ↑
(Potassium zincate)

Aluminium and Zinc have unique nature; They react with acids and can even react with hot concentrated alkalis to form hydrogen and a soluble salt. Salt s (oxides and hydroxide) of these metals are Amphoteric.

Question 19.
Complete and balance the following reactions.
(a) Na + H2O →_____________ +___________
(b) Ca + H2O →_____________ +___________
(c) Mg + H2O →_____________ +___________
(d) Zn + H2O →_____________ +___________
(c) Fe + H2O →_____________ +___________
(d) Zn + HCl →_____________ +___________
(e) Al + H2SO4 →_____________ +___________
(f) Fe + HCl →_____________ +___________
(g) Zn + NaOH →_____________ +___________
(h) Al + KOH + H2O→_____________ +___________

Solution:

(a) 2Na + 2H2O → 2NaOH + H2
(b) Ca + 2H2O → Ca(OH)+ H2
(c) Mg + H2O → MgO + H2
(d) Zn + H2O → ZnO + H2
(e) 3Fe + 4H2O → Fe3O4 + 4H2
(f) Zn + 2HCl → ZnCl2 + H2
(g) 2Al + 3H2SO → Al2(SO4)3 + 3H2
(h) Fe + 2HCl → FeCl2 + H2
(i) Zn + 2NaOH → Na2ZnO2 + H2
(j) 2Al + 2KOH + 2H2O → 2KAlO2 + 3H2

Solution 20.

(a) Lead reacts with dilute sulphuric acid and hydrochloric acid and forms an insoluble coating of lead sulphate and lead chloride respectively. Hence, further reaction is prevented.

(b) When potassium and sodium react with dilute sulphuric acid, H2SO4 or dilute HCl, the reaction is highly explosive and practically not feasible.

Solution 21.

(a) Sodium hydroxide + zinc → hydrogen + Sodium zincate
(b) Calcium + water calcium → hydroxide + Hydrogen

Solution 22.

  1. Hydrogen gas is collected by the downward displacement of water. This is because-
    • It is virtually insoluble in water.
    • It forms an explosive mixture with air and therefore, cannot be collected by downward displacement of air, even though it is lighter than the air.
  2. A candle when brought near the mouth of the jar containing hydrogen gas starts burning but the candle is extinguished when pushed inside the jar because, Because hydrogen is a combustible gas but a non-supporter of combustion i.e. it burns itself but does not allow substances to burn in it.
  3. An oxy-hydrogen flame is used for welding and cutting metals because, when a mixture, of hydrogen and oxygen is burnt, temperatures as high as 2500°C is produced. This forms the basis for its use in oxy-hydrogen flame used in cutting and welding of flames.
  4. Apparatus for the laboratory preparation of hydrogen should be air tight and away from a naked flamebecause, mixture of hydrogen and air explodes violently when brought near a flame.

Question 23.
a. Helium is preferred to hydrogen for filling balloons because it is:
i. lighter than air
ii. almost as light as hydrogen
iii. non-combustible
iv. inflammable
b. Reacting with water, an active metal produces
i. oxygen
ii. nitric acid
iii. a base
iv. none of these
c. A metal oxide that is reduced by hydrogen is
i. Al2O3
ii. CuO
iii. CaO
iv. Na2O
d. Which of the following statements about hydrogen is incorrect?
i. It is an inflammable gas
ii. It is the lightest gas.
iii. It is not easily liquefied
iv. It is a strong oxidizing agent.
e. For the reaction PbO + H2→ Pb + H2O, which of the following statements is wrong?
i. H2 is the reducing agent.
ii. PbO is the oxidizing agent.
iii. PbO is oxidized to Pb.
iv. H2 is oxidized to H2O.
f. Which metal gives hydrogen with all of the following: water, acids, alkalis?
i. Fe
ii. Zn
iii. Mg
iv. Pb
g. Which of the following metals does not give hydrogen with acids?
i. Iron
ii. Copper
iii.Lead
iv. Zinc

Solution:
(a) (iii) non-combustible
(b) (iii) a base (c) (iii) CaO
(d) (iv) It is strong oxidising agent.
(e) (iii) PbO is oxidised to Pb
(f) (ii) Zn
(g) (ii) Copper

Question 11.
Choose terms from the options given in brackets to complete these sentences.
a. When CuO reacts with hydrogen,………………… is reduced and ……………….is oxidized to ………………… .
(CuO, H2,Cu,H2O)
b. Hydrogen is ………………… soluble in water.
(sparingly, highly, moderately)
c. Metals like …………….. , ……………… and ……………… give H2 with steam.
(iron, magnesium, aluminium, sodium , calcium)
d. Sodium ………………. reacts smoothly with cold water.
(metal, amalgam, in the molten state)
e. A metal …………….. hydrogen in the activity series gives hydrogen with …………… acid or …………… acid.
(above, below, dilute hydrochloric, concentrated hydrochloric, dilute sulphuric).

Solution 24.

(a) CuO, H2O
(b) sparingly
(c) amalgam
(d) iron, magnesium, aluminium
(e) above, dilute hydrochloric, dilute sulphuric.

Question 25.
Correct the following statements:

  1. Hydrogen is separated from CO by passing the mixture through caustic potash solution.
  2. All metals react with acids to give hydrogen.
  3. Hydrogen is dried by passing it through conc. H2SO4.
  4. Very dilute nitric acid reacts with iron to produce hydrogen.
  5. Conc. H2SO4 reacts with zinc to liberate hydrogen.

Solution:

(a) Hydrogen is used as a fuel in the form of coal gas, water gas and liquid hydrogen.
(b) All metals above hydrogen in reactivity series react with acids to give hydrogen.
(c) Metals like palladium or platinum or nickel absorb hydrogen at room temperature.
(d) The reaction of hydrogen with oxygen is explosive with pop sound.
(e) Concentrated sulphuric acid reacts with zinc to liberate sulphur dioxide.

More Resources for Selina Concise Class 9 ICSE Solutions

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure

ICSE SolutionsSelina ICSE Solutions
APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 4 Pressure in Fluids and Atmospheric Pressure. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 4 Pressure in Fluids and Atmospheric Pressure

Exercise 4(A)

Solution 1S.

Thrust is the force acting normally on a surface.
Its S.I. unit is ‘newton’.

Solution 2S.

Pressure is the thrust per unit area of the surface.
Its S.I. unit is ‘newton per metre2‘ or ‘pascal’.

Solution 3S.

(a) Pressure is measured in ‘bar’.
(b) 1 bar = 105 pascal.

Solution 4S.

One pascal is the pressure exerted on a surface of area 1 m2 by a force of 1N acting normally on it.

Solution 5S.

Thrust is a vector quantity.

Solution 6S.

Pressure is a scalar quantity.

Solution 7S.

Thrust is the force applied on a surface in a perpendicular direction and it is a vector quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.

Solution 8S.

Pressure exerted by thrust is inversely proportional to area of surface on which it acts. Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it.
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that sand, our body does not sink into the sand. In both the cases, the thrust exerted on the sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust acts on a large area and when we stand, the same thrust acts on a small area.

Solution 9S.

The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and it can be driven into with less effort.

Solution 10S.

(a) It is easier to cut with a sharp knife because even a small thrust causes great pressure at the edges and cutting can be done with less effort.
(b) Wide wooden sleepers are placed below the railway tracks so that the pressure exerted by the rails on the ground becomes less.

Solution 11S.

A substance which can flow is called a fluid.

Solution 12S.

Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the fluid is called fluid pressure.

Solution 13S.

A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a fluid exerts pressure at all points in all directions.

Solution 14S.

Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a series of holes in the wall of the vessel anywhere below the free surface of the liquid. The water spurts out through each hole. This shows that the liquid exerts pressure at each point on the wall of the bottle.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 3
Liquid exerts pressure at all points in all directions

Solution 15S.

Pressure at a point in a liquid depends upon the following three factors:

  1. Depth of the point below the free surface.
  2. Density of liquid.
  3. Acceleration due to gravity.

Solution 16S.

P = Po + hρg
Here, P = Pressure exerted at a point in the liquid
Po = Atmospheric pressure
h = Depth of the point below the free surface
ρ = Density of the liquid
g = Acceleration due to gravity

Solution 17S.

Consider a vessel containing a liquid of density ρ. Let the liquid be stationary. In order to calculate pressure at a depth, consider a horizontal circular surface PQ of area A at a depth h below the free surface XY of the liquid. The pressure on the surface PQ will be due to the thrust of the liquid contained in cylinder PQRS of height h with PQ as its base and top face RS lying on the frees surface XY of the liquid.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 2
Total thrust exerted on the surface PQ
= Weight of the liquid column PQRS
= Volume of liquid column PQRS x density x g
= (Area of base PQ x height) x density x g
= (A x h) x ρ x g

This thrust is exerted on the surface PQ of area A. Therefore, pressure is given as shown below.
P = Thrust on surface / Area of surface
P = Ah ρg / A = hρg
Thus, Pressure = depth x density of liquid x acceleration due to gravity

Solution 18S.

Due to dissolved salts, density of sea water is more than the density of river water, so pressure at a certain depth in sea water is more than that at the same depth in river water.

Solution 19S.

(a) P2 = P1 + h ρ g,
(b) P2 > P1

Solution 20S.

The reason is that when the bubble is at the bottom of the lake, total pressure exerted on it is the atmospheric pressure plus the pressure due to water column. As the gas bubble rises, due to decrease in depth the pressure due to water column decreases. By Boyle’s law, PV = constant, so the volume of bubble increases due to decrease in pressure, i.e., the bubble grows in size.

Solution 21S.

The pressure exerted by a liquid increases with its depth. Thus as depth increases, more and more pressure is exerted by water on wall of the dam. A thicker wall is required to withstand greater pressure, therefore, the thickness of the wall of dam increases towards the bottom.

Solution 22S.

The sea divers need special protective suit to wear because in deep sea, the total pressure exerted on the diver’s body is much more than his blood pressure. To withstand it, he needs to wear a special protective suit.

Solution 23S.

Laws of liquid pressure:

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

Solution 24S.

The liquid from hole B reaches a greater distance on the horizontal surface than that from hole A.
This explains that liquid pressure at a point increases with the depth of point from the free surface.

Solution 25S.

(i) As the diver moves to a greater depth, pressure exerted by sea water on him also increases.
(ii) When the diver moves horizontally, his depth from the free surface remains constant and hence the pressure on him remains unchanged.

Solution 26S.

Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Solution 27S.

Two applications of Pascal’s law:

  1. Hydraulic press
  2. Hydraulic jack

Solution 28S.

The principle of a hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.
Hydraulic press and hydraulic brakes work on this principle.

Solution 29S.

Hydraulic press works on principle of hydraulic machine.
It states that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.
Use: It is used for squeezing oil out of linseed and cotton seeds.

Solution 30S.

(i) X : Press Plunger; Y: Pump Plunger

(ii) When the lever is moved down, valve B closes and valve A opens, so the water from cylinder P is forced into the cylinder Q.

(iii) Valve B closes due to an increase in pressure in cylinder P. This pressure is transmitted to the connecting pipe and when the pressure in connecting pipe becomes greater than the pressure in the cylinder Q, valve A opens up.

(iv) When the release valve is opened, the ram (or press) plunger Q gets lowered and water of the cylinder Q runs out in the reservoir.

Solution 31S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 1
Working: When handle H of the lever is pressed down by applying an effort, the valve V opens because of increase in pressure in cylinder P. The liquid runs out from the cylinder P to the cylinder Q. As a result, the piston B rises up and it raises the car placed on the platform. When the car reaches the desired height, the handle H of the lever is no longer pressed. The valve V gets closed (since the pressure on the either side of the valve becomes same) so that the liquid may not run back from the cylinder Q to cylinder P.

Solution 32S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 4

Working: To apply brakes, the foot pedal is pressed due to which pressure is exerted on the liquid in the master cylinder P, so liquid runs out from the master cylinder P to the wheel cylinder Q. As a result, the pressure is transmitted equally and undiminished through the liquid to the pistons B1and B2 of the wheel cylinder. Therefore, the pistons B1 and B2 get pushed outwards and brake shoes get pressed against the rim of the wheel due to which the motion of the vehicle retards. Due to transmission of pressure through the liquid, equal pressure is exerted on all the wheels of the vehicle connected to the pipe line R.

On releasing the pressure on the pedal, the liquid runs back from the wheel cylinder Q to the master cylinder P and the spring pulls the break shoes to their original position and forces the pistons B1 and B2 to return back into the wheel cylinder Q. Thus, the brakes get released.

Solution 33S.

(a) h ρ g (b) same (c) the same (d) directly proportional (e) directly proportional.

Solution 1M.

Pa

Solution 2M.

h ρ g

Solution 3M.

P1 < P2

Solution 4M.

P2 – P1 = h ρ g

Solution 1N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 5

Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 6

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 7

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 8

Solution 5N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 9

Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 10

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 11

Solution 8N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 12

Solution 9N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 13

Solution 10N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 14

Solution 11N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 15

Solution 12N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 16

Solution 13N.

Data is incomplete

Solution 14N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 17

Exercise 4(B)

Solution 1S.

The thrust exerted per unit area of the earth surface due to column of air, is called the atmospheric pressure on the earth surface.

Solution 2S.

1.013 x 10 5 pascal

Solution 3S.

Atmospheric pressure is measured in ‘torr’.
1 torr = 1 mm of Hg.

Solution 4S.

At normal temperature and pressure, the barometric height is 0.76 m of Hg at sea level which is taken as one atmosphere.
1 atmosphere = 0.76 m of Hg = 1.013 x 105 pascal

Solution 5S.

We do not feel uneasy under enormous pressure of the atmosphere above as well as around us because of the pressure of our blood, known as blood pressure, is slightly more than the atmospheric pressure. Thus, our blood pressure balances the atmospheric pressure.

Solution 6S.

Experiment to demonstrate that air exerts pressure:

Take a thin can fitted with an airtight stopper. The stopper is removed and a small quantity of water is boiled in the can. Gradually the steam occupies the entire space of can by expelling the air from it [Fig (a)]. Then stopper is then tightly replaced and simultaneously the flame beneath the can is removed. Cold water is then poured over the can.

It is observed that the can collapses inwards as shown in fig (b).

The reason is that the pressure due to steam inside the can is same as the air pressure outside the can [Fig (a)]. However, on pouring cold water over the can, fitted with a stopper [fig (b)], the steam inside the can condenses producing water and water vapour at very low pressure. Thus, the air pressure outside the can becomes more than the vapour pressure inside the closed can.

Consequently, the excess atmospheric pressure outside the can causes it to collapse inwards.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 18

Solution 7S.

(i) When air is removed from the balloon, the pressure inside the balloon (which was due to air in it) is much less than the atmospheric pressure outside and hence the balloon collapses.

(ii) Water is held inside the dropper against the atmospheric pressure because the pressure due to height column of liquid inside the dropper is less than the atmospheric pressure. By pressing the dropper we increase the pressure inside the dropper and when it becomes greater than the atmospheric pressure the liquid comes out of the dropper.

(iii) There is no air inside a completely filled and sealed can. When a single hole is made to drain out the oil from the can, some of the oil will come out and due to that the volume of air above the oil will increase and hence the pressure of air will decrease. But if two holes are made on the top cover of the can, air outside the can will enter it through one hole and exert atmospheric pressure on the oil from inside along with the pressure due to oil column, and it will come out of the can from the other hole.

Solution 8S.

When syringe is kept with its opening just inside a liquid and its plunger is pulled up in the barrel, the pressure of air inside the barrel below the plunger becomes much less than the atmospheric pressure acting on the liquid. As a result, the atmospheric pressure forces the liquid to rise up in the syringe.

Solution 9S.

In a water pump, on pulling the piston up, the pressure of air inside the siphon decreases and the atmospheric pressure on the water outside increases. As a result, the atmospheric pressure pushes the water up in pump.

Solution 10S.

(a) Pressure increases inside the bell jar.
(b) Pressure decreases inside the balloon.

Solution 11S.

A barometer is used to measure atmospheric pressure.

Solution 12S.

A barometer is an instrument which is used to measure the atmospheric pressure.

Construction of a simple barometer:

A simple mercury barometer can be made with a clear, dry, thick-walled glass tube about 1 metre ling. The glass tube is sealed at one end and is filled with mercury completely. While filling the tube with mercury care has to be taken so that there are no air bubbles present in the mercury column. Close the open end with thumb and turn the tube upside down carefully over a trough containing mercury. Dip the open end under the mercury level in the trough and remove the thumb.

The mercury level in the tube falls until it is about 76 cm (h =760 mm) vertically above the mercury level. It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. The empty space above the mercury column is called the ‘Torricellian vacuum’.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 19

Solution 13S.

In given figure, at all points such as C on the surface of mercury in trough, only the atmospheric pressure acts. When the mercury level in the tube becomes stationary, the pressure inside tube at the point A, which is at the level of point C, must be same as that at the point C. The pressure at point A is due to the weight (or thrust) of the mercury column AB above it. Thus, the vertical height of the mercury column from the mercury surface in trough to the level in tube is a measure of the atmospheric pressure.
The vertical of the mercury column in it (i.e., AB = h) is called the barometric height.
Had the pressure at points A and C be not equal, the level of mercury in the tube would not have been stationary.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 20

Solution 14S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 21

Solution 15S.

It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. Hence, barometric height is used as unit to express the atmospheric pressure.

Solution 16S.

The atmospheric pressure at a place is 76 cm of Hg means at normal temperature and pressure, the height of the mercury column supported by the atmospheric pressure is 76 cm.
76 cm of Hg = 1.013 x 105 pascal

Solution 17S.

The space above mercury is a vacuum. This empty space is called ‘Torricellian vacuum’.

This can be shown by tilting the tube till the mercury fills the tube completely. Again when the mercury column becomes stationary, the empty space is created above the mercury column. If somehow air enters into the empty space or a drop of water gets into the tube, it will immediately vaporize and the air will exert pressure on mercury column due to which the barometric height will decrease.

Solution 18S.

(a) The barometric height remains unaffected.
(b) The barometric height remains unaffected.
(c) The barometric height decreases.

Solution 19S.

Two uses of barometer:

  1. To measure the atmospheric pressure.
  2. For weather forecasting

Solution 20S.

Two advantages of using mercury as barometric liquid:

  1. The density of mercury is greater than that of all the liquids, so only 0.76m height of mercury column is needed to balance the normal atmospheric pressure.
  2. The mercury neither wets nor sticks to the glass tube therefore it gives the correct reading.

Solution 21S.

Water is not a suitable barometric liquid because:

  1. The vapour pressure of water is high, so its vapours in the vacuum space will make the reading inaccurate.
  2. Water sticks with the glass tube and wets it, so the reading becomes inaccurate.

Solution 22S.

In a simple barometer, there is no protection for the glass tube but in Fortin’s barometer, this defect has been removed by enclosing the glass tube in a brass case.
In a simple barometer, a scale cannot be fixed with the tube (or it cannot be marked on the tube) to measure the atmospheric pressure but Fortin’s barometer is provided with a vernier calipers to measure the accurate reading.

Solution 23S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 22
To measure the atmospheric pressure, first the leather cup is raised up or lowered down with the help of the screw S so that the ivory pointer I just touches the mercury level in the glass vessel. The position of the mercury level in the barometer tube is noted with the help of main scale and the vernier scale. The sum of the vernier scale reading to the main scale reading gives the barometric height.

Solution 24S.

A barometer calibrated to read directly the atmospheric pressure is called an aneroid barometer. It has no liquid, it is light and portable.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 23

Construction: Figure above shows the main parts of an aneroid barometer. It consists of a metallic box B which is partially evacuated. The top D of the box is springy and corrugated in form of a diaphragm as shown. At the middle of diaphragm, there is a thin rod L toothed at its upper end. The teeth of rod fit well into the teeth of a wheel S attached with a pointer P which can slide over a circular scale. The circular scale is initially calibrated with a standard barometer so as to read the atmospheric pressure directly in terms of the barometric height.

Working: When atmospheric pressure increases, it presses the diaphragm D and the rod L gets depressed. The wheel S rotates clockwise and pointer P moves to the right on the circular scale. When atmospheric pressure decreases, the diaphragm D bulges out due to which the rod L moves up and the wheel S rotates anti-clockwise. Consequently, the pointer moves to the left.

Solution 25S.

Aneroid barometer has no liquid and it is portable. It is calibrated to read directly the atmospheric pressure.

Solution 26S.

(i) In a mine, reading of a barometer increases.
(ii) On hills, reading of barometer decreases.

Solution 27S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 24

Solution 28S.

Factors that affect the atmospheric pressure are:

  1. Height of air column
  2. Density of air

Solution 29S.

A fountain pen filled with ink contains some air at a pressure equal to atmospheric pressure on earth’s surface. When pen is taken to an altitude, atmospheric pressure is low so the excess pressure inside the rubber tube forces the ink to leak out.

Solution 30S.

On mountains, the atmospheric pressure is quite low. As such, nose bleeding may occur due to excess pressure of blood over the atmospheric pressure.

Solution 31S.

An altimeter is a device used in aircraft to measure its altitude.

Principle: Atmospheric pressure decreases with the increase in height above the sea level; therefore, a barometer measuring the atmospheric pressure can be used to determine the altitude of a place above the sea level.

The scale of altimeter is graduated with height increasing towards left because the atmospheric pressure decreases with increase of height above the sea level.

Solution 32S.

(a) It indicates that the moisture is increasing i.e., there is a possibility of rain.
(b) It indicates the coming of a storm or cyclone.
(c) It indicates that the moisture is decreasing i.e., it indicates dry weather.

Solution 1M.

1 torr = 1 mm of Hg

Solution 2M.

76 cm of Hg

Solution 3M.

Solution 1N.

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Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 26

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 27
Assumption: Atmospheric pressure falls linearly with ascent.

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 28

Solution 5N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 29

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Selina Concise Physics Class 9 ICSE Solutions Reflection of Light

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light

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Selina ICSE Solutions for Class 9 Physics Chapter 7 Reflection of Light

Exercise 7(A)

Solution 1S.

The return of light into the same medium after striking a surface is called reflection.

Solution 2S.

Black silvered surface reflects most of the light incident on it.

Solution 3S.

(a) Plane mirror: Plane mirror is a highly polished and smooth reflecting surface made from a clear plane glass sheet, usually thin and silvered with suitable reflecting abrasive (for example, mercury) on one side. Once this pasting is done, then the glass becomes opaque but due to the reflecting property of the abrasive, the plane glass sheet becomes a plane glass reflector or a plane glass mirror.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 1
(b) Incident ray: The light ray striking a reflecting surface is called the incident ray.

(c) Reflected ray: The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.

(d) Angle of incidence: The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. It is denoted by the letter i.

(e) Angle of reflection: The angle which the reflected ray makes with the normal at the point of incidence is called the angle of reflection. It is denoted by the letter r.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 2

Solution 4S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 3
Regular reflection occurs when a beam of light falls on a smooth and polished surface and irregular reflection occurs when a beam of light falls on a rough surface. Since the surface is uneven, from different points light rays get reflected in different directions and give rise to irregular reflection.

Solution 5S.

Reflection of light from a plane mirror is regular reflection and reflection of light from plane sheet of paper is irregular reflection of light.

Solution 6S.

Laws of reflection:

  1. The angle of incidence is equal to the angle of reflection.
  2. The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.

Solution 7S.

Laws of reflection:

  1. The angle of incidence is equal to the angle of reflection.
  2. The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.

Experiment to verify the laws of reflection:

Fix a white sheet of paper on a drawing board and draw a line MMas shown in figure. On this line, take a point O nearly at the middle of it and draw a line OA such that ∠MOA is less than 90o. Then draw a normal ON on line MM1 at the point O, and place a small plane mirror vertical by means of a stand with its silvered surface along MM1.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 4

Next fix two pins P and Q at some distance (≈5 cm) apart vertically on line OA, on the board. Keeping eye on the other side of normal (but on the same side of mirror), see clearly images P’ and Q’ of the pins P and Q. Next fix a pin R such that it is in line with the images of pins P and Q as observed in the mirror. Next, fix one more pin S such that the pin S is in line with the pin R as well as images P’ and Q’ of pins P and Q.

Draw small circles on paper around the positions of pins as shown in figure. Remove the pins and draw a line OB joining the pin points S and R, which meets the surface of mirror at O. The angles AON and BON are measured and recorded.

The experiment is then repeated for the angle of incidence ∠AON equal to 40o, 50o, 60o.
From results, it is observed that angle of incidence is equal to the angle of reflection. This verifies the first law of reflection.

The experiment has been performed on a flat drawing board, with mirror normal to the plane of board on which white sheet of paper is being fixed. Since the lower tips of all the pins also lie on the same plane (i.e., the plane of paper), it proves the second law of reflection.

Solution 8S.

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Solution 9S.

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Solution 10S.

(a) Angle of incidence = 90– 30o = 60o
(b) Angle between the incident ray and reflected ray = Angle of incidence + Angle of reflection
Angle of reflection = Angle of incidence = 60o
Therefore, Angle between the incident ray and reflected ray = 60o + 60o = 120o

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 7

Solution 11S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 8

Solution 12S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 9

Solution 13S.

(a) Three characteristics of image formed by plane mirror:

  1. Image formed in erect (upright)
  2. Image formed is virtual
  3. Image formed is of the same size as the object

(b) The image is situated at the same perpendicular distance behind the mirror as the object in front of it.

Solution 14S.

Real ImageVirtual image
1. A real image is formed due to actual intersection of the reflected rays.1. A virtual image is formed when the reflected rays meet if they are produced backwards.
2. A real image can be obtained on a screen.2. A virtual image cannot be obtained on a screen.
3. A real image is inverted with respect to the object.3. A virtual image is erect with respect to the object.

Solution 15S.

The interchange of the left and right sides in the image of an object in a plane mirror is called lateral inversion.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 10
Figure above shows the image formation of a letter P in a plane mirror.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 43
Solution 16S.

The letters on the front of a ambulance are written laterally inverted, so that the driver of the vehicle moving ahead of the ambulance reads these words laterally inverted as AMBULANCE, in his rear view mirror, and gices side to pass the ambulance first.

Solution 17S.

Due to lateral inversion, , it becomes difficult to read the image of the text of a page formed due to reflection by a plane mirror.

Solution 1M.

i = r

Solution 2M.

Erect and of same size

Solution 3M.

virtual with lateral inversion

Solution 1N.

Angle of incidence (i) + Angle of reflection(r) = 90
But, as per the laws of reflection, i = r
Therefore, 2 i = 90o
Or, i = r = 45o

Solution 2N.

Distance between man and his image = 6m
Distance between man and mirror + distance between mirror and image = 6m
But, Distance between man and mirror (object distance) = distance between mirror and image (image distance)
Therefore, distance of man from mirror = 6/2 = 3m

Solution 3N.

(a) Image of the insect is formed 1m behind the mirror.
(b) Distance between the insect and his image = 1 + 1 = 2 m

Solution 4N.

Initially, distance of the object from the mirror = 60 cm.
Therefore, image is formed at a distance 60 cm from the mirror, behind it.
Thus, initial distance between the object and image = 60 + 60 = 120 cm
If the mirror is moved 25 cm away from the object,
The new distance of the object from the mirror = 60 + 25 = 85 cm
The new image is now at a distance 85 cm from the mirror behind it.
Thus, new distance of the image from the object = 85 + 85 = 170 cm
Taking the position of the object as reference point, the distance between the two positions of the image = new distance of image from the object – initial distance of the image from the object
= (170 – 120) cm = 50 cm
Thus, the image shifts 50 cm away.

Solution 5N.

Distance between man and chart = 3m
Distance between man and mirror = 2m
Therefore, distance between chart and mirror = 5 m
Now, final image is formed on the mirror, which is at a distance of 2 m from the man, therefore, the chart as seen by patient is (5m + 2m =) 7m away.

Exercise 7(B)

Solution 1S.

If two mirrors make an angle θ with each other and object is placed in between the two mirrors, the number of images formed is n or (n – 1) depending upon n = 360o / θo is odd or even.

(a) If n = 360o / θo is odd,
(i) The number of images formed is n, when the object is placed asymmetrically between the mirrors.
(ii) The number of images formed is n-1, when the object is placed symmetrically between the mirrors.

(b) If n = 360o / θo is even, the number of images is always n-1.

Solution 2S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 44

Solution 3S.

For two mirrors kept perpendicular to each other, three images are formed for an object kept in between them.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 11

Solution 4S.

For two mirrors kept parallel to each other, an infinite number of images are formed for an object kept in between them.Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 12

Solution 5S.

Two uses of plane mirror:

  1. In barber’s shop for seeing the hairs at the back of head, two mirrors facing each other are fixed on opposite walls at the front and back of the viewer.
  2. In solar heating devices such as a solar cooker, solar water heater, etc., a plane mirror is used to reflect the incident light rays from sun on the substance to be heated.

Solution 1M.

5

Solution 2M.

In a barber’s shop, two plane mirrors are placed parallel to each other.

Solution 1N.

(a) Angle between the mirrors, θ = 90o
Now, n = 360o / θo = 360o / 90= 4, which is even.
Hence number of images formed will be (n-1); i.e., 4-1 = 3 images

(b) Angle between the mirrors, θ = 60o
Now, n = 360o / θo = 360o / 60= 6, which is even.
Hence number of images formed will be (n-1); i.e., 6-1 = 5 images

Solution 2N.

Angle between the mirrors, θ = 50o
Now, n = 360o / θo = 360o / 50= 7.2  7, which is odd.
(i) When placed asymmetrically, number of images formed will be n, i.e. 7.
(ii) When placed symmetrically, number of images formed will be (n-1); i.e. 7-1 = 6 images

Exercise 7(C)

Solution 1S.

A reflecting surface which is a part of a sphere is called a spherical mirror.

Solution 2S.

Two kinds of spherical mirrors are concave and convex.

Distinction between concave and convex mirror: A concave mirror’s bulging surface is silvered and reflection takes place from the hollow surface but a convex mirror’s inner surface is silvered and reflection takes place from the bulging surface.

Solution 3S.

Pole: The geometric centre of the spherical surface of mirror is called the pole of mirror.
Principal axis: It is the straight line joining the pole of the mirror to its centre of curvature.
Centre of curvature: The centre of curvature of a mirror is the centre of the sphere of which the mirror is a part.

Solution 4S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 13

Solution 5S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 14

Solution 6S.

Focus of a concave mirror: The focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to principal axis, pass after reflection from the mirror.
Focal length of a concave mirror: The distance of the focus from the pole of the concave mirror is called its focal length.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 15

Solution 7S.

Focus of a convex mirror: The focus of a convex mirror is a point on the principal axis from which, the light rays incident parallel to principal axis, appear to come, after reflection from the mirror.
Focal length of a convex mirror: The distance of the focus from the pole of the convex mirror is called its focal length.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 16

Solution 8S.

Incident ray is directed towards the centre of curvature because the ray is normal to the spherical mirror, so ∠i = ∠r = O.

Solution 9S.

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Solution 10S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 18

Solution 11S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 19

Solution 12S.

Two convenient rays that are chosen to construct the image by a spherical mirror for a given object:

  1. A ray passing through the centre of curvature: A ray of light passing through the centre of curvature of a concave mirror or a ray directed in the direction of centre of curvature of a convex mirror is reflected back along the same path after reflection.
    Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 20
  2. A ray parallel to the principal axis: A ray of light parallel to the principal axis, after reflection pass through the principal focus in case of a concave mirror or appears to diverge from it in case of convex mirror.
    Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 21

Solution 13S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 22

Solution 14S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 23

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 24
The image formed is virtual, erect and magnified.

Solution 17S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 25
The image formed is real, inverted and magnified.

Solution 18S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 26
The image formed is virtual, erect and diminished.

Solution 19S.

Convex mirror always produces erect and virtual images. The images formed are diminished, i.e. the size of the image is shorter than the size of the object.

Solution 20S.

(a) If the object is placed between the pole and focus of a concave mirror, the image formed is magnified and erect.
(b) The image is virtual.

Solution 21S.

(a) If the object is placed at the centre of curvature of a concave mirror, the image formed is of same size.
(b) The image formed is real and inverted.

Solution 22S.

(a) An image which can be obtained on a screen is called a real image.
(b) A concave mirror can be used to obtain a real image of an object.
(c) No, it does not form real image for all locations of the object.

Solution 23S.

When an object is moved from infinity towards the pole of mirror, the image formed moves away from the mirror. The image formed is real and inverted.

Solution 24S.

In a convex mirror, the image formed is always virtual, upright and diminished. It is always situated between its pole and focus, irrespective of the distance of object in front of the mirror.

Solution 25S.

(a) Concave, (b) Concave, (c) Convex and (d) Concave

Solution 26S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 27

Solution 27S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 28

Solution 28S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 29

Solution 29S.

The image formed by a convex mirror is always between pole and focus. Hence, the maximum distance that can be obtained in convex mirror is the focal length. For this case the object has to be at infinity.

Solution 30S.

The maximum distance that can be obtained in concave mirror is infinity. For this case the object has to be at focus.

Solution 31S.

To distinguish between a plane mirror, concave mirror and convex mirror, the given mirror is held near the face and image is seen. There can be following three cases:

Case (i): If the image is upright, of same size and it does not change in size by moving the mirror towards or away from the face, the mirror is plane.
(ii) If the image is upright and magnified, and increases in size on moving the mirror away, the mirror is concave.
(iii) If the image is upright and diminished and decreases in size on moving the mirror away, the mirror is convex.

Solution 32S.

Two uses of concave mirror:

  1. It is used as a shaving mirror.
  2. It is used as reflector in torch, head light of automobiles etc.

Solution 33S.

(a) Concave mirror
(b) Concave mirror

Solution 34S.

(a) The person’s face is between the pole and focus of the mirror.
(b) The image formed is erect, virtual and magnified.

Solution 35S.

A convex mirror is preferred as a rear view mirror because it has a wider field of view as compared to a plane mirror of same size.

Solution 36S.

A convex mirror diverges the incident beam and always forms a virtual, small and erect image between its pole and focus. Thus, a driver can see all the traffic approaching from behind. This fact enables the driver to use it as a rear view in vehicles to see all the traffic approaching from behind.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 30

Solution 1M.

Retraces its path

Solution 2M.

Erect and diminished

Solution 3M.

Concave mirror

Solution 1N.

Focal length = ½ (Radius of curvature)
Or, f = 40/2 = 20 cm

Solution 2N.

Radius of curvature = 2 x focal length
Or, R = 2f = 2 x 10 = 20 cm

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 31
The image is 30 cm in front of the mirror, 3 cm high, real, inverted and magnified.

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 32
The image is 6 cm behind the mirror.
Yes the image is magnified.

Solution 5N.

The size of the image is equal to the size of the object if the object is placed at the centre of curvature of a concave mirror.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 33
Hence, the object should be placed at 50 cm.

Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 45
The position of the object is 12 cm in front of the mirror.
Its size is 1 cm.

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 35

Solution 8N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 36
A ray passing parallel to the principal axis passes through the focal point after reflection. Hence, the focal length is 12 cm.

Solution 9N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 37

Solution 10N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 38

Solution 11N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 39

Solution 12N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 40

Solution 13N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 41

Solution 14N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 42

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Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution

Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution

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Selina ICSE Solutions for Class 9 Chemistry Chapter 8 Atmospheric Pollution

Page No: 128

Question 1.
Define the following terms:
pollution
pollutant
air pollution
Solution:
Pollution may be defined as contamination of air, water or soil by undesirable amounts of materials or heat and is caused by the concentration of substances which have harmful effects.

Toxic and otherwise harmful substances which have an undesirable impact on different components of the environment and life forms are known as pollutants.

Air pollution means degradation of air quality due to concentration of harmful contaminants which affect human, plant and animal lives.

Question 2.
Name any four gaseous pollutants.
Solution:
Sulphur dioxide, hydrogen sulphide, carbon monoxide and hydrocarbons.

Question 3.
Name the compounds of sulphur that cause air pollution. Also state the harmful effects of sulphur compounds.
Solution:
Compounds of sulphur like sulphur dioxide, sulphur trioxide and hydrogen sulphide are pollutants.

Harmful effects of oxides of sulphur:

  • It causes headache, vomiting and even death due to respiratory failure.
  • It destroys vegetation and weakens building materials/constructions.
  • It mixes with smoke and fog to form smog, which is very harmful.

It is oxidised by atmospheric oxygen into sulphur trioxide (SO3) which combines with water to form sulphuricacid.
2SO2(g) + O2(g) →2SO3(g)
SO3(g) + H2O(l) → H2SO4(l)

Question 4.
State:
natural sources of air pollution
man-made sources of air pollution
Solution:
Volcanoes, decaying vegetation, forest fires and dust storms.
Automobiles, factories, industrial processes and decay of crop residue in rural areas.

Question 5.
a. How do oxides of nitrogen enter the atmosphere?
b. What are their harmful effects?
Solution:
Nitric oxide (NO) and nitrogen dioxide (NO2) enter the atmosphere in the following ways:

  1. On burning of fuels in furnaces, temperature increases. At high temperature, nitrogen and oxygen present in air combine to form oxides of nitrogen.
  2. When fuel burns in an internal combustion engine, oxides of nitrogen are produced, and they enter the atmosphere as exhaust gases from automobile engines.
  3. Nitric acid is formed by the reaction between atmospheric nitrogen and oxygen in the presence of electric discharge, which occurs during thunderstorms when there is lightning.
  4. Nitric oxide further reacts with atmospheric oxygen and ozone to form nitrogen dioxide.

Harmful effects of the oxide of nitrogen

  • Nitrogen dioxide is very harmful to plants and animals.
  • It causes irritation in the mucous membrane.
  • Large concentrations of NOmay cause serious lung diseases.
  • Nitrogen dioxide causes serious injury to vegetation; it damages plant leaves.
  • In sunlight, nitrogen dioxide oxidises hydrocarbons to form photochemical smog. Photochemical smog causes eye irritation, asthma attacks and nasal and throat infections.

Question 6.
State the origin and health impact of smog.
Solution:
Smog is a dark, thick, dust and soot-laden fog pollutant which is a combination of oxides of nitrogen and sulphur and of partially oxidised hydrocarbons and their derivatives produced by industries and automobiles.
Smog is noxious and irritating. It reduces visibility, induces respiratory troubles and can cause death by suffocation.

Question 7.
What are the harmful effects of oxides of sulphur?
Solution:
Harmful effects of oxides of sulphur:

  1. It causes headache, vomiting and even death due to respiratory failure.
  2. It destroys vegetation and weakens building materials/constructions.
  3. It mixes with smoke and fog to form smog, which is very harmful.

It is oxidised by atmospheric oxygen into sulphur trioxide (SO3) which combines with water to form sulphuricacid.

2SO2(g) + O2(g) →2SO3(g)
SO3(g) + H2O(l) → H2SO4(l)

Question 8.
State the main sources and effects of carbon monoxide.
Solution:
Carbon monoxide is formed by incomplete combustion of fuels in homes, factories and automobiles.

Effects of carbon monoxide are as follows:

  • It is a highly poisonous gas.
  • It reduces the oxygen-carrying capacity of blood by an amount equivalent to the amount of haemoglobin converted to carboxyhaemoglobin.
    Haemoglobin + CO → Carboxyhaemoglobin
  • Because heart and brain are the two tissues most sensitive to oxygen depletion, they show the most serious effects of carbon monoxide exposure.
  • In high concentrations, carbon monoxide may kill by paralysing normal brain action.

Question 9.
Give the mechanism of the action of carbon monoxide.
Solution:

  • It is a highly poisonous gas.
  • When inhaled, it passes through the lungs directly into the blood stream. There it combines with haemoglobin, the substance which carries oxygen to body tissues. Because haemoglobin binds with carbon monoxide 200 times more strongly than oxygen, even low concentrations of carbon monoxide in air have magnified effects on the body.
  • It reduces the oxygen-carrying capacity of blood by an amount equivalent to the amount of haemoglobin converted to carboxyhaemoglobin.
    Haemoglobin + CO → Carboxyhaemoglobin

Question 10.
How can we control carbon monoxide poisoning?
Solution:
Carbon monoxide pollution can be controlled in the following ways:

  1. By switching over from internal combustion engines to electrically powered cars.
    Many pollution control devices are now installed in cars. Most of these devices help reduce pollution by burning gasoline completely. Complete combustion of gasoline produces only carbon dioxide and water vapour.
    2C8H18 + 5O2 → 16CO2 + 18H2O
  2. By using substitute fuels for gasoline: Natural gas in both compressed (CNG) and liquefied forms (LNG) is now increasingly being used as fuel. Alcohols are other feasible substitutes.
  3. By using catalytic convertors: Nitrogen oxide is reduced to nitrogen and oxygen in the presence of finely divided platinum or palladium as a catalyst.
    Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 1

Page No: 130

Question 1.
Why does rain water have pH less than 7?
Solution:
Carbon dioxide reacts with water to form weak carbonic acid which is slightly acidic having pH about 5.6.
Hence, the pH of rain water usually ranges between 5.6 and 3.5; at times, it can be as low as 2.

Question 2.
pH of acid rain is sometimes as low as 2. Explain.
Solution:Normal rain is only slightly acidic having pH about 5.6.
This is because carbon dioxide reacts with it to form weak carbonic acid.
CO+ H2O → H2CO3
pH of acid rain usually ranges between 5.6 and 3.5; at times, it can be as low as 2.

Question 3.
Explain the formation of acid rain due to:
1. Oxides of sulphur
2. Oxides of nitrogen
Solution:

  • Sulphur is a non-metallic element found in coal and fuel oil. When these fuels are burned, sulphur combines with oxygen in air to form its gaseous oxides, sulphur dioxide (SO2) and sulphur trioxide (SO3).
    S + O2 → SO2
    2SO2 + O2 → 2SO3
  • Sulphur dioxide and sulphur trioxide react with water to form H2SO4 which is the main cause of acid rain.
    2SO2 + O2 + 2H2O → 2H2SO4
    SO3 + H2O → H2SO4
  • Nitric acid is formed by the combination of nitrogen and oxygen. Nitrogen and oxygen combine in the presence of thunder and lightning. Oxides of nitrogen are also produced by internal combustion engines.
    N2 + O→ 2NO
  • Nitrogen oxide then gets oxidised in the atmosphere to nitrogen dioxide.
    2NO +O2 → 2NO2
  • Nitrogen dioxide combines with water to form a mixture of nitrous acid and nitric acid.
    2NO2 + H2O → HNO2 + HNO3

Question 4.
What are the causes of acid rain?
Solution:
The main causes of acid rain are the formation of mineral acids such as carbonic acid, nitric acid and sulphuric acid during rains.

Question 5.
Give the impact of acid rain:
1. on plants
2. on soil
3. on water bodies
Solution:
Acid rain causes loss of nutrients from plants, thus damaging their leaves.
It removes calcium and potassium both basic ingredients of soil, thus making it lose its fertility which ultimately damages forests.
Acid rain has serious ecological impacts as it affects water bodies too. The water of lakes and rivers is gradually becoming acidic due to acid rain which is affecting aquatic life.

Question 6.
How does a scrubber help in reducing the formation of acid rain?
Solution:
A scrubber can also be used to reduce the formation of acid rain. It is a device which absorbs gaseous pollutants. It is used for removing sulphur dioxide from a smoke stack, and usually consists of a fine spray of water and gas rising from the stack, which is passed through the scrubber where water absorbs sulphur dioxide.

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Page No: 131

Question 1.
What do you understand by Green House effect?
Solution:
Heating of the Earth and its environment due to solar radiation trapped by carbon dioxide and water vapour in the atmosphere is called greenhouse effect.

Question 2.
What are green house gases? How are they responsible for global warming?
Solution:
Gases which contribute to the greenhouse effect are called greenhouse gases. These gases are carbon dioxide, water vapour, oxides of nitrogen, methane, ozone and chlorofluorocarbons. Sunlight reaching Earth consists of three types of radiation-UV radiation, visible radiation and IR radiation. As sunlight passes through the atmosphere, most UV radiation is absorbed by ozone; 30% of IR radiation reaches the Earth’s surface, heating it up. As the Earth’s surface becomes hot, it starts emitting radiation with less energy than the incoming radiation and thus with longer wavelength. Some emitted IR radiation escapes from the Earth’s surface and some are absorbed by CO2, thus remaining on the Earth. Trapped radiation warms the Earth’s surface and lower layers of the atmosphere.

Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 3
Question 3.
State the sources and effects of the following gases:
1. Carbon dioxide
2. Methane
3. Water vapour
Solution:
Sources of carbon dioxide:

  • Burning of fossil fuels such as coal, natural gas and petroleum
  • Industrial processes such as manufacture of lime and those in fermentation units
  • Biological decay of plants
  • Respiration by animals, human beings and plants

Effects of carbon dioxide:

  • Greenhouse effect and global warming

Sources of methane:

  • Anaerobic decomposition of organic matter in soil, water and sediments
  • Incomplete combustion of fossil fuels

Effects of methane:

  • Greenhouse effect and global warming

Sources of water vapour:

  • Burning of hydrocarbons
  • Evaporation and transpiration

Effects of water vapour:

  • Greenhouse effect and global warming

Question 4.
State the ways of reducing the presence of green house gases.
Solution:
Ways of reducing the presence of greenhouse gases:

  • Minimise the use of automobiles: Depending on the situation, one can use a bicycle, the public transport system and car pools.
  • Plant more trees to increase green cover.
  • Avoid burning of dry leaves and wood.
  • Avoid smoking. It is illegal to smoke in public places and work places, because smoke is harmful not only for the one who is smoking but also for others sitting nearby.
  • Help people in understanding global warming; most people are unaware of it.

Question 5.
State the effects of green house gases on the atmosphere.
Solution:
Effects of global warming

  • Rise in sea level: Due to global warming, glaciers and polar ice caps have started to melt, and gradually this may lead to an increase in the sea level. This will in turn flood several coastal areas in countries such as India, Bangladesh, the Netherlands and the Maldives.
  • Global warming will cause more water to evaporate from water bodies, thus forming more water vapour. Because water vapour also contributes to the greenhouse effect, global warming will further increase.
  • Global warming can lead to changes in the rain pattern and thus shift in crop zones. For example, wheat-producing zones will shift from Russia and Canada to the less fertile polar regions.
  • Change in rain pattern due to global warming will also affect trees and plants in forests which are natural habitats of wild life. With destruction of forests, many species of wild life will also begin to die out.

Question 6.
State the role of a green house in growing plants.
Solution:
A greenhouse collects light and converts it to heat. It also stores thermal energy and helps moderate temperature and produces a controlled environment for plants to grow and thrive. It even offers protection from wind, rain, snow and other weather elements and protects fruits from invading pests and animals.

Question 7.
Our atmosphere acts as a green house. Explain.
Solution:
Our atmosphere contains greenhouse gases such as CO2, water vapour, O3, CH4, oxides of nitrogen and CFCs and allows the sunrays to come in. Sunlight reaching the Earth consists of three types of radiation-UV radiation, visible radiation and IR radiation. As sunlight passes through the atmosphere, most UV radiation is absorbed by ozone; 30% of IR radiation reaches the Earth’s surface, heating it up. As the Earth’s surface becomes hot, it starts emitting radiation with less energy than the incoming radiation and thus with longer wavelength. Some emitted IR radiation escapes from the Earth’s surface and some are absorbed by CO2, thus remaining on the Earth. Trapped radiation warms the Earth’s surface and lower layers of the atmosphere.

Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 3

Question 8.
How can we reduce global warming?
Solution:

Ways of reducing global warming:

  • Minimise the use of automobiles: Depending on the situation, one can use a bicycle, the public transport system and car pools.
  • Plant more trees to increase green cover.
  • Avoid burning of dry leaves and wood.
  • Avoid smoking. It is illegal to smoke in public places and work places, because smoke is harmful not only for the one who is smoking but also for others sitting nearby.
  • Help people in understanding global warming; most people are unaware of it.

Page No: 134

Question 1.
What is a pollutant?
Solution:
The toxic substances that have an undesirable impact on different components of the environment and are injurious to life and property are known as pollutants.

Question 2.
What is the effect of the following pollutants on living beings (one in each case)?
Fluorides
Smoke particles
Lead
Mercury compounds
Smog
Nitrogen oxide
Solution:
Effect of Pollutant on living beings:

  • Fluorides: Effects teeth and bones.
  • Smoke Particles: Cause asthma and lung diseases.
  • Lead: Damages the nervous and digestive systems and can cause cancer .
  • Mercury compounds: They cause disease like Minamata commonly found in fishermen.
  • Smog: It reduces visibility and induces respiratory troubles.
  • Nitrogen Oxide: Causes death of many plants.

Question 3.
What is air pollution? How does this pollution take place?
Solution:
Air pollution: Deterioration of air quality around us is called air pollution. It is defined as the presence of a contaminant in the atmosphere in a concentration large enough to injure human, plant and animal life.

Air pollution takes place due to the presence of gaseous pollutants like oxides of sulphur, hydrocarbons, smoke, oxides of carbon, oxides of nitrogen, dust, particulate pollutants like mist, spray and fume.

Question 4.
What are the components of clean, dry air?
Solution:
Components of air is:

Pure Air componentsBy Volume percentageConcentration parts per million (ppm)
Nitrogen78.09780,900
Oxgyen20.94209,400
Inert Gases
Argon
0.939300
Neon18
Helium5
Krypton1
Xenon1
Carbon-dioxide0.03315
Methane1
Hydrogen0.5
Natural pollutants
Oxides of nitrogen
Ozone

0.52
0.02

Question 5.
Name some particulate pollutants.
Solution:
Particulate Pollutant are dust, smoke, mist, spray and fume.

Question 6.
Why is cigarette-smoking harmful?
Solution:
Cigarette smoking is harmful not only for one who is smoking but also for sitting nearby and so one should avoid it. Tobacco smoke causes lung cancer and asthma.

Question 7.
What is smog? State its damaging effects.
Solution:
Smog: A smog is a pollutant which is a combination of oxides of sulphur and nitrogen, partially oxidized hydrocarbons and their derivatives produced by industries and automobiles from a dark, thick dust and soot laden fog known as smog.

Damaging Effect: Smog is noxious and irritating. It reduces visibility, induce respiratory troubles and can cause death by suffocation.
Photochemical Smog damages the tissues of certain plants and even decreases the yield of citrus fruits and grapes.

Question 8.
What do you understand by ppm?
Solution:
ppm means parts per million. That is, share in 10, 00000.

Question 9.
Describe the major air pollutants. How does carbon monoxide pollute our environment?
Solution:
Major air pollutants are: Large amounts of Carbon monoxide, Sulphur dioxide, H2S, Chlorine, HCl, Hydrocarbons and particulates. Particulate matter like sand, dust etc. Secondary pollutant like (PAN) peroxyaryl.

Question 10.
How do you propose to control:
a. carbon monoxide emission
b. SOx emission
Solution:
Control of: (i) Carbon monoxide, CO emission:
Emission of CO can be controlled by :

  • Switching over from internal combustion engine to electrically powered cars. (b) Using alcohols, CNG, LNG in place of gasoline.
  • By using Catalytic Mufflers or Convertors.
  • Using pollution control devices to burn gasoline completely.
  • Using lead free petrol.
  • By using catalytic convertors
    Nitrogen oxide is reduced to nitrogen and oxygen in the presence of finely divided platinum or palladium as catalyst.
    2NO → N2 + O2
    2NO2 → N2 + 2O2
  • Carbon monoxide changes to carbon dioxide in the presence of finely divided platinum as catalyst.
    CO → CO2 + H2O

(ii) (SOx) oxides of Sulphur emission: Oxides of sulphur (SO, SO) emission can be reduced by-
(a) Using coal or oil that has low sulphur content. (b) By using Scrubber, a device that absorbs gaseous pollutants.

Question 11.
Give the composition, causes and effects of acid rain.
Solution:
Acid rain

Factories in big cities release nitrogen dioxide and sulphur dioxide as their wastes. These gases dissolve in rainwater during rains and form nitrous acid and sulphurous acid. As the rain falls, these acids come down to the ground as an acid rain.

The normal rain is slightly acidic having a pH about 56 as carbon dioxide gas reacts with it to form a weak carbonic acid.
CO2 + H2O → H2CO3
(Carbonic acid)

The pH of acid rain ranges between 56 – 35 and in some cases pH can go even lower than 2.
The two forms of deposition of acid rains are:

  • Dry deposits-Particles containing sulphates and nitrates
  • Wet deposits-dew, rain, fog, smoke

Formation of acid rain

  • Acid rain refers to rain which has pH less than 5.6. It is mainly caused by atmospheric pollutants.
  • Natural sources: Bacterial decomposition, forest fires, volcanic eruptions.
  • Man made sources: Industries and smelting plants, automobile exhausts, power plants etc.
  • Oxides of nitrogen and sulphur interact with water vapour in presence of sunlight in the atmosphere to form nitric acid and sulphuric acid mist respectively. This mist remains as vapours at high temperatures and condenses at low temperatures. These acids mix with rain (snow or fog) and fall down on the Earth resulting in acid rain.

Causes of acid rain

  • The formation of mineral acids like carbonic acid, nitric acid and sulphuric acid is the main cause of acid rain.

Formation of Nitric acid and Nitrous acid

  • Nitrogen and oxygen (that is oxides of nitrogen) combines in the presence of thunder and lightning to form nitric acid.
  • They are also produced by internal combustion engines (automobile engines). This then gets oxidized in the atmosphere to nitrogen dioxide. Nitrogen dioxide combines with water to form a mixture of nitrous acid and nitric acid.

N2 + O2 → 2NO
(Nitrogen oxide)

2NO + O2 → 2NO2
(Nitrogen dioxide)

2NO2 + H2O → HNO2 + HNO3
(Nitrous acid) (Nitric acid)

Formation of Sulphuric acid and Sulphurous acid

Impurities in the coal: Coal used in power plants contains upto 4% sulphur. On combustion it forms pollutant sulphur dioxide (i.e, oxides of sulphur).
S + O2 → SO2
(Sulphur dioxide)

Sulphur dioxide reacts with water vapour to form sulphurous acid.
SO+ H2O → H2SO3
(Sulphurous acid)

Sulphur dioxide can also be oxidized to sulphur trioxide.
2SO2 + O2 → 2SO3
(Sulphur trioxide)

Sulphur trioxide reacts with water vapour to form sulphuric acid.
SO3 + H2O H2SO4
(Sulphuric acid)

Impact of Acid rain

Changes the acidity of soil: The acids present in the acid rain like, nitric acid, nitrous acid and sulphuric acid, sulphurous acid increases the acidity of soil. It removes calcium and potassium minerals i.e., basic ingredients from the soil losing their fertility.

The hydrogen ions H+ which are added to the soil, when acid rain falls on the Earth interact chemically with existing soil minerals.
Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 4

Affects water bodies and marine organisms: The water of lakes and rivers is becoming acidic, which may no longer support aquatic life.

Material damage: It increases corrosion of metals, disintegrates paper and leather. Weakens building materials such as statues, marbles, sculptures, limestone, slate, mortar etc. These materials become pitted and weakened mechanically. The Taj Mahal in India faces this problem.
CaCO3 + H2SO4 → CaSO4 + CO2 + H2O
CaCO3 + 2HNO→ Ca(NO3)2 + CO2 + H2O

Impact on living things: It damages forests. Acid rain gets absorbed by plants, animals directly or indirectly toxicity enters food chain affecting humans. They can affect a person’s breathing, at sufficiently high concentrations. Sulphur dioxide irritates the upper respiratory tract, which serve to expel soot particles and dust in the inhaled air. At even lower concentrations, it does still greater harm by injuring lung tissues.

Question 12.
Explain the effect of sulphur dioxide on the atmosphere.
Solution:
Harmful effects of oxides of sulphur:

  • It causes headache, vomiting and even death due to respiratory failure.
  • It destroys vegetation and weakens building materials/constructions.
  • It mixes with smoke and fog to form smog, which is very harmful.
  • It is oxidised by atmospheric oxygen into sulphur trioxide (SO3) which combines with water to form sulphuric acid (H2SO4). Sulphuric acid is the cause of acid rain.
    2SO2(g)  + O2(g) → 2SO3(g)
    SO3(g) + H2O →H2SO4(aq)

Question 13.
Explain the formation of ozone in the atmosphere.
Solution:
In the atmosphere ozone is formed by the action of ultraviolet rays of the sun on oxygen.
3O2 (g) → 2O3(g)

The high energy UV radiations break oxygen molecules into oxygen atoms.
O2 + Far UV → O + O
Oxygen molecule Oxygen atoms

Oxygen atom reacts with oxygen molecule to form ozone.
O + O2 → O3
Atom Molecule Ozone

The Net reaction is:
3O2 + Far UV → 2O3

Question 14.
What is the function of ozone in the atmosphere?
Solution:
It is formed by the action of ultraviolet rays of the Sun on oxygen.
O3 → O + O2

Ozone layer acts as a blanket in the atmosphere 16 km height above the Earths surface.
It absorbs harmful ultraviolet rays (UV radiations) coming from the Sun and prevents them to reach the surface of the Earth.

Ultraviolet rays have very harmful effects on living things. It causes skin cancer. It destroys many organic species which are necessary for life.

Thus it protects the life on earth from harmful effects of Ultra Violet Rays. Which can cause (a) Skin cancer (b) destroy many organic species necessary for life.

Question 15.
State the chemicals responsible for ozone layer destruction.
Solution:
Chemicals responsible for ozone destruction free radical chlorine (Cl) and nitrioxide (NO) are responsible for ozone depletion i.e. react with O3Free radical chlorine (Cl) is produced by UV rays from chlorofluoro carbons enter the atmosphere because of excessive use as solvents, Aerosol, Spray, Propellants, Refrigerants and blowing agents for plastic foams.

Chemicals responsible for the depletion of ozone layer

Fuel of planes: Burning of fuels of planes emits large quantity of nitric oxide and other gases in the atmosphere. Nitric oxide reacts with ozone and form nitrogen dioxide and nitrogen trioxide. This causes depletion of ozone.

NO(g) + O3(g) → NO2(g) + O2(g)
(Nitrogen dioxide)

NO2(g) + O3(g) → NO3(g) + O2(g)
(Nitrogen trioxide)

Excessive use of chlrofluro carbon: It is released by refrigerators and air conditioning systems.
It causes reduction of ozone layer that protects us from harmful ultraviolet rays (UV radiations) of the Sun.

The chlorofluro carbons are decomposed by the ultraviolet rays to highly reactive chlorine which is produced in the atomic form.
Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 5
The free radical [Cl] reacts with ozone to form chlorine monoxide.

Cl + O3 → ClO + O2
(Chlorine monoxide)

This causes depletion of ozone layer. Chlorine monoxide then reacts with atomic oxygen to produce more chlorine free radicals.

ClO + O → Cl + O2
(Free radical)
Again this free radical destroys ozone and the process continues giving rise to depletion of ozone layer.

Question 16.
Name any two:
natural sources of atmospheric pollution.
gases which are responsible for the formation of acid rain.
Solution:
Natural sources of atmospheric pollution:

  1. Decay of plants and animals
  2. Disintegration of rocks and soil
  3. SO2 and NO2 are gases responsible for acid rain

Question 17.
Explain the term ‘global warming’. State two ways by which global warming can be reduced.
Solution:
Global warming is the increase in temperature of Earth due to enhanced concentration of greenhouse gases (CFCs) in the atmosphere.

Two ways to reduce global warming:

  • Plant more trees to increase green cover
  • Minimise the use of automobiles

Question 18.
State two effects of ozone depletion.
Solution:
Effects of ozone depletion:

  1. UV rays of the Sun reach Earth and cause sun burn, premature ageing of the skin and skin cancer.
  2. UV radiation can also damage several parts of the eyes, including the lens, cornea, retina and conjunctiva.

Question 19.
What is the cause of acid rain? Give any two impacts of acid rain.
Solution:
Causes:

Sulphur and nitrogen oxides are emitted by burning fossil fuels. Such smoke and gases entering the atmosphere make a dilute soup of sulphuric and nitric acids. This falls on the land surface in the form of acid rain damaging the things on Earth.

Impacts:

  1. Acid rain accelerates the decay of building materials and paints, including buildings, statues and sculptures which are part of our nation’s culture and heritage.
  2. Acid rain causes respiratory problems in humans, especially for people suffering from asthma. It may cause throat irritation, dry cough and severe headache.

Question 20.
Explain the methods of preventing acid rain.
Solution:
Methods of preventing acid rain:

  1. By using coal or oil which has low sulphur content. This reduces the emission of oxides of sulphur and nitrogen responsible for acid rain.
  2. By using a scrubber, a device which absorbs gaseous pollutants.

Question 21.
State an advantage of CNG (Compressed Natural Gas).
Solution:
Using CNG causes less pollution. It does not contain lead, and it has low maintenance cost.

Question 22.
State how CFC break ozone layer.
Solution:
Depletion of O3 by CFC:

CFC is broken by UV rays of sunlight to produce [Cl] atom or free radical [Cl] which is highly reactive in the atomic form and it forms ClO(g) with O2.

CFCl3 → CFCl2 + Cl(atom)
Cl(g) + O3(g) → ClO(g) + O2(g)

This depletes ozone.

ClO further produces more[Cl] free radical and destroys more of O3,thereby resulting in ozone depletion.
ClO(g) + O(g)→ Cl(g) + O2(g)

Question 23.
Describe the methods of saving ozone layer.
Solution:
Methods to protect the ozone layer:

  1. Using alternative products such as HCFCs (hydrochlorofluorocarbons)
  2. Montreal Protocol, an international treaty, helps prevent ozone depletion.

Question 24.
Fill in the blanks:
The pollutants such as NO2, SO2 and SO3 dissolved in the moisture of air are the cause of ____________________.
Excessive release of carbon dioxide in the atmosphere is the cause of __________ effect which produces global warming.
Ozone layer prevents the harmful ________ radiation of the sun to reach the earth.
Decrease of the concentration of ozone in the stratosphere is the cause of formation of __________ holes.
Ozone depletion is mainly caused by the active __________ atoms generated from CFC in the presence of UV radiation.
Solution:

  1. acid rain
  2. greenhouse
  3. ultraviolet
  4. ozone
  5. chlorine

Question 25.
Select the correct answer:
a. Excessive release of carbon dioxide in the atmosphere is the cause of
i. Depletion of ozone
ii. formation of polar vartex
iii. global warming
iv. formation of smog
b. Inhalation of air polluted with carbon monoxide is dangerous because:
i. CO combines with O2 dissolved in blood.
ii. CO combines with haemoglobin of blood.
iii. CO removes water from the body and causes dehydration.
iv. CO causes coagulation of proteins in the body
c. Decrease of amount of ozone in stratosphere is called depletion of zone and it is caused by
i. UV radiations of sun
ii. Use of CFC compounds
iii. excessive use of detergents
iv. Use of polychlorinated biphenyls
Solution:

  1. global warming
  2. CO combines with haemoglobin of blood.
  3. Use of CFC compounds

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Selina Concise Biology Class 9 ICSE Solutions Respiration in Plants

Selina Concise Biology Class 9 ICSE Solutions Respiration in Plants

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Biology Chapter 7 Respiration in Plants. You can download the Selina Concise Biology ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Biology for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Biology Chapter 7 Respiration in Plants

Exercise 1

Solution A.

  1. (c) in which glucose is broken down into pyruvate
  2. (b) Stomata and lenticels
  3. (d) Baker’s yeast

Solution B.1.

Respiration is an essential life process for all living organisms. Therefore, plants also respire all day and all night.

Solution B.2.

(a) ATP
(b) Respiration
(c) Fungus
(d) Glycolysis
(e) Oxygen
(f) Soda lime

Solution B.3.

(a) False. Aerobic respiration of one mole of glucose yields 38 ATP.
(b) False. Anaerobic respiration in plants yields ethanol.
(c) True
(d) False. All leaves of a green plant normally respire aerobically at night.

Solution C.1.

The energy liberated in respiration is stored in the form of ATP inside the cells. Some part of it is lost as heat to the surroundings.

Solution C.2.

During day time, both photosynthesis and respiration takes place in green plants. Evolution of COis an indicator to demonstrate respiration in living organisms. CO2 produced during respiration in plants gets utilized during photosynthesis and thus, there is no evolution of CO2. Therefore, it is difficult to demonstrate respiration in green plants as there is no evolution of CO2 during day time.

Solution C.3.

Respiration is said to be the reverse of photosynthesis due to the following reasons:

  1. In respiration, the organic food is broken down into its inorganic compounds, i.e., CO2 and H2O, while in photosynthesis the organic food is synthesized from its inorganic components, i.e., CO2 and H2O.
  2. In respiration, CO2 is given out, while in photosynthesis CO2 is consumed.
  3. In respiration, O2 is consumed, while in photosynthesis O2 is released or evolved.
  4. In respiration, energy is liberated, while in photosynthesis energy is absorbed.

Solution C.4.

illing makes the soil porous and airy. The underground parts of the plants get sufficient amount of oxygen for respiration. In this way, tilling helps the crops to grow faster.

Solution C.5.

ATP: Adenosine triphosphate
ADP: Adenosine diphosphate

Solution C.6.

Cell respiration cannot occur in any organism at a temperature of about 65°C because the enzymes involved in respiration become inactivated at high temperatures.

Solution C.7.

(a) Lenticels
(b) Cytoplasm
(c) Glucose
(d) Photosynthesis, respiration
(e) Pyrogallate of potash
(f) KOH (caustic potash)

Solution D.1.

Oxidation of organic food particularly carbohydrates in living cells to release energy is called respiration.

Similarities between respiration and burning:

  1. Both require oxygen
  2. Both produce energy
  3. Both result in the formation of CO2 and water

Differences between respiration and burning: 

RespirationBurning
1. Occurs in a series of chemical steps1. Occurs in a single step
2. Carried out by enzymes2. Carried out by heat
3. Biochemical process3. Physico-chemical process
4. Energy is liberated in the form of ATP and some heat4. Energy is liberated in the form of heat and light
5. No light energy is produced5. Light energy is produced
6. Cellular process6. Non-cellular process
7. Occurs at body temperature7. Occurs at high temperature (at ignition point)
8. No supply of heat energy is required8. Supply of heat energy is required
9. The organic compound is oxidized to carbon dioxide and water.9. The organic compound initially chars and later burns, producing a flame.

Solution D.2.

Difference between aerobic and anaerobic respiration in plants: 

Aerobic respiration in plantsAnaerobic respiration in plants
1. Also called oxybiotic respiration1. Also called anoxybiotic respiration
2. Proceeds in the presence of oxygen2. Proceeds in the absence of oxygen
3. Occurs in mitochondria3. Occurs in cytoplasm
4. Complete breakdown of glucose4. Incomplete breakdown of glucose
5. End-products are carbon dioxide and water5. End-products are ethyl alcohol and carbon dioxide
6. Large quantity of energy is liberated (38ATP) from one mole of glucose6. Small quantity of energy is liberated (2ATP) from one mole of glucose
7. Occurs normally throughout the life7. Occurs temporarily for short periods

Solution D.3.

(a) Experiment demonstrating that germinating seeds produce heat.

  1. Take two thermos flasks ‘A’ and ‘B’.
  2. Take about 200 bean or pea seeds and soak them in water for more than 24 hours.
  3. Divide the seeds into two equal groups. Kill one group of seeds by boiling them and then, wash them with dilute formalin to prevent bacterial decay. Place the live germinating seeds in flask A and the killed one in flask B.
  4. Insert a thermometer in each of the flasks and plug their mouths with cotton wool. Note the initial reading in the thermometer.
  5. After few hours, the thermometer in flask A will show a higher reading indicating that the germinating seeds produce heat. There will not be any rise in the temperature of flask B.
    Selina Concise Biology Class 9 ICSE Solutions Respiration in Plants 1

(b) Experiment demonstrating that germinating seeds give out carbon dioxide.

  1. Take two flasks ‘A’ and’ B’.
  2. Place some wet cotton wool at the bottom of both the flasks.
  3. Place some soaked seeds of pea in flask A and an equal number of boiled or dead seeds in flask B. Add some carbolic acid to flask B to prevent the growth of bacteria on dead seeds.
  4. Cork the flasks and keep them undisturbed for few days.
  5. After few days, the seeds in flask A will have germinated and the seeds in flask B do not show any signs of germination.
  6. The gases in each of the flasks are then tested by removing the cork and tilting the flask over a test tube containing limewater and then shaking up the test tube.
  7. The expected CO2 present in flask A will turn limewater milky showing that germinating seeds give out CO2, while the gas in flask B will show no effect on limewater.
    Selina Concise Biology Class 9 ICSE Solutions Respiration in Plants 2

(c) Experiment demonstrating that germinating seeds can respire even in total absence of air.

  1. Take few soaked and peeled off pea seeds and push them into the mouth of a test tube filled with mercury and inverted in a beaker of mercury. The seeds will float to the top and will be completely surrounded by mercury.
  2. After few days, the level of mercury falls and the presence of some gas in the test tube can be detected.
  3. Add some KOH to the test tube. The gas present in the test tube will be absorbed and the level of mercury will again rise showing that the gas was CO2. Hence, we can prove that germinating seeds respire even in the total absence of air.
    Selina Concise Biology Class 9 ICSE Solutions Respiration in Plants 3

Solution D.4.

(a) Lenticels help the stem to participate in respiration.
(b) Stomata help the leaves to participate in respiration.
(c) Root hairs help the roots to participate in respiration.

Solution E.1.

(a) Five flasks (A-E) are having different plant parts for observation.

(b) Limewater absorbs CO2 present in the flasks. If CO2 is produced, limewater turns milky.

(c) All the five flasks (A-E) containing plant parts will show respiration and thus, the tubes containing limewater will turn milky. The flask F does not contain any plant part. So, there is no respiration and hence, there is no change.

(d) Set-up F acts as a control for the experiment, where all the conditions are similar except for the plant part, which is missing.

(e) It can be concluded from this experiment that CO2 is produced during respiration.

Solution E.2.

(i) (a) 6H2O (b) C2H5OH
(ii) (a) Aerobic respiration; (b) Anaerobic respiration
(iii) Reaction (a) can be completed only in the presence of oxygen, while reaction (b) can be completed in the absence of oxygen.

Solution E.3.

(a) Potassium hydroxide solution is kept in test tube X and Y to check the presence of CO2, as KOH can absorb CO2, if it is produced.

(b) When the process of respiration takes place in test tube X, the O2 present in the test tube is consumed. The CO2 produced due to respiration is absorbed by the KOH present in the test tube. Hence, a space is developed due to consumption of oxygen and hence, the coloured water rises in tubing 1.

(c) In test tube Y, boiled peas soaked in disinfectant are kept because the boiled seeds are dead and cannot respire. Therefore, the process of respiration does not take place in test tube Y.

(d) Respiration is the cause of the above rise.

(e) The catabolic process of releasing energy from simple sugars for carrying out life processes is called respiration.

Solution E.4.

(a) The experiment was set up to study that CO2 is given off from living plant material which performs respiration.

(b) Soda lime is kept in bottle ‘A’ so as to check the (presence of) passage of CO2 present in bottle ‘B’. This is because soda lime has a tendency to absorb CO2 present in air.

The limewater present in bottle ‘B’ shows that no CO2 is entering bottle C with the air passing through bottle C.

(c) The bottle ‘D’ having limewater will show that CO2 is produced due to respiration taking place in bottle ‘C’ having germinating seeds in it. Due to the CO2 produced in bottle C, the limewater turns milky, which passes through bottle ‘D’.

(d) C6H12O6 + 6O2 6CO2 + 6H2O + 38ATP (Energy)

(e) The bottle ‘C’ when covered with black cloth will have no scope for photosynthesis. Hence, only the process of respiration can be observed.

(f) By introducing a thermometer in bottle ‘C’, the change in temperature is checked or observed. If respiration takes place, there will be a rise in temperature and if there is no rise in temperature, no respiration occurs. Therefore, a three-holed rubber stopper and a thermometer were introduced.

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