Selina Concise Physics Class 8 ICSE Solutions – Energy

Selina Concise Physics Class 8 ICSE Solutions – Energy

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 4 Energy. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions For Class 8 Physics Chapter – 4 – Energy

  • When we are pushing a wall we are not doing any work as the position of wall is not change i.e. wall has not moved in the direction of force.
  • WORK: “is said to be done if on applying force on a body, the body moves (or changes it position) from it place in the direction of force. W = F × d
    Or
    “Work is said to be done by a force applied on a body, if it changes its size or shape.”
  •  FACTORS AFFECTING THE AMOUNT OF WORK DONE : W = F × d
    (i) Magnitude of force applied.
    (ii) Distance moved by the body in the direction of force.
    UNIT OF WORK : W = F × d
    s.i  unit W= 1N × 1m = Nm = joule (J)
    1 kgf = 9.8 N is force on 1 kg ∴ F = mg
    Work done – 1 kgf × m = 9.8 N m = 9.8 J = 10 J nearly
  • A cooli standing with a box on his head, does no work as distance moved is zero. ~
  •  A cooli with a box on his head and walking is doing no work as force is acting vertically downward and direction of motion is at right angle.
  •  ENERGY: “is capacity of doing work.”
    Or
    “The work done on a body in changing its state is called energy.”
    s.1. unit of energy = S.I. unit of work = (J)
  •  JOULE: “A body is said to possess a energy of one joule. If a force of 1 Newton moves the body by a distance of 1 metre in the direction of force.”
  • MECHANICAL ENERGY: “The energy possessed by a body due to its state of rest or state of motion is called mechanical energy.
  •  Potential energy and kinetic energy are mechanical energies.
  •  POTENTIAL Energy (P.E.) : “Is energy possessed by body due to its state of rest or position.” P.E. = mgh
  •  KINETIC Energy (K.E.) : “Is energy possessed by body due to its motion.”
    K.E. = 1/2 M V2
  • GRAVITATIONAL POTENTIAL ENERGY: “When a stone or water is raised (lifted) from ground to a height, work is done against the force of gravity. This work is stored in the stone or water in the form of GRAVITATIONAL POTENTIAL ENERGY.”
  • A stretched bow, due to change in position possesses potential energy. When stretched bow is relreased the arrow comes in motion and due to motion possesses the kinetic energy and hits the body on which it strikes.
  •  When a body at a hight, it possess P.E. = mgh. When it falls, height decreases and speed increases
    ∴ its P.E. decreases and K.E. increases.
  • Powder : “Rate of doing work”. P = W/t
    Selina Concise Physics Class 8 ICSE Solutions Chapter 4 Energy 1

Test your self

A.Objective Questions

1. Write true or false for each statement

(a) A coolie does no work against the force of gravity while carrying a luggage on a road.
Answer. True.

(b) The energy stored in water of a dam is the kinetic energy.
Answer. False.
The energy stored in water of a dam is the potential energy.

(c) The energy of a flying kite is kinetic energy.
Answer. True.

(d) Work done by a boy depends on the time in which he does work.
Answer. False.

(e) Power spent by a body depends on the time for which it does work.
Answer. True.

2. Fill in the blanks

(a) Work is said to be done by a forte only when the body moves.
(b) Work done = Force × distance moved in direction of force.
(c) The energy of a body is its capacity to do work.
(d) The S.I. unit of energy is joule.
(e) The potential energy of a body is due to its state of rest or position and kinetic energy of body is due to its state of motion.
(f) Gravitational potential energy U = mass × force of gravity on unit mass × height.
(g) Kinetic energy = 1/2 × mass × (speed)2
(h) Power P=work done/time taken.
(i) The S . i.  unit of power is watt
(j) I H.P. = 746 W

3. Match the following
Selina Concise Physics Class 8 ICSE Solutions Chapter 4 Energy 2

4. Select the correct alternative

(a) The S.I. unit of work is

  1. second
  2. metre
  3. joule
  4. newton

Answer:
joule

(b) No work is done by a force if the body

  1. moves in direction of force
  2. does not move
  3. moves in opposite direction
  4. none of the these

Answer:
does not move

(c) Two coolies A and B do some work in time 1 minute and 2 minute respectively. The power spent is

  1. same by both coolies
  2. is more by coolie A than by B
  3. is less by coolie A than by B
  4. nothing can be said.

Answer:
is more by coolie A than by B

(d) The expression of power P is

  1. P = mgh
  2. P = P = 1/2 Mv2
  3. P = F × d
  4. P = F × d/t

Answer:
P = F × d/t

(e) I H.P. ¡s equal to

  1. 1 W
  2.  1 J
  3.  746 J
  4.  746 W

Answer:
 746 W

(f) When a boy doubles his speed, his kinetic energy becomes

  1. half
  2. double
  3. four times
  4. no change

Answer:
four times

(g) A boy lifts a luggage from height 2 m to 4 m. The potential energy will become

  1. half
  2. double
  3. one-third
  4. one-fourth

Answer:
double

Selina Concise Physics Class 8 ICSE Solutions – Heat Transfer

Selina Concise Physics Class 8 ICSE Solutions – Heat Transfer

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 6 Heat Transfer. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Physics Chapter 6 Heat Transfer

  •  Heat is a form of energy. When two bodies are in contact heat flows from body at higher temperature to body at lower temperature till the lower temperature of both is same.
  •  When a body is heated, its molecules move faster about their means position and kinetic energy increases and with fall in temperature their K.E. decreases.
  •  When a substance is heated
    (i) It expands i.e. a change in size takes place
    (ii) Change in temperature takes place.
    (iii) Change in state takes place.
  • CHANGE OF STATE : “The process of change from one state to another at a constant temperature is called change of state.”
  • Solid on heating changes into LIQUID. LIQUID on absorbing heat changes to VAPOURS some SOLIDS on heating DIRECTLY change in vapours called SUBLIMATION. Substance is called SUBLIMATE.
    SOLIDIFICATION on cooling when a vapours change into SOLID. GAS OR VAPOURS on cooling \(\xrightarrow { Condensation }\) changes to LIQUID also called LIQUIFACTION.
  •  MELTING: Change of solid into liquid at constant temperature. FUSION ⇒ FREEZING is change of LIQUID into SOLID at constant temperature and change of solid into liquid at a constant _ temperature is called FUSION.
  •  EVAPORATION: “Change liquid to gas at ALLTEMPERATURES” It is surface phenomenon. “
  • VAPOURIZATION : “Change of liquid into vapours at fixed temperature”.
  •  METING POINT: “Is the temperature at which a solid starts melting and remains constant till the whole of solid melts.”
    M.P. is same as freezing point.
    M.P. of ice is 0°C or freezing point, of water is 0°C.
  •  BOILING POINT: “Is the temperature of a liquid at which it start, boiling i.e. change into vapours or gaseous state.”
    B .P. of pure water is 100°C.
  • ABSOLUTE ZERO: “The temperature at which molecular motion completely ceases.”
  •  FACTORS EFFECTING THE RATE OF EVAPORATION :
    (i) Temperature: Increases with increase in temperature
    (ii) S.A.: Increases with increase in S.A.
    (iii) BLOWING AIR—Renewal of air increases evaporation.
    (iv) NATURE—Some liquids like spirit, alcohol, petrol evaporate easily.
  •  EVAPOURATION → produces coolness, BOILING produces Hotness.
  •  LINEAR EXPANSION: When a solid rod (metal) is heated change in length takes place, which depends upon
    (i) original length (L0)
    (ii) Increase in temperature
    (iii) Material of rod.
    Let L0 be the original length at 0°C, when heated to T°C final length becomes L
    Increase in length (Lt – L0) a L0 (T – 0)
    Or
    Coefficient of linear expension a which depends upon material of rod.
    Lt– L0 = L0 α T
    α = Lt – L/ LT = increase in length / original length × Rise in temperature
  •  When a metal plate is heated, change in area takes place and the expansion is called SUPERFICIAL expansion.
  •  When a solid of volume vis heated change in volume called cubical expansion takes place.
  • α : β : γ = 1 : 2 : 3

Test your self

A. Objective Questions

1. Write true or false for each statement

(a) Evaporation is rapid on a wet day.
Answer. False.

(b) Evaporation takes place only from the surface of liquid.
Answer. True.

(c) All molecules of a liquid take part in the process of evaporation.
Answer. False.

(d) Temperature of a liquid rises during boiling or vaporization
Answer. False.

(e) All molecules of a liquid take part in boiling.
Answer. True.

(f) Boiling is a rapid phenomenon.
Answer. True.

(g) All solids expand by the same amount when heated to the same rise in temperature.
Answer. False.

(h) Telephone wires are kept tight between the two poles in winter.
Answer. True.

(i) Equal volumes of different liquids expand by the different amount when they are heated to the same rise in temperature.
Answer. True.

(j) Solids expand the least and gases expand the most on being heated.
Answer. True.

(k) A mercury thermometer makes use of the property of expansion of liquids on heating.
Answer. True.

(l) Kerosene contracts on heating.
Answer. False.

2. Fill in the blanks 

(a) Boiling occurs at a fixed temperature.
(b) Evaporation takes place at all temperature.
(c) The molecules of liquid absorb heat from surroundings in evaporation.
(d) Heat is absorbed during boiling.
(e) Cooling is produced in evaporation.
(f) A longer rod expands more than a shorter rod on being heated to the same temperature.
(g) Liquids expand more than the solids.
(h) Gases expand more than the liquids.
(i) Alcohol expands more than water.
(j) Iron expands less than copper.

3. Match the following
Selina Concise Physics Class 8 ICSE Solutions Chapter 6 Heat Transfer 1

Selina Concise Physics Class 8 ICSE Solutions Chapter 6 Heat Transfer 2

4. Select the correct alternative 

(a) In evaporation

  1. all molecules of liquid begin to escape out
  2.  only the molecules at the surface escape out
  3.  the temperature of liquid rises by absorbing heat from surroundings.
  4.  the molecules get attracted within the liquid.

(b) The rate of evaporation of a liquid increases when :

  1.  temperature of liquid falls
  2.  liquid is poured in a vessel of less surface area
  3.  air is blown above the surface of liquid
  4.  humidity increases.

(c) During boiling or vaporization

  1.  all molecules take part
  2.  temperature rises
  3.  no heat is absorbed
  4.  the average kinetic energy of molecules increases.

(d) The boiling point of a liquid is increased by

  1.  increasing the volume of liquid
  2.  increasing the pressure, on liquid
  3.  adding ice to the liquid
  4.  decreasing pressure on liquid.

(e) Two rods A and B of the same metal, but of length 1 m and 2 m respectively, are heated from 0°C to 100°C. Then

  1.  both the rods A and B elongate the same
  2.  the rod A elongates more than the rod B
  3.  the rod B elongates more than the rod A
  4.  the rod A elongates, but the rod B contracts.

(f) Two rods A and B of the same metal, same length, but one solid and the other hollow, are heated to the same rise in temperature.
Then

  1.  the solid rod A expands more than the hollow rod B
  2.  the hollow rod B expands more than the solid rod A
  3.  the hollow rod B contracts, but the solid rod A expands
  4.  both the rods A and B expand the same.

(g) A given volume of alcohol and the same volume of water are heated from the room temperature to the same temperature then.

  1.  alcohol contracts, but water expands
  2.  water contracts, but alcohol expands
  3.  water expands more than alcohol
  4.  alcohol expands more than water.

(h) The increase in length of a metal rod depends on

  1.  the initial length of the rod only
  2.  the rise in temperature only
  3.  the material of rod only
  4.  all the above three factors.

(i) The correct statement is

  1.  Iron rims are cooled before they are placed on the cart wheels.
  2.  A glass stopper gets tighten on warming the neck of the bottle.
  3.  Telephone wires sag in winter, but become tight in summer.
  4.  A little space is left between two rails on a railway track.

Selina Concise Physics Class 8 ICSE Solutions – Electricity

Selina Concise Physics Class 8 ICSE Solutions – Electricity

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 8 Electricity. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Physics Chapter 8  Electricity

  •  ELECTRICITY: “Is the rate of flow of electrons”. i = \(\overset { q }{ \underset { t }{ – } } \)
  •  To keep electrons move, potential difference is needed. This is done by a cell or battery.
  •  Potential difference “is the amount of work done in moving a unit positive charge from one point to other.”
    Potential difference V = Work done (W) / charge moved (Q)
    Or W = QV but charge = it
    Hence, W = VIt or Electrical energy = VIt
  •  Power is “Rate of doing work”
    P = W/t = VIt/t = VI
    Power is measured in watt or J S-1
    1 Watt = 1 Volt × 1 Ampere
  • S.I. unit of charge is coulomb (C).
    S.I. unit of current is Ampere (A).
    S.I. unit of P.D. is volt.
    S.I. unit of electrical energy is Joule (J) and of power is watt (W)
    1 kWh-3600000J = 3.6 x 106 J
  •  ELECTRIC power is generated at the GENERATING STATION at 11000 volt, or 11 kV as these stations are at very far off place from areas where it is to be used. The voltage (A.C.) is of 50 HZ frequency.
  • AT GRID SUB-STATION this alternating current (A.C) voltage is stepped up from 11 kV to 132 kV to minise the loss of energy in transmission line wires.
  • At MAIN-SUB-STATION this voltage is stepped down from 132 kV to 33 kV and transmitted to city SUB-STATION.
  •  At CITY SUB STATION, it is further stepped down from 33 kV to 220 V for supply to hourses for consumers.
  •  Colour coding: Live wire — Red or Brown
    Neutral—Black or light blue
    Earth wire — Green or yellow
  •  1 kWh = 1 unit: Power Rating on an appliance 100 W – 220 V means the appliance when worked on a 220 V will consume 100 W electricity power
  • OVER LOADING: is the condition of Electric circuit, when it draws more current than it is designed for i. e. when a number of appliances are switched on at a time i.e. geyser, A.C. Electric motor etc. or a large number of plugs are put in the same socket.
  •  EARTHING: is done in a house near the kWh meter. Earthing is a safety device which puts the appliance at zero potential.
  •  SHORT CIRCUITING: If the insulation on the wire of cable used f in the wiring (or used with an appliance) breaks. The LIVE WIRE
    COMES IN CONTACT WITH THE NEUTRAL WIRE, this result in SHORT CIRCUITING
  •  FUSE: “Is a device used to limit the current in an electric circuit”. The use of fuse protects the appliance in circuit from being damaged Fuse is always connected in live wire. A fuse wire should have
    (i) High resistance
    (ii) Low melting point.
    These days miniature circuit breakers (MCB) are used. It is AUTOMATIC breaker, when current flowing excess.
  •  Appliances in a house are connected in parallel.

Test yourself

A. Objective Questions

1. Write true or false for each statement:

(a) A fuse wire has a high melting point.
Answer. False.

(b) Flow of protons constitutes electric current.
Answer. False.

(c) Silver is an insulator of electricity.
Answer. False.

(d) S.I. unit and commercial unit of electrical energy are same.
Answer. False.

(e) Overloading of electric current in circuits can lead to short circuiting.
Answer. True.

(f) Our body can pass electricity through it.
Answer. True.

(g) All metals are insulators of electricity.
Answer. False.

(h) The earth wire protects us from an electric shock.
Answer. True.

(i) A switch should not be touched with wet hands.
Answer. True.

(j) AH electrical appliances in a household circuit work at the same voltage.
Answer. True.

(k) In a cable, the green wire is the live wire.
Answer. False.

(l) A fuse is connected to the live wire.
Answer. True.

(m) A switch is connected to the neutral wire.
Answer. False.

2. Fill in the blanks

(a) The unit in which we pay the cost of electricity is kWh.
(b) The electrical energy consumed in a house is measured by kWh meter.
(c) In a household electrical circuit, the appliance are connected in parallel with the mains.
(d) A switch is connected to the live wire.
(e) The red colour insulated wire in a cable is the live wire.
(f) One kilowatt hour is equal to 3.6 x 106 joule.
(g) A fuse wire should have low melting point.

3. Match the following
Selina Concise Physics Class 8 ICSE Solutions Chapter 8 Electricity 1

Selina Concise Physics Class 8 ICSE Solutions Chapter 8 Electricity 2

4. Select the correct alternative

(a) All wires used in electric circuits should be covered with

  1.  colouring material
  2.  conducting material
  3.  an insulating material
  4.  none of the above

(b) Electric work done per unit time is

  1.  electrical energy
  2.  electric current
  3.  electric voltage
  4.  electrical power

(c) One kilowatt ¡s equal to

  1.  100 watt
  2.  1000 watt
  3.  10 watt
  4.  none of these

(d) Fuse wire is an alloy of

  1.  tin-lead
  2.  copper-lead
  3.  tin-copper
  4.  lead-silver

(e) A fuse wire should have

  1.  a low melting point
  2.  high melting point
  3.  very high melting point
  4.  none of the above

(f) When switch of an electric appliance is put off, it disconnects

  1.  the live wire
  2.  the neutral wire
  3.  the earth wire
  4.  the live and the neutral wire

(g) The purpose of an electric meter in a house is

  1.  to give the cost of electricity directly
  2.  to give the consumption of electrical energy
  3.  to safeguard the circuit from short circuiting
  4.  to put on or off the mains.

(h) If out of the two lighted bulbs in a room, one bulb suddenly fuses, then

  1.  other bulb will glow more
  2.  other bulb will glow less
  3.  other bulb will also fuse
  4.  other bulb will remain lighted unaffected.

Selina Concise Physics Class 8 ICSE Solutions – Force and Pressure

Selina Concise Physics Class 8 ICSE Solutions – Force and Pressure

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 3 Force and Pressure. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions For Class 8 Physics Chapter – 3 – Force and Pressure

  • FORCE : “Is the cause which changes the state of a body (rest or
    state of motion) or it changes the size or shape of a body”.

Weight of a body → The force with which a body is attracted towards the centre of earth mg = force of gravity.

  • A force does not change the mass of the body that is why mass of a
    body on earth and moon in same but weight → force exerted on
    body is different.
  •  Force cannot be seen but it is felt.
  • → represents force, length of arrow gives magnitude and arrow points the direction.
  •  S.I. unit of force is Newton (N).
  • Newton: “Is that much force, which when acting on a body of mass 1 kg produces in it (increases) a speed of 1 M s-1 in the direction of its motion.
  • 1 kgf = 9.8 N = 10 N (nearly)
  • RIGID body: “When a force is applied on a body and inter-spacing between its constituent particles do not change is called RIGID body” force can cause only the motion in it.
  •  NON-RIGID body: “When force applied changes inter-spacing.” Force causes both change in its size (shape) and the motion in body.
  • TURNING EFFECT: “When force is applied on a pivoted (at a point) body, it can turn it and turning of body about point of rotation is called TURNING EFFECT.” or Moment of force.
    This is measured as:
    TURNING EFFECT = MOMENT OF FORCE
    Force x perpendicular distance from point of rotation.
    Selina Concise Physics Class 8 ICSE Solutions Chapter 3 Force and Pressure 1
    Moment of force = F x OP
    S.I. unit of moment of force=N × m=Nm
  • THRUST: “Force acting normally on a surface.” Smaller the area of surface, larger is thrust.
  • PRESSURE : “Thrust per unit area.p = Thrust/area = F/A S.I unit of area A pressure is Nm-2 or pascal (Pa)
  •  If Thrust is measured in kgf and area in Cm2, then pressure is expressed as kgf Cm-2.
    ATMOSPHERIC Pressure: 1 atm = 76 cm of mercury column 1 atm = 1.013 x 105 Pa
  •  FACTORS AFFECTING THE PRESSURE : P = F/A
    (i) Area: Greater the area, lesser is pressure and lesser area, greater is pressure.
    (ii) Magnitude of thrust acting: greater thrust, greater pressure.
  • Factors Affecting LIQUID PRESSURE = hdg
    (i) High of liquid column: increases with height
    (ii) Density of liquid: increases with density of liquid.
    (iii) Gravity constant.
  • ATMOSPHERIC PRESSURE : “Pressure exerted by the air of atmosphere around us.”

STANDARD ATMOSPHERIC PRESSURE 1 Atm = 76 cm of Hg column = 1.013 x 10Pa

 

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) The S.I. unit of force is kgf.
Answer. False.
The S.I. unit of force is newton.

(b) A force always produces both the linear and turning motions.
Answer. False.

(c) Moment of force = force × perpendicular distance of force –  from the pivoted point.
Answer. True.

(d) Less force is needed when applied at a farther distance from the pivoted point.
Answer. True.

(e) For a given thrust, pressure is more on a surface of large j area.
Answer. False.
For a given thrust, pressure is less on a surface of large area.

(f) The pressure on a surface increases with an increase in the thrust on the surface.
Answer. True.

(g) A man exerts same pressure on the ground whether he is standing or he is lying.
Answer. False.
A man exerts different pressure on the ground whether he is  standing or he is lying.

(h) It is easier to hammer a blunt nail into a piece of wood than a sharply pointed nail.
Answer. False.
It is not easier to hammer a blunt nail into a piece of wood than a sharply pointed nail.

(i) The S.I. unit of pressure is pascal.
Answer. True.

(j) Water in a lake exerts pressure only at its bottom.
Answer. False.

(k) A liquid exerts pressure in all directions.
Answer. True.

(l) Gases exert pressure in all directions.
Answer. True.

(m) The atmospheric pressure is nearly 105 Pa.
Answer. True.

(n) Higher we go, greater is the air pressure.
Answer. False.

2. Fill in the blanks

(a) 1 kgf = 10 N (nearly).
(b) Moment of force = force × distance of force from the point of turning
(c) In a door, handle is provided farthest from the hinges.
(d) The unit of thrust is newton .
(e) Thrust is the normal force acting on a surface.
(f) Pressure is the thrust acting on a surface of unit area.
(g) The unit of pressure is pascal
(h) Pressure is reduced if area of surface increases.
(i) Pressure in a liquid increases with the depth.
(j) The atmospheric pressure on earth surface is nearly 105 Pa.

3. Match the following

Selina Concise Physics Class 8 ICSE Solutions Chapter 3 Force and Pressure 2

4. Select the correct alternative

(a) SI. unit of moment of force ¡s

  1. N
  2.  N cm
  3.  kgfm
  4.  N m

(b) To obtain a given moment of force for turning a body, the force needed can be decreased by

  1. applying the force at the pivoted point
  2.  applying the force very close to the pivoted point
  3.  applying the force farthest from the pivoted point
  4.  none of the above

(c) The unit of thrust is

  1.  kgf
  2.  kg
  3.  g
  4. m s-1

(d) The unit of pressure is

  1.  N × m
  2.  kgf
  3.  N m-2
  4.  kgf m2

(e) The pressure and thrust are related as

  1.  Pressure = Thrust
  2.  Pressure = Thrust x Area
  3.  Pressure = Thrust / Area,
  4.  Pressure = Area / Thrust

(f) A body weighing 5 kgf, placed on a surface of area 0.1 m2, exerts a thrust on the surface equal to

  1.  50 kgf
  2.  5 kgf
  3.  50 kgf  m-2
  4.  5 kgf  m-2

P.Q. A body weighing 5 kgf, placed on a surface of area 0.1 m2, exerts a pressure on the surface equal to

  1.  50 kgf
  2.  5 kgf
  3.  50 kgf m-2
  4.  5 kgf m-2

(g) The feet of lizards act like

  1.  moving pads
  2.  drilling pads
  3.  suction pads
  4.  none of the above

(h) Pressure exerted by a liquid is due to its

  1.  weight
  2.  mass
  3. volume
  4.  area

(i) Pressure inside a liquid increases with :

  1.  increase in depth
  2.  decrease in depth
  3.  decrease in density
  4.  none of the above

(j) The atmospheric pressure at sea level is nearly

  1.  10 Pa
  2.  100,000 Pa
  3.  100 Pa
  4.  10,000 Pa

(k) Nose bleeding may occur at a high altitude because

  1.  the atmospheric pressure decreases
  2.  the oxygen content of atmosphere decreases
  3.  the atmospheric pressure increasess
  4.  there are strong air currents at the high altitude

Selina Concise Physics Class 8 ICSE Solutions – Physical Quantities and Measurement

Selina Concise Physics Class 8 ICSE Solutions – Physical Quantities and Measurement

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Selina Concise ICSE Solutions For class 8 Physics chapter 2 – Physical Quantities and Measurement

  • MASS Is the quantity of matter contained in a body.
    VOLUME is the space occupied by body.
  • Equal mass of IRON and cotton, iron will have less volume than cotton.
  • Equal volume of Iron and cotton, the mass of iron is more than mass of cotton, because iron denser than cotton.
  • DENSITY “Is ratio of mass of substance to volume of substance”
    D = M/V = KG/M3 The SI. unit of density is kg M-3
  • Density of a substance does not change with change in shape or size.
  • When a substance is heated it expands and volume increases. Hence density decreases.
    Water has maximum density at 4°C i.e. density of water increases from 0°C to 4°C and decreases above 4°C.
  • Volume of substance is measured by formula V = L × B × H or 4/3 or by measuring cylinder.
  • Mass is measured by beam balance or spring balance.
  • RELATIVE DENSITY of substance is the density compare with water i.e. How many times the substance is DENSER than water. Since density of water is 1  Gcm-3, so density of a substance in Gcm-3 = relative density of substance.
    S.I. unit of R.D. > has no units — since it is the ratio of same
    quantities.
  • If a substance has density more than liquid it SINKS in the liquid and if the density of substance is LESS than liquid it floats on liquid.
  • BUOYANT FORCE “The force exerted by liquid acting vertically
    upward on a body and is equal to the weight of liquid displaced by its immersed part.”
  • Weight of body Acting vertically downward. This force has the tendency to sink the body.
  •  LAW OF FLOATATION “When a body floats in a liquid, the weight of the liquid displaced by its immersed part is equal to the total weight of the body.” While floating
    wt. of floating body W=wt. of liquid displaced by its immersed part FB i.e. Apparent wt. of floating body is zero.
    Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 1

Density of body is greater than density of liquid. The body sinks.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 2

Density of body is equal to the density of liquid. The body float where ever it is left in liquid.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 3

Density of body is less than density of liquid. The body rises to the surface and floats.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) Equal volumes of the two different substances have equal masses.
Answer False.
Equal volumes of the two different substances have different masses.

(b) The density of a piece of brass will change by changing its size or shape.
Answer False.

(c) The density of a liquid decreases with increase in its temperature.
Answer True.

(d) Relative density of water is 1.0.
Answer True.

(e) Relative density of a substance is expressed in g cm-3.
Answer False.
Relative density of a substance has no units.

(f) When a body is immersed in a liquid, the buoyant force experienced by the body is equal to the volume of the liquid displaced by it.
Answer False.
The buoyant force is equal to the weight of the liquid displaced by the immersed part of body.

(g) A body experiences the same buoyant force while floating in watr or alcohol.
Answer True.

(h) A body experiences the same buoyant force when it floats or sinks in water.
Answer False.

(i) A body floats in a liquid when its weight becomes equal to the weight of the liquid displaced by its submerged part. .
Answer True.

(j) A body while floating, sinks deeper in a liquid of low density than in a liquid of high density.
Answer True.

2. Fill in the blanks

(a) 1 kg is the mass of 1000 ml of water at 4°C.
(b) Mass = density x volume.
(c) The S.I. unit of density is Kg m-3
(d) Density of water is 1000 Kg m-3.
(e) 1 g cm-3 = 1000 Kg m-3.
(f) The density of a body which sinks in water is more than 1000 Kg m-3.
(g) Abody sinks in a liquid A, butt floats in a liquid B. The density of liquid A is less than the density of liquid B.
(h) A body X sinks in water, but a body Y floats on water. The density of the body X is more than the density of body Y.
(i) The buoyant force experienced by a body when floating in salt¬water is equal to or same that of when floating in pure water.
(j) The weight of a body floating in a liquid is zero.

3. Match the following

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 4

4. Select the correct alternative 

(a) The correct relation is

  1. Density = Mass x Volume
  2. Mass = Density x Volume
  3.  Volume = Density x Mass
  4. Density = Mass + Volume

(b) The relative density of alcohol is 0.8. Its density is

  1.  0.8
  2. 800 kg nr3
  3.  800 g cm-3
  4. 0.8 kg m-3

(c) A block of wood of density 0.8 g cm-3 has a volume of 60 cm3. The mass of block is

  1.  60.8 g
  2.  75 g
  3. 48 g
  4. 0.013 g

(d) The density of aluminium is 2.7 g cm-3 and that of brass 8.4 g cm’3. The correct statement is

  1.  Equal masses of aluminium and brass have equal volumes
  2. The mass of a certain volume of brass is more than the mass of equal volume of aluminium.
  3.  The volume of a certain mass of brass is more than the volume of equal mass of aluminium.
  4.  Equal volumes of aluminium and brass have equal masses.

(e) A density bottle has a marking 25 mL on it. It means that:

  1.  the mass of density bottle is 25 g
  2. the density bottle will store 25 ml of any liquid in it
  3.  the density bottle will store 25 ml of water, but more volume of liquid denser than water.
  4.  the density bottle will store 25 ml of water, but more volume of a liquid lighter than water.

(f) The correct statement is

  1.  The buoyant force on a body is equal to the volume of the liquid displaced by it ‘
  2. The buoyant force on a body is equal to the volume of the body
  3. The buoyant force on a body is equal to the weight of the liquid displaced by it
  4.  The buoyant force on a body is always equal to the weight of the body.

(g) A piece of wood floats on water. The buoyant force on wood will be

  1.  zero
  2.  more than the weight of the wood piece
  3. equal to the weight of the wood piece
  4. less than the weight of the wood piece.

(h) The weight of a body is more than the buoyant force experienced by it, due to a liquid. The body will

  1. sink
  2.  float with its some part outside the liquid
  3.  float just below the surface of liquid
  4. float with whole of its volume above the surface of liquid.

B. Short/Long Ans Questions 

Question 1.
Define the term density of a substance.
Answer:
Density of a substance is defined as “Mass per Unit volume”.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 5

Question 2.
Name the S.I. unit of density. How is it related to g Cm-3 ?
Answer:
S.I. unit of density is kg M-3 In C.GS. system unit of mass is g and unit of volume is Cm3, so CGS unit of density is g Cm-3 (gram per cubic centimetre)
Relationship between S.I. and C.GS. units

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 6

Question 3.
The density of brass is 8.4 g cm-3. What do you mean by this statement ?
Answer:
This statement meAns one cubic centimetre volume of brass has mass of 8.4 g.

Question 4.
Arrange the following substances in order of their increasing density:
Iron, Cork, Brass, Water, Mercury.
Answer:
Cork, Water, Iron, Brass, Mercury.

Question 5.
How does the density of a liquid (or gas) vary with temperature?
Answer:
Most of the liquids increase in volume with increase in temperature, but water shows anomalous behaviour. Water has maximum volume at 4°C and maximum density at 4°C.
Actually, when volume increases density decreases and when volume decreases the density increases.
But water when cooled from a high temperature, contracts upto 4°C because volume decreases and expands when cooled further below 4°C and hence density of water increases when it is cooled upto 4°C while decreases when cooled further below 4°C. In other words, the density of water is maximum at 4°C equal to 1 g Cm-3 or lOOO kg m-3.

Question 6.
A given quantity of a liquid is heated. Which of the following quantity will vary and how ?
(a) mass, (b) volume and (c) density
Answer:
When a given quantity of liquid is heated
(a) Mass : does not change.
(b) Volume: changes and increases with rise in temperature.
(c) Density : Changes and decreases.
Density = Mass / volume

Question 7.
Describe an experiment to determine the density of the material of a coin.
Answer:
Density = Mass / volume
To find the density of the material of a coin, we need to find its (i) mass—by common beam balance and (ii) Its volume by measuring cylinder.
Measure the mass of coin.
EXPERIMENT – Let the mass of coin shown by beam balance = M (gram) = 50 g (ray)
Measure the vol. of coin.
Initial volume of water = V1 = 40 ml (say)
Final volume of water
When coin is added in the cylinder=V2 = 50 ml (say)
Then vol. of coin = V2 – V1 = 50 – 40 = 10 ml

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 7

Question 8.
Describe an experiment to determine the density of a liquid.
Answer:
To determine the density of a liquid D = M / V
We need to find (i) the vol. of liquid say milk, (ii) mass of liquid.
EXPERIMENT:
(i) To find the mass of milk:
wt. of empty 100 c.c beaker = Mg = 70 g (say)
Fill the beaker (half) with milk and weigh again=M2 g = 116 g (say)
(ii) To find the vol. of milk:
TrAnsfer this milk into measuring cylinder and note the volume V = 40 c.c (say)

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 8
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 9

Question 9.
What is a density bottle ? How is it used to find the density of a liquid ?
Answer:
DENSITY bottle is a small glass bottle having a glass stopper at its neck. The bottle can store a fixed volume of a liquid. Generally the volume of bottle is 25 ml or 50 ml. Stopper has a narrow hole through it. When bottle is filled with liquid and stopper is inserted, THE EXCESS LIQUID RISES THROUGH THE HOLE and drains out. Thus the bottle will contain the same volume of liquid each time when it is filled. It is used to determine the density of a liquid.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 10

 

Question 10.
Define the term relative density of a substance.
Answer:
RELATIVE DENSITY: “is the ratio of density of a substance to the density of water at 4° C.”
Or
RELATIVE DENSITY “is theratio of mass of the substance to the mass of an equal volume of water at 4° C.”

Question 11.
What is the unit of relative density ?
Answer:
UNIT OF RELATIVE DENSITY: No units since it is a pure ratio.

Question 12.
Distinguish between density and relative density.
Answer:
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 11

Question 13.
Explain the meaning of the statement ‘relative density of aluminium is 2.7’ ?
Answer:
The statement ‘Relative density of aluminium is 2.7’ meAns .
A piece of aluminium of any volume has mass 2.7 times that of an equal volume of water.
i.e. Aluminium is 2.7 times heavier than water.

Question 14.
How does the density of a body and that of a liquid determine whether the body will float or sink into that liquid ?
Answer:
If the density of a body is LESS than the density of LIQUID, the body will FLOAT on the surface of liquid.
If the density of a body is MORE than the density of liquid, the body will SINK in a liquid.

Question 15.
A cork piece floats on water surface while an iron nail sinks in it. Explain the reason.
Answer:
CORK floats on water meAns density of cork is LESS than density of water.
IRON nail: Sinks in water meAns density of iron nail is MORE than density of water.

Question 16.
Which of the following will sink or float on water ? (Densityof water = 1 g Cm-3)
(a) body A having density 500 kg m-3
(b) body B having density 2520 kg m-3
(c) body C having density 1100 kg m-3
(d) body D having density 0.85 g m-3
Answer:
Density of water = 1 g Cm-3
(a) Density of body A = 500 kg m-3 = 500 × = 0.5 = 0.5 g Cm-3
Density of body A ¡s less than density of water hence A will float on water
(b) Density of body B = 2520 kg m-3 = 2520 × 1/1000 = 2.52 g Cm-3
Density of body B is more than density of water and hence B will SiNK in water
(c) Density of body C = 1100kg m-3 = 1100 × 1/1000 = 1.1 g Cm-3
is greater than water.
Hence, body C will sink in water.
(d) Density of body D = 0.85 g Cm-3 < 1.0 g Cm-3
Density of body D is less than the density of water hence body D will FLOAT on water

Question 17.
What is the iaw of floatation ?
Answer:
When a body floats in a liquid, the weight of the liquid displaced by its immersed part is equal to the total weight of the body. This is the law of floatation, i.e. while floating. Weight of the floating body = Weight of the liquid displaced by its immersed part.

Question 18.
The density of water is 1.0 g Cm-3. The density of iron is 7.8 × 10″3 g Cm-3. The density of mercury is 13.6 g Cm-3.
Ans the following:
(a) Will a piece of iron float or sink in water ?
(b) Will a piece of iron float or sink in mercury ?
Answer:
Density of water 1.0 g Cm-3
(a) Density of piece of iron = 7.8 × 10-3 g Cm-3
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 12
∴ Density of piece of iron is LESS than density of water.
Hence, piece of iron will FLOAT in water.
(b) Density of piece of iron = 7.8 × 10-3
Density of mercury is 13.6 × 10-3 g Cm-3
Since 7.8 × 10-3 < 13.6 × 10-3
∴ Density of piece of iron is LESS than density of mercury
∴ Piece of iron will FLOAT in mercury

Question 19.
The diagram given below show a body floating in three different liquids. A, B and C at different levels.
(a) In which liquid does the body experience the greatest buoyant force ?
(b) Which liquid has the least density ?
(c) Which liquid has the highest density ?

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 13Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 14
Answer:
(a) Buoyant force is same in each case as the wt. of body is same in each case and Buoyant force is equal to the weight of liquid displaced by the immersed part of body which balances the wt. of body.
(b) The liquid A has the least density as body immerses the maximum.
(c) Liquid C has the highest density as the body immerses the least.

Question 20.
For a floating body, how is its weight related to the buoyant force ?
Answer:
When a body floats in a liquid. The weight of the liquid displaced by its immersed part is equal to the total weight of the body.

Question 21.
Why does a piece of ice float on water ?
Answer:
FLOATATION OF ICE ON WATER : Density of 0.9 g Cm-3 is less than density of water 1 g Cm-3. Hence, ice floats on water.

Question 22.
Explain why an iron needle sinks in water, but a ship made of iron floats on water.
Answer:
Density of iron is more than density of water, ∴ weight of iron nail is more than wt. of water displaced by it and nail SINKS. While shape of iron ship is made in such a way that it displaces MORE WEIGHT OF WATER than its own weight. Secondly the ship is HOLLOW and THE EMPTY SPACE contains AIR which makes the AVERAGE DENSITY OF SHIP LESS THAN THAT OF WATER and hence ship floats on water.

Question 23.
It is easier to swim in sea water than in river water. Explain the reason.
Answer:
Density of sea water is greater than density of river water, [because of impurities]
(i) In each case the weight of water displaced will be equal to the weight of the man.
∴ Ratio of weight of sea water and river water displaced by man is 1: 1.
(ii) With smaller portion of man’s body submerged in sea water, the wt. of sea water displaced is equal to the total weight of body. While to displace the same weight of river water, a larger portion of the body will have to be submerged ¡n water.
∴ It is easier for man to swim in sea water.

Question 24.
Icebergs floating on sea water are dangerous for ships. Explain the reason.
Answer:
ICEBERGS are very dangerous for ships as ICEBERGS are huge masses of ice floating in sea [density of ice being 0.917 g Cm-3]
with about 9/10 portion below water and only 1/10 portion of it above surface of water.

Question 25.
Explain why it is easier to lift a stone under water than in air.
Answer:
In water, the stone experience a buoyant force which counter balances the weight of the stone acting downward and this makes the stone lighter and thus easier to lift the stone in water.

Question 26.
What is a submarine ? How can it be made to’dive in water and come to the surface of water.
Answer:
SUBMARINE: Submarine is a water-tight boat which can travel under water like a ship. It is providgd with water tanks. When submarine is to dive, water is filled in water tanks and it is made heavier and average density of submarine becomes greater than the density of sea water and it sinks. To make the submarine rise to the surface of water, water tanks are emptied and average density.of submarine becomes less than the density of sea water and it rises to surface of water.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 15

While submarine is underwater soldiers can see the enemy activities through periscope.

Question 27.
A balloon filled with hydrogen rises in air. Explain the reason.
Answer:
A balloon filled with hydrogen rises to a certain height as it displaces more wt. of air than wt. of balloon but as it rises higher density of air DECREASES there and upthrust becomes less and ultimately upthrust becomes equal to the weight of balloon and balloon stops rising further.

C. Numericals

Question 1.
The density of air is 1.28 g/Iitre. Express it in:
(a) g cm3 (b) kg m
Answer:
(a) The density of air is I .28g/litre
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 16
Question 2.
The dimensions of a hail are 10 m × 7 m × 5 m. If the density of air is 1.11 kg m-3, find the mãss of air in the hail.
Answer:
The dimensions of hall 10m × 7m × 5m
i.e. V350 m3
Density of air(D)= 1.11 kg m-3
M = V × D 350 ×  1.11 =388.5 kg

Question 3.
The density of aluminium is 2.7 g cm3. Express it in kg m-3
Answer:
Density of aluminium = 2.7 g/Cm3

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 17
Question 4.
The density of alcohol is 600 kg m-3. Express it in g Cm-3.
Answer:
Density of alcohol is = 600 kg/m-3

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 18
Question 5.
A piece of zinc of mass 438.6 g has a volume of 86 Cm3. Calculate the density of zinc.
Answer:

Mass of Zinc (M) = 438.6 g
Volume V = 86 Cm3
Density (D) = ?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 19
Question 6.
A piece of wood of mass 150 g has a volume of 200 Cm3. Find the density of wood ¡n
(a) C.GS. unit, (b) S.l. unit
Answer:
(a) Mass of wood (M) = 150 g
Volume of wood (V) = 200 Cm3
Density (D) =?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 20
(b) In S.I. system = 0.75 × 1000 750 kg/ m3

Question 7.
Calculate the volume of wood of mass 6000 kg if the density of wood is 0.8 g Cm-3
Answer:
Volume of wood (V) = ?
Mass of wood (M) = 6000 kg
Density of wood D = 0.8 g/ Cm3
D=O.8g/Cm3=o.8 × IOOO = 800kg /m3
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 21

Question 8.
Calculate the density of solid from the following data :
(a) Mass of solid = 72 g
(b) Initial volume of water in measuring cylinder = 24 ml
(c) Final volume of water when solid is completely immersed in water = 42 ml
Answer:
Mass of solid (M) = 72 g
Intial volume of water V1 = 24 ml
Final volume of water V2 = 42 ml
Volume of solid (V) = V2 – V1 = 42 – 24 = 18 Cm3
Density of solid (D) = ?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 22
Question 9.
The mass of an empty density bottle is 21.8 g, when filled completely with water is 41.8 g and when filled completely with liquid it is 40.6 g. Find :
(a) the volume of density bottle
(b) the relative density of liquid
Answer:
Density of water is 1 g Cm3
∴ Volume of density bottle = weight of water in grams completely filling the bottle
(a) Volume of density bottle:
Mass of empty density bottle = M1 =21.8 g
Mass of bottle + water = M2  41.8 g
∴ Mass of water completely fih1ig the density bottle = M2 — M1
=41.8 —21.8
20g
But 1 g of water has volume = 1 cc
∴ Volume of bottle (density bottle) = volume of water =20 c.c. =20 ml
(b) The relative density of liquid:
Mass of 20 c.c. of liquid = (mass of density bottle + mass of 20 c.c of liquid- mass of density bottle)
= 40.6—21.8
= 18.8 g
Mass of 20 C.C of water = 20g
Relative density of liquid
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 23
Question 10.
From the following observations, calculate the density and relative density of a brine solution. Mass of empty density
bottle = 22 g
Mass of bottle + water = 50 g
Mass of bottle + brine solution = 54 g

Answer:
Mass of empty bottle, M1 = 22 g
Mass of bottle + water, M2 =50 g
Mass of bottle + brine solution, M3 =54 g
Mass of water = M2 — M1 =50—22=28 g
Mass of brine solution = M3 — M1 54—22 = 32 g
Density of brine solution = Mass of brine solution / Mass of water
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 24
Question 11.
The mass of an empty density bottlfe is 30 g, it is 75 g when filled completely with water and 65 g when filled completely with a liquid. Find :
(a) volume of density bottle,
(b) density of liquid, and
(c) relative density of liquid.
Answer:
Mass of empty density bottle (M1) =30 g
Mass of bottle + Water (M2) 75 g
Mass of liquid + Liquidx (M3)= 65 g
Mass ofwater=M2—M1=75—30=45 g
(a) Volume of density bottle = Mass of water 45 g
(b) Density of Iiquid x = ?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 25
(c) Mass of water in the density bottle =75 — 30 = 45 g
∴ Volume of water in density bottle = 45 cc
and mass of equal volume of liquid in density bottle 65—30 = 35g
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 26

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Practicing OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(f) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Question 1.
Find the square root of the following complex numbers.
(i) 3 + 4i
(ii) – 8 + 6i
(iii) – 40 – 42i
(iv) i
(v) \(\left(\frac{2+3 i}{5-4 i}+\frac{2-3 i}{5-4 i}\right)\)
Solution:
(i) \(\sqrt{3+4 i}\) = x + iy; where x, y ∈ R
On squaring both sides ; we have
3 + 4i = (x + iy)²
⇒ 3 + 4i = x² – y² + 2ixy
On comparing real and imaginary parts on both sides ; we have
x² – y² = 3 …(1)
and 2xy = 4 …(2)
∴ x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{3^2+4^2}\) = 5 … (3)
On adding (1) and (3); we have
2x² = 8 ⇒ x² = 4 ⇒ x = ± 2
eqn. (3) – eqn. (1) gives ;
2y² = 2 ⇒ y = ± 1
Since xy be +ve
∴ x and y both are of same sign
Hence x = 2, y = 1 or x = – 2, y = – 1
∴ \(\sqrt{3+4 i}\) = 2 + i or – (2 + i)

(ii) \(\sqrt{- 8 + 6 i}\) = x + iy where x, y ∈ R
On squaring both sides ; we have
– 8 + 6i = (x + iy)² = x² – y² + 2ixy
On ωmparing real and imaginary parts on both sides, we get
x² – y² = – 8 … (1)
and 2xy = 6 … (2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{(-8)^2+6^2}\)
= \(\sqrt{64+36}\) = 10 … (3)
On adding (1) and (3) ; we have
2x² = 2 ⇒ x = ± 1
eqn. (3) – eqn. (1) gives ;
2y² = 18 ⇒ y² = 9 ⇒ y = ± 3
Since xy be +ve ∴ both x and y are of same sign.
∴ x = 1; y = 3 or x = – 1, y = – 3
Thus \(\sqrt{-8+6 i}\) = 1 + 3i or – (1 + 3i)

(iii) \(\sqrt{-40-42 i}\) = x – iy where x, y ∈ R
On squaring both sides ; we have
– 40 – 42i = (x – iy)² = x² – y² – 2ixy
On comparing real and imaginary parts on both sides ; we have
– 40 = x² – y² …(1)
and 2xy = 42 …(2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{(-40)^2+42^2}\)
= \(\sqrt{1600+1764}\)
= \(\sqrt{3364}\) = 58 … (3)
On adding (1) and (3) ; we have
2x² =18 ⇒ x = ± 3
eqn. (3) – eqn. (1) gives ;
2y² = 98 ⇒ y² = 49 ⇒ y = ± 7
Since xy be of +ve sign ∴ both x and y are of same sign
i.e. when x = 3, y = 7 or x = – 3, y = – 7
Thus, \(\sqrt{-40-42 i}\)
= 3 – 7i or – 3 + 7i
= ±(3 – 7i)

(iv) Let \(\sqrt{i}\) = x + iy;
On squaring; we have i = (x + iy)² = x² – y² + 2ixy
On comparing real and imaginary parts on both sides, we have
x² – y² = 0 …(1)
and 2xy = 1 …(2)
∴ x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{0^2+1^2}\) = 1 … (3)
On adding eqn. (1) and eqn. (3); we have
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f) 1

Question 2.
If ω is a cube root of unity, then
(i) ω + ω² = ….
(ii) 1 + ω = …
(iii) 1 + ω³ = ….
(iv) ω³ = ….
Solution:
Now ω be the cube root of unity
∴ ω = \(\frac{-1+\sqrt{3} i}{2}\) and ω² = \(\frac{-1-\sqrt{3} i}{2}\)

(i) 1 + ω² = \(\frac{-1+\sqrt{3} i}{2}+\frac{-1-\sqrt{3} i}{2}\)
= \(\frac { -2 }{ 2 }\) = – 1

(ii) 1 + ω = 1 + \(\frac{-1+\sqrt{3} i}{2}=\frac{2-1+\sqrt{3} i}{2}=\frac{1+\sqrt{3} i}{2}\) = – ω²

(iii) 1 + ω² = 1 + \(\frac{-1-\sqrt{3} i}{2}=\frac{2-1-\sqrt{3} i}{2}=\frac{1-\sqrt{3} i}{2}=-\left(\frac{-1+\sqrt{3} i}{2}\right)\) = – ω

(iv) ω³ = ω². ω = \(\left(\frac{-1-\sqrt{3} i}{2}\right)\left(\frac{-1+\sqrt{3} i}{2}\right)=\frac{(-1)^2-(\sqrt{3} i)^2}{4}=\frac{1+3}{4}\) = 1

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Question 3.
If 1, ω, ω² are three cube roots of unity, prove that
(i) (1 + ω²)4 = ω
(ii) (1 + ω – ω²)³ = (1 – ω + ω²)³ = – 8
(iii) (1 – ω) (1 – ω²) = 3
(iv) \(\frac{1}{1+\omega}+\frac{1}{1+\omega^2}\) = 1
Solution:
(i) (1 + ω²)4 = (- ω)4= ω³ . ω = ω [∵ 1 + ω + ω² = 0 and ω³ = 1]

(ii) (1 + ω – ω²)² = (- ω² – ω²)³ = (- 2ω²)³ = – 8ω6 = – 8 (ω³)² = – 8
and (1 – ω + ω²)³ = (- ω – ω)³ = (- 2ω)³ = – 8ω³ = – 8 x 1 = – 8 [∵ 1 + ω + ω² = 0 and ω³ = 1]

(iii) (1 – ω) (1 – ω²) = 1 – ω – ω² + ω³ = 1 – (ω + ω²) + 1 [∵ ω³ = 1]
= 1 – (- 1) + 1 [∵ 1 + ω + ω² = 0]
= 1 + 1 + 1 = 3

(iv) \(\frac{1}{1+\omega}+\frac{1}{1+\omega^2}=\frac{1}{-\omega^2}+\frac{1}{-\omega}\) [∵ 1 + ω + ω² = 0]
= \(\frac{1+\omega}{-\omega^2}=\frac{-\omega^2}{-\omega^2}\) = 1

Question 4.
(i) (1 – ω – ω²)6 = 64
(ii) (1 + ω – ω²) (1 – ω + ω²) = 4
Solution:
(i) (1 – ω – ω²)6 = [1 – (ω + ω²)]6 = [1 – (- 1)]6 = 26 = 64 [∵ 1 + ω + ω² = 0]

(ii) (1 + ω – ω²) (1 – ω + ω²) = (- ω² – ω²) (- ω – ω) [∵ 1 + ω + ω² = 0]
= (- 2ω²) (- 2ω)
= 4ω³
= 4 x 1 = 4

Question 5.
(3 + 5ω + 3ω²)6 = (3 + 5ω² + 3ω)6 = 64
Solution:
(3 + 5ω + 3ω²)6 = (3 + 5ω² + 3ω)6
= [3 (- ω) + 5ω]6 [∵ 1 + ω + ω² = 0]
= (2ω)6 = 64 (ω³)² = 64
and (3 + 5ω² + 3ω)6 = [3 (1 + ω) + 5ω²]6 = [3 (- ω²) + 5ω²]6 [∵ 1 + ω + ω² = 0]
= (2ω²)6 = 64ω12 = 64 (ω³)4 = 64 [∵ ω³ = 1]

Question 6.
ω28 + ω29 + 1 = 0
Solution:
ω28 + ω29 + 1 = (ω³)9 . ω + (ω³)9 . ω² + 1
= 19 . ω + 19 . ω² + 1 [∵ ω³ = 1]
= ω + ω² + 1 = 0

Question 7.
Prove that \(\left(\frac{-1+i \sqrt{3}}{2}\right)^n+\left(\frac{-1-i \sqrt{3}}{2}\right)^n\) is equal to 2 if n be a multiple of 3 and is equal to – 1 if n be any other integer.
Or
If 1, ω, ω² are the cube roots of unity, prove that ωn + ω2n = 2 or – 1 acωrding as n is a multiple of 3 or any other integer.
Solution:
We know that cube root of unity are 1, ω, ω²
where ω = \(\frac{-1+\sqrt{3} i}{2}\) and ω² = \(\frac{-1-\sqrt{3} i}{2}\)
∴ \(\left(\frac{-1+i \sqrt{3}}{2}\right)^n+\left(\frac{-1-i \sqrt{3}}{2}\right)^n=\omega^n+\left(\omega^2\right)^n=\omega^n+\omega^{2 n}\)

Case-I : When n be a multiple of 3
∴ n = 3k
∴ ωn + ω2n = ω3k + ω6k = (ω³) + (ω³)2k = 1k + 12k =1 + 1 = 2

Case-II : When n be not a multiple of 3
∴ n = 3k + 1, 3k + 2

Subcase – I : When n = 3k + 1
ωn + ω2n = ω3k+1 + ω2(3k+1) = ω3k . ω + ω6k . ω²
= 1 . ω + 1 . ω² = ω + ω² = – 1 [∵ ω + ω² +1 = 0 and ω3k = (ω³)k = 1]

Subcase-II : When n = 3k+2
∴ ωn + ω2n = ω3k+1 + ω2(3k+2) = ω3k. ω² + ω6k + ω4 = (ω³)k ω² + (ω³)2k . ω³ . ω
= ω² + ω = – 1 [∵ (ω³)k = 1k = 1]
Thus, ωn + ω2n = 2 or – 1
according as n is a multiple of 3 or any other integer.

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Prove the following:

Question 8.
(1 – ω + ω²) (1 + ω – ω²) (1 – ω – ω²) = 8.
Solution:
(1 – ω + ω²) (1 + ω – ω²) (1 – ω – ω²) = (1 + ω² – ω) (1 + ω – ω²) (1 – (ω + ω²))
= (- ω – ω) (- ω² – ω²) (1 – (- 1)) [∵ 1 + ω + ω² = 0]
= (- 2ω) (- 2ω²)² = 8ω³ = 8 [∵ ω³ = 1]

Question 9.
(i) (1 + ω) (1 + ω²) (1 + ω4) (1 + ω8) …. to 2n factors = 1
(ii) (1 – ω + ω²) (1 – ω² + ω4) (1 – ω4 + ω8)…… to 2n factors = 22n.
Solution:
(i) L.H.S = (1 + ω) (1 + ω²) (1 + ω4) (1 + ω8) …. to 2n factors
= (1 + ω) (1 + ω²) (1 + ω³ . ω) (1 + ω6 . ω²) … 2n factors
= (1 + ω) (1 + ω²) (1 + ω) (1 + ω²) …. 2n factors
= (1 + ω)n (1 + ω²)n = [(1 + ω) (1 + ω2)]n
= [1 + ω + ω² + ω³]n = [0 + 1]n = 1n = 1 [∵ 1 + ω + ω² = 0 and ω³ = 1]

(ii) (1 – ω + ω²) (1 – ω² + ω4) (1 – ω4 + ω8) to 2n factors
= (1 – ω + ω²) (1 – ω² + ω³ . ω) (1 – ω³ . ω + ω6 . ω²) to 2n factors
= (1 – ω + ω²) (1 – ω² + ω) (1 – ω + ω²)…. 2n factors [∵ ω³ = 1]
= [(1 – ω + ω²) …. n factors] [(1 – ω² + ω) …. n factors
= (- 2ω)n (- 2ω²)n [∵ 1 + ω + ω² = 0]
= 2n (ω . ω²)n (- 1)2n
= 22n (ω³)n
= 22n1n
= 22n

Question 10.
\(\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}\) = ω
Solution:
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f) 2

Question 11.
\(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}\) = – 1
Solution:
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f) 3

Question 12.
If ω is a cube root of unity and n is a positive integer which is not a multiple of 3, then show that (1 + ωn + ω2n) = 0.
Solution:
When n is not a multiple of 3
∴ n be of the form 3k + 1 or 3k + 2
Case – I. When n = 3k + 1
1 + ωn + ω2n = 1 + ω3k+1 + ω6k+2 = 1 + (ω³)k ω + (ω³)2k . ω² = 1 + 1k . ω + 12k ω²
= 1 + ω + ω² = 0

Case-II. When n = 3k+2
1 + ωn + ω2n = 1 + ω3k+2 + ω2(3k+2)= 1 + ω3k . ω² + ω6k . ω4
= 1 + (ω³)k . ω² + (ω³)2k . ω³ω = 1 + ω² + ω = 0 [∵ ω³ = 1]

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Question 13.
Show that (x + ωy + ω²z) (x + ω²y + ωz) = x² + y² + z² – yz – zx – xy.
Solution:
L.H.S = (x + ωy + ω²z) (x + ω²y + ωz)
= x² + (ω² + ω) xy + xz (ω + ω²) + ω³y² + yz (ω² + ω4) + ω³z²
= x² – xy + xz(- 1) + y² + yz (- 1) + z² [∵ 1 + ω + ω² = 1]
= x² + y² + z² – xy – yz – zx = R.H.S

Question 14.
Show that x³ + y³ = (x + y) (ωx + ω²y) (ω²x + ωy).
Solution:
R.H.S = (x + y) (ωx + ω²y) (ω²x + ω²y) = (x + y) (ω³x² + ω²xy + ω4xy + ω³y²)
= (x + y) [1 . x² + xy (ω² + ω³ . ω) + 1 . y²] [∵ ω³ = 1]
= (x + y) (x² + xy (ω² + ω) + y²) = (x + y) (x² – xy + y²) [∵ 1 + ω + ω² = 1]
= x³ + y³ = L.H.S

Question 15.
If 1, ω, ω² are cube roots of unity, prove that 1, ω² are vertices of an equilateral triangle.
Solution:
Since 1, ω and ω² are the cube root of unity.
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f) 4
Thus, 1, ω and ω² are the vertices of an equilateral triangle.

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Well-structured OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(e) facilitate a deeper understanding of mathematical principles.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Question 1.
Illustrate in the complex plane, the set of points satisfying the following conditions. Explain your answer:
(i) |z| ≤ 3
(ii) arg (z – 2) = \(\frac { π }{ 3 }\)
(iii) | i – 1 – 2z | > 9
Solution:
(i) Let z = x + iy
∴ | z | ≤ 3 ⇒ | x + iy ≤ 3
⇒ x² + y² ≤ 9
which represents the set of points in interior and on the circle with centre (0, 0) and radius 3.
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 1

(ii) arg (z – 2) = \(\frac { π }{ 3 }\)
⇒ arg (x + iy – 2) = \(\frac { π }{ 3 }\)
⇒ tan-1\(\left(\frac{y}{x-2}\right)=\frac{\pi}{3}\)
⇒ \(\frac{y}{x-2}=\tan \frac{\pi}{3}\) = \(\sqrt{3}\) ⇒ y = \(\sqrt{3}\)(x – 2)
which represents the set of points on line which intersects x-axis at (2, 0) and making an angle of 60° with +ve direction of x-axis.
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 2

(iii) Given | i – 1 – 2z | > 9
| i – 1 – 2 (x + iy) | > 9, where z = x + iy
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 3

Question 2.
Illustrate and explain the region of the Argand’s plane represented by the inequality |z + i| ≥ |z + 2|.
Solution:
Let z = x + iy
given | z + i | ≥ | z + 2 |
⇒ | x + iy + i | ≥ | x + iy + 2 |
⇒ |x + i (y + 1) | ≥ |x + 2 + iy |
⇒ \(\sqrt{x^2+(y+1)^2}\) ≥ \(\sqrt{(x+2)^2+y^2}\)
On squaring both sides ; we have
x² + (y + 1)² ≥ (x + 2)² + y²
⇒ x² + y² + 2y + 1 ≥ x² + 4x + 4 + y²
⇒ 2y + 1 ≥ 4x + 4
⇒ y ≥ 2x + \(\frac { 3 }{ 2 }\)
which represents the set of points in the region lies above the line PQ intersects coordinate axes at (0, \(\frac { 3 }{ 2 }\)) and (- \(\frac { 3 }{ 4 }\), 0)
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 4

Question 3.
Illustrate and explain the set of points z in the Argand diagram, which represents | z – z1 | ≤ 3 where z1 = 3 – 2i
Solution:
Given | z – z1| ≤ 3 ;
where z1 = 3 – 2i
⇒ | z – (3 – 2i) | ≤ 3
⇒ | x + iy – 3 + 2i | ≤ 3
⇒ | (x – 3) + i (y + 2) | ≤ 3
⇒ (x – 3)² + (y + 2)² ≤ 9
which represents the set of points in the interior and on the circle with centre (3, – 2) and radius 3.

Question 4.
If z = x + yi and ω = \(\frac{(1-z i)}{z-i}\) then
| ω | = 1 implies that in the complex plane
(a) z lies on the imaginary axis
(b) z lies on the real axis
(c) z lies on the unit circle
(d) None of these
Solution:
Given z = x + iy and w = \(\frac{1-z i}{z-i}\)
and |w| = 1 ⇒ \(\left|\frac{1-z i}{z-i}\right|\) = 1
⇒ |1 – zi | = |z – i| [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
⇒ | 1 – (x + iy) i | = | x + iy – i |
⇒ |1 – ix + y| = |x + i(y – 1)|
⇒ 1(1 + y) – ix| = | x + | 0 – 1) |
⇒ \(\sqrt{(1+y)^2+(-x)^2}=\sqrt{x^2+(y-1)^2}\)
On squaring both sides ; we have
(1 + y)² + x² = x² + (y – 1)²
⇒ y² + 2y + 1 – x² = x²+y² – 2y + 1
⇒ 4y = 0
⇒ y = 0
which is the eqn. of x-axis. z lies on real axis i.e. x-axis

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Question 5.
Find the locus of a complex number z such that arg \(\left(\frac{z-2}{z+2}\right)\) = \(\frac { π }{ 3 }\).
Solution:
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 5

Question 6.
If the amplitude of z – 2 – 3i is \(\frac { π }{ 4 }\), then find the locus of z = x +yi.
Solution:
Given z = x + iy
and amp (z – 2 – 3i) = \(\frac { π }{ 4 }\)
⇒ amp (x + iy – 2 – 3 z) = \(\frac { π }{ 4 }\)
⇒ amp [(x – 2) + i (y – 3)] = \(\frac { π }{ 4 }\)
⇒ tan-1\(\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4}\)
⇒ \(\frac{y-3}{x-2}=\tan \frac{\pi}{4}=1\)
⇒ y – 3 = x – 2
⇒ x – y + 1 = 0
Hence locus of z represents a straight line intersecting coordinate axes at (- 1, 0) and (0, 1).

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Question 7.
Find the locus of z if
ω = \(\frac{z}{z-\frac{1}{3} i}\) = 1.
Solution:
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 6

Question 8.
A variable complex number z is such that the amplitude of \(\frac{z-1}{z+1}\) is always equal to \(\frac { π }{ 4 }\).
Illustrate the locus of z in the Argand plane.
Solution:
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 7

Question 9.
Find the radius and centre of the circle z\(\bar { z }\)+ (1 – i) z + (1 + i) \(\bar { z }\) – 7 =0.
Solution:
Given eqn. of circle be z\(\bar { z }\)+(1 – i)z + (1 + i)\(\bar { z }\) – 7 = 0 …(1)
where z = x + iy and \(\bar { z }\) = x – iy
∴ eqn. (1) becomes ;
⇒ (x + iy) (x – iy) + (1 – i) (x + iy) + (i + i) (x – iy) – 7 = 0
⇒ x² + y² + x + iy – ix + y + x – iy + ix + y – 7 = 0
⇒ x² + y² + 2x + 2y – 7 = 0
⇒ (x² + 2x) + (y² + 2y) – 7 = 0
⇒ (x² + 2x + 1) + (y² + 2y + 1) – 9 = 0
⇒ (x + 1)² + (y + 1)² = 9
which represents a circle with centre (- 1, – 1) and radius 3.

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Question 10.
What is the region represented by the inequality 3 < | z – 2 – 3i|< 4 in the Argand plane.
Solution:
Given 3 < |z – 2 – 3i| < 4;
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e) 8
where z = x + iy
⇒ 3 < | x + iy – 2 – 3i| < 4
⇒ 3 < | (x – 2) + | (y – 3) | < 4
⇒ 3 < \(\sqrt{(x-2)^2+(y-3)^2}\) < 4
⇒ 9 < (x – 2)² + (y – 3)² < 4
which represents the set of points lies in the region between two concentric circles with centre (2, 3) and radius 2 and 3.

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

Students can track their progress and improvement through regular use of OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(d).

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

Find the modulus and amplitude of the following complex numbers and hence express them into polar form.

Question 1.
\(\sqrt{3}\) + i
Solution:
Let z = \(\sqrt{3}\) + i = x + iy
Here x = \(\sqrt{3}\) > 0 and y = 1 > 0
∴ point (x, y) i.e. (-\(\sqrt{3}\), 1)
lies in first quadrant.
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d) 1

Question 2.
– \(\sqrt{3}\) + i
Solution:
Let z = – \(\sqrt{3}\) + i …(1)
put – \(\sqrt{3}\) = r cos θ …(2)
1 = r sin θ …(3)
On squaring and adding (1) and (2);
3 + 1 = r² (cos² θ + sin² θ)
⇒ r² = 4
⇒ r = 2 (r > 0)
On dividing (3) by (2); we have tan θ
tan θ = – \(\frac{1}{\sqrt{3}}\) = – tan \(\frac { π }{ 6 }\) = tan (π – \(\frac { π }{ 6 }\))
[∵ cos θ < 0 and sin θ > 0
∴ θ lies in 2nd quad]
⇒ θ = \(\frac { 5π }{ 6 }\)
⇒ arg (z) = \(\frac { 5π }{ 6 }\)
∴ from eqn. (1) ; we have
z = r [cos θ + i sin θ]
= 2 [cos \(\frac { 5π }{ 6 }\) + i sin \(\frac { 5π }{ 6 }\) ]

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

Question 3.
– 2 + 2\(\sqrt{3}\) i
Solution:
Let z = – 2 + 2\(\sqrt{3}\) i … (1)
put – 2 = r cos θ …(2)
2\(\sqrt{3}\) = r sin θ …(3)
since cos θ < 0 and sin θ > 0 ∴ θ lies in 2nd quadrant.
On squaring and adding (2) and (3); we have
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d) 2
which is the required polar form.

Question 4.
– 1 – i
Solution:
Let z = – 1 – i = x + iy
Here x = – 1 < 0 and y = – 1 < 0
∴ point (x, y) i.e. (- 1, – 1) lies in IIIrd quadrant.
∴ tan θ = \(\left|\frac{{Im}(z)}{{Re}(z)}\right|=\left|\frac{y}{x}\right|=\left|\frac{-1}{-1}\right|\) = 1
⇒ α = \(\frac { π }{ 4 }\)
∴ arg (z) = – (π – α) = – (π – \(\frac { π }{ 4 }\)) = – \(\frac { 3π }{ 4 }\)
and r = \(\sqrt{x^2+y^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}\)
Thus required polar form of z is given by
z = r [cos (arg z) + z sin (arg z)]
⇒ z = \(\sqrt{2}\left[\cos \left(\frac{-3 \pi}{4}\right)+i \sin \left(\frac{-3 \pi}{4}\right)\right]\)

Question 5.
– 2i
Solution:
Let z = – 2i × x + iy
∴ |z| = |- 2i| = 2
where x – 0 ; y = – 2
∴ point (x, y) i.e. (0, – 2) lies in IVth quadrant.
∴ tan θ = \(\left|\frac{{Im}(z)}{{Re}(z)}\right|=\left|\frac{-2}{0}\right|\) → ∞ ⇒ α = \(\frac { π }{ 2 }\)
∴ θ = arg (z) = – α = – \(\frac { π }{ 2 }\)
Thus polar form of z is given by z = | z | (cos θ + z sin θ)
= 2 \(\left[\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\right]\)

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

Question 6.
– 1 – \(\sqrt{3}\) i
Solution:
Let z = – 1 – \(\sqrt{3}\) i …(1)
put – 1 = r cos θ …(2)
– \(\sqrt{3}\) = r sin θ …(3)
Here cos θ, sin θ < 0
∴ θ lies in 3rd quadrant. On squaring and adding eqn. (2) and eqn. (3); we have
r² (cos² θ + sin² θ) = (- 1)² + (- \(\sqrt{3}\))² = 1 + 3 = 4
r² = 4 ⇒ r = 2 (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = \(\frac{-\sqrt{3}}{1}\) = \(\sqrt{3}\)
= \(\tan \left(\frac{\pi}{3}\right)=\tan \left[-\left(\pi-\frac{\pi}{3}\right)\right]\)
⇒ θ = – \(\frac { 2π }{ 3 }\)
∴ from (1) ; z = r [cos θ + i sin θ]
i.e. z = 2\(\left[\cos \left(-\frac{2 \pi}{3}\right)+i \sin \left(-\frac{2 \pi}{3}\right)\right]\)

Question 7.
– 2
Solution:
Let z = – 2 = x + iy
Here x = – 2 and y = 0
∴ (- 2, 0) lies on negative x-axis.
∴ | z | = | – 2 | = 2
∴ arg (z) = π ;
∴ z = 2 [cos π + i sin π] be the required polar form.

Question 8.
\(\frac{(1+i)^{13}}{(1-i)^7}\)
Solution:
Let
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d) 3
Further x = – 8 ; y = 0 i.e. point (- 8, 0) lies on negative real axis.
∴ arg (z) = π
∴ z = 8 [cos π + i sin π] be the required polar form

Question 9.
(3 + i) (4 + i)
Solution:
Let z = (3+ i) (4 + i)
= 12 + 3i + 4i – 1 = 11 + 7i …(1)
∴ | z | = \(\sqrt{11^2+7^2}\)
= \(\sqrt{121+49}=\sqrt{170}\)
put 11 = r cos θ …(2)
7 = r sin θ …(3)
On dividing (3) by (2); we have
tan θ = \(\frac { 7 }{ 11 }\)
⇒ θ = tan-1\(\frac { 7 }{ 11 }\) = tan-1(0.636)
⇒ θ = 32°29′
[since cos θ, sin θ > 0
∴ θ lies in first quadrant]
∴ arg (z) = θ = 32° 29′
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = \(\sqrt{170}\) [cos (32° 29′) + i sin (32° 29′)]
[On squaring and adding eqn. (1) and (2); r = \(\sqrt{170}\) ]

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

Question 10.
\(\frac{(1+i)(2+i)}{(3+i)}\)
Solution:
Let z = \(\frac{(1+i)(2+i)}{3+i}=\frac{2+i+2 i-1}{3+i}\) = \(\frac{1+3 i}{3+i} \times \frac{3-i}{3-i}\)
⇒ z = \(\frac{3-i+9 i+3}{3^2-i^2}\)
⇒ z = \(\frac{6+8 i}{9+1}=\frac{3+4 i}{5}=\frac{3}{5}+\frac{4}{5} i\) … (1)
∴ | z | = \(\frac{\sqrt{3^2+4^2}}{5}=\frac{5}{5}\) = 1
Put \(\frac { 3 }{ 5 }\) = r cos θ … (2)
and \(\frac { 4 }{ 5 }\) = r sin θ … (2)
On squaring and adding eqn. (2) and (3);
(\(\frac { 3 }{ 5 }\))² = r² (cos² θ + sin² θ)
⇒ r² = \(\frac{9+16}{25}\) 1 (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = \(\frac{4}{5} \times \frac{5}{3}=\frac{4}{3}\) = 1.3333
⇒ θ = tan-1 (1.3333) = 53° 8′ [∵ θ lies in first quadrant]
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = 1 [cos (53° 8′) + i sin (53° 8′)]
which is the required polar form.

Question 11.
\(\frac{5-i}{2-3 i}\)
Solution:
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d) 4

Question 12.
\(\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}\)
Solution:
Let z = \(\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}\)
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d) 5
⇒ α = 18° 12′ and arg (z) = a = 18° 12′
∴ required polar form of z be given by z = | z | [cos α + i sin α]
⇒ z = \(\sqrt{\frac{41}{85}}\) [cos (18° 12′)+ i sin (18° 12′)]

Question 13.
Change the following complex numbers into polar form.
(i) – 4 + 4\(\sqrt{3}\) i
(ii) \(\frac{1+3 i}{1-2 i}\)
(iii) \(\frac{1+2 i}{1-(1-i)^2}\)
(iv) \(\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}\)
Solution:
(i) Let z = – 4 + 4\(\sqrt{3}\)i …(1)
put – 4 = r cos θ …(2)
4\(\sqrt{3}\) = r sin θ …(3)
On squaring and adding eqn. (2) and eqn. (3); we have
r² = (- 4)² + (4\(\sqrt{3}\))² = 16 + 48 = 64
⇒ r = 8 (∵ r > 0)
since cos θ < 0 and sin θ > 0 ∴ θ lies in 2nd quadrant.
On dividing eqn. (3) by eqn. (2); we have
tan θ = – \(\sqrt{3}\) = – tan\(\frac { π }{ 3 }\) = tan(π – \(\frac { π }{ 3 }\))
= tan\(\frac { 2π }{ 3 }\)
∴ θ = \(\frac { 2π }{ 3 }\)
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = 8\(\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right]\)
which is the required polar form

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

(ii) Let z = \(\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}\)
⇒ z = \(\frac{1+2 i+3 i-6}{1^2-(2 i)^2}=\frac{5 i-5}{1+4}\) = i – 1
⇒ z = – 1 + i …(1)
put – 1 = r cos θ …(2)
and 1 = r sin θ …(3)
since cos θ < 0 and sin θ > 0
∴ θ lies in 2nd quadrant.
On squaring and adding eqn. (1) and (2); we have
r² (cos² θ + sin² θ) = (- 1)² + (1)²
⇒ r² = 2 ⇒ r = \(\sqrt{2}\) (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = – 1 = – tan \(\frac { π }{ 4 }\) = tan (π – \(\frac { π }{ 4 }\))
⇒ θ = π – \(\frac { π }{ 4 }\) = \(\frac { 3π }{ 4 }\)
∴ from (1); z = r [cos θ + 2 sin θ]
⇒ z = \(\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]\)
which is the required polar form.

(iii) Let z = \(\frac{1+2 i}{1-(1-i)^2}=\frac{1+2 i}{1-(1-1-2 i)}\)
⇒ z = \(\frac{1+2 i}{1+2 i}\) = x + iy
Here x = 1 ; y = 0
∴ point (x, y) i.e. (1,0)
lies on positive real axes.
∴ z = 1 = 1 + i0 = cos 0° + 2 sin 0°
which is the required polar form.

(iv) Let z = \(\frac{1+7 i}{(2-i)^2}=\frac{1+7 i}{4-1-4 i}\) = \(\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}\)
⇒ z = \(\frac{3+4 i+21 i-28}{3^2-(4 i)^2}=\frac{-25+25 i}{9+16}\)
⇒ z = – 1 + i … (1)
put – 1 = r cos θ …(2)
and 1 = r sin θ … (3)
since cos θ < 0 and sin θ > 0 ∴ lies in 2nd quadrant.
On squaring and adding eqn. (1) and (2); we have
r² (cos² θ + sin² θ) = (- l)² + (l)²
⇒ r² = 2 ⇒ r = \(\sqrt{2}\) (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = – 1 = – tan \(\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right)\)
⇒ θ = π – \(\frac { π }{ 4 }\) = \(\frac { π }{ 4 }\)
∴ from (1) ; z = r [cos θ + i sin θ]
⇒ z = \(\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]\)
which is the required polar form.

Question 14.
Given the complex number z = \(\frac{-1+\sqrt{3} i}{2}\) and w = \(\frac{-1-\sqrt{3} i}{2}\) (where i = \(\sqrt{-1}\))
(i) Prove that each of these complex numbers is the square of the other.
(ii) Calculate the modulus and argument of w and z.
(iii) Calculate the modulus and argument of \(\frac { w }{ z }\).
(iv) Represent z and w accurately on the complex plane.
Solution:
Given z = \(\frac{-1+\sqrt{3} i}{2}\) and w = \(\frac{-1-\sqrt{3} i}{2}\)
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d) 6
Thus each of the complex numbers is the square of the other.

(ii) z = – \(\frac{1}{2}+\frac{\sqrt{3}}{2}\) = x + iy
OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(d) 7

(iv) Clearly the complex number z is represented by point P\(\frac{-1-\sqrt{3} i}{2}\) and the complex number is represented by point Q\(\frac{-1-\sqrt{3} i}{2}\) in complex plane.

Selina Concise Physics Class 8 ICSE Solutions – Light Energy

Selina Concise Physics Class 8 ICSE Solutions – Light Energy

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 5 Light Energy. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Physics Chapter 5 Light Energy

  • Light is a form of energy which produces in us the sensation of sight i.e. we can see objects only when light falls on them and then reflected into our eye.
  •  Velocity of light in air or in vacuum is 300000 km per second.
    Or
    3 x 10ms-1
  • As light passes into different mediums its speed changes and depends upon the density of medium i.e. it decreases with increase in density i.e. it is 2.25 × 108 m/s in water and 2 x 108 ms-1 in glass as water is
    denser than air ( \(_{ w }^{ a }{ \mu }\) = 1.33 ) and glass is still optically denser than water
    ( \(_{ g }^{ a }{ \mu }\) =1.5 ) i.e. slower in water and still slower in glass.
  • Light travels in a straight line.
  • As light travels from one transparent medium to other transparent medium and falls oblique at another medium, its path changes and this change in path is called REFRACTION OF LIGHT.
  •  When ray of light travels from RARER (less-denser) to DENSER medium, it bends TOWARD the normal AND when it travels from a DENSER to a RARER medium it bends away from NORMAL
  •  ANGLE of INCIDENCE : “The angle which incident ray makes with normal”. “∠i”
  •  ANGLE OF REFRACTION: “The angle which refracted ray makes with normal” “ ∠r ”
    ∠i is not equal to ∠r
  •  LAWS OF REFRACTION or SNELL’S LAWS OF REFRACTION:
    (i) Incident ray, normal at the point of incidence and Refracted ray all lie in the same plane.
    (ii) Ratio of sine of angle of incidence to the sine of angle of refraction is constant.
  • Selina Concise Physics Class 8 ICSE Solutions Chapter 5 Light Energy 1
  •  EFFECTS OF REFRACTION :
    (i) A coin placed in water appears to be raised.
    (ii) Swimming pool seen from above appears SHADOW.
    (iii) A pencil in water appears to be bent.
    (iv) MIRAGE in desert, EARLY Sunrise, LATE SUN set are all due to REFRACTION of light.
  • White light is a band of seven colours-VIBGYOR. Speed of all colours of the white light in AIR or VACUUM is same, but different different transparent mediums.
  •  In glass or water Speed of VIOLET colour is MINIMUM and speed of RED light is MAXIMUM
  • Refractive index of medium is minimum for VIOLET light and R.I. of medium is maximum for red light.
  • DISPERSION: “The splitting (breaking) of white light into seven colours is called DISPERSION OF LIGHT.
  •  CAUSE OF DISPERSION: Speed of different colours is different in glass or water and different colours get separated from each other on refraction at second surface of glass prism.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) Water is optically denser than glass.
Answer. False.
Water is optically denser than air.

(b) A ray of light when passes from glass to air, bends towards the normal.
Answer. False.

(c) The speed of light is more in glass than in water.
Answer. False.

(d) The depth of a pond when seen from above appears to be less.
Answer. True.

(e) Light travels at a lower speed in water than in air.
Answer. True.

(f) Light travels in the same straight line path while passing through different media.
Answer. False.

(g) The angle formed between the normal and the refracted ray is known as the angle of incidence.
Answer. False.

(h) At the point of incidence, a line drawn at right angles to the surface, separating the two media, is called the normal.
Answer. True.

(i) Image is formed by a mirror due to refraction of light.
Answer. False.

(j) Rays of light incident parallel to the principal axis pass through the focus after reflection from a concave mirror.
Answer. True.

(k) A convex mirror is used as a shaving mirror.
Answer. False.

(l) The focal length of a convex mirror is equal to its radius of curvature.
Answer. False.

(m) A concave mirror converges the light-rays, but a convex mirror diverges them.
Answer. True.

(n) A virtual image formed by a spherical mirror is always erect and situated behind the mirror.
Answer. True.

2. Fill in the blanks

(a) Water is optically denser than air.
(b) Air is optically rarer than glass.
(c) When a ray of light travels from water to air, it bends away from the normal.
(d) When a ray of light travels from air to glass, it bends towards the normal.
(e) When white light passes through a prism, it disperses
(f) The splitting of white light into its constituent colours is called dispersion.
(g) A concave mirror is obtained on silvering the outer surface of a part of a hollow glass sphere.
(h) Radius of curvature of a spherical mirror is two times its focal length.
(i) The angle of incidence for a ray of light passing through the centre of curvature of a spherical mirror is
(j) A convex mirror always forms a virtual image.
(k) A concave mirror forms a virtual image for an object placed between pole and focus.

 

3. Match the following
Selina Concise Physics Class 8 ICSE Solutions Chapter 5 Light Energy 2

4. Select the correct alternative

(a) The speed of light in air or vacuum is

  1. 3 × 10M s-1
  2.  2.25 × 108 m s-1
  3.  332 ms-1
  4.  2.0 × 108 ms-1

(b) A ray of light moving from an optically rarer to a denser medium

  1.  bends away from the normal
  2.  bends towards the normal
  3.  remains undeviated
  4.  none of the above

(c) The angle between the normal and refracted ray is called

  1.  angle of deviation
  2.  angle of incidence
  3.  angle of refraction
  4.  angle of emergence.

(d) The property of splitting of white light into its seven constituent colours is known as

  1.  rectilinear propagation
  2.  refraction
  3.  reflection
  4.  dispersion

(e) The seven colours in the spectrum of sunlight in order, are represented as :

  1.  VIBGYOR
  2.  VIGYBOR
  3.  BIVGYOR
  4.  RYOBIVG

(f) A ray of light passing through centre of curvature of a spherical mirror, after reflection

  1. passes through the focus
  2.  passes through the pole
  3.  becomes parallel to the principal axis
  4.  retraces its own path.

(g) If the radius of curvature of a concave mirror is 20 cm, its focal length is:

  1.  10 cm
  2.  20 cm
  3.  40 cm
  4.  80 cm

(h) The image formed by a convex mirror is

  1.  erect and diminished
  2.  erect and enlarged
  3.  inverted and diminished
  4.  inverted and enlarged.

(i) The image formed by a concave mirror is of the same size as the object, if the object is placed

  1. at the focus
  2. between the pole and focus
  3.  between the focus and centre of curvature
  4.  at the centre of curvature.

(j) A convex mirror is used

  1.  as a shaving mirror
  2.  as a head mirror by a dentist
  3.  as a rear view mirror by a driver
  4.  as a reflector in torch.

Selina Concise Physics Class 7 ICSE Solutions – Heat

Selina Concise Physics Class 7 ICSE Solutions – Heat

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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

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Selina Concise ICSE Solutions for Class 7 Physics Chapter 5 Heat

  • Points to Remember
  •  Heat is a form of energy that leads to the sensations of hotness or coldness.
  •  Temperature is the degree of hotness and coldness of a body.
  •  Thermometer is used to measure temperature.
  •  The S.I. unit of temperature is °C.
  •  The most common liquid for a thermometer is mercury.
  •  The main sources of heat are (i) Fire (ii) Sun (iii) Electricity.
  •  Those substances which can easily catch fire are called inflammable substances.
  •  Those substances which are fire resistant are called non-inflammable substances.
  •  The fixed temperature at which freezing of liquid occurs is known as freezing point.
  •  The temperature at which vapourisation occurs is known as the boiling point.
  •  Substances through which heat is easily conducted are called good conductors e.g. silver, gold, copper etc.
  •  Substances through which heat is not easily conducted are called Insultors.
  •  Radiation is the process of transfer of heat from a hot body to a cold body without affecting the intermediate medium.

Test Yourself

A. Objective Questions 

1. Write true or false for each statement

(a) On touching a lump of ice, we feel cold because some heat passes from our body to the ice.
Answer. True.

(b) Heat flows from a body at a high temperature to a body at a low temperature when they are kept in contact. .
Answer. True.

(c) All solids expand by the same amount when heated to the same rise in temperature.
Answer. False.

(d) Telephone wires are kept tight between the two poles in summer.
Answer. False.

(e) Equal volumes of different liquids expand by different amounts when they are heated to the same rise in temperature.
Answer. True.

(f) Solids expand the least and gases expand the most on being heated.
Answer. True.

(g) A mercury thermometer makes use of the property of expansion of liquids on heating.
Answer. True.

(h) Kerosene contracts on heating.
Answer. False.

(i) Water is a bad conductor of heat.
Answer. True.

(j) Medium is necessary for the transfer of heat by radiation.
Answer. False.

(k) Land and sea breezes are convection currents of cold and warm air.
Answer. True.

(l) Liquids are heated by conduction and radiation.
Answer. False.

(m) Black surfaces are the poor absorbers of heat radiations.
Answer. False.

2. Fill in the blanks

(a) Heat is a form of energy.
(b) Temperature determines the degree of hotness or coldness of a body.
(c) On heating a body, its temperature rises.
(d) We use a thermometer for measuring the temperature of a body.
(e) The S.I. unit of temperature is kelvin.
(f) In a thermometer, the commonly used liquid is mercury.
(g) The temperature of a normal human body is 37 °C.
(h) A person is said to have fever if his body temperature is more than 98.6
(i) A hot metallic piece is placed in tap water contained in a bucket. Heat will flow from metallic piece to water.
(j) The temperature of boiling water is 100°C.
(k) Liquids expand more than the solids.
(l) Gases expand more than the liquids.
(m) Heat transfer in solids is by conduction.
(n) Heat transfer in liquids and gases is by convection.
(o)Metals are conductors of heat.
(p) Still air is an insulator of heat.
(q) Black and dull surfaces are good absorbers of heat.

3. Match the following

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 1

4. Select the correct alternative

(a) If we add a lump of ice to a tumbler containing water,

  1. heat flows from water to ice
  2.  heat flows from ice to water
  3.  heat flows from water to ice if water is more
  4.  heat flows from ice to water if ice is more

(b) The temperature of pure melting ice is

  1.  0°C
  2.  100°C
  3.  95°C
  4.  98.6°F

(c) A thermometer uses

  1.  water
  2. mercury
  3.  air
  4.  none of the above

(d) Which of the statement is correct

  1.  Iron rims are cooled before they are placed on cart wheels
  2.  A glass stopper gets tight on warming the neck of the bottle
  3.  Telephone wires sag in winter, but become tight in summer
  4. A little space is left between two rails on a railway track

(e) Heat in a liquid is transferred by

  1.  conduction
  2. convection
  3.  radiation
  4.  conduction and radiation

(f) In the process of convection, heat travels

  1.  sideways
  2.  downwards
  3. upwards
  4.  in all directions

(g) The vacuum kept in between the walls of a thermos flask reduces the heat transfer by

  1.  conduction only
  2.  convection only
  3.  radiation only
  4. conduction and convection

B. Short/Long Answer Questions

Question 1.
What is heat ? State its S.I. unit.
Answer:
Heat is a form of energy which flows. It is the energy of motion of molecules constituting the body.
The unit of heat is same as that of energy, The S.I. unit of heat is joule (abbreviated as J) and other common units of heat are calorie and kilo calorie, where 1 k cal = 1000 cal.

Question 2.
What is meant by the term temperature.
Answer:
Temperature is a quantity which tells the thermal state of a body (i.e. the degree of hotness or coldness). It determines the direction of flow of heat when the two bodies at different temperatures are placed in contact.

Question 3.
State the three units of temperature.
Answer:
The S.I. unit of temperature is kelvin or K. The other most common unit of temperature is degree Celsius (°C) and degree Fahrenheit (°F).

Question 4.
Name the instrument used to measure the temperature of a body.
Answer:
To measure the temperature of a body with the help of a thermometer.

Question 5.
Name two scales of temperature. How are they inter-related?
Answer:
Two scales of temperature are
(i) Celsius (ii) Fahrenheit
Relation:

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 2

Question 6.
How is the size of a degree defined on a Celsius scale ?
Answer:
The interval between the ice point and steam point divided by 100 (hundred) equal parts is called a degree on the Celsius scale.

Question 7.
How is the size of a degree defined on a Fahrenheit scale?
Answer:
The interval between the ice point and steam point divided into 180 equal parts is called a degree on the Fahrenheit scale.

Question 8.
State the temperature of (i) ice point and (ii) steam point, on the Celsius scale.
Answer:
(i) Ice point. Is the the mark on Celsius scale at which ice melts. Ice point on the Celsius scale is 0°C.
(ii) Steam point. On the Celsius scale is the mark at which water changes into steam at normal atmospheric pressure. On Celsius scale it is 100°C.

Question 9.
Write down the temperature of (i) lower fixed point, and (ii) upper fixed point, on the Fahrenheit scale.
Answer:
Lower fixed point: On the Falirenheit scale is the mark at which pure ice melts. It is 32°F on Fahrenheit scale.
Upper fixed point: On the Fahrenheit scale is the mark at which water starts changing into steam at normal atmospheric pressure. It is 212°F.

Question 10.
What is the Celsius scale of temperature ?
Answer:
Celsius scale is that which has ice point as 0°C and steam point marked as 100°C.

Question 11.
What is the Fahrenheit scale of temperature ?
Answer:
Fahrenheit scale is that which has ice point as 32°F and the steam point marked as 100°C.

Question 12.
What is the Kelvin scale of temperature ?
Answer:
On Kelvin scale of temperature zero mark is when no molecular motion occurs. Ice point is at 273 and steam point is at 373 K. Thus 0 K = – 273°C and one degree on Kelvin scale is same as one degree on Celsius scale.

Question 13.
The fig. shows a glass tumbler containing hot milk which is placed in a tub of cold water. State the direction in which heat will flow.

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 3
Answer:
When we bring two objects of different temperature together, energy will always be transferred from hotter to the cooler object.
Here, also heat will flow from hot milk tumbler to tub of cold water.

Question 14.
Draw a neat labelled diagram of a laboratory thermometer.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 4

Question 15.
Write down the body temperature of a healthy person.
Answer:
The temperature of a healthy persons is 98.6 degrees fahrenheit or 37.0 degree Celsius or 310 k.

Question 16.
What do you understand by thermal expansion of a substance ?
Answer:
The expansion of a substance when, heated, is called thermal expansion.
Or
Thermal expansion is the tendency of matter to change .in shape, area and volume in response to a change in temperature.

Question 17.
Name two substances which expand on heating.
Answer:
Mercury and Aluminium wire.

Question 18.
Why do telephone wires sag in summer ?
Answer:
The telephone wires will sag in summer due to expansions and will break in winter due to contraction.
Therefore, while putting up the wires between the poles, care is taken that in summer they are kept slightly loose so that they may not break in winter due to contraction.
While in winter they are kept light so that they may not sag too much in summer due to expansion.

Question 19.
Iron rims are heated before they are fixed on the wooden wheels. Explain the reason.
Answer:
The wooden wheels of a bullock-cart are fitted with iron tyres. To ensure a tight fit, the tyre is made slightly smaller in diameter than the wheel. The tyre is first heated due to which it expands. The heated tyre is then fitted on the wheel. When the tyre cools, it contracts and makes a tight fit on the wheel.

Question 20.
Why are gaps left between successive rails on a railway track ?
Answer:
The rails of railway track are made of steel. While laying the railway track, a small gap is left between the two successive length of rails. The reason is that the rails expand in summer. The gap is provided to allow for this expansion. If no gap is left, the expansion in summer will cause the rails to bend sideways. This may result in a train accidents.

Question 21.
A glass stopper stuck in the neck of a bottle can be removed by pouring hot water on the neck of the bottle. Explain why ?
Answer:
When hot water is poured over the neck of the bottle, it expands. As a result the stopper gets loosened and can be removed easily.

Question 22.
Why is a cement floor laid in small pieces with gaps in between?
Answer:
The floor is laid in small pieces with gaps in between to allow for the expansion during summer. However glass strips can be placed in the gaps.

Question 23.
One end of a steel girder in a bridge is not fixed, but is kept on roliers. Give the reason.
Answer:
In the construction of a bridge, steel girders are used. One end of the girder is fixed into the concrete or brick pillars and its other end is not fixed, but it is placed on rollers. The reason is that if there is any rise (or fall) in temperature of atmosphere, the girder can freely expand (or contract) without affecting the pillars.

Question 24.
Describe one experiment to show that liquids expand on heating.
Answer:
(i) Take an empty bottle with a tight fitting cork having a hole drilled in its middle, a drinking straw, two bricks, a wire guaze and a burner.
(ii) Fill the bottle completely with water and add few drops of ink in it to make it coloured.
(iii) Fix the cork in the mouth of the bottle and pass the drinking straw through the cork. Put some molten wax around the hole so as to avoid the leakage of water.
(iv) Pour some more water into the drinking straw so that water level in the straw can be seen. Mark the water level in the straw as shown in Figure.

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 5
(v) Place the bottle on the wire gauze kept over the two bricks as shown in Figure. Then heat the bottle by means of a burner.
(vi) Look at the level of water in the straw.
You will notice that as the water is heated more and more, the level of water in the drinking straw rises. This shows that water expands on heating.

Question 25.
State one application of thermal expansion of liquids.
Answer:
Mercury is a metal found in liquid state. It expands more and uniformly over a wide range of temperature. So mercury is used as thermometric liquid.

Question 26.
Describe an experiment to show that air expands on heating.
Answer:
(i) Take an empty bottle. Actually the empty bottle contains air. Attach a rubber balloon to its neck as shown in Figure. Initially, the balloon is deflated.
(ii) Place the bottle in a water bath containing boiling water. After some time you will notice that the balloon gets inflated as shown in Figure. The reason is that the air inside the bottle expands on heating and it fills the balloon.

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 6
(iii) Take the bottle out of the water bath and 7 allow it to cool by itself. We will notice that the balloon gets deflated and it collapses. This is because the air inside the balloon and the bottle, has contracted on cooling. The air from balloon passes to the bottle, so the balloon gets deflated.

Question 27.
An empty glass bottle is fitted with a narrow tube at its mouth. The open end of the tube is kept in a beaker containing water. When the bottle is heated, bubbles of air are seen escaping into the water. Explain the reason.
Answer:
When the bottle is heated, bubbles of air are seen escaping into the water. This happens because the air present in glass bottle expands on heating and tries to escape out through the tube into the water.

Question 28.
State which expands more, when heated to the same temperature : solid, liquid or gas ?
Answer:
Gases expand much more than the liquids and the solids. Like liquids, the gases do not have a definite shape, so they also have only the cubical expansion.

Question 29.
Name the three modes of transfer of heat.
Answer:
There are three modes of transfer of heat (i) Conduction (ii) Convection (iii) Radiation.
(i) Conduction “is that mode of transfer of heat, when heat travels from hot end to cold end from particle to particle of the medium, without actual movement of particles.”
(ii) Convection. “Is a process of transfer of heat by actual move-ment of the medium particles.”
(iii) Radiation. “Is that mode of transfer of heat in which heat directly passes from one body to the other body without heating the medium.”

Question 30.
Name the mode of transfer of heat in the following :
(a) solid,
(b) liquid,
(c) gas
(d) vacuum
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 7

Question 31.
What are the good and bad conductors of heat ? Give two examples of each.
Answer:
Good conductors. “The substances through which heat is easily conducted are called good conductors of heat.”
Example : Copper, iron.
Bad conductors. “The substances through which heat is not conducted easily are called bad conductors of heat or poor conductors of heat.”
Example : Wood, cloth.

Question 32.
Name a liquid which is a good conductor of heat.
Answer:
Mercury is good conductor of heat.

Question 33.
Name a solid which is a good conductor of heat.
Answer:
Aluminium is a good conductor of heat.

Question 34.
Select good and bad conductors of heat from the following :
copper, mercury, wood, iron, air, saw-dust, cardboard, silver, plastic, wool.
Answer:
Good conductors — Mercury, copper, silver, iron.
Bad conductors — Wood, air, saw dust, plastic, wool, cardboard.

Question 35.
Why is an oven made of double walls with the space in between filled with cork ?
Answer:
An oven is made of double walls and the space between them is filled with wool, cork etc. because the wool and cork are the insulator of heat. They prevent the heat of the oven to escape.

Question 36.
Why do we use cooking utensils made up of copper.
Answer:
Cooking utensils are made of metals such as copper, aluminium, brass, steel etc., so that heat is easily conducted through the base to their contents. But they are provided with handles of bad conductors (such as ebonite or wood) to hold them easily as handles will not conduct heat from the utensil to our hand.

Question 37.
Why is a tea kettle provided with an ebonite handle ?
Answer:
Tea kettles are provided with wooden or ebonite handles. The wood or the ebonite being the insulators of heat, does not pass heat from the utensils to our hand. Thus, we can hold the hot utensils or pans comfortably by their handles.

Question 38.
In summer, ice is kept wrapped in a gunny bag. Explain the reason.
Answer:
In summer, the ice is kept wrapped in a gunny bag or it is covered with saw dust. The air filled in the fine pores of the gunny bag or saw dust, is the insulator of heat. The air does not allow heat from outside to pass through it to the ice. Thus, the ice is prevented from melting rapidly.

Question 39.
Explain why
(a) we wear woolen clothes in winter.
(b) the water pipes are covered with cotton during very cold weather.
Answer:
(a) Woolen clothes have fine pores filled with air. Wool and air both are bad conductors of heat. Therefore in winter, we wear woolen clothes as they check the conduction of heat from the body to the surroundings and thus keeps the body warm.
(b) During very cold weather, the water pipes are covered with cotton. The cotton has air trapped in its fine pores. The cotton and air are the insulators of heat. They do not pass heat from water inside the pipes to the outside atmosphere. Thus, cotton prevents the water in the pipes from freezing.

Question 40.
Why are quilts filled with fluffy cotton ?
Answer:
Quilts are filled with fluffy cotton. Air is trapped in the fine pores of cotton. Cotton and air are the insulators of heat. They check heat from our body to escape and thus keep us warm.
The newly made quilts are warmer than the old ones because in the old quilts, there is no air trapped in the cotton.

Question 41.
State the direction of heat transfer by way of convection.
Answer:
By the process of convection, heat is always transferred vertically upwards. The reason is that the medium particles near the source of heat absorb heat from the source and they start moving faster. As a result, the medium at this place becomes less dense so it rises up and the medium from above being denser, moves down to take its place. Thus, current is set up in the medium which is called a convection current. The current continues till the entire medium acquires the same temperature.

Question 42.
Why is a ventilator provided in a room ?
Answer:
Ventilators and windows are provided in rooms for proper ventilation. The reason is that when we breathe out in a room, the air in the room becomes warm and impure. The warm air is less dense i.e. lighter, so it rises up and moves out through the ventilators. Then the cold fresh air comes in the room through the windows to take its place. Thus the continuous circulation of fresh air keeps the air in the room fresh.

Question 43.
Why are chimneys provided over furnace in factories ?
Answer:
Chimneys are provided over the furnace in factories. This is because the hot gases coming out of the furnace are less dense than the air. They rise up through the chimney. The smoke, fumes etc. around the furnace rush in so as to take their place and they are sucked out. Thus, the chimney helps to remove the undesired fumes, smoke etc. from the premises.

Question 44.
What are the land and sea breezes ? Explain their formation.
Answer:
LAND BREEZE : Blowing of breeze (air) from land towards sea is called land breeze.
During night land and sea water both lose heat. Specific heat capacity of land being very low as compared to that of sea water, land loses heat energy fast and cools more rapidly as compared to sea. Sea water being at higher temperature, the air above it becomes lighter and rise up. Air from land being at higher pressure. So air from land starts blowing towards sea and gives rise to Land Breeze.

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 8
SEA BREEZE : Blowing of breeze (cold air) from sea towards land during the day is called the SEA BREEZE. During day time land and sea both are heated equally by the sun, but land has very low specific heat capacity as compared to sea, is heated up more quickly. Thus air above land due to heat becomes lighter and rises up. Thus pressure decreases and cold and humid air above the sea starts blowing towards land, thereby giving rise to SEABREEZE.

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 9

Question 45.
Why is the freezing chest in a refrigerator fitted near its top?
Answer:
Freezing chest in a refrigerator is fitted near the top, because it cools the remaining space of the refrigerator by convection current. Air near the top comes in contact with the freezing chest gets cooled, becomes denser and therefore descends while the hot air from the lower part rises and hence convection currents produced cool the whole space inside.

Question 46.
Explain briefly the process of heat transfer by radiation.
Answer:
RADIATION. “The transfer of heat energy from a hot body to cold body directly, without heating the medium between two bodies is called RADIATION.”
The radiant heat or thermal radiation is of the form of ELECTROMAGNETIC WAVES. These waves can travel even in vacuum in all directions in straight line with the speed of light. They do not heat the medium through which they pass. Heat radiations are also called INFRA-RED RADIATIONS because the wavelength of heat radiations is longer than that of visible light. These radiations can cause heating effect only if they are absorbed by some material.

Question 47.
Give one example of heat transfer by radiation.
Answer:
When we sit in the sun, we feel warm. We cannot get heat from the sun by the process of conduction or convection because most of the space between the sun and the earth is a vacuum and both of these modes of heat transfer require medium. Hence, one must be getting heat from the sun by the mode of radiation.

Question 48.
Why do we prefer to wear white or light coloured clothes in summer and black or dark coloured clothes in winter ?
Answer:
We prefer to wear white clothes in summer. The reason is that the white clothes reflects most of the sun’s heat and absorb very little of the sun’s heat, thus they keep our body cool.
We prefer to wear black and dark coloured clothes in winter. The reason is that the black or dark colour clothes absorb most of the sun’s heat and keep our body warm.

Question 49.
The bottom of a cooking utensil is painted black. Give the reason.
Answer:
The bottom part of the cooking utensil or pan is painted black. The reason is that the black surface absorbs more heat and so the contents of utensil or pan get cooked rapidly if its bottom part is painted black.

Question 50.
Draw a labelled diagram of a thermo flask. Explain how the transfer of heat by conduction, convection and radiation is reduced to a minimum in it.
Answer:
Heat transfer is minimised because of:
(1) The vacuum between the two walls, rubber, glass, cork and air do not allow the loss of heat by conduction.
(2) Cork in the neck of flask and the cup over it prevent loss of heat by convection.
(3) Heat cannot be lost by conduction or convection because of vacuum between the two walls.
(4) Heat loss is also minimised by radiation, by making outer surface of inner wall and inner surface of outer wall silvered. The inner wall is a BAD RADIATOR and the outer wall is a GOOD REFLECTOR of radiation.

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 10

C. Numericals

Question 1.
The temperature of a body rises by 1°C. What is the corresponding rise on the (a) Fahrenheit scale (b) Kelvin scale?
Answer:
(a) Since 100 divisions on Celsius scale =180 divisions on the Fahrenheit scale 1 division on Celsius scale
∴ 1 division on Celsius scale
= 1.80 / 1.00 × 1
= 1.8 divisions in the Fahrenheit scale.
For 1°C rise corresponding rise in Fahrenheit = 1.8°F
(b) Since 100 divisions in the Celsius scale = 100 divisions in the Kelvin scale
1 division on Celsius scale = 100 / 100 × 1
= 1 division on Kelvin scale
For 1°C rise corresponding rise in Kelvin is 1 K.

Question 2.
The temperature rises by 18°F. What is the rise on the Celsius scale ?
Answer:
Since 100 divisions on the Celsius scale =180 divisions on the Fahrenheit scale
∴ 18 divisions on Fahrenheit scale.

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 11

Question 3.
Convert 5°F to the Celsius scale.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 12

Question 4.
Convert 40°C to the (a) Fahrenheit scale (b) Kelvin Scale.
Answer:
(a) Fahrenheit scale
C = 40°C
Substitute value of C = 40° in below equation

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 13

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 14

Question 5.
Convert – 40°F to the Celsius scale.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 5 Heat 15

Selina Concise Chemistry Class 7 ICSE Solutions – Elements, Compounds and Mixtures

Selina Concise Chemistry Class 7 ICSE Solutions – Elements, Compounds and Mixtures

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Chemistry Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 3 Elements, Compounds and Mixtures

Points to Remember :

  1. Every substance is made up of very tiny particles, called molecules. Molecules are formed from even smaller particles called atoms.
  2. Element— (a) Element is the simplest pure substance. It cannot be divided further into simpler substances by any chemical method, e.g. oxygen, hydrogen, sulphur, etc.
    (b) At present 116 elements are known, of which 92 are natural elements.
  3. Based on their properties, elements are classified into : metals, non-metals, metalloids, noble gases.
  4. Metals are ductile, malleable, good conductors of heat and electricity, high melting and boiling points. Metals are sonorous, e.g., Iron, Gold, Silver, etc.
  5. Non-metals are solids and brittle in nature, bad conductor of heat and electricity (exception Graphite) low melting and boiling points, e.g. sulphur, carbon, hydrogen, etc.
  6. Metalloids— These elements show properties of both metals and non-metals. They are hard solids, e.g. Boron, Silicon, Arsenic.
  7. Inert or noble gases— These elements do not react chemically with other elements or compounds are called noble (Inert) gases, e.g., helium, neon, argon, etc.
  8. Symbols of Elements— Each element is denoted by a symbol usually to first letter.
    Examples : Oxygen by O Hydrogen by H.
  9. Atom— “An Atom is the smallest particle of an element that can take part in a chemical reaction but may or may not have independent existence.”
    The atom of an element exhibits all the properties of that element.
  10. Molecule— A molecule is the smallest particle of a pure substance of element or compound which has independent existence. It exhibits all the properties of pure substance.
  11. Atomicity— The number of atoms of an element that join together to form a molecule of that element is known as the atomicity.
  12. Molecular Formula— of an element is the symbolic representation of its molecule. It indicates the number of atoms present in it. e.g. Magnesium oxide – MgO.

EXERCISE – I

Question 1.
Write the symbols of helium, silver, krypton, antimony, barium.
Answer:

Element                   Symbol
Helium                         He
Silver                            Ag
Krypton                       Kr
Antimony                    Sb
Barium                        Ba

Question 2.
Write the names of following elements Na, C, Kr, U, Ra, Fe, Co.
Answer:

Symbol           Element
Na                        Sodium
C                          Carbon
Kr                        Krypton
U                         Uranium
Ra                        Radium
Fe                           Iron
Co                        Cobalt

Question 3.
Define :

  1. Elements : An element is the basic form of matter that cannot be broken down into simpler substances by chemical reactions.
  2. Compounds : A compound is a pure substance formed by the chemical combination of two or more elements in a fixed ratio by mass.

Question 4.
Name the main metal present in the following :
Answer:

(a) Haemoglobin                                    Iron
(b) Chalk                                               Calcium
(c) Chlorophyll                                   Magnesium
(d) Chocolate wrappers                    Aluminium

Question 5.
Give four examples of non-metallic elements.
Answer:
Examples : Hydrogen, oxygen, nitrogen, carbon, chlorine, sulphur, phosphorus, etc.

Question 6.
What do you understand by :
Answer:

  1. Metalloids : Metalloids are those substances which have some properties of metals and some of non-metals e.g. boron, silicon.
  2. Noble gases : Noble gases are those which do not react chemically with other elements or compounds e.g. helium, neon, etc.

Question 7.
Select elements and compounds from the following list: Iron, plaster of paris, chalk, common salt, copper, aluminium, calcium oxide, cane sugar, carbon, silica, sodium sulphate, uranium, potassium carbonate, silver, carbon dioxide.
Answer:
Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 3 Elements, Compounds and Mixtures-7

EXERCISE – II

Question 1.
State four difference between compounds and mixtures.
Answer:

Compound Mixture
1. A compound is a pure substance. 1. A mixture is an impure substance.
2. Compounds are always homogeneous. 2. Mixtures may be homogeneous or heterogeneous.
3. A compound has a fixed composition, i.e., it is formed when two or more pure substances chemically combine in a definite ratio by mass. 3. A mixture has no fixed composition, i.e., it is formed by mixing two or more substances in any ratio without any chemical reaction.
4. Formation of a compound involves change in energy. 4. Formation of a mixture does not involve any change in energy.
5. Compounds have specific set of properties. 5. Mixtures do not have any specific set of properties.
6. Components of compounds can be separated only by complex chemical processes. 6. Components of mixtures can be separated by simple physical methods.

 Question 2.
What are the characteristic properties of a pure substance? Why do we need them?
Answer:
Pure substance : Pure substances have a definite set of properties such as boiling point, melting point, density, etc. They are all homogeneous i.e., their composition is uniform throughout the bulk. Both elements and compounds are pure substances.
Pure substances are needed to :

  1. Manufacture medicines.
  2. To prepare chemicals in industry.
  3. For scientific purposes.
  4. To maintain the good health of human beings.

Question 3.
Give two examples for each of the following :
(a) Solid + Solid mixture
(b) Solid + Liquid mixture
(c) Liquid + Liquid mixture
Answer:
(a) Solid + Solid mixture :Sand and sugar,

  • Sand and stone,
  • sand and sugar.

(b) Solid + Liquid mixture :

  • Sand and water,
  • Charcoal and water.

(c) Liquid + Liquid mixture :

  • Oil in water,
  • Alcohol and water.

Question 4.
Define :

  1. Evaporation : Is the process ~of converting a liquid into its vapours state either by exposing it to air or by heating.
  2. Filtration : The process of separating solid particles from liquid by allowing it to pass through a filter paper is called filtration.
  3. Sublimation : The process in which a solid changes directly into its vapours on heating is called sublimation.
  4. Distillation : Distillation is the method of getting a pure liquid from a solution by evaporating and then condensing the vapours.
  5. Miscible liquids : Homogeneous liquid-liquid mixtures are called miscible liquids.
  6. Immiscible liquids : Heterogeneous liquid-liquid mixtures are called immiscible liquids.

Question 5.
Name the process by which the components of following mixtures can be separated.

  1. Iron and sulphur
  2. Ammonium chloride and sand
  3. Common salt from sea water
  4. Chaff and grain
  5. Water and mustard oil
  6. Sugar and water
  7. Cream from milk

Answer:

  1. Magnetic separation.
  2. Sublimation.
  3. Evaporation.
  4. Winnowing separates chaff (lighter) from heavier grains in two different heaps.
  5. Mustard oil and water is liquid-liquid immiscible mixture and is separated by separating funnel. Water being the heavier forms the lower layer.
  6. By evaporation in this process of converting a liquid into its vapour state by heating. Liquid is heated and water evaporate and sugar is obtained.
  7. Centrifugation.

Question 6.
How will you separate a mixture of common salt, chalk powder and powdered camphor? Explain.
Answer:
Comphor with sublimation. Chalk powder by Alteration then the residual left is common salt.

Question 7.
How is distillation more advantageous than evaporation?
Answer:
The advantage of distillation is that both components of the
solid and liquid mixture are obtained. Whereas in evaporation only solid is obtained.

Question 8.

  1. What is chromatography?
  2. Why is it named so?
  3. What are the advantages of chromatography?
  4. Name the simplest type of chromatography?
  5. On what principle is this method based?
  6. What is meant by stationary phase and mobile phase in chromatography?

Answer:

  1. The process of separating different dissolved constituents of a mixture by their absorption on an appropriate material is called chromatography.
  2. It is named so, because earlier it was used to separate mixtures containing coloured components only but these days this technique is applied to colourless substances too.
  3. Advantages of chromatography :
    (i) A very small quantity of the substance can be separated.
    (ii) Components with very similar physical and chemical properties can be separated.
    (iii) It identifies the different constituents of a mixture.
    (iv) lt also helps in quantitative estimation of components of a mixture.
  4. The simplest type of chromatography is “Paper chromatography”.
  5. Chromotography is based on differential affinities of compounds towards two phases i.e. stationary and mobile phase.
  6. The filter paper acts as “stationary phase” while the solvent act as “mobile phase”.

Question 9.
On what principle are the following methods of separation based? Give one example of a mixture for each of the methods mentioned in which they are used
Answer:

  1. Sublimation : Change of solid into vapours directly on heating and change of vapours into solid again on
    Example : Salt from ammonium chloride.
  2. Filtration : The process of separating insoluble solid particles from a liquid by allowing it to pass through a filter is called Alteration. These filters allow liquids to pass through them but not solids. The insoluble solid left on the filter is called the residue, while the liquid which passes through the filter is called the filtrate. Mixtures like chalk and water, clay and water, tea and tea leaves, sawdust and water, etc., are separated by this method.
  3. Sedimentation and decantation : The settling down of suspended, insoluble, heavy, solid particles in a solid- liquid mixture when left undistrubed is called sedimentation.
    The solid which settles at the bottom is called sediment while the clear liquid above it is called supernatant liquid.
    The process of pouring out the clear liquid, without disturbing the sediment, is called decantation.
    Example : A mixture of sand and water.
  4. Solvent extraction method : This method is used when one of the solid components is soluble in a liquid.
    Example : A mixture of sand and salt can be separated by this method. Salt gets dissolved in water while sand settles down in the container. The salt solution is then decanted. Salt is separated from the solution by evaporation. In this way, they can be separated.
  5. Magnetic separation : This method is used when one of the components of the mixture is iron. Iron gets attracted towards a magnet and hence can be separated. Mixtures of iron and sulphur, iron and sand, etc., can be separated by moving a magnet over them. Iron gets attached to the magnet and is separated.
  6. By using a separating funnel : It is a simple device used to separate the components of a liquid-liquid heterogeneous mixture.
    Example : Kerosene oil and water. The mixture is placed in a separating funnel and allowed to stand for sometime. The components form two clear layers. Water being heavier forms the lower layer and oil being lighter forms the upper layer. When the stopper of the funnel is opened, the heavier liquid trickles out slowly and is collected in a vessel. The stopper is closed when the bottom layer is entirely removed the funnel. In this way, the two liquids are separated.
  7. Fractional distillation : The process of distillation is used for separating the components of a homogeneous liquid-liquid mixture, like water and alcohol. This is based on the fact that alcohol boils at a lower temperature than water. The vapour of alcohol are collected and cooled while water is left behind in the original vessel. Thus, two liquids having different boiling points can be separated by distillation provided that difference in their boiling points must be 25 °C or more.

OBJECTIVE TYPE QUESTIONS

Question 1.
Fill in the blanks:
Answer:

  1. Elements are made up of same kind of atoms.
  2. Elements and compounds are pure substances.
  3. In a mixture the substances are not combined chemically.
  4. Clay is separated from water by the method called loading and decantation.
  5. Crystallisation is a process to obtain a very pure form of a solid dissolved in a liquid.
  6. Camphor and ammonium chloride can sublimate.

Question 2.
Give one word answers for the following :
Answer:

  1. The solid particles which remain on the filter paper after the filtration residue.
  2. The liquid which evaporates and then condenses during the process of distillation distillate.
  3. The process of transferring the clean liquid after the solid settles at the bottom of the container decantation.
  4. The process by which two miscible liquids are separated fractional distillation.

MULTIPLE CHOICE QUESTIONS

Select the correct alterative from the choices given for the following statements:
Question 1.
A pure liquid is obtained from a solution by :
Answer:

  1. evaporation
  2. distillation
  3. Alteration
  4.  crystallisation

Question 2.
Components of crude petroleum can be separated by :
Answer:

  1. distillation
  2. evaporation
  3. filtration
  4. fractional distillation

Question 3.
Example of a homogeneous mixture is :
Answer:

  1. tap water 
  2. distilled water
  3.  sand and water
  4. water and oil

Question 4.
In chromatography the filter paper is :
Answer:

  1.  stationary phase 
  2. mobile phase
  3. mixture
  4. none of the above

Question 5.
A set of mixture is :
Answer:

  1.  ink, honey, icecream, milk
  2. tapwater, gold, common salt, alloy
  3.  milk, brass, silver, honey
  4. butter, petroleum, tapwater, iron

 

Selina Concise Physics Class 7 ICSE Solutions – Physical Quantities and Measurement

Selina Concise Physics Class 7 ICSE Solutions – Physical Quantities and Measurement

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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

Selina Class 7 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 7 Physics Chapter 1 Physical Quantities and Measurement

  • Points to Remember
  • The mass of a body is the quantity of matter contained in a body regardless of its volume or any force acting on it.
  •  The weight of a body is the force with which every body is attracted towards its centre.
  •  The unit of mass in S.I. system is kilogram (kg). Higher units of mass are quintal and metric tonne.
  •  The weight of body changes with acceleration due to gravity.
  •  Weight is zero at the centre of the earth.
  •  Mass per unit volume of a substance is called density of the body.
  •  The unit of density in S.I. system is kg m-3 and gcm-3 in C.G.S. system.
  •  The density in S.I. system is 1000 × numerical value in C.G.S. system.
  •  The density of liquids and gases decreases or increases with the rise or fall in temperature.
  • The cycle of upward and downward movements of the fluid form currents in the medium which are known as convectional currents.

Test Yourself

A. Objective Questions 

1. Write true or false for each statement

(a) The S.I. unit of volume is litre.
Answer. False.
The S.I. unit of volume is cubic metre.

(b) A measuring beaker of capacity 200 ml can measure only the volume. 200 ml of a liquid.
Answer. True.

(c) cm2 is a smaller unit of area than m2.
Answer. True.

(d) Equal volumes of two different substances have equal masses.
Answer. False.
Equalvolumes of two different substances have different masses.

(e) The S.I. unit of density is g cm-3.
Answer. False.
The S.I. unit of density is Kg m-3.

(f) 1 g cm-3 = 1000 kg m-3.
Answer. True.

(g) The density of water is maximum at 4°C.
Answer. True.

(h) The speed 5 ms-1 is less than 25 km h-1.
Answer. True.

(i) The S.I. unit of speed is ms-1.
Answer. True

2. Fill in the blanks

(a) l m3106 cm3
(b) The volume of an irregular solid is determined by the method of displacement of liquid.
(c) Volume of a cube = (one side)
(d) The area of an irregular lamina is measured by using a grapl paper.
(e) Mass = density × volume.
(f) The S.I. unit of density is kg m-3.
(g) 1 g cm-3 = 1000 kg m-3.
(h) 36 km h-1 = 10 ms-1.
(i) Distance travelled d = speed v × time t.

3. Match the following

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 1

4. Select the correct alternative 

(a) One litre is equal to :

  1.  1 cm-3
  2. 1 m3
  3.  10-3  cm3
  4. 10-3 m3

(b) A metallic piece displaces water of volume 15 ml. The volume of piece is :

  1.  15 cm3
  2.  15 m3
  3.  15 × 103 cm3
  4.  15 × 103 cm3

(c) A piece of paper of dimensions 1.5 m x 20 cm has area :

  1.  30 m2
  2.  300 cm2
  3.  0.3 m2
  4.  3000 m3

(d) The correct relation is :

  1.  d = M × V
  2.  M = d × Y  
  3.  V = d × M
  4.  d = M + V

(e) The density of alcohol is 0.8 g cm-3. In S.I. unit, it will be :

  1.  0.8 kg m-3
  2.  0.0008 kg m-3
  3.  800 kg m-3
  4.  8 x 103 kg m-3

(f) The density of aluminium is 2.7 g cm-3 and of brass is 8.4 g cm-3. For the same mass, the volume of:

  1.  both will be same
  2.  aluminium will be less than that of brass
  3.  aluminium will be more than that of brass
  4.  nothing can be said.

(g) A block of wood of density 0.8 g cm-3 has a volume of 60 cm3. The mass of block will be :

  1.  60.8 g
  2.  75 g
  3.  48 g
  4.  0.013 g

(h) The correct relation for speed is

  1.  Speed = distance x time
  2.  speed = distance / time
  3.  speed = time / distance
  4.  speed = 1 / distance x time

(i) A boy travels a distance 150 m in 1 minute. His speed is

  1.  150 m s-1
  2.  2.5 m s-1
  3.  25 m s-1
  4.  9 m s-1

B. Short/Long Answer Questions

Question 1.
Define the term volume of an object.
Answer:
The space occupied by an object is called its volume.

Question 2.
State and define the S.I. unit of volume.
Answer:
S.I. unit of volume – The S.I. unit of volume is cubic metre. In short form, it is written as m3.
One cubic metre is the volume of a cube of each side 1 metre as shown in figure below i.e., 1 m3 = 1 m × 1 m × 1 m.

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 2

Question 3.
State two smaller units of volume. How are they related to the S.I. unit?
Answer:
A smaller unit of volume is cubic centimetre (symbol cm3) and cubic decimetre (symbol 1 dm3). One cubic centimetre is the volume of a cube of each side 1 centimetre, i.e.,
1 cm3 = 1 cm × 1 cm × 1 cm.
Relationship between m3 and cm3
l m= lm × lm × lm
= 100 cm × 100 cm × 100 cm
= 10,00,000 cm3 = 106 cm3.
Relationship between m3 and dm3
l m3 = lm × 1 m × 1 m .
= 10 dm × 10 dm × 10 dm
= 1000 dm
= 103 dm
Note 1 m = 10 dm

Question 4.
How will you determine the volume of a cuboid ? Write the formula you will use.
Answer:
Volume of a cuboid = length × breadth × height.

Question 5.
Name two devices which are used to measure the volume of an object. Draw their neat diagrams.
Answer:
Two devices that are used to measure the volume of an object are :
(i) Measuring cylinder and
(ii) Measuring beaker.

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 3

Question 6.
How can you determine the volume of an irregular solid (say a piece of brass) ? Describe in steps with neat diagrams.
Answer:
To measure the volume of a piece of stone.
Take a piece of brass, a measuring cylinder, fine thread of sufficient length and some water.
Place a measuring cylinder on a flat horizontal surface and fill it partially with water. Note the reading of the water level very carefully. Now tie the piece of brass with a thread and dip it completely into water. We see that the level of water rises. Note the reading of the new water level.

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 4
The difference in the two levels of water gives the volume of the piece of brass
Initial level of water = 60 ml
Level of water when brass is immersed = 80 ml
∴ Volume of water displaced = 80 ml – 60 ml = 20 ml
∴ Volume of the piece of brass = 20 cm3
Note : 1 ml = 1 cm3

Question 7.
You are required to take out 200 ml of milk from a bucket full of milk. How will you do it ?
Answer:
By using the measuring beaker A measuring beaker is used to measure a fixed volume of liquid from a large volume. Suppose it is required to measure 200 ml of milk from the milk contained in a bucket. For this, take the measuring beaker of capacity 200 ml. Wash it and dry it. Then, immerse the measuring beaker well inside the milk contained in the bucket so that the beaker gets completely filled with the milk.
Take out the measuring beaker from the bucket gently so that no milk splashes out and then pour the milk from the measuring beaker into the another empty vessel.

Question 8.
Describe the method in steps to find the area of an irregular lamina using a graph paper.
Answer:
Method to find the area of an irregular lamina using a graph paper : First, place the lamina over a graph paper and draw its boundary line on the graph paper with a pencil. Then remove the lamina and count and note the number of complete squares as well as the number of squares more than half within the boundary line (only the squares less than half, are left while counting). The area of lamina is equal to the sum of the area of complete squares and the area of squares more than half. Let n be the total number of complete and more than half or half squares within the boundary of lamina. Since area of one big square is 1 cm × 1cm = 1 cm2, so the area of lamina will be n x

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 5

Question 9.
Define the term density of a substance.
Answer:
The density of a substance is defined as the mass of a unit volumx of that substance.

Question 10.
State the S.I. and C.G.S. units of density. How are they inte related ?
Answer:
The S.I. unit of mass is kilogram (symbol kg) and of volume is
cubic metre (symbol m3). Therefore S.I. unit of density is kg/m3
or kg m-3.
The C.G.S. unit of mass is gram (symbol g) and of volume is cubic centimetre (symbol cm3). Therefore the C.G.S. unit of
density is g/cm3 or g cm-3.

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 6

Question 11.
‘The density of brass is 8.4 g cm’3’. What do you mean by the statement ?
Answer:
Density of brass is 8.4 g cm-3. This means that unit volume of brass contain 8.4 g mass.

Question 12.
Arrange the following substances in order of their increasing density:
(a) iron
(b) cork
(c) brass
(d) water
(e) mercury
Answer:
b<a<c<d<e

Question 13.
How does the density of water changes when :
(a) it is heated from 0°C to 4°C,
(b) it is heated from 4°C to 10°C ?
Answer:
(a) Water contracts on heating from 0°C to 4°C and expands on heating above 4°C.
(b) The density of water is maximum at 4°C. It decreases when it is cooled from 4°C to 0°C or it is heated above 4°C.

Question 14.
Write the density of water at 4°C.
Answer:
The density of water at 4°C is 1.0 g cm-3, or 1,000 kg m-3

Question 15.
Explain the meaning of the term speed.
Answer:
The distance covered or travelled by a body in one second is called the speed of the body, i.e.

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 7

Question 16.
Write the S.I. unit of speed.
Answer:
The S.I. unit of speed is metre/second or metre per second. Its symbol is m s-1.

Question 17.
A car travels with a speed 12 m s”1, while a scooter travels with a speed 36 km h-1. Which of the two travels faster ?
Answer:
Speed of car = 12 m s-1
Speed of scooter = 36 km h-1
here, 1 km = 1000 m
1 hr = 3600 sec

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 8
∴ Speed of car is more. Car travels faster than scooter.

C. Numericals

Question 1.
The length, breadth and height of a water tank are 5 m, 2.5 m and 1.25 m respectively. Calculate the capacity of the water tank in (a) m3 (b) litre.
Answer:
Given,
Length (1) = 5m
Breadth (b) = 2.5 m
and Height (h) = 1.25 m

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 9

Question 2.
A solid silver piece is immersed in water contained in a measuring cylinder. The level of water rises from 50 ml to 62 ml. Find the volume of silver piece.
Answer:
Given, initial level of water .v1 = 50 ml
Final level of water v2 = 62 ml
Volume of silver piece V = v2 – v1
= 62 ml – 50 ml
= 12 ml or 12 cm3

Question 3.
Find the volume of a liquid present in a dish of dimensions 10 cm x 10 cm x 5 cm.
Answer:
Volume of water = Length × breadth × height
= 10 cm × 10 cm × 5 cm
= 500 cm3 or 500 ml.

Question 4.
A rectangular field is of length 60 m and breadth 35 m. Find the area of the field.
Answer:
Length of a rectangular field = 60 m
Breadth of rectangular field = 35 m
∴ Area = 60 m × 35 m
= 2100 m2

Question 5.
Find the approximate area of an irregular lamina of which boundary line is drawn on the graph paper shown in fig. 1.16. below.

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 10
Answer:
From figure, the number of complete squares = 11
The number of squares more than half = 9
∴ Total number of squares = 11 + 9 = 20
∴ Area of the 1 square = 1 cm × 1cm = 1 cm2
∴ Area of 20 squares = 20 × 1 cm2 = 20 cm2
∴ Approximate area of irregular lamina = 20 cm2

Question 6.
A piece of brass of volume 30 cm3 has a mass of 252 g. Find the density of brass in (i) g cm-3, (ii) kg m-3.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 11

Question 7.
The mass of an iron ball is 312 g. The density of iron is 7.8 g cm-3. Find the volume of the ball.
Answer:

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Question 8.
A cork has a volume 25 cm3. The density of cork is 0.25 g cm-3. Find the mass of cork.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 14

Question 9.
The mass of 5 litre of water is 5 kg. Find the density of water in g cm-3.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 15

Question 10.
A cubical tank of side 1 m is filled with 800 kg of a liquid. Find: (i) the volume of tank, (ii) the density of liquid in kg m-3.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 16

Question 11.
A block of iron has dimensions 2 m × 0.5 m × 0.25 m. The density of iron is 7.8 g cm-3. Find the mass of block.
Answer:
Given, l = 2m
b = 0.5 m

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 17

Question 12.
The mass of a lead piece is 115 g. When it is immersed into a measuring cylinder, the water level rises from 20 ml mark to 30 ml mark.
Find:
(i) the volume of the lead piece,
(ii) the density of the lead in kg m-3.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 18

Question 13.
The density of copper is 8.9 g cm-3. What will be its density in kg m-3 ?
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 19

Question 14.
A car travels a distance of 15 km in 20 minute. Find the speed of the car in (i) km h-1, (ii) m s-1.
Answer:
Distance travelled by car =15 km
Time taken = 20 minutes
(i) Speed of car in km h-1
Convert 20 minutes to hour

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Question 15.
How long a train will take to travel a distance of 200 km with a speed of 60 km h-1 ?
Answer:
Distance covered by train = 200 km
Speed of train = 60 km h-1

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Question 16.
A boy travels with a speed of 10 m s-1 for 30 minute. How much distance does he travel ?
Answer:
Speed of boy = 10 m s-1
Time taken = 30 minutes
speed = distance travelled / time taken
Distance travelled = Speed × Time taken
Convert 30 minutes to seconds
1 minute = 60 sec
30 minute 60 × 30 = 1800 seconds
Putting the value of speed and time we get
Distance travelled = 10 ms-1 × (1800 sec) = 18000 m
= 18000 metre or 18 km Ans.

Question 17.
Express 36 km h-1 in m s-1
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 23

Question 18.
Express 15 m s-1 in km h-1.
Answer:

Selina Concise Physics Class 7 ICSE Solutions Chapter 1 Physical Quantities and Measurement 24

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