Dattu – ML Aggarwal Solutions

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 12 Symmetry

August 11, 2022August 30, 2018 by Dattu

ML Aggarwal Class 6 Solutions Chapter 12 Symmetry

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 12 Symmetry

ML Aggarwal Solutions ICSE Solutions Selina ICSE Solutions

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 12 Symmetry Exercise 12.1

Solution 01:

(i) One
(ii) None
(iii) One
(iv) One
(v) None
(vi) None
(vii) One
(viii) One
(ix) Three

Solution 02:

(i) One
(ii) None
(iii) One
(iv) One
(v) None
(vi) None

Solution 03:
(i) One
(ii) None
(iii) One
(iv) One

Solution 04:

(i) One
(ii) None
(iii) One
(iv) One

Solution 05:

(i) One
(ii) None
(iii) One
(iv) One
(v) None
(vi) None

Solution 06:

J, S, L and K have no line of symmetry.

Solution 07:

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 12 Symmetry Exercise 12.2

Solution 01:
(i)

(ii)

(iii)

Solution 02:
(i)

(ii)

(iii)

Solution 03:

Solution 04:

Solution 05:

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes

August 11, 2022August 28, 2018 by Dattu

ML Aggarwal Class 6 Solutions Chapter 11 Understanding Symmetrical Shapes

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes

ML Aggarwal Solutions ICSE Solutions Selina ICSE Solutions

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.1

Solution 01:
It gives accurate measurement and avoids error due to thickness of ruler or positioning of eye (due to angular viewing)

Solution 02:
By measuring the lengths of the given figure

(i) AB = CD
(ii) BC < AB
(iii) AC = BD
(iv) CD < BD

Solution 03:
Given that AC = 10 CM, AB = 6 CM and BC = 4 CM
By constructing line segment by the given data, the model drawn as below.

Point B lies in between A and C

Solution 04:

By measuring the Lengths of line segments in the above figure
AB = 3 CM
BC = 1.5 CM
(i) It can be observed that AC = AB + BC [i.e. 4.5 CM = 3 CM + 1.5 CM]
(ii) AC – BC = AB 4.5 CM – 1.5 CM = 3 CM by measurement fount that AB = 3 CM, so AC – BC = AB.

Solution 05:
By measuring the lengths of the given figure.

Given data
AB = 1.9 CM
BC = 0.7 CM
CD = 1.9 CM
AD = 4.5 CM
(i) AC + BD = 2.6 CM + 2.6 CM = 5.2 CM AD + BC = 4.5 CM + 0.7 CM = 5.2 CM Hence, AC + BD = AD +BC.
(ii) AB + CD = 1.9 CM + 1.9 CM = 3.8 CM AD – BC = 4.5 CM – 0.7 CM = 3.8 CM Hence, AC + BD = AD +BC.

Solution 06:
By measuring the lengths of the given triangle ABC as below

AB = 2.6 CM, AC = 3.8 CM and BC = 3.8 CM.
(i) AB + BC = 2.6 CM + 3.8 CM = 6.4 CM AC = 3.8 CM Hence, AB + BC > AC.
(ii) BC + AC = 3.8 CM + 3.8 CM = 7.6 CM AB = 2.6 CM Hence, AB > BC + AC.
(iii) AC + AB = 3.8 CM + 2.6 CM = 5.4 CM BC = 3.8 CM Hence, AC + AB > BC.

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.2

Solution 01:
(i) When the hour hand moves from 4 to 10 clockwise, fraction of revolution turned = ½ Number of right angles turned = 2.

(ii) When the hour hand moves from 2 to 5 clockwise, fraction of revolution turned = ¼. Number of right angles turned = 1.

(iii) When the hour hand moves from 7 to 10 clockwise, fraction of revolution turned = ¼. Number of right angles turned = 1.

(iv) When the hour hand moves from 8 to 5 clockwise, fraction of revolution turned = ¾. Number of right angles turned = 3.

(v) When the hour hand moves from 11 to 5 clockwise, fraction of revolution turned = ½ Number of right angles turned = 2.

(vi) When the hour hand moves from 6 to 3 clockwise, Fraction of revolution turned = ¾. Number of right angles turned = 3.

Solution 02:
(i) When the hour hand moves from 10 and makes half revolution, clockwise it will stop at 4.

(ii) When the hour hand moves from 4 and makes 1/4 revolution, clockwise it will stop at 7.

(iii) When the hour hand moves from 4 and makes 3/4 revolution, clockwise it will stop at 1.

Solution 03:
(i) When the hour hand moves from 6 and turns through 1 right angle, clockwise it will stop at 9.

(ii) When the hour hand moves from 8 and turns through 2 right angles, clockwise it will stop at 2.

(iii) When the hour hand moves from 10 and turns through 3 right angles, clockwise it will stop at 7.

(iv) When the hour hand moves from 7 and turns through 2 straight angles, clockwise it will stop at 7.

Solution 04:
(i) While turning from north to south Fraction of a revolution = ¾. Number of right angles = 3.

(ii) While turning from south to east Fraction of a revolution = 1/4. Number of right angles = 1.

(iii) While turning from east to west (clockwise). Fraction of a revolution = 1/2. Number of right angles = 2.

Solution 05:
(i) Straight angle – (c) Half of a revolution
(ii) Right angle – (d) One fourth of a revolution
(iii) Complete angle – (f) One complete revolution
(iv) Acute angle – (b) Less than one fourth of a revolution
(v) Obtuse angle – (e) Between ¼ and ½ of a revolution
(vi) Reflex angle – (a) More than half of a revolution

Solution 06:
(i) Acute angle
(ii) Obtuse angle
(iii) Right Angle
(iv) Straight angle
(v) Reflex angle
(vi) Reflex angle
(vii) Acute angle
(viii) Obtuse angle

Solution 07:
(i) Angle a and Angle c are acute, Angle b is obtuse
(ii) Angle x and Angle z are Obtuse, Angle y is acute
(iii) Angle p is obtuse, Angle q and Angle s are acute and Angle r is reflex.

Solution 08:

By measuring the protractor marked angles are as follows
(i) 62^o
(ii) 116^o
(iii) 121^o

Solution 09:

By measuring the protractor marked angles are as follows
(i) 315^o
(ii) 235^o

Solution 10:
In the clock the angle between every numeric is 30^o i.e. angle between 1 and 2 is 30^o, 2 and 3 is 30^o and 4 and 6 is 30^o x 2 = 60^o

Similarly,
(i) Angle between the hands of the clock – 60^o

(ii) Angle between the hands of the clock – 30^o

(iii) Angle between the hands of the clock – 150^o

Solution 011:

Smaller angle formed by the hour and minutes hands of a clock at 7’O clock is 150o [30o x 5 = 150o] (Type of Angle – Obtuse angle) as shown in the above model
Other Angle = 360^o – 150^o = 210^o (Type of Angle – Reflex angle)

Solution 12:
One is a 30^o – 60^o – 90^o set square; the other is a 45^o – 45^o -90^o set square. The angle of measure 90o is common between them.

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.3

Solution 01:
Two straight line are called perpendicular lines if they intersect at right angles.
In the given models (i), (iii) and (iv) are perpendicular lines.

Solution 02:
(i) Yes, CE = EG; E is the midpoint of CG
(ii) Yes, PF Line bisect segment BH – E is the midpoint of BH and Line P bisects line segment BH.
(iii) Line segment DF, Line segment BH
(iv) All are true (AC > FG, CD = GH and BC < EG)

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.4

Solution 01:

(i) Two sides are equal – Isosceles triangle
(ii) Three sides are different – Scalene triangle
(iii) Three sides are equal – Equilateral triangle

Solution 02:

(i) Angle is 90^o – Right angled triangle
(ii) Angle is more than 90^o – Obtuse angled triangle
(iii) Angle is less than 90^o – acute angled triangle

Solution 03:

(i) Angle is less than 90^o – acute angled triangle and two sides are in equal in length- Isosceles triangle.
(ii) Angle is 90^o – right angled triangle and three sides are in not equal in length- scalene triangle.
(iii) Angle is more than 90^o – Obtuse angled triangle and two sides are in equal in length- Isosceles triangle.
(iv) Angle is 90^o – right angled triangle and two sides are equal in length- Isosceles triangle.
(v) Angle is less than 90^o – acute angled triangle and three sides are in equal in length- Equilateral triangle.
(vi) Angle is more than 90^o – Obtuse angled triangle and three sides are in not equal in length- scalene triangle.

Solution 04:
(i) Three sides of equal length – (e) Equilateral
(ii) 2 Sides of length – (g) Isosceles
(iii) All sides of different length – (a) Scalene
(iv) 3 acute angles – (f) Acute angled
(v) 1 right angle – (d) Right Angled
(vi) 1 Obtuse angle- C) Obtuse Angled
(vii) 1 Right angle with two sides of equal length – (b)Right angled Isosceles

Solution 05:
(i) False
(ii) True
(iii) True
(iv) False
(v) False
(vi) False
(vii) True
(viii) False

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.5

Solution 01:
(i) True
(ii) True
(iii) True
(iv) True
(v) False
(vi) False
(vii) False
Solution 02:
(i) Not a polygon, because it is not a closed curve
(ii) Polygon, because it is a simple closed curve made up entirely of line segments
(iii) Not a polygon, because it is not a simple curve
(iv) Not a polygon, because it is not made up of entirely line segments.

Solution 03:
(i) Pentagon

(ii) Quadrilateral

(iii) Hexagon

(iv) Octagon

Solution 04:
ABCDE is a regular pentagon and diagonals as in the below figure.

Solution 05:

Let ABCDEF be regular hexagon then
(i) Triangle ABC is an Isosceles triangle.
(ii) Triangle CEF is a right angled triangle.

Solution 06:
ABCD is a regular quadrilateral – Square.

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.6

Solution 01:
(i) Cuboid
(ii) Cuboid
(iii) Cuboid
(iv) Cylinder
(v) Cube
(vi) Sphere

Solution 02:
(i) Cone

(ii) Sphere

(iii) Cube

(iv) Pyramid

(v) Cylinder

(vi) Cuboid

Solution 03:
(i) A cube has 6 square faces, 12 edges and 8 vertices.
(ii) A triangular prism has 2 triangular faces, 3 rectangular faces, 9 edges and 6 vertices.
(iii) A triangular pyramid has 4 faces, 6 edges and 4 vertices.

Frank ICSE Solutions for Class 9 Physics Chapter 5 – Heat Solutions

August 11, 2022November 10, 2017 by Dattu

Frank ICSE Solutions for Class 9 Physics Chapter 5 – Heat Solutions

PAGE NO: 191.
Solution – 01.
Heat is defined as a form of energy which flows from one point to another on account of temperature difference.
Solution – 01.

Yes, heat is a form of energy

Solution – 02.

Joule is the SI unit of heat.

Solution – 03.

One calorie is defined as the quantity of heat required to raise the temperature of 1 gram of water through 1oC.

Solution – 04.

1 calorie = 4.2 joules.

Solution – 05.

Temperature is the degree of hotness or coldness of a body compared to other bodies around it.
SI unit of temperature is Kelvin (K).

Solution – 06.

We feel cold on touching ice because heat flows from our warm hands to cold ice. Due to this flow of heat from hand to ice, the temperature of our hand falls. This is why we feel cold.

Solution – 07.

We feel cold on touching ice because heat flows from our warm hands to cold ice. Due to this flow of heat from hand to ice, the temperature of our hand falls. This is why we feel cold.

Solution – 08.

Solution – 09.

Heat flows from a body at a higher temperature to a body of lower temperature.

Solution – 10.

Yes, heat is the cause of temperature because temperature of a body rises when the heat flows into the body.

Solution – 11.

Heat changes the temperature of a body due to flow of heat in or out of the given body.

Solution – 12.

Calorie. Because 1 calorie = 4.2 joules.

Solution – 13.

No, the exact relation is as given
1 calorie = 4.2 joules.

Solution – 14.

Yes, because the heat flow is only due to temperature difference between the temperature of two bodies.

PAGE NO: 229.
Solution – 01.

Temperature is the degree of hotness or coldness of a body compared to other bodies around it.
SI unit of temperature is Kelvin (K)

Solution – 02.

Normal temperature of human body is 370C.

Solution – 03.

To convert 200c into0F
Toc /100 = (Tof -32)/180
20/100 = (Tof – 32)/180
20 X 180/100 = Tof -32
Tof = 36+32 = 680F

Solution – 04.

Upper fixed point on the Celsius scale is 1000C.

Solution – 05.

Tk = 80 K
Tc = Tk -273
Tc = 80 – 273
Tc = -1530C

Solution – 06.

SI unit of latent heat is Joule per kg (J/kg).

Solution – 07.

Relative humidity is defined as the amount of water vapour in the air compared to the amount needed for saturation.

Solution – 08.

Coefficient of Linear expansion is equal to the change in length of a rod of length 1m when its temperature rises by 10C. Its SI unit is oc-1.

Solution – 09.

Celsius was the scientist who discovered the first thermometer in 1710.

Solution – 10.

According to principle of calorimetry of mixtures,
Heat gained = Heat lost

Solution – 11.

SI unit of coefficient of cubical expansion is oc-1.

Solution – 12.

Two uses of bimetallic strip are
(a). As thermostat in electric iron
(b). As balance wheel in watches

Solution – 13.

Telephone wires sag in summer because due to heat of the sun, the wire expands and increases in length, thus they sag in summer.

Solution – 14.

Solution – 15.

Not all substances expand on heating. Some examples of substances which do not expand on heating are plastics, polythene and rubber.

Solution – 16.

Evaporation is the phenomenon of a change of a liquid into vapour without raising the temperature. Evaporation needs energy for phase change from liquid to gases. As water evaporates off your skin, it absorbs energy(heat) from the body to make the phase change to gas thus cooling the body.

Solution – 17.

Factors affecting evaporation are
(a). Humidity- more the humidity less is the evaporation
(b). surface area- more the surface area more is the evaporation
(c). wind- more the wind more is the evaporation
(d). temperature- more the temperature more is the evaporation

Solution – 18.

The cold air that blows from land towards sea during night, is called land breeze
The cold air that blows from the sea towards the land during the day is known as the sea breeze. These breezes are the examples of natural convection current.

Solution – 19.

No, the conduction is not possible in gases. Gases are bad conductors.

Solution – 20.

No, conduction is not possible in vacuum.

Solution – 21.

The velocity of thermal radiations is equal to the speed of light i.e. 3 x 108 m/s.

PAGE NO: 230.
Solution – 22.

We wear woolen clothes in winter because woolen clothes have tiny pores and air is trapped in these pores and being a bad conductor, the trapped air obstructs the flow of body heat to the surroundings.

Solution – 23.

A newly made quilt is warmer than an old one because the cotton in the old quilt gets compressed and very little air will remain trapped in it, hence heat insulation is quite poor.

Solution – 24.

In cold countries, water pipes are covered with poor conductors because poor conductor prevents water from freezing and thus prevent these pipes from bursting.

Solution – 25.

Three devices used to detect heat radiations are
(a). Blackened bulb thermometer
(b). Differential air thermo scope
(c). Thermopile

Solution – 26.

The increase in size of a body on heating is called thermal expansion.

Solution – 27.

Linear expansion is the increase in length of a solid on heating.

Solution – 28.

Coefficient of Linear expansion is equal to the change in length of a rod of length 1m when its temperature rises by 10C.

Solution – 29.

A bimetallic strip consists of two metal strips- one with high coefficient of expansion and the other with low coefficient of expansion.

Solution – 30.

SI unit of coefficient of linear expansion is oc-1.

Solution – 31.

Water is the substance which contracts, when heated from 00C to 40C.

Solution – 32.

Coefficient of volume expansion is equal to the change in volume of a rod of volume 1m3 when its temperature rises by 1oC.

Solution – 33.

SI unit of coefficient of volume expansion is oC-1.

Solution – 34.

Two uses of bimetallic strip are
(a). As thermostat in electric iron
(b). As balance wheel in watches

Solution – 35.

We should heat the neck of the bottle because due to heating the neck will expand and loosen the stopper stuck in the neck. In this way, we can easily remove the stopper from the bottle.

Solution – 36.

When hot water is poured into a thick glass tumbler, it generally cracks because on pouring hot water in the tumbler the inner surface heats up and expands more as compared to its outer surface. This unequal expansion between the two surfaces causes a strain and the tumbler cracks.

Solution – 37.

A substance is made up of molecules arranged in a lattice. On heating, the molecules vibrate faster in the lattice and bump into each other harder. So the distance between the molecules increases thus expanding lattice. Thus, the substances expand on heating.

Solution – 38.

There are three types of thermal expansion
(a). Linear expansion
(b). Superficial expansion
(c). Cubical expansion

Solution – 39.

Gaps are left in the railway tracks because the tracks gets heated during the day and as a result they increase in length. If the gaps are not provided, the railway line would buckle outward and may cause derailment.

Solution – 40.

The beams of the bridges expand maximum during the summer days and contract maximum during the winter nights. If the beams are fixed at both ends on the pillars, they may develop crack due to expansion and contraction. To avoid this, beams are made to rest on rollers on the pillars to provide space for expansion.

Solution – 43.

Solution – 44.

A ventilator is provided in a room because it helps in removing the hot air from the room and allows the fresh and cold air to come in.

Solution – 45.

No, it is not possible to heat a liquid or gas from above because the transfer of heat through convection takes place vertically upwards in liquids and gases. So if they are heated from above, the liquid or gas at the top will only be heated because most liquids and gases are themselves bad conductor of heat so they cannot conduct heat from top layer to the bottom layer.

Solution – 46.

(a). Water is heated generally from below because water itself is a bad conductor of heat and the transfer of heat through convection take place vertically upwards.
(b). Land becomes warmer than water during the day because water has more specific heat capacity so it absorbs the heat and heats up slowly but on the other hand land has less specific heat and it heats up faster than water.

Solution – 47.

Main characteristics of thermometric substance are
(a). The substance should have high coefficient of expansion so that it is sensitive to the smallest change in temperature
(b). The substance should have uniform expansion all over its entire volume
(c). The substance should have minimum specific heat so that it absorbs minimum heat from the body under measurement.

Solution – 48.

Wood is an insulator of heat.

Solution – 49.

(a). In cold countries, windows are provided with two glass panes because in between these two glass panes, a thin layer of air is present: air being a bad conductor obstructs the conduction of heat from the room to outside.
(b). 1 calorie = 4.2 joules
(c). Yes, it is possible to boil water in a thin paper cup because when heated the heat in the paper cup is transferred to the water through convection and paper cup doesn’t get sufficient heat to get burnt

Solution – 50.

Thermometer works on the principle that substances expand on heating and contract on cooling. So we use a thermometric substance which expands and contracts uniformly.

Solution – 51.

Advantages of mercury and alcohol as thermometric liquid are
(a). They both are good conductors of heat.
(b). They have high coefficient of expansion thus are sensitive to the smallest change in temperature
(c). Their freezing points are very low and boiling point is high in case of mercury
Disadvantages
(d). Alcohol is transparent and this makes hard to read the thermometer.
(e). It does not have uniform expansion.
(f). Mercury is less sensitive than alcohol as its coefficient of expansion is less than alcohol.
(g). Alcohol is a volatile liquid.

Solution – 52.

Lower point of a thermometer is the temperature at which ice starts melting at normal atmospheric pressure i.e. 0oC
Upper point of a thermometer is the temperature at which water just starts boiling at normal atmospheric pressure i.e. 100oC.

Solution – 53.

Solution – 54.

(i). Laboratory thermometer is used to measure and observe the temperature of various chemical reactions
(ii). Clinical thermometer is used to measure human body temperature
(iii). Six’s maximum and minimum thermometer is used in meteorology and horticulture.

Solution – 55.

Solution – 56.

The temperature that is common in both clinical and Fahrenheit scale is -40oC
Derivation is as follows
Let the temperature be x
C/100= (F – 32)/180
x/100 = (x-32)/180
x X 180/100 = x-32
9/5x = x-32
-4/5x =32
X = – 40

Solution – 57.

(a). 60oC
60/100 = (F-32) / 180
F = 6 X 18 +32
= 110oF
(b). 100oC
100/100 = (F – 32) /180
F = 180 X 1 + 32
= 212oF
(c). -40oC
-40/100 = (F – 32) /180
F = -4 X 18 + 32
= 40oF
(d).85oC
85/100 = (F – 32) /180
F = 85 X 18/10 +32
= 185oF

Solution – 58.

(a). 104oF
C = (F – 32) X 100/180
C = 72 X 100/180
= 40 oC
(b). 95oF
C = (F – 32) X 100/180
= 63 X 10/18
= 35oC
(c). 113oF
C = (F – 32) X 100/180
= 81 X 10/18
= 45oC
(d). 32oF
C = (F – 32) X 100/180
= 0 X 10/18
= 0oC

Solution – 59.

Solution – 60.

Three modes of heat transfer are
(a). Conduction involves the transfer of heat from the hot end to the cold end from particle to particle of the medium.
(b). Convection is the transfer of heat from one body to another by actual movement of the particles of the medium
(c). Radiation is the transfer of heat from one body to another without the need of an intervening material medium

Solution – 62.

Solution – 63.

A wooden knob and a metal latch are both being at same temperature but it feels colder to touch the latch because metal is a good conductor and as soon as we touch it heat from our hand flows to the latch and we feel cold while on the other hand wood is a bad conductor of heat, heat of our hand does not flow into it therefore it does not feel cold.

Solution – 64.

The flask consists of double walled glass container with vacuum between the walls A and B to prevent heat loss due to conduction and convection as vacuum is the excellent insulator .to prevent heat loss by radiation, the inner side of the wall A and outer side of wall B is silvered. It has a narrow mouth which is closed by a non-conducting rubber stopper.

Solution – 65.

The spiral starts moving because due to the flame of the candle the spiral heats up and expands. While expanding, the spiral tries to create space for the extension in length and an outward pull is created which causes the spiral to move.

Solution – 66.

(i). In winters, the human body covered with a blanket keeps warm because the blanket has air trapped in it which provide heat insulation to the body from the surroundings and keep us warm
(ii). It is better to use two thin blankets to keep the body warm rather than using a single blanket of equal thickness because in between the two thin blankets there is more air trapped than in the single blanket of equal thickness so using two thin blankets better heat insulation is provided to the body from the surroundings and keep us warm
(iii). In winter the birds fluff their feathers in order to trap air in their feathers so that the air provides heat insulation to their body from the surroundings and keep them warm and save them from winter.
(iv). Old quilts are less warmer than new ones because the cotton in the old quilt gets compressed and very little air will remain trapped in it, hence heat insulation is quite poor
(v). People wear light colured clothes in winter because these clothes reflect most of the sun’s radiations and absorb only a little of them. Therefore, they keep themselves cool.

PAGE NO: 231.
Solution – 67.

Transformation of Sun’s energy in sun-eco system through a food chain is called energy flow.

Solution – 68.

Solution – 69.

Any energy transfer is not 100% because energy is lost to the surroundings in the form of heat, friction losses during the transfer of energy. Therefore complete energy is not transferred.

Solution – 70.

Bio gas is produced by the action of bacteria on decaying organic matter. The primary source of bio gas in villages is dung of cow, or buffalo. The bio gas is mostly methane which can be used as a chief source of light and heat energy.

Solution – 73.

Solution – 74.

PAGE NO: 232.
Solution – 76.

Water is not used as a thermometric liquid because It has low coefficient of expansion so it is less sensitive to temperature changes. Moreover, It is transparent thus making it difficult to read the thermometer and water evaporates with time thus producing error and also the freezing and boiling points are also low.

Solution – 77.

The sensitivity of a thermometer can be increased by using a substance having high coefficient of expansion and uniform expansion so that its expands with the slightest change in temperature.

Solution – 78.

(i). When hot water is poured into a thick glass tumbler, it generally cracks because on pouring hot water in the tumbler the inner surface heats up and expands more as compared to its outer surface. This unequal expansion between the two surfaces causes a strain and the tumbler cracks.
(ii). Pyrex glass tumbler does not crack on adding hot water because Pyrex glass has low coefficient of expansion. It does not expand less when hot water is added to the tumbler.

Solution – 79.

Solution – 80.

Solution – 81.

Solution – 82.

Temperature in oC = 1oC
C/100 = (F-32)/180
F = 1×18/10 + 32
= 33.8oF

Solution – 83.

Lower fixed point = 10oC
Upper fixed point =130oC
Range of thermometer= 130oC – 10oC
= 120oC
No of divisions = 100
So least count = 120/100 = 1.2oC
On actual thermometer 40oC would have 40 divisions
So, on this thermometer it would show
= 40 x LC = 48oC

Solution – 84.

The green house is referred to a glass house. The heat enters the house but cannot escape out, because the glass reflects the heat back to the inside of the house. This makes glass house warmer than the outside environment. This phenomenon is called green house effect.

Solution – 85.

Global warming occurs due to the presence of carbon di oxide, CFCs, methane in the atmosphere. Carbon dioxide acts as a transparent gas to incoming shortwave radiations which the earth re-radiates into space. It, therefore traps the outgoing radiations thus warming lower atmosphere of the earth thereby causing global warming.

Solution – 86.

Harmful effects of global warming are
(a). The atmospheric temperature of earth would increase thereby making it difficult for a living being to survive
(b). It would melt down the polar caps thus increasing the size of the ocean and leading to floods, tsunami, etc.
(c). The increase in temperature would affect climate and rainfall thus affecting flora and fauna.
(d). Human beings would be vulnerable to diseases as microbes would get warmth to grow.

Solution – 87.

The temperature in a green house rises because heat enters the house through the glass but cannot escape out, because the glass reflects the heat back to the inside of the house. This makes glass house warmer than the outside environment.

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Frank ICSE Solutions for Class 9 Physics Chapter 4 – Fluids Solutions

August 11, 2022November 10, 2017 by Dattu

Frank ICSE Solutions for Class 9 Physics Chapter 4- Fluids Solutions Solutions

Frank ICSE Solutions for Class 9 Physics Chapter 4 – Fluids Solutions

PAGE NO: 157.
Solution – 01.

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

Solution – 02.

Pressure is given by
P = h Xp Xg.
Where h is height of liquid column, p is density of liquid, g is acceleration due to gravity.
Density of mercury is = 1.36 X104 kg/m3.
h= height of mercury column which is given = 75 cm = 0.75 m.
So pressure = 0.75 X 1.36 X104X9.8 = 9.996X104Nm-2.

Solution – 03.

Pressure is a scalar physical quantity.

Solution – 04.

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

Solution – 05.

The force acting normally on a surface is known as thrust.
SI unit of thrust is N.

Solution – 06.

Solution – 07.

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

Solution – 08.

Pressure is the physical quantity which is measured in bar.

Solution – 09.

Thrust is a vector quantity.

Solution – 10.

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

Solution – 11.

Solution – 12.

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

Solution – 13.

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

Solution – 14.

Solution – 15.

The pressure at a point in a liquid depends upon on the following three factors:
(i) It depends on the point below the free surface (h).
(ii) It depends on density of liquid (p).
(iii) It depends upon acceleration due to gravity (g) of the place.

Solution – 16.

Solution – 17.

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

Solution – 18.

The laws of liquid pressure are
(i) Pressure inside the liquid increases with the depth from the free surface of the liquid.
(ii) Pressure is same at all points on a horizontal plane, in case of a stationary liquid.
(iii) Pressure is same in all directions about a point inside the liquid.
(iv) Pressure at same depth is different in different liquids. It increases with the increase in the density of the liquid.
(v) A liquid will always seek its own level.

Solution – 19.

Solution – 20.

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

Solution – 21.

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

Solution – 22.

Solution – 23.

Manometer is a simple pressure gauge that measures differences in pressure at the two ends of the apparatus.
Manometer is a U shaped tube containing water whose one limb is dipped in vessel and vessel is tightly covered with plastic sheet. U shaped tube has two limbs one towards the vessel and other is opened to atmosphere.
Now if level of water toward atmospheric open limb is more than level of water in limb towards apparatus end then liquid is said to be at higher pressure than atmosphere. And if level of water toward atmospheric open limb is less than level of water in limb towards apparatus end then liquid is said to be at lower pressure than atmospheric pressure.

PAGE NO: 158.
Solution – 24.

Solution – 25.

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

Solution – 28.

Pascal’s law states that pressure applied to an enclosed liquid, is transmitted equally to every part of the liquid or in other words when pressure is applied at a point in a confined fluid, it is transmitted undiminished and equally in all directions throughout the liquid.
Hydraulic machines such as hydraulic press, hydraulic brakes and hydraulic jack are application of pascal’s law.

Solution – 29.

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

Solution – 30.

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

Solution – 31.

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

Solution – 32.

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

Solution – 33.

Mercury is used in barometer because
(i) It can be obtained in pure form.
(ii) It does not vaporizeat ordinary temperatures.
(iii) Its density is high and hence the length of the mercury column supported by atmospheric pressure is 76 cm which is practically possible.

PAGE NO: 173.
Solution – 01.

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

Solution – 02.

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

Solution – 03.

Buoyant force act on a body in upward direction.

Solution – 04.

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

Solution – 05.

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

Solution – 06.

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

Solution – 07.

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

Solution – 08.

Upthrust or buoyant force depends on the following factors:
(i) Volume of body submerged in the liquid.
(ii) Density of the liquid.
(iii) Acceleration due to gravity.

Solution – 09.

Solution – 10.

Weight of the body in air = 300 gf.
Apparent Weight of the completely immersed body in water = 280 gf.
(i) Loss in weight of the body = Weight of body in air – apparent weight of immersed body.
Loss in weight = 300 gf – -280 gf = 20 gf.
(ii) As upthrust on the body = loss in weight
(iii) So uptrust = 20 gf.

Solution – 11.

Edge of metal cube = 5 cm.
Density of the metal cube = 9 gcm-3 = 9 X 103 kgm-3.
Volume of the metal cube = 125 cm3 = 125X10-6 m3.
Mass of the metal cube =9 X 103X125X10-6 = 1125 X10-3 =1.125 kg.
Weight of the liquid = mass X gravity = 1.125 X10 = 11.25 N.
Density of liquid = 1.2 gcm-3= 1.2 X103 kgm-3.
Upthrust of the liquid = V X p Xg.
Upthrust = 125X10-6 X1.2 X103 X10 = 1.5 N.
Apparent weight of the body = weight of liquid – upthrust
Apparent weight = 11.25 N – 1.5 N = 9.75 N
Tension in the string is equal to the apparent weight of the body
So, tension in string would be 9.75 N.

Solution – 12.

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

Solution – 13.

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

Solution – 14.

Solution – 15.

PAGE NO: 174.
Solution – 16.

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

Solution – 17.

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

Solution – 18.

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

Solution – 19.

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

Solution – 20.

(i) SI unit of buoyant force is N.
(ii) SI unit of density is Kgm-3.
(iii) SI unit of weight of body is N.
(iv) Relative density is a pure ratio it has no dimension.

Solution – 21.

Density of iron is more than the density of water so it sinks down in the water but in case of ship, it is design in such a manner that it encloses large quantity of air in air tight bags and in rooms and corridors which makes the average density of ship less than that of water and ship floats on the surface of the water.

Solution – 22.

The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
(Density of floating body / Density of liquid) = fraction submerged.
The Fraction of ice submerged in water remain same as density of ice and water remain same during melting. As ice melts some volume of ice decrease and convert into water and volume of water increase by same amount. So, level of water remains same during melting.

Solution – 23.

The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
(Density of floating body / Density of liquid) = fraction submerged.
Height of wooden piece = 15 cm.
Height of wooden piece sinks in water = 10 cm.
Fraction of wooden piece submerged in water = 10/15 = 0.67.
As liquid is water so ratio of Density of wooden by density of water gives relative density of floating wooden piece. So, relative density of wooden block is 0.67.
Height of wooden piece = 15 cm.
Height of wooden piece sinks in spirit = 12 cm.
Fraction of wooden piece submerged in water = 12/15 = 0.8.
We know density of wooden piece = 0.67
(Density of floating body / Density of liquid) = fraction submerged.
Density of liquid/spirit = (Density of floating body /fraction submerged)
Density of liquid/spirit = 0.67/0.8 = 0.83.
Relative density of spirit is 0.83.

Solution – 24.

(i) When a body is completely immersed in water then it displaces equal volume of water to its own weight. Volume of body of man is same in both river and sea so weight of water of sea displaced by him is equal to the weight of water of river displaced by him. And ratio of weights would be 1:1.
(ii) Sea water contains mineral salts and density of sea water increase due to presence of these. As density of sea water is more than the normal water so it apply more buoyant force than usual one and a person find it easy to swim in sea water.

Solution – 25.

The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
(Density of floating body / Density of liquid) = fraction submerged.
Fraction of wooden piece submerged in water = 2/3 = 0.67.
As liquid is water so ratio of Density of wooden by density of water gives relative density of floating wooden piece.
So, relative density of wooden block is 0.67.
Density of water in SI system = 1000 Kg m-3.
Density of wood=relative density�density of water =0.67X1000 Kg m-3 =670 kgm.

Fraction of wooden piece submerged in oil = 3/4 = 0.75.
We know density of wooden piece = 0.67
(Density of floating body / Density of liquid) = fraction submerged.
Relative Density of oil = (Relative Density of wooden block/fraction submerged)
Density of oil = 0.67/0.75 = 0.893.
Density of water in SI system = 1000 Kg m-3.
Density of oil =relative density X density of water =0.893 X1000 Kg m-3 =893 kgm-3.

Solution – 26.

Relative density of Ice = 0.92
Relative density of sea water = 1.025
Let total volume of iceberg = X cm3.
Volume of iceberg above water = 800 cm3.
Volume of iceberg in submerged in the water = (X – 800) cm3.
Fraction of iceberg submerged = (X- 800)/X
Now we know that fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
(Density of ice / Density of sea water) = fraction submerged
0.92/1.025 = (X-800)/X
0.8975 X = X – 800
X – 0.8975 X = 800
0.1025 X = 800
X = 800/0.1025 = 7804.8 cm3.
Total volume of iceberg = 7804.8 cm3.

Solution – 27.

Relative density of wax = 0.95
Relative density of brine = 1.1
(Density of wax/ Density of brine) = fraction submerged
0.95/1.1 = fraction of volume submerged
Fraction of volume submerged = 0.86

Solution – 28.

Relative density of Ice = 0.9 cm
Relative density of sea water = 1.1 cm
(Density of ice / Density of sea water) = fraction submerged of iceberg
0.9/1.1 = fraction of iceberg submerged
Fraction of iceberg submerged = 9/11.

Solution – 30.

Lactometer is commonly used for testing the purity of milk.

Solution – 31.

Density of water at 4oc in SI system is = 1000 Kgm-3.

Solution – 32.

Side of wooden cube = 10 cm.
Volume of wooden cube = 10 X 10 X 10 = 1000 cm3.
Mass of wooden cube = 700 g.
Density of wooden cube = mass/volume = 700/1000 = 0.7 gcm-3.
Density of water = 1 gcm-3.
(Density of floating body / Density of liquid) = fraction submerged
0.7/1 =fraction submerged
Fraction of wooden cube submerged in water = 0.7
Height of wooden cube = 10 cm
Part of wooden cube which is submerged = 10 X 0.7 = 7 cm
So, wooden cube will float in water with 3 cm height above the water surface.

Solution – 33.

Volume of wooden block = 0.032 m3.
Mass of wooden block = 24 Kg.
Density of wooden block = mass/volume = 24/0.032 = 750 Kgm-3.
Density of water = 1000 Kgm-3.
(Density of floating body / Density of liquid) = fraction submerged
750/1000 =fraction submerged
Fraction of wooden block submerged in water = 0.75
Total volume of wooden block = 0.032 m3.
Part of volume of wooden block which is submerged = 0.032 X 0.75 = 0.024 m3.

Solution – 34.

Relative density = density of substance /density of water at 40C.
As relative density of platinum is 21.50, this means platinum is 21.5 times denser than water at 4oC.

PAGE NO: 175.
Solution – 35.

Density of mercury = 13600 Kgm-3.
Density of water at 4oC = 1000 kg m-3.
Relative density = density of substance /density of water at 40C.
Relative density of mercury = 13600 Kgm-3/1000 kg m-3 = 13.6.

Solution – 36.

volume of body = 100 cm3.
Weight of body = 1 kgf = 1000 gf
Mass of body= 1000 gm.
Density of liquid = 1000 gm/100cm3 = 10 gcm3.
Density of water at 4o = 1gcm-3.
Relative density = density of substance /density of water at 40 C
Relative density = 10 gcm3 /1 gcm3 = 10
Mass of body= 1000 gm.
Density of water = 1 gcm-3
Acceleration due to gravity = 10 ms-2.
Upthrust = V X p Xg.
Upthrust = 100 X 1 Xf = 100 gf.
Resultant weight of the body = weight – upthrust = 1000 gf – 100 gf = 900 gf.

Solution – 37.

When a body is completely immersed in water then it displaces equal volume of water to its own weight.
So, volume of body = 20000 cm3.
Mass of body = 70 kg = 70000 gm
Density of body = mass /volume = 70000/20000= 3.5 gm cm-3.
Density of water in C.G.S system = 1g cm-3.
Relative density of body = density of body /density of water =3.5 gm cm-3/1g cm-3.
Relative density = 3.5.

Solution – 38.

Relative density = density of mercury /density of water.
Density of mercury = relative density X density of water.
Relative density = 13.6.
Density of water in C.G.S system = 1g cm-3.
So, density of mercury in C.G.S system = 13.6 X1 = 13.6 gcm-3.
Density of water in SI system = 1000 Kg m-3.
So, density of mercury in SI system = 13.6 X1000 = 13.6 X103 Kgcm-3.

Solution – 39.

Density of iron is = 7.8 X 103 Kg m-3.
Density of water at 4oC = 103 Kg m-3.
Relative density of a substance is the ratio of the density of the substance to the density of water at 4oC.
So, relative density of iron is = 7.8 X 103 Kg m-3/103 Kg m-3 = 7.8

Solution – 40.

(i) Mass of a metallic piece remains unchanged with increase in temperature.
(ii) Volume of metallic piece increases with increase in temperature.
(iii) Density of metallic piece decreases with increases in temperature.

Solution – 41.

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

Solution – 42.

PAGE NO: 177.
Solution – 01.

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

Solution – 02.

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

Solution – 03.

Pressure is a scalar quantity.

Solution – 04.

Thrust is a vector quantity.

Solution – 05.

SI unit of density is Kgm-3.

Solution – 06.

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

Solution – 07.

Solution – 08.

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

Solution – 09.

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

Solution – 10.

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

Solution – 11.

Yes, all liquid exert pressure.

Solution – 12.

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

Solution – 13.

Pascal’s law is principle of hydraulic machines.

Solution – 14.

Brahma press depends upon Pascal’s law.

Solution – 15.

(i) A hydraulic press can be used for extracting juice of sugarcane, sugar beet etc.
(ii) A hydraulic press can be used for pressing cotton bales, quilts, books etc.

Solution – 16.

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

Solution – 17.

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

Solution – 18.

Barometer is used for measuring the atmospheric pressure.

Solution – 19.

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

Solution – 20.

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

PAGE NO: 178.
Solution – 21.

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

Solution – 23.

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

Solution – 24.

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

Solution – 25.

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

Solution – 27.

Factors which affect the atmospheric pressure as we go up are
(i) Weight of gaseous column.
(ii) Density of gaseous column.

Solution – 28.

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

Solution – 29.

Barometer is a device used for measuring atmospheric pressure.
Simple barometer has two main defects
(i) It is not suitable for making accurate measurement of atmospheric pressure as any change in the level of mercury in the tube changes the level of the free surface of mercury is trough and fixed scale cannot be used with it.
(ii) Simple barometer is not portable. So, it cannot be used by airmen, navigators, mountaineers, who need a portable barometer.

Solution – 31.

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

Solution – 32.

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

Solution – 33.

Weight of solid in air = 2.10 N
Relative density of solid = 8.4
Now, Relative density = weight of solid in air/ loss of weight of solid in water.
Loss of weight of solid in water = weight of solid in air/ Relative density.
Loss of weight of solid in water = 2.10/8.4 = 0.25 N.
Weight of solid in water = weight in air – loss of weight in water
Weight of solid in water = 2.10 – 0.25 =1.85 N.

Relative density of liquid =1.2
We know
Relative density of liquid = Loss of weight of solid in liquid/loss of weight of solid in water.
Loss of weight of solid in liquid = Relative density X loss of weight of solid in water.
Loss of weight of solid in liquid = 1.2 X 0.25 = 0.3 N.
Weight of solid in liquid = weight of solid in air – loss of weight of solid in liquid.
Weight of solid in liquid = 2.10 – 0.3 = 1.8 N.

Solution – 34.

Density of iron is 7800Kgm-3.
This means a cube of iron having side 1m would weigh 7800 Kg.
Density of water at 4oC is 1000 Kgm-3.

Solution – 35.

Relative density of body = 0.52
Density of water at 4oC = 1000 Kgm-3.
Density of body = 0.52 X 1000 Kgm-3= 520 Kgm-3
We know density = mass X volume.
Mass = density X volume
Mass = 520 X 2 =1040 Kg.
Mass of given body is 1040 Kg.

Solution – 36.

Piece of metal weighs in air = 44.5 f
Piece of metal weighs in liquid = 39.5 f.
Loss of weight of metal in liquid = 44.5 – 39.5 = 5f.
Relative density = weight of solid in air/ loss of weight of solid in water.
Relative density of liquid =44.5f/5f =8.9
Relative density of liquid = 8.9

Solution – 37.

Solution – 38.

Principle of floatation states that a floatating body displaces an amount of fluid equal to its own weight.
Ship is designed in such a manner that it encloses large quantity of air in air tight bags and in rooms and corridors which makes the average density of ship less than that of water.

Solution – 40.

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

Solution – 41.

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

Solution – 43.

Solution – 44.

(i) A balloon filled with hydrogen has low density than air so it rises over the air but as height increases density of air decreases and at a certain height the density of hydrogen in balloon and density of air become equal. And as there is no density difference there is no pressure difference also and hence balloon stops rising further.
(ii) Density of egg is greater than fresh water so it sinks in fresh water but due to addition of salt density of water increases which makes the density of salt water greater than egg and hence floats in a strong solution of salt.
(iii) The bottom of the hydrometer is made heavier by loading it with lead shots so that it floats vertically with some of its portion outside the surface of water in the jar.
(iv) Relative density of Ice is = 0.9 cm-3
Relative density of sea water is = 1 cm-3
(Density of ice / Density of sea water) = fraction submerged of iceberg
0.9/1 = fraction of iceberg submerged
Fraction of iceberg submerged = 9/10.
Thus in colder countries where there are icebergs in oceans, only about 1/10 is seen above water and the remaining water 9/10 remain submerged. Hence, there is danger of these icebergs to the ships sailing in these oceans.

Solution – 45.

Solution – 46.

Solution – 47.

Solution – 48.

Solution – 49.

Solution – 50.

Solution – 51.

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Frank ICSE Solutions for Class 9 Physics Chapter 3 – Laws of Motion Solutions

August 11, 2022November 1, 2017 by Dattu

Frank ICSE Solutions for Class 9 Physics Chapter 3 – Laws of Motion Solutions

Frank ICSE Solutions for Class 9 Physics Chapter 3 – Laws of Motion Solutions

PAGE NO: 113.
Solution – 01.
The property by which a body neither changes its present state of rest or of uniform motion in a straight line nor tends to change the present state,is known as inertia.

Solution – 02.

A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.

Solution – 03.

The greater is the MASS , the greater is the inertia of the object.

Solution – 04.

An object possess two kind of inertia, inertia of rest and inertia of motion.A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.

Solution – 05.

1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it. 1 newton would produce acceleration of 1 ms-2 in a mass of 1 kg.

Solution – 06.

The acceleration produced bya force in an object is directly proportional to the applied FORCE and inversely proportional to the MASS of the object.

Solution – 07.

SI unit of force is Newton (N).

Solution – 08.

Acceleration is the physical quantity associated with N kg-1.

Solution – 09.

1 N = 105 Dyne.

Solution – 10.

As mass of loaded van is greater than sports car so it would require more force to stop.

Solution – 11.

We know force = mass X acceleration.
a= f/m = 12 N / 4 kg. = 3 ms-2
so acceleration of the body would be 3 ms-2.

Solution – 12.

SI unit of force is Newton whereas CGS unit of force is dyne.
1 newton / 1dyne = 105.

Solution – 13.

SI unit of momentum is kgms-1.

Solution – 14.

Momentum is defined as the amount of motion contained in the body. It is given by the product of the mass of the body and its velocity.

Solution – 15.

Momentum is the physical quantity associated with the motion of the body.

Solution – 16.

Momentum is possessed by bodies in MOTION.

Solution – 17.

A fast pitched soft ball has more momentum.

Solution – 18.

SI unit of momentum is kgms-1 and CGS unit of momentum is g cms-1.
And their ratio is = 1000 X 100 g ms-1= 1:10.

Solution – 19.

A body at rest has zero momentum as its velocity is zero.

Solution – 20.

According to Newton’s third law, for every action there is always an equal and opposite reaction.

Solution – 21

When a force acts on a body then this is called an action.

Solution – 22.

No, action and reaction never act on a same body they always act simultaneously on two different bodies.

Solution – 23.

2nd law of motion gives the definition of force.

Solution – 24.

Newton’s third law explains this statement.

Solution – 25.

Force is a vector quantity.

Solution – 26.

This means these forces are balanced forces.

Solution – 27.

Passengers tend to fall sideways when the bus takes a sharp turn due to the inertiaof direction.

Solution – 28.

Passengers are thrown in the forward direction as the running bus stops suddenly because due to their inertia of motion, their upper body continues to be in the state of motion even though the lowerbody comes to rest when the bus stops

Solution – 29.

Passengers tends to fall in backward direction when bus starts suddenly because due to their inertia of rest, as soon as the bus starts, their lower body comes in motion but the upper body continues to be in the state of rest.

Solution – 30.

No, internal forces cannot change the velocity of a body.

Solution – 31.

When a hanging carpet is beaten using a stick, the dust particles will start coming out of the carpet because the part of the carpet where the stick strikes, immediately comes in motion while the dust particle sticking to the carpet remains at rest . Hence a part of the carpet moves ahead alongwith the stick, and the dust particles fall down due to the earth’s pull.

Solution – 32.

When we shake the branches of a tree, the fruits and leaves remain in state of rest while branches comes in rest so fruits and leaves are detached from the tree.

Solution – 33.

We know force = mass X acceleration
F1 = 10 X 5 = 50 dyne.
F2 = 20 X 2 = 40 dyne.
So first body require more force

Solution – 34.

PAGE NO: 114.
Solution – 35.

initial velocity of the object = 0 ms-1
Acceleration of the object = 8 ms-2.
Time = 5 s.
Distance covered would be S = ut + 1/2 at2.
S = 1/2 X 8 X 5 X5 = 100 m.

Solution – 36.

Initial velocity of the truck = 0 ms-1
Distance covered by truck = 100 m
Time taken to cover this distance = 10 s.
We know Distance covered would be S = ut + 1/2 at2.
100 =1/2 Xa X100
a= 2 ms-2.
Mass of truck = 5 metric tons = 5000 kg.
Force acted on truck = mass X acceleration
Force = 5000 X 2 = 10000 N.

Solution – 37.

Momentum is used for quantifying the motion of body.

Solution – 38.

When we fire a gun, a force is exerted in the forward in the forward direction as the bullets comes out; in reaction to which an equal and opposite force is act in the backward direction and hence, we feel a backward jerk on the shoulder.

Solution – 39.

A person applies force on water in backward direction and water according to third law of motion water apply an equal and opposite force in forward direction which helps a person to swim.

Solution – 40.

Newton’s third law of motion is involved in the working of a jet plane.

Solution – 41.

Yes, a rocket can propel itself in a vacuum once it is given initial velocity.

Solution – 42.

Action is equal and opposite to reaction but they act on different bodies and object moves as movement requires an unbalanced force and these are provided once inertia is overcome.

PAGE NO: 125.
Solution – 01.

Sir Isaac Newton stated the law of gravitation.

Solution – 02.

Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.

Solution – 03.

Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

Solution – 04.

Acceleration due to gravity is the acceleration experienced by a body during free fall.

Solution – 05.

g = GM/R2

Solution – 06.

We know that law of gravitation.
F = G ( m1 X m2)/R2.
Here G is universal constant and is called constant of gravitation. It doesnot depend upon on the value of m1,m2 or R.
Its value is same between any two objects in the universe.

Solution – 07.

SI unit of constant of gravitation is Nm2kg-2.

Solution – 08.

we know that law of gravitation.
F = G ( m1 X m2)/R2.
(a) If distance between them is halved then put R = R/2.
F = 4 X G( m1 X m2)/ R2.
F1 = 4 F.
(b) If distance between them is doubled then put R = 2R.
F = G( m1 X m2)/ 4R2.
F1 = F/4.
(c) If distance between them is made four times then put R = 4R.
F = G( m1 X m2)/16 R2.
F1 = F/16.
(d) If distance between them is infinite then put R = infinite.
F = G( m1 X m2)/ R2.
F1 = 0.
(e) If distance between them is almost zero then put R = 0.
F = G( m1 X m2)/ 0.
F1 = infinite.

Solution – 09.

All objects in the universe attract each other along the line joining their CENTRES.

Solution – 10.

The force of attraction between any two material objects is called FORCE OF GRAVITATION.

Solution – 11.

The gravitational force of the earth is called earth’s GRAVITY.

Solution – 12.

The Gravity is a particular case of GRAVITATIONAL FORCE OF EARTH.

Solution – 13.

The value of G is extremely SMALL.

Solution – 14.

Yes the law of gravitation is also applicable in case of the sun and moon.

Solution – 15.

we know that law of gravitation.
F = G ( m1Xm2)/R2.
Mass of earth = 6X1024kg.
Mass of the person = 100 kg.
G = 6.7 X10-11 Nm2kg-2.
Radius of earth = 6.4 X 1014.
F = (6.7 X10-11X 100 X 6 X1014 )/ (6.4 X6.4 X1012) = 981.4N
Force of gravity due to earth acting on a 100 kg person is 981.4 N.

Solution – 16.

Objects fall towards the earth due to force of gravitation.

Solution – 17.

Because the masses of persons are not large enough to overcome the value of small constant of gravitation so the force of gravitation is very small and negligible to feel.

Solution – 18.

Initial speed of ball is = 4.9 ms-1.
Acceleration due to gravity = -9.8 ms-2.
(a) We know v2 – u2 =2as
At highest point final velocity is zero so
0 – 4.9 X 4.9 = 2 X (-9.8) S
S = 1.125 m
(b) We know v = u + at
0 = 4.9 – 9.8 t
T = 0.5 sec.
(c) for highest point initial velocity is zero
Acceleration due to gravity is = 9.8 ms-2.
Final velocity at ground is v
V2 – 0 = 2 X9.8 X 1.125
V = 4.9 ms-1.
Time taken to reach ground from highest point
V = u + at
4.9 = 0 + 9.8 t
T = 4.9/9.8 = 0.5 sec.
So time of ascent is equal to time descent.

Solution – 19.

g = GM/R2.

Solution – 20.

Value of the g at the surface of the earth is 9.8ms-2

Solution – 21.

Mass of the body is constant at all positions so mass will not change. But weight will change as gravity on the surface of earth is almost 6 times than on the surface of the moon, so its weight will increase almost 6 times on the surface of earth.

Solution – 22.

We will weigh more on the surface of the earth.

Solution – 23.

Beam balance is used to measure the mass of a body.

Solution – 24.

Spring scale is used to measure the weight of a body.

Solution – 25.

The weight is greater at the poles than the equator.

Solution – 26.

Newton 1N = 9.8 kgwt.

Solution – 27.

We will weigh more on earth surface as value of g is greater on earth surface.

Solution – 28.

No, the force of gravitation between two objects does not depend on the medium between them.

Solution – 29.

we know that law of gravitation.
F = G ( m1Xm2)/R2.
Now m1 = 2 m1
m2 = 2 m2
R = 2 R
F1 = G ( 2m1 X2 m2)/4R2.
F1 = F
So force between them remains same.

PAGE NO: 126.
Solution – 30.

Yes, in absence of gravity all freely falling body have same force acting on them.

Solution – 31.

g= GM/R2
it means acceleration due to gravity is directly proportional to the mass of body and inversely proportional to the square of distance between earth and object.

Solution – 32.

Yes a body falling freely near the earth surface has a constant acceleration.

Solution – 33.

As we know
g= 1/R2
so value of g is more at poles than equator so value of g is maximum near a camp site in Antarctica as this lie on the pole.

Solution – 34.

PAGE NO: 128.
Solution – 01.

Force is that external agency which tends to change the state of rest or the state of motion of a body.

Solution – 02.

1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.

Solution – 03.

Newton is the SI unit of force whereas dyne is the CGS unit of force.
1 N = 105 dyne.

Solution – 04.

No, force is a vector quantity.

Solution – 05.

A force can produce MOTION in an objectat rest. It can ACCELERATE an object and can change its DIRECTION of motion.

Solution – 06.

(a) force changes the shape of skin.
(b) force produces stretching in the rubber.
(c) force provides retardation to the car and finally stops the car.
(d) force decreases the momentum of ball and finally stops the ball.

Solution – 07.

No, every force does not produce motion in every type of body.

Solution – 08.

The amount of inertia of a body depends on its MASS.

Solution – 09.

You can change the direction in which an object is moving by APPLYING FORCE ON IT.

Solution – 10.

A man riding on a car has INERTIA of motion.

Solution – 11.

When a body is at rest , it will continue to remain at rest unless some external force is applied to change its state of rest. This property of body is called inertia of rest.

Solution – 12.

(i) Weight of the book is action and normal force applied by table on book is reaction.
(ii) Force applied by man on ground is action and force of friction is the reaction.
(iii) Force applied by hammer on nail is action and normal force applied by nail on hammer is reaction to this force.
(iv) Firing of bullet is the action and recoiling of gun is the reaction.
(v) Force applied by us on wall is action and opposite force applied by wall on us or we can say that resistance of wall to our force is reaction.

Solution – 13.

A book lying on a table will remain placed on table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.This is an example of inertia of motion.

Solution – 14.

Unbalance external force causes motion in the body.

Solution – 15.

Linear Momentum is defined as the amount of motion contained in a body. It is given by the product of the mass of the body and its velocity.

Solution – 16.

SI unit of momentum is kgms-1.

Solution – 17.

According to Newton’s first law force is that external agency which tends to change the state of rest or the state of motion of a body.

Solution – 18.

According to Newton’s first law, everybody continues in its state of rest or in uniform motion in a straight line unless compelled by some external force to act otherwise.

PAGE NO: 129.
Solution – 19.

Out of all these, as mass of truck is greatest and mass is measure of inertia so a truck has maximum inertia.

Solution – 20.

It is advantageous to run before taking a long jump because after running we get motion of inertia which helps in long jumping.

Solution – 21.

Ball moving on a table top stops eventually due to force of friction between the ball surface and table surface.

Solution – 22.

Force is equal to the rate of change of linear momentum.

Solution – 23.

According to newton second law of motion, when a force acts on a body, the rate of change in momentum of a body is equal to the product of mass of the body and acceleration produced in it.
Yes, Newton’s first law is contained in the second law as if force is zero then acceleration would be zero which means body would remain in its state of rest or in state of constant motion.

Solution – 24.

1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.
1 newton / 1dyne = 105.

Solution – 25.

1 newton = 1 kg X 1 ms-1 = 1000 g X 100 cms-1 = 105 cms-1.
1 dyne 1 g X 1 cms-1 = 1cms-1.
So 1 newton = 105 dyne

Solution – 26.

No, the body will not move as the two forces are equal and opposite and they constitute balanced forces.

Solution – 27.

As these forces are balanced so they will not affect the motion and motion of the body will remain unaffected.

Solution – 28.

According to Newton’s third law, for every action there is always an equal and opposite reaction. Rocket works on the same principle. The exhaust gases produced as the result of the combustion of the fuel are forced out at one end of the rocket. As a reaction , the main rocket moves in the opposite direction.

Solution – 29.

According to Newton’s third law, every action has equal and opposite reaction so force exerted by the wall on the boy is 30 N.

Solution – 30.

Newton stated the law of inertia.

Solution – 31.

Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
Law of gravitation is called universal because it applies to all bodies of universe.

Solution – 32.

Gravity is the force of attraction betwen the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

Solution – 33.

Person will weigh more at Delhi as we know that gravity decreases with increase in height. Now as Shimla is at a height from Delhi so weight is less in Shimla and more in Delhi.

Solution – 34.

Spring scale is used to measure the weight of a body.

Solution – 35.

Gravity is another kind of FORCE. It exerts all through the UNIVERSE. The sun’s gravity keeps the PLANETS in their orbits. Gravity can only be felt with very large MASS.

Solution – 36.

(i) Objects fall on the earth due to gravitational force between the earth and object.
(ii) Atmosphere doesnot escape because molecules of atmosphere are attracted by earth due to gravitational force of earth.
(iii) A moon rocket needs to reach a certain velocity because during its motion earth attracts the rocket towards it by its gravitational force.

Solution – 37.

‘g’ is acceleration due to earth’s gravity and ‘G’ is universal gravitational constant.

Solution – 38.

Free fall means motion of a body under the gravity of earth only.

Solution – 39.

Yes, we have a gravitational force of attraction between us and a book. But our mass is very small so the force between us and book is very small almost negligible.

Solution – 40.

Yes, the force of gravitation of earth affects the motion of moon, because moon is revolving around earth and centripetal force for this revolution is provided by earth’s gravitation.

Solution – 41.

Inertial mass is measure of inertia of the object. According to second law of motion F = m X a
m= F/a and this mass is called as inertial mass.
Newton law of gravitation gives another definition of mass.
F = (G m1m2)/R2
Thus m2 is the mass of the body by which another body of mass m1 attracts it towards it by law of gravitation. This mass is called gravitational mass.

Solution – 42.

Newton law of gravitation is that Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
(i) Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

(ii) ‘g’ is acceleration due to earth’s gravity and ‘G’ is universal gravitational constant.

Solution – 43.

Yes, it is true that apple attracts the earth towards it with same force but the mass of earth is so huge that acceleration produced in it due to this force is very much small and negligible to notice.

Solution – 44.

we know that law of gravitation is
F = G ( m1Xm2)/R2
Here the F is force due to attraction and this force is equal to weight of the body m2g.
So m2g = G ( m1Xm2)/R2
g= (G Xm2)/R2.
Here R is the distance between earth centre and the object centre. Now if we go on higher altitude say ‘H’ then this R would increase to (R + H)
And value of gravity at height H becomes
g’= (G�m2)/( R +H)2.
As denominator increases so g’ would be less than g and hence we can say that gravity decreases on higher altitudes.

Solution – 45.

(i) The force exerted by the block on is the weight of box and that is equal to 20N.
(ii) The force exerted by string on block is equal to the tension in the string and this is also equal to the 20N.

Solution – 46.

we know F = m X a
m= F/a
so we can calculate mass of each body
Mass of body 1 m1 = 4/8 = 0.5 kg.
Mass of body 2 m2 = 4/20 = 0.2 kg.
Total mass when two masses are tied together M = 0.5 + 0.2 = 0.7 kg.
Now as force is acting on total mass so acceleration produced is
a= 4/0.7 = 5.71 ms-2.

PAGE NO: 130.
Solution – 47.

Solution – 48.

Initial speed of body = 5 ms-1
Final speed of body = 8 ms-1
Time taken to acquire this speed = 2 s.
Acceleration of body = ( v- u)/t
a= (8- 5)/2 = 1.5 ms-2.
Force applied on body = 0.9 N.
we know F = m X a.
m = f/a = 0.9/1.5 = 0.6 kg
mass of the body is 600 gm.

Solution – 49.

Solution – 50.

The force that acts on a body for a very short time but produces a large change in its momentum, is known as impulsive force.

Solution – 51.

initial velocity of body = 0 ms-1.
Final velocity of body = 100 ms-1.
Mass of body = 20 kg.
Force applied = 100N.
We know that
F X t = m (v – u)
100 t = 20 (100 -0)
T = 2000/100 = 20 s.

Solution – 52.

SI unit of retardation is ms-2.

Solution – 53.

Force applied is equal to the product of mass and acceleration produced in the body.
F = mass X acceleration.

Solution – 54.

According to Newton’s second law of motion, when a force acts on a body, the rate of change in momentum of a body equals the product of mass of the body and acceleration produced in it due to that force, provided the mass remains constant.
Mass of body = 400 g = 0.4kg
Force applied on body = 0.02 N
Acceleration = force/mass = 0.02/0.4 = 0.05 ms-2.

Solution – 55.

Linear Momentum is defined as the physical quantity which is associated with bodies in linear motion. It is given by the product of the mass of the body and its velocity.
Mass of body = 1 kg
Acceleration produced = 10 ms-2.
Force applied would be = 1 X 10 N = 10 N.
Mass of second body = 4 kg.
As same force has to be applied on second body so force = 10N.
Acceleration produced is = F/M =10/4 = 2.5 ms-2.

Solution – 56.

Mass of P is m1= m.
Velocity of P is v1 =2 v
Mass of Q is m2 = 2m
Velocity of Q is v2 = v.
(i) inertia of P/inertia of Q = m1/m2 = 1/2.
So ratio of inertia of two bodies is 1:2.
(ii) Momentum of P/momentum of Q = m1v1/m2v2 = 1
So ratio of momentum of two bodies is 1:1.
(iii) As force required to stop them is equal to change in their momentum from moving to rest.
So ratio would be same as the ratio of their momentum i.e 1: 1.

Solution – 57.

According to newton second law
F = m X a
a= (v – u)/t.
F = m(v -u)/t
F = (mv – mu)/t
As F= m X a
ma = (mv – mu)/t
so rate of change of momentum = mass X acceleration.
This relation holds good when mass remains constant during motion.

Solution – 58.

Solution – 59.

According to newton third law, for every action there is always an equal and opposite reaction.
To demonstrate newton third law blow a balloon and hold its neck tightly facing downwards. When we release the balloon, the balloon will moves up instead of falling to the ground. As air is releasing from bottom of balloon and this air apply equal and opposite force to the balloon and this force helps balloon to move upwards.

Solution – 60.

time for which force is applied = 0.1 s.
Mass of body = 2 kg
Initial velocity of body = 0 ms-1
Final velocity of body = 2 ms-1.
We know FX t = m (v – u)
F X 0.1 = 2 (2 – 0)
F = 4 /0.1 = 40 N.

Solution – 61.

mass of ball = 500g = 0.5 kg.
Initial speed of the ball = 30 ms-1
Final speed of ball = 0 ms-1
Time taken by player to stop the ball = 0.03 s.
We know FX t = m (v – u)
F X 0.03 = 0.5 (0 – 30)
F = – 1.5 / 0.03 = – 500 N
(- ) sign shows that player has to apply force in opposite direction of the motion of the ball.

Solution – 62.

Time for which force is applied =0.1 s.
Mass of the body = 3.2 kg.
Initial speed of body = 0 ms-1
After removal of forces body covers a distance of 3m in 1 second so final speed of body = 3/1 = 3ms-1.
We know FX t = m (v – u)
F X 0.1 = 3.2 (3 -0)
F = 9.6/0.1 = 96 N.
So applied force is 96 N.

Solution – 63.

PAGE NO: 131.
Solution – 64.

Time for which force is applied =3 s.
Mass of the body = 2 kg.
Initial speed of body = 0 ms-1
Force applied = 10 N.
(i) We know F X t = m (v – u)
10 X3 = 2 (v- 0)
v = 15 ms-1.
Final velocity is 15 ms-1.
(ii) As m(v – u) is change in momentum and this is equal to the F X t so change in momentum is equal to the 30 kgms-1.

Solution – 65.

(i) We always prefer to land on sand instead of hard floor while taking a high jump because sand increases the time of contact.As FX t = m ( v – u ) and our change in momentum is constant so if time increases then force experienced would decrease.
(ii) Again while catching a fast moving ball, we always pull our hands backwards to increase reaction time so force experienced would decrease.

Solution – 66.

Height of cliff = 98 m.
Initial velocity of stone = 0 ms-1.
Acceleration due to gravity = 9.8 ms-2.

(i) We know H = ut + 1/2 gt2.
98 = 1/2 X 9.8X t2.
t2 = 98X2/9.8 = 20
t= 4.47 sec.
(ii) Final velocity when it strikes the ground
V2 – u2 = 2 g H
V2 = 2X9.8X98
V2= 1920
V= 44.6 ms-1.

Solution – 67.

Initial speed of ball is = 9.8 ms-1.
Acceleration due to gravity = -9.8 ms-2.
Final speed at maximum height = 0 ms-1.
We know v = u + at
0 = 9.8 – 9.8 t
T = 1 sec.
We know v2 – u2 =2as
At highest point final velocity is zero so
0 – 9.8 X 9.8 = 2 X (-9.8) S
S = 4.9 m.
for highest point initial velocity is zero
Acceleration due to gravity is = 9.8 ms-2.
Final velocity at ground is v
V2 – 0 = 2 X9.8 X 4.9
V = 9.8 ms-1.
Time taken to reach ground from highest point
V = u + at
9.8 = 0 + 9.8 t
T = 9.8/9.8 = 1 sec.
Total time = time of ascent + time of descent.
Total of flight = 1+ 1 = 2 seconds.

Solution – 68.

Initial speed of ball = 10 ms-1.
Acceleration due to gravity on ball = – 9.8 ms-2
We know that from first equation of motion
v= u + gt.
After 1 sec
v= 10 – 9.8 X1
v= 0.2 ms-1
so velocity after 1 sec would be 0.2 ms-1.

Velocity after 2 seconds
v= 10 – 9.8X2 = 10 – 19.6 = -9.6 ms-1.
Here negative sign shows that velocity is in downward direction and magnitude is 9.6 ms-1.

Solution – 69.

Maximum Height attained by ball = 19.6 m
Let initial speed of ball = u ms-1.
Acceleration applied on ball due to gravity = -9.8 ms-2.
Final speed of ball at maximum height = 0 ms-1.
We know that from second equation of motion
V2 – u2 = 2as
0 -u2 = 2 X(-9.8)X19.6
u2 = 19.6 X 19.6
u= 19.6 ms-1
so initial speed of ball to attain maximum height of 19.6 m should be 19.6ms-1.

Solution – 70.

Height of tower = 98 m
Acceleration due to gravity on stone = 9.8 ms-2.
Initial speed of ball= 0 ms-1.
Let initial speed of second stone is v ms-1.
We know from second equation of motion
S = ut + 1/2 a Xt2.
98 = 0 + 1/2 X9.8Xt2.
t2 = 20
t= 4.47 sec.
As second stone is thrown 1 sec later so time taken by second body to cover distance of 98 m is = 4.47 – 1 = 3.47sec.
So again put t= 3.47 sec and S = 98 m in second equation of motion we get
98 = vX3.47 + 1/2 X9.8X3.47X3.47.
98 = 3.47Xv + 59
3.47X v = 98 – 59
v= 39/3.47 = 11.23 ms-1.
Initial speed of second stone should be 11.23 ms-1.

Solution – 71.

Mass of object 1 m1 = 200 mg = 200 �10-6 kg = 2 �10-4 kg.
Mass of object 2 m2= 200 mg = 200 �10-6 kg = 2 �10-4 kg.
Distance between the two objects = 1 mm = 10-3 m
We know law of gravitation is
F = G ( m1�m2)/R2
F = (6.67 � 10-11�2�10-4�2�10-4)/(10-3�10-3)
F = 6.67 �2�2�10-11-4-4+3+3
F = 26.68 �10-13 N
So these two objects would experience a force of 26.68 �10-13 N.

Solution – 72.

Radius of earth = 6.38 �103 km = 6.38�106 m
G = 6.67 � 10-11
Acceleration due to gravity = 9.8 ms-2.
We know that
g= (G � M)/R2.
9.8 = (6.67 �10-11�M)/ ( 6.38�106�6.38 �106)
9.8 �6.38�6.38�1012 = 6.67 �10-11� M
398.9 �1012 = 6.67 �10-11� M
M = 398.9 �1012/6.67 �10-11
M = 59 � 1023 kg
So mass of earth is 59 �1023 kg.

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Frank ICSE Solutions for Class 9 Physics Chapter 2 – Motion in One Dimensiont Solutions

August 11, 2022October 29, 2017 by Dattu

Frank ICSE Solutions for Class 9 Physics Chapter 2 – Motion in One Dimensiont Solutions

Frank ICSE Solutions for Class 9 Physics Chapter 2 – Motion in One Dimensiont Solutions

PAGE NO: 61.
Solution – 01.

A body is said to be in state of rest if it does not change its position with respect to its surrounding objects with time.

Solution – 02.

A vector quantity is that physical quantity which is represented by both magnitude and direction.

Solution – 03.

No mass is not a vector quantity.

Solution – 04.

A vector is represented by an arrow. The length of the arrow represents the magnitude of vector quantity and arrow head gives the direction of vector quantity.

Solution – 05.

If a book is lying in almirah then it is at rest.

Solution – 06.

A body is said to be in motion when it change its position with respect to its surrounding objects with time.

Solution – 07.

Yes rest and motion are relative to each other.

Solution – 08.

Out of Force and Energy, Force is a vector quantity.

Solution – 09.

Examples of scalars are distance and mass.

Solution – 10.

Out of these positions, (i) and (ii) positions of body lie on same straight line as direction of these two are same.

Solution – 11.

A vector quantity is represented only when its magnitude and direction are specified so this quantity is a vector quantity.

PAGE NO: 62.
Solution – 12.

Passengers sitting in a train are at rest with respect to each other.

Solution – 13.

Yes we are at rest as well as motion because we are at rest with respect to a observer which is itself at rest and we are in motion with respect to a observer which is in motion.

Solution – 14.

The platform is in motion with respect to train. As train is moving with respect to platform so platform would also look in motion with respect to train.

Solution – 15.

Solution – 16.

Solution – 17.

The physical quantity representing the magnitude and its direction is a vector quantity.

Solution – 18.

(i) Yes we can add two scalars.
(ii) Yes we can add two vectors.
(iii) Yes we can multiply two scalars.
(iv) No we cannot add a scalar quantity to a vector quantity.
(v) Yes we can subtract two scalars.
(vi) No we cannot subtract a scalar quantity from a vector quantity. Reverse is also not true.
(vii) Yes we can multiply vectors.

Solution – 19.

The actual length of the path covered by a moving object irrespective of its direction of motion is called the distance travelled by the object.

Solution – 20.

No the distance covered by a body cannot be less than the magnitude of its displacement.

Solution – 21.

The displacement of a moving body is defined as the change in its position along a particular direction

Solution – 22.

SI unit for measurement of distance and diplacement is metre denoted by m.

Solution – 23.

Solution – 24.

Yes a body can have negative displacement.

Solution – 25.

If a body is moving in a straight line then the displacement of a body is equal to the distance travelled by it.

Solution – 26.

(i) Distance is a scalar quantity whereas displacement is a vector quantity.
(ii) Distance is always positive but displacement can be negative,zero or positive.

Solution – 27.

Solution – 28.

Solution – 29.

Solution – 30.

Solution – 31.

PAGE NO: 79.
Solution – 01.

Speed of a body can be defined as distance covered by the body in unit time.

Solution – 02.

Average speed of a body can be defined as ratio of total distance covered by a body In total time.

Solution – 03.

Both speed and average speed have same unit and that is ms-1.

Solution – 04.

No speed and average speed of a body have different meaning.

Solution – 05.

60 km/hr can be converted into m/s to compare with 15m/s.
60 km/hr = (60 X 1000)/3600 = 16.66 m/s. so speed 60 km/hr is greater.

Solution – 06.

20m/s can be converted inti km/hr as
20 m/s = (20 X 3600)/1000 = 72 km/hr.

Solution – 07.

Solution – 08.

SI unit of velocity is ms-1.

Solution – 09.

No as their direction is different they don’t have same velocity.

Solution – 10.

we convert all the speeds in m/s to compare them.
36 km/hr = (36 X 1000)/3600 = 10m/s.
2 km/min = (2 X 1000)/60 = 33.3 m/s.
7 m/s = 7 m/s.
So increasing order of speed is 7m/s < 10m/s <33m/s. Solution – 11.

let total distance be S.
Boy covers distance S/2 with speed u then time taken by him to cover this distance would be T1 =S/2u.
Again boy covers rest of the distance S/2 with speed v then time taken by him to cover this distance would be T2 =S/2v.
So total time taken by boy to cover the distance S is T = T1 + T2.
Total time T= S/2 (1/u +1/v) = s(u+v)/2uv.
And average speed = S/T = 2uv/(u+v).

Solution – 12.

Yes uniform speed and constant speed have same meaning.

Solution – 13.

let S be the distance between P and Q.
Body covers forward journey distance S (P to Q) with speed u then time taken by him to cover this distance would be T1 =S/u.
Again body covers backward journey distance S (Q to P) with speed v then time taken by him to cover this distance would be T2 =S/v.
So total time taken by body to cover the distance S is T = T1 + T2.
Total time T= S (1/u +1/v) = s(u+v)/uv.
And average speed = 2S/T = 2uv/(u+v).

Solution – 14.

As body goes from P to Q and then return back to P so the displacement of the body would be zero and hence average velocity would also be zero.

Solution – 15.

let distance between P and Q is S.
Speed of car while travelling from P to Q is 20 m/s.
Let car take time T1 to travel from P to Q then T1= S/20.
Speed of car while travelling from Q to P is 30 m/s.
Let car take time T2 to travel from Q to P then T2= S/30.
Total time = T1 + T2 = S/20 +S/30 =S/12.
So average speed of journey = total distance/ total time = 2S/(S/12) = 24 m/s.
Average speed of journey is 24 m/s.

Solution – 16.

Speed is a scalar quantity whereas velocity is a vector quantity. So speed doesn’t have its direction and velocity has a particular direction.

Solution – 17

Speed and velocity of a moving body become equal when the body moves in a straight line.

Solution – 18.

When the velocity of a moving body doesn’t change with time then the velocity of the body is said to be constant or uniform.Yes uniform velocity and constant velocity are one and the same thing.

Solution – 19.

Acceleration of a body is rate of change of its velocity with respect to the time.

Solution – 20.

SI unit of acceleration is ms-2.

PAGE NO: 80.
Solution – 21.

If the acceleration of a moving body is constant in magnitude and direction then the path of the body must not be a straight line because in circular motion also acceleration of a body is constant in magnitude and always directed towards the centre.
So the path of the body may be a straight line and may be a circular one.

Solution – 22.

No the relation S = v X t cannot used to find the total distance covered by a body moving with non-uniform speed.

Solution – 23.

Yes area under a speed time graph in a given interval gives the total distance covered by a body.

Solution – 24.

Yes the motion is uniform and the uniform speed is given by area under speed time graph divided by time interval.
So speed = 500/20 =25 m/s.

Solution – 25.

Positive acceleration corresponds to situation when velocity is continuously increasing with respect to the time.

Solution – 26.

Negative acceleration corresponds to situation when velocity is continuously decereasing with respect to the time.

Solution – 27.

If a body falls towards earth then it would have a positive acceleration.

Solution – 28.

If a body has acceleration of 8.5 ms-2 then it means its velocity is increasing at a rate of 8.5 ms-1 per second.

Solution – 29.

SI unit of retardation is ms-2.

Solution – 30.

first convert 60 km/h in m/s.
60 km/hr =(60 X 1000)/3600 = 16.7 m/s.
This is initial velocity of car i.e u = 16.7 m/s.
As car stops in 10 seconds so final velocity is =0 m/s.
So acceleration = (v-u)/t = (0-16.7)/10 = -1.67 ms-2.
Acceleration of car is = -1.67 ms-2.

Solution – 31.

-30 m/s is speed.

Solution – 32.

Velocity corresponds to the rate of change of displacement.

Solution – 33.

No the speed of a body cannot be negative.

Solution – 34.

A flying bird most likely to have a non uniform velocity.

Solution – 35.

Let initial velocity be u.
Final velocity is v= 0 m/s.
Time taken by body to come to rest = 10 sec.
Retardation =2.5 ms-2.
We know v = u +at.
Then u = v – at.
u = 0 – (-2.5 X 10) = 25 m/s.
So initial velocity of the body is 25 m/s.

Solution – 36.

Equation of motion gives us the picture of motion of moving body.

Solution – 37.

First equation of motion is v = u + at.
Second equation of motion is s= ut + 1/2a t2.
Third eqution of motion is v2 – u2 =2as.

Solution – 38.

Four variables are present in each equation of motion.

Solution – 39.

Four variables are present in each equation of motion and if any of three is known to us then fourth can be easily find with the help of these equation of motion.

Solution – 40.

SI unit of acceleration and retardation is ms-2.

Solution – 41.

Distance is the physical quantity which is equal to the area under speed-time graph.

Solution – 42.

A uniformly accelerated motion is one in which speed is constantly increasing or decreasing with time while a non uniform motion is one in which speed is not constantly changing with time.

Solution – 43.

No we cannot use this relation for a body moving with uniform acceleration.

Solution – 44.

Slope of a graph is given as rate of change of y coordinates to the x coordinate. In speed time graph speed is on the y axis and time is on the x axis. And we define acceleration as rate of change of speed with respect to time. So slope of a speed time graph gives acceleration.

Solution – 45.

(i) Motion of blades of an electric fan.
(ii) Motion of moon around earth.

Solution – 46.

A straight line curve on speed time graph indicates that acceleration of the body is uniform and a zigzag or curved line indicates that acceleration of a body is not uniform.

Solution – 47.

Two quantities are directly proportional to each other.

Solution – 48.

As we distance = speed X time.
Speed = 42 km/hr.
Time = 10 m = 1/6 hr.
Distance = 42 X 1/6 =7 km.

Solution – 49

Initial velocity u =10 ms-1.
Acceleration a = 2 ms-2.
Time t = 10 s.
By using first equation of motion
V = u + at.
V = 10 + 2 X 10.
V (final velocity) = 30 ms-1.

Solution – 50.

Initial velocity u = 10 km/hr. = (10 X 1000)/3600 = 8.33 ms-1.
Final velocity = 64 km/hr = (64 X 1000)/3600 = 17.77 ms-1.
Time = 10 s.
Acceleration = (v-u)/t = (17.77- 8.33)/10 = 9.44/10 = 0.94 ms-2.

PAGE NO: 81.
Solution – 51.

Solution – 52.

No a body cannot have a speed negative.

Solution – 53.

No2 distance covered by body during nth second is not more than the distance covered in n seconds.

Solution – 54.

Solution – 55.

Solution – 56.

If speed time graph is moving upward then the body is accelerating and if it is starting from origin then it means the body has initial velocity =0.

Solution – 57.

Speed time graph is moving upward then the body is accelerating and if it is not starting from origin then it means the body has some initial velocity.

Solution – 58.

Solution – 59.

Solution – 60.

Solution – 61.

Solution – 62.

PAGE NO: 83.
Solution – 01.

Displacement and velocity are two examples of vectors.

Solution – 02.

SI unit of retardation is ms-2.

Solution – 03.

Velocity is the physical quantity associated with the rate of change of displacement with time.

Solution – 04.

Solution – 05.

There are three types of rectilinear motion Translational , vibrational and rotational.

Solution – 06.

A body is said to have a uniform velocity if it covers equal displacement in equal interval of time.

Solution – 07.

Acceleration is a vector quantity.

Solution – 08.

Slope of speed time graph represents acceleration.

Solution – 09.

If a stone is dropped from a certain height then it undergoes non uniform velocity motion.

Solution – 10.

This means the body has a positive acceleration.

Solution – 11.

Solution – 12.

(i) No a body with constant acceleration cannot have a zero velocity.
(ii) No a body with an acceleration in vertical direction cannot move horizontally.
(iii) No in an accelerated motion a body cannot have a constant velocity.

Solution – 13.

Solution – 14.

(i) In displacement-time graph a straight line parallel to time axis shows that body is at rest position.
(ii) In displacement-time graph a straight line inclined to the time axis with an acute angle means body is moving with a positive velocity.

Solution – 15.

No a accelerating body cannot have constant speed.

Solution – 16.

(i) In displacement-time graph a straight line shows body is at rest if it is parallel to time axis and shows a body is moving with uniform velocity if it is inclined to x axis.
(ii) In velocity-time graph a straight line shows body is moving with uniform constant velocity if it is parallel to x axis and shows body is moving constant acceleration of it is inclined to x axis.

Solution – 17.

Average speed during different time intervals for a uniform motion is same.

Solution – 18.

Velocity of a stone thrown vertically upward at its maximum height is Zero.

Solution – 19.

Velocity of a stone thrown vertically upwards decrease because acceleration due to gravity is acting on downward direction.

Solution – 20.

Linear velocity would be equal to linear speed if body is moving in a straight line.

Solution – 21.

Solution – 22.

PAGE NO: 84.
Solution – 23.

Solution – 24.

During circular motion
(a) Speed remains constant.
(b) Velocity changes continuously.

Solution – 25.

The statement is not correct , the correct statement is “the earth is moves round the sun with constant speed”.

Solution – 26.

As in circular motion direction changes continuously with motion so after two complete revolutions we can say that direction has changed infinite times.

Solution – 27.

As after completing 3 revolution in circular motion the displacement is = 0. so the ratio of distance covered to the displacement is infinite.

Solution – 28.

The graph becomes straight line with positive slope with time axis and represents almost a constant acceleration.

Solution – 29.

Retardation is negative of acceleration so retardation the body is +3.4 ms-2.

Solution – 30.

Bus is moving with initial velocity of u = 60 km/hr.
60 km/hr = ( 60 X 1000)/3600 = u = 16.66 ms-1.
Reaction time = t =1/15 sec.
Distance would the bus had moved before pressing the bus would be = u X t.
S = 16.66 X 1/15 = 1.1 m.
Now if the driver is intoxiacated then reaction time would be t = 1/2 seconds.
So S becomes S = u X t = 16.66 X 1/2 = 8.33m.

Solution – 31.

Time difference of 0.1 s denotes the time taken by sound to go from device to wall and back to wall. As the distance between wall and device is 15 m so total distance covered by sound is 2 X 15 m =30 m.
So speed of sound is = total distance covered/time taken = 30/0.1 =300 ms-1.
So speed of sound is 300 ms-1.

Solution – 32.

Solution – 33.

Solution – 34.

slope of velocity time graph represents acceleration of the body.

Solution – 35.

PAGE NO: 85.
Solution – 36.

Solution – 37.

Solution – 38.

Solution – 39.

Solution – 40.

Solution – 41.

let total distance be S.
Boy covers distance S/2 with speed A then time taken by him to cover this distance would be T1 =S/2A.
Again boy covers rest of the distance S/2 with speed B then time taken by him to cover this distance would be T2 =S/2B.
So total time taken by boy to cover the distance S is T = T1 + T2.
Total time T= S/2 (1/A +1/B) = s(A+B)/2AB.
And average speed = S/T = 2AB/(A+B).

Solution – 42.

Car travls 30 km distance with speed 60 km/hr
Time taken by car to travel this distance = 30/60 = 0.5 hr.
Car travels another distance of 30 km with speed of 20 km/hr.
Time taken by car to travel this distance = 30/20 = 1.5 hr.
Total time taken = 1.5 + 0.5 = 2 hr.
Total distance = 30+ 30 = 60 km.
Average speed of car = 60/2 = 30 km/hr.

Solution – 43.

Train travels first 40 km at speed of 30 km/hr.
Time taken by train to cover this distance is = distance/speed = 40/30 = 4/3 hr.
Let speed of train to cover next 80 km is v .
Then time taken by train to cover these 80 km is 80/v.
Total time becomes T = 4/3 +80/v = ( 4v + 240)/3v.
Total distance= 120 km.
Average speed = 60 km/h (given)
However average is given by = total distance /total time.
So (120 X 3 X v)/(4v +240) = 60
360 v = 240v +14400
120v = 14400
v= 14400/120 =120 km/hr.
so train has to cover those 80 km at a speed of 120 km/hr.

Solution – 44.

Solution – 45.

Solution – 46.

Solution – 47.

Initial velocity of car u = 18 km/hr.
Final velocity of car v= 36 km/hr.
Time taken by body = 15 min. = 1/4 hr
Acceleration of car a = ( v- u )/t = (36 – 18 ) X 4 = 72 kmh-2.

Solution – 48.

Initial speed of car u = 50 km/h.
Final speed of car v = 55 km/h.
Time taken by car to attain this speed is = 1 sec. = 1/3600 hr.
Acceleration of the car is = (55 – 50 )X 3600 = 18000 kmh-2.

Solution – 49.

(a) 7200 km/h2 = ( 7200 X 1000)/(3600 X 3600) = 5/9 ms-2.
(b) 1/36 m/s2 = (1 X3600 X 3600)/(36 X1000) = 3600 kmh-2.

Solution – 50.

initial velocity u = 20 m/s.
Acceleration = 5 m/s2.
T = 2 s.
We know v= u + at.
v= 20 + 5X2= 30 m/s.

Solution – 51.

acceleration of the car = 10 ms-2.
Initial velocity u = 10 m/s.
Final velocity v = 30 m/s.
We v = u + at.
T = (v- u)/a
T = (30 – 10 )/10 = 2 sec.
Time taken by car is 2 sec.

PAGE NO: 86.
Solution – 52.

Solution – 53.

Solution – 54.

Let total distance be S.
Body covers distance S/2 with speed 40 ms-1 then time taken by him to cover this distance would be T1 =S/2 X 40.
Again body covers rest of the distance S/2 with speed 60 ms-1 then time taken by him to cover this distance would be T2 =S/2 X 60.
So total time taken by body to cover the distance S is T = T1 + T2.
Total time T= S/2 (1/40 +1/60) = s(40+60)/2 X 40 X 60 = s/48.
And average speed = S/T = 48 ms-1.
So average speed is 48 ms-1.

Solution – 55.

As displacement for the motion from A to B and B to A is zero so the average velocity of the body would be zero.

Solution – 56.

Initial velocity of body u = 0.5 ms-1.
Final velocity of the body v = 0 ms-1 as body comes to rest finally.
Retardation of body = 0.05 ms-2.
We know that v = u + at.
0 = 0.5 – 0.05t
T = 0.5/0.05 = 10 sec.

Solution – 57.

Initial speed of train = 90 km/hr
Speed of train imn m/s = ( 90 X 1000 )/3600 = 25 m/s.
Retardation of the train = 2.5 ms-2.
Final speed of train at platform = 0 m/s.
We know that v2 – u2 = 2as.
0 – 25 X25 = 2 X (-2.5) X s
S = 625/5 = 125 m.
So driver should apply the brakes 125 m before the platform.

Solution – 58.

Train travels first 30 km at speed of 30 km/hr.
Time taken by train to cover this distance is = distance/speed = 30/30 = 1 hr.
Let speed of train to cover next 90 km is v .
Then time taken by train to cover these 90 km is 90/v.
Total time becomes T = 1 +90/v = ( v + 90)/v.
Total distance= 120 km.
Average speed = 60 km/h (given)
However average is given by = total distance /total time.
So (120 � v)/(v +90) = 60
120 v = 60v +5400
60v = 5400
v= 5400/60 =90 km/hr.
so train has to cover those 90 km at a speed of 90 km/hr.

Solution – 59.

speed of train = 30 km/hr.
Speed in m/s = ( 30 X1000 )/3600 = 50/6 m/s.
Lenth of train = 50 m.
Let lenth of bridge be s metre.
Train has to cover total distance of 50 +s to cross that bridge.
Time taken by train to cover this distance = 36 sec.
So as time taken = total distance /total time taken.
36 = ( 50 +s ) X 6/ 50.
1800 = 300 + 6s
6s = 1500.
S = 1500/6 = 250m
Length of bridge is 250 m.

Solution – 60.

Solution – 61.

PAGE NO: 87.
Solution – 62.

Solution – 63.

(i) No vehicle is moving with uniform velocity.
(ii) Vehicle B is moving with constant acceleration.
(iii) At 6 seconds both vehicles would meet.
(iv) Velocity of both the vehicles is 60 m/s when they meet.
(v) Vehicle B is ahead at the end of 7th sec and by 70 m.

Solution – 64.

The given question is wrong as distance can never DECREASE with progress of time.

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Frank ICSE Solutions for Class 9 Physics Chapter 1 – Measurement Solutions

August 11, 2022October 24, 2017 by Dattu

Frank ICSE Solutions for Class 9 Physics Chapter 1 – Measurement Solutions

Frank ICSE Solutions for Class 9 Physics Chapter 1 – Measurement Solutions

PAGE NO: 15.
Solution – 01.

Measurement is an act or the result of comparison of a quantity whose magnitude is unknown with a predefined standard.

Solution – 02.

The physical quantities like mass, length and time which do not depend on each other are known as fundamental quantities.

Solution – 03.

Length, mass, time are the three fundamental quantities.

Solution – 04.

Unit is a standard quantity of the same kind with which a physical quantity is compared for measuring it.

Solution – 05.

A standard metreis equal to 1650763.73 wavelengths in vacuum, of the radiation from krypton isotope of mass 86

Solution – 06.

Three systems of unit are
(a) C.G.S system – fundamental unit of length is centimetre(cm), of mass is gram(gm), of time is second(s).
(b) F.P.S system- fundamental unit of length is foot(ft), of mass is pound(lb), of time is second(s).
(c) M.K.S system- fundamental unit of length is metre(m), of mass is kilogram(kg), of time is second(s).

Solution – 07.

The SI unit of mass is Kilogram. One standard kilogram is equal to the mass of a cylinder of nearly same height and diameter and made up of platinum and iridium alloy.
Three units of length greater than a metre are

Solution – 08.

(a). Decameter = 10 metre
(b). Hectometer = 100 metre
(c). Kilometer = 1000 metre

Solution – 09.

Solution – 10.

Light year is defined as the distance travelled by light in vacuum in one year.

Solution – 11.

Two units of length smaller than a metre are
(a). Decimeter = 0.1 metre
(b). Centimeter = 0.01 metre

Solution – 12.

Leap year because it is a unit of time.

Solution – 13.

Order of magnitude of a physical quantity is defined as its magnitude in powers of ten when that physical quantity is expressed in powers of ten with one digit towards the left decimal.
For example, volume= 52.37 m3 then the order of magnitude is 102m3.

Solution – 14.

No, micron is not same as millimeter because micron is equal to 10-6metre while a millimeter is equal to 10-3metre.

Solution – 15.

Solution – 16.

Solution – 17.

A leap year refers to a year in which February has 29 days and the total days in the year are 366 days.

PAGE NO: 16.
Solution – 18.

Solution – 19.

Solution – 20.

PAGE NO: 28.
Solution – 01.

When one complete rotation is given to the screw hand, it moves forward or backward by a distance is called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = distance travelled by screw in n rotations/n rotations

Solution – 02.

No, least count is not same as pitch because least count is found by dividing pitch by number of divisions on the circular scale.

Solution – 03.

Two uses of vernier caliper are
(a). Measuring the internal diameter of a tube or a cylinder.
(b). Measuring the length of an object.

Solution – 04.

Two limitations of metre rule
(a). There comes an error of parallax due to thickness of the metre rule.
(b). We cannot use metre rule for measuring small thickness.

Solution – 05.

When one complete rotation is given to the screw hand, it moves forward or backward by a distance called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = distance travelled by screw in n rotations/n rotations
Least count refers to the smallest reading that can be accurately measured while using an instrument. The least count is the value of one division on its scale.

Solution – 06.

Initial level of water in cylinder = 30 ml
Level of water in cylinder after immersing piece of copper = 50 ml
Volume of copper piece = 50-30 = 20 ml

PAGE NO: 29.
Solution – 07.

Solution – 08.

The ratchet is used in a screw gauge to hold the object under measurement gently between the studs.

Solution – 09.

If the zero of the circular scale does not coincide with the zero of the main scale (pitch scale), this is known as zero error. There are two types of zero error –
(a). If the zero of the circular scale remains below the line of graduation then it is called positive zero error
(b). If the zero of the circular scale lies above the line of graduation then it is called negative zero error
For positive zero error correction, the zero error should always be subtracted from the observed reading
For negative zero error correction, the zero error must be added to the observed reading.

Solution – 10.

Two scales in a screw gauge are
(a). A linear scale called the main scale graduated in half millimeters
(b). A circular scale divided into 50 or 100 equal parts.

Solution – 11.

Due to constant use, there is space for the play of screw gauge but gradually this space increases with the use or wear and tear, so that when the screw is moved by rotating it in some direction, it slips in the nut and does not cover any linear distance for some rotation of the screw head. The error due to this is known as backlash error.
It is avoided by turning the screw always in the same direction.

Solution – 12.

Following procedure is used to measure the diameter of a wire
(a). Calculate the least count and zero error of the screw gauge.
(b). Place the wire in between the studs. Turn the ratchet clockwise so as to hold the wire gently in between the studs. Record the main scale reading.
(c). Now record the division of circular scale that coincides with the base line of main scale. This circular scale division multiplied by least count will give circular scale reading.
(d). The observed diameter is obtained by adding the circular scale reading to the main scale reading. Subtract the zero error if any, with its proper sign, from the observed diameter to get the true diameter.

Solution – 13.

Solution – 14.

Screw gauge measures a small length to a high accuracy because it has the lowest least count among the given three instruments. And low least count means high accuracy

Solution – 15.

Solution – 16.

Solution – 17.

Solution – 18.

If the zero of the circular scale remains below the line of graduation then it is called positive zero error. When there is positive zero error, then the instrument reads more than the actual reading. Therefore in order to get the correct reading, the zero error should always be subtracted from the observed reading.

Solution – 19.

Pitch of the screw gauge = 0.5mm = 0.05 cm
Circular scale divisions = 100
Least Count of screw gauge = pitch of the gauge/circular scale divisions
= 0.05/100
= 0.0005cm

Solution – 20.

If the zero of the circular scale lies above the line of graduation then it is called negative zero error. When there is negative zero error, then the instrument reads less than the actual reading. Therefore in order to get the correct reading, the zero error should always be added to the observed reading.

Solution – 21.

(a). False, because the accuracy is higher in case of screw gauge due to lower least count value of 0.01mm
(b). True
(c). False, because its least count is limited to 0.1 cm. thus this length can be measured with an instrument of least count of 0.001 cm i.e. screw gauge
(d). False, the ratchet is used to hold the object under measurement gently between the studs.
(e). True

Solution – 22.

The space occupied by a body is known as its volume. SI unit of volume is cubic metre (m3)

Solution – 23.

Yes, 1 cm3 is same as 1 ml.

Solution – 24.

1 m3 = 1000 litre
1 litre = 1/1000 m3
= 0.001 m3

PAGE NO: 30.
Solution – 25.

Solution – 26.

SI unit of volume is cubic metre or metre3 (m3).
The relation between liter and metre3
1 metre3 = 1000 liter
Solution – 27.

Pitch of the screw = 0.5 mm
Least count = 0.001 mm
Number of divisions = pitch/least count
= 0.5/0.001
= 500

Solution – 28.

Solution – 29.

Precautions to be taken while measuring volume of a solid lighter than water using displacement method
(a). The sinker should be insoluble in water
(b). The sinker should have a high density than water.
(c). Lower meniscus should be read to note down the readings and error due to parallax should be avoided.

Solution – 30.

Measurement of volume of an irregular solid soluble in water using a graduated cylinder.
(a). In this case, kerosene or any liquid whose density is lighter than water and in which the solid is not soluble is used.
(b). Fill the graduated cylinder with the liquid.
(c). Record the lower meniscus of liquid and let the value be V1.
(d). Tie the solid whose volume is to be measured to a strong string and lower it into the water gently.
(e). Note the reading carefully and let the value be V2
(f). Volume of the solid, V = V2 – V1

PAGE NO: 38.
Solution – 01.

Solution – 02.

A seconds pendulum is a pendulum which takes 2 seconds to complete one oscillation. The length of seconds pendulum, where g = 9.8ms-2, is nearly 1 m.

Solution – 03.

A stopwatch is used to measure short intervals of time.

Solution – 04.

SI unit of frequency is hertz (Hz).

Solution – 05.

When a pendulum completes one oscillation in one second, then the frequency is one hertz.

Solution – 06.

The time period, T and frequency of oscillation, f are related as,
T = 1/f or f = 1/T

Solution – 07.

One complete to and fro motion of a pendulum about its mean position is known as oscillation. Amplitude is the magnitude of the maximum displacement of the bob from the mean position on either side when an oscillation takes place.

Solution – 08.

SI unit of amplitude is metre (m).

Solution – 09.

A seconds pendulum is a pendulum which takes 2 seconds to complete one oscillation. The length of seconds pendulum, where g = 9.8ms-2, is nearly 1 m.

Solution – 10.

Solution – 11.

Solution – 12.

When a pendulum is taken from earth to moon surface, its time period will increase because the acceleration due to gravity on moon is less than that on earth and the time period depends inversely on square root of acceleration due to gravity.

Solution – 13.

If time period of a pendulum becomes infinite, the pendulum will not oscillate at all as pendulum will take infinite time to complete one oscillation.

Solution – 14.

Effective length of a simple pendulum is the distance of the point of oscillation (i.e. the centre of the gravity of bob) from the point of suspension

Solution – 15.

Solution – 16.

Solution – 17.

Solution – 18.

The time period of a pendulum is independent of mass of the bob.

Solution – 19.

Solution – 20.

The quantity of matter contained Mass of a body can be measured by using a beam balance. in a body is called its mass. Mass is always constant for a given body.

Solution – 21.

A beam balance works on the principle of moments. According to the principle of moments, under equilibrium condition, the clockwise moment due to the body on one side of beam equals the anti clockwise moment due to standard weights on the other side of beam.

Solution – 22.

Precautions to be taken to measure the mass of a body using beam balance are
(a). The beam must be gently lowered before adding or removing weights from the pan.
(b). The weights should not be carried with bare hands to avoid the change in weights due to moisture and dust particles from the surrounding.
(c). The lever should be turned gently, in order to prevent knife edges from chipping.
(d). Never keep the wet or hot objects on the pan.
(e). The weights should be placed into weight box after use.
(f). Whenever you are near the actual weight, you should carefully try the weights in the descending order.

Solution – 23.

SI units of time and mass are second (s) and kilogram (kg) respectively.

Solution – 24.

Conditions for a beam balance to be true are
(a). Both the pans must be of equal weights.
(b). Both the arms must be of equal lengths.

PAGE NO: 44.
Solution – 01.

Least count of an instrument refers to the smallest reading that can be accurately measured while using the instrument. For an instrument provided with a scale the least count is the value of one division on its scale.

Solution – 02.

Maximum possible error is 0.1 cm.

Solution – 03.

Slope of a graph indentifies the proportional relationship between the quantities plotted.

Solution – 04.

Solution – 05.

Solution – 06.

Accuracy is the extent to which a reported measurement approaches the true value of the quantity measured. This extent is usually described by the least count of the instrument and since the least count for a given instrument is limited hence, the accuracy of the instrument is limited.

Solution – 07.

Two types of error in a measurement are
(a). Random errors-these errors are due to various factors. In a number of observations we get different readings every time.
These errors can be minimized by taking observations a large number of times and taking the arithmetic mean of the readings.
(b). Gross error- these errors are due to carelessness of the observer like parallax, improper setting of the instrument.
These errors can be minimized only when the observer is careful in setting up of instrument and taking readings.

Solution – 08.

3000g is the most accurate measurement because it has maximum number of significant figures = 4.

Solution – 09.

Basically there is no difference between the quantity being measured but there is a difference of significant figures in the measurement.
(a). Number of significant figures = 3
(b). Number of significant figures = 4
(c). Number of significant figures = 5
Since (c) part has maximum number of significant figures = 5, therefore it is most accurate among the given three.

PAGE NO: 46.
Solution – 01.

Unit is a standard quantity of the same kind with which a physical quantity is compared for measuring it.

Solution – 02.

The units which can neither be derived from one another, nor can they be further resolved into other units are known as fundamental units.

Solution – 03.

The units which can be expressed in terms of fundamental units of mass, length and time are known as derived units.

Solution – 04.

A standard metre is equal to 1650763.31 wavelengths in vacuum, of the radiation from krypton isotope of mass 86.

Solution – 05.

One standard kilogram is equal to the mass of a cylinder of nearly same height and diameter and made up of platinum and iridium alloy.

Solution – 06.

SI unit of electric current is Ampere (A).

PAGE NO: 47.
Solution – 07.

Light year is defined as the distance travelled by light in vacuum in one year.

Solution – 08.

1 Parsec is bigger because 1 Parsec is 3.26 times a light year

Solution – 09.

1 Fermi is smaller because 1 Fermi is 10-15 m while 1 micron is 10-6 m.

Solution – 10.

Parsec refers to the distance at which an arc of length equal to 1 astronomical unit subtends an angle of one second at a point.
No, parsec is not same as astronomical unit (A.U.).
1 Parsec = 2 X 105 A.U.

Solution – 11.

Least count of a vernier caliper used in laboratory is 0.1mm = 0.01cm

Solution – 12.

Vernier caliper is an instrument used for measuring small lengths of solid objects where an ordinary scale cannot be applied. We can measure the length accurately up to the order of 10-2 cm, 10-3 cm depending upon the vernier used. Therefore a vernier caliper is important to measure the fraction of a smallest division of a measuring scale which otherwise could not be done by the judgment of the eye.

Solution – 13.

Least count of an instrument refers to the smallest reading that can be accurately measured while using the instrument. For an instrument provided with a scale the least count is the value of one division on its scale.

Solution – 14.

No, we cannot measure the thickness of a paper with vernier caliper as its least count is only 0.1mm. We should use screw gauge instead as its least count is 0.01 mm as the thickness of the paper is in the range of 10-2 mm.

Solution – 15.

If the zero of the one scale (vernier scale or circular scale of screw gauge) does not coincide with the zero of the main scale, this is known as zero scale, zero error arises. There are two types of zero error –
(a) If the zero of the scale remains below the line of graduation of the main scale then it is called positive zero error
(b) If the zero of the scale lies above the line of graduation of the main scale then it is called negative zero error

Solution – 16.

Screw gauge consists essentially of a screw with a uniform pitch which moves in a nut, thus it is named as screw gauge because the major working part is a screw.

Solution – 17.

Solution – 18.

Solution – 19.

Material used for making screw gauge is stainless steel to avoid expansion and contraction due to change in weather as stainless steel absorbs a little heat.

Solution – 20.

When one complete rotation is given to the screw hand, it moves forward or backward by a distance is called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = distance traveled by screw in n rotations/n rotations

Solution – 21.

If the zero of the circular scale lies above the line of graduation then it is called negative zero error. When there is negative zero error, then the instrument reads less than the actual reading. Therefore in order to get the correct reading, the zero error should always be added from the observed reading.

Solution – 22.

Due to constant use, there is space for the play of screw gauge but gradually this space increases with the use or wear and tear, so that when the screw is moved by rotating it in some direction, it slips in the nut and does not cover any linear distance for some rotation of the screw head. The error due to this is known as backlash error.
It is avoided by turning the screw always in the same direction

Solution – 23.

A screw are threaded to twist in, when turned with a screw driver while nails are smooth to slide in straight when pounded with hammer.

Solution – 24.

Screw has two types of motions: linear and circular motions.

Solution – 25.

Unit of Least count of an instrument is cm.

Solution – 26.

1 micron = 10-6 m.

Solution – 27.

A physical balance works on the principle of moments. According to the principle of moments, under equilibrium condition, the clockwise moment due to the body on one side of beam equals the anti clockwise moment due to standard weights on the other side of beam.

Solution – 28.

1 light year = 9.46 X 1015 m

Solution – 29.

Solution – 30.

Solution – 31.

Yes, the vibration is same as the oscillation.

Solution – 32.

The time period, T and frequency of oscillation, f are related as,
T = 1/ for f = 1/T

Solution – 33.

An ideal pendulum is a simple pendulum consists a heavy mass (called the bob) considered as a point mass suspended by a thread which is considered to be mass less and inextensible or non-elastic, from a fixed point or rigid support and in which there is no friction between the support and the string

Solution – 34.

Wall clock with a pendulum will run at a faster rate in winter as it pendulum rod get shorter and the pendulum will swing at a faster rate thus the clock would run faster in winters.

Solution – 35.

Measurement is needed for precise description of any phenomenon happening in the world. For example, if a body is freely falling down to the ground, to understand this phenomenon we must know its velocity, time it will take to reach the ground , etc and to get answer to all our questions we need measurement.

Solution – 36.

Solution – 37.

Solution – 38.

The maintenance of standard units is essential because any variation in these standards would lead to wrong measurements, misleading results and confusing generalizations. The standards are preserved in such a way that they do not undergo any change with the change in temperature, pressure, humidity and other environmental changes.

Solution – 39.

Main characteristics of a standard unit are as follows
(a). It must be well defined.
(b). It must be of proper size. Very small or large size may cause inconvenience.
(c). It should be easily accessible
(d). It must be reproducible at all places without any difficulty.
(e). It must be accurately defined and must not change with time, place and physical conditions such as pressure, humidity, etc.
(f). It must be widely acceptable all over the world.

Solution – 40.

The units which can neither be derived from one another, nor can they be further resolved into other units are known as fundamental units. Some of the fundamental units are metre (length), kilogram (mass), second (time), Kelvin (temperature), ampere (current), etc.

Solution – 41.

Solution 41 page 47″>

Solution – 42.

Solution – 43.

PAGE NO: 48.
Solution – 44.

If the zero of the circular scale does not coincide with the zero of the main scale (pitch scale) when the end of the movable screw is brought in contact with the fixed end then the screw gauge is said to have a zero error.

Solution – 45.

In this case, the zero error is positive
Least count of screw gauge = 0.01 mm
Thus, zero error = 0 + 4 X L.C. = 0.04 mm

Solution – 46.
In this case, the zero error is negative
Least count of screw gauge = 0.01 mm
Thus, zero error = (50-47) X L.C.
= 3 X 0.01
= 0.03 mm

Solution – 47.

No, we cannot measure the diameter of a wire by wrapping it around a pencil because it is not very accurate. We can use screw gauge for this purpose as it can measure the diameter correct up to 1/100 of millimeter or even less.

Solution – 48.

Solution – 49.

Solution – 50.

Number of threads =20
Distance covered in 20 threads = 10 mm
Pitch of the screw gauge = 10/20 =0.5 mm
No of divisions on circular scale = 50
Least count = pitch/no of divisions =
= 0.01 mm

Solution – 51.

(i) Oscillation – One complete to and fro motion of a pendulum about its mean position is known as oscillation.
(ii) Time period – The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum.
(iii) Frequency -the number of oscillation made by the pendulum in one second is called frequency. Its SI unit is Hertz (Hz).
(iv) Amplitude – Amplitude is the magnitude of the maximum deviation of the bob from the mean position on either side when an oscillation takes place.

Solution – 52.

Solution – 53.

Mass of the metal = 540g
Volume = 200cm3
Density = mass of metal/ volume
= 540 /200 = 2.70 g/cm3

Solution – 54.

Mass of copper = 540 g
Density of copper = 9 g/cm3
Volume of copper used in the alloy = mass of copper / density
= 540/9 = 60 cm3
Mass of iron = 240 g
Density of iron = 8 g/cm3
Volume of iron used in the alloy = mass of iron / density
= 240/8 = 30 cm3
Total mass of the alloy = 540 + 240 = 780 g
Total volume of the alloy = 60 + 30 = 90
Density of the alloy = mass of the alloy / density of the alloy = 780 / 90 = 8.67 g/cm3

Solution – 55.

PAGE NO: 49.
Solution – 56.

Solution – 57.

For measuring the length of an object using a vernier calipers, these steps are followed :-
(a). First of all we find the least count and zero error of the vernier calipers.
(b). Place the object whose length is to be measured below the lower jaws and move the jaw till it touches the object. Record the main reading.
(c). Note the division on the vernier scale that coincides with some division of the main scale. Multiply this number of vernier division with least count. This is vernier scale reading.
(d). Record the observed length by adding the main scale reading and the vernier scale reading. Also, subtract zero error with its proper sign, if any, from the observed length to find the true length of the object.

Solution – 58.

Solution – 59.

Following procedure is used to measure the diameter of a wire
(a). Calculate the least count and zero error of the screw gauge.
(b). Place the wire in between the studs. Turn the ratchet clockwise so as to hold the wire gently in between the studs. Record the main scale reading.
(c). Now record the division of circular scale that coincides with the base line of main scale. This circular scale division multiplied by least count will give circular scale reading.
(d). The observed diameter is obtained by adding the circular scale reading to the main scale reading. Subtract the zero error if any, with its proper sign, from the observed diameter to get the true diameter.

Solution – 60.

In order to measure the length of an object using a metre rule, the metre rule must be placed with its marking close to the object, such that the zero marking on the scale coincides with one end of the object. Then the reading on the scale corresponding to the other end of the object will give the length of the object.
Precautions to be taken for measuring the length of the object, the eye must be kept vertically above the end of the object to avoid parallax and the corresponding marking along the line should be carefully read.
The meter scale can measure up to an accuracy of 1mm or 0.1 cm

Solution – 61.

Solution – 62.

Solution 62 page 49″>

Solution – 63.

Solution – 64.

Solution – 65.

Solution – 66.

(i) Oscillation – One complete to and fro motion of a pendulum about its mean position is known as oscillation.
(ii) Amplitude – Amplitude is the magnitude of the maximum deviation of the bob from the mean position on either side when an oscillation takes place.
(iii) Frequency – the number of oscillation made by the pendulum in one second is called frequency. Its SI unit is Hertz (Hz).
(iv) Time period – The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum.

Solution – 67.

PAGE NO: 50
Solution – 68.

Solution – 69.

To measure mass of a body using a physical balance
(i). Before starting, bring the plumb line just above the pointed projection by adjusting the leveling screws at the base. The beam is then gently raised using the lever. And it should be ensured that the pointer swings equally on both sides of the zero mark of the scale.
(ii). Now lower the beam gently and given body is kept on left pan.
(iii). Next, place some weight on the right pan form the weight box using the forceps.
(iv). Now the lever is turned towards right so that the beam rises and the power begins to swing to pointer swing on either side. It must be carefully noted that the side to which the pointer moves more, denotes lesser mass on that side.
(v). Go on adjusting the standard weights till the pointer swings equally on both sides of the zero mark.
(vi). At this stage, the total mass of weights on the right pan gives the mass of the body.
Three precautions to be taken to measure the mass of a body using beam balance are
(a). The beam must be gently lowered before adding or removing weights from the pan.
(b). The weights should not be carried with bare hands to avoid the change in weights due to moisture and dust particles from the surrounding
(c). Whenever you are near the actual weight, you should carefully try the weights in the descending order.
Conditions for a beam balance to be true are
(vii). Both the pans must be of equal weights.
(viii). Both the arms must be of equal lengths

Solution – 70.

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Frank ICSE Solutions for Class 10 Maths Equation of A Straight Line Ex 13.1

Answer 1.

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Frank ICSE Solutions for Class 10 Maths Equation of A Straight Line Ex 13.2

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