Category: Frank ICSE

Frank ICSE Solutions for Class 9 Physics – Fluids

PAGE NO: 157
Solution 1:
The thrust on the unit surface is known as pressure. The SI unit of pressure is Nm^-2.

Solution 2:
Pressure is given by
P = h X_p X_g.
Where h is height of liquid column, p is density of liquid, g is acceleration due to gravity.
Density of mercury is = 1.36 x 10⁴ kg/m³.
h= height of mercury column which is given = 75 cm = 0.75 m.
So pressure = 0.75 x 1.36 x 104 x 9.8 = 9.996 x 10⁴Nm^-2.

Solution 3:
Pressure is a scalar physical quantity.

Solution 4:
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 5:
The force acting normally on a surface is known as thrust.
SI unit of thrust is N.

Solution 6:

Solution 7:
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 8:
Pressure is the physical quantity which is measured in bar.

Solution 9:
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:

• It depends on the point below the free surface (h).
• It depends on density of liquid (p).
• 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

• Pressure inside the liquid increases with the depth from the free surface of the liquid.
• Pressure is same at all points on a horizontal plane, in case of a stationary liquid.
• Pressure is same in all directions about a point inside the liquid.
• Pressure at same depth is different in different liquids. It increases with the increase in the density of the liquid.
• 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

• It can be obtained in pure form.
• It does not vaporizeat ordinary temperatures.
• 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 1:
All liquid exerts an upward force on the body placed in it. This Phenomenon is called buoyancy.

Solution 2:
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 3:
Buoyant force act on a body in upward direction.

Solution 4:
Upthrust is defined as the upward force on the object provided by the liquid because the object has displaced some of the fluid.

Solution 5:
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 6:
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 7:
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 8:
Upthrust or buoyant force depends on the following factors:

• Volume of body submerged in the liquid.
• Density of the liquid.
• Acceleration due to gravity.

Solution 9:

Solution 10:
Weight of the body in air = 300 gf.
Apparent Weight of the completely immersed body in water = 280 gf.

• 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.
• As upthrust on the body = loss in weight
• 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 = 125 x 10^-6 m³.
Mass of the metal cube =9 x 10³ x 125 x 10^-6 = 1125 x 10^-3 =1.125 kg.
Weight of the liquid = mass x gravity = 1.125 x 10 = 11.25 N.
Density of liquid = 1.2 gcm^-3= 1.2 x 10³ kgm^-3.
Upthrust of the liquid = V X_p X_g.
Upthrust = 125 x 10^-6 x 1.2 x 10³ x 10 = 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 4⁰C.

Solution 20:

• SI unit of buoyant force is N.
• SI unit of density is Kgm^-3.
• SI unit of weight of body is N.
• 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:

• 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.
• 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 x density of water =0.67 x 1000 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 x 1000 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 4⁰c 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 4⁰C.

PAGE NO : 175
Solution 35:
Density of mercury = 13600 Kgm^-3.
Density of water at 4⁰C = 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 X_g.
Upthrust = 100 x 1 gf = 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 x 1 = 13.6 gcm^-3.
Density of water in SI system = 1000 Kg m^-3.
So, density of mercury in SI system = 13.6 x 1000 = 13.6 x 10³ Kgcm^-3.

Solution 39:
Density of iron is = 7.8 x 10³ Kg m^-3.
Density of water at 4⁰C = 10³ Kg m^-3.
Relative density of a substance is the ratio of the density of the substance to the density of water at 4⁰C.
So, relative density of iron is = 7.8 x 10³ Kg m^-3/103 Kg m^-3 = 7.8

Solution 40:

• Mass of a metallic piece remains unchanged with increase in temperature.
• Volume of metallic piece increases with increase in temperature.
• 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 1:
All liquid exerts a upward force on the body placed in it. This Phenomenon is called buoyancy.

Solution 2:
The upward force which any liquid exerts upon a body placed in it is called the upthrust.

Solution 3:
Pressure is a scalar quantity.

Solution 4:
Thrust is a vector quantity.

Solution 5:
SI unit of density is Kgm^-3.

Solution 6:
The relative density of a substance is the ratio of the density of the substance to the density of water at 4⁰C.

Solution 7:

Solution 8:
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 9:
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:

• A hydraulic press can be used for extracting juice of sugarcane, sugar beet etc.
• 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

• Weight of gaseous column.
• 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

• 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.
• 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 4⁰C is 1000 Kgm^-3.

Solution 35:
Relative density of body = 0.52
Density of water at 4⁰C = 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:

• 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.
• 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.
• 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.
• 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:

PhysicsChemistryBiologyMaths

Frank ICSE Solutions for Class 9 Physics – Laws of Motion

PAGE NO: 113
Solution 1:
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 2:
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 3:
The greater is the MASS , the greater is the inertia of the object

Solution 4:
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 5:
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 6:
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 7:
SI unit of force is Newton (N).

Solution 8:
Acceleration is the physical quantity associated with N kg^-1.

Solution 9:
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 gms^-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 x 5 = 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 x a x 100
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 1:
Sir Isaac Newton stated the law of gravitation.

Solution 2:
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 3:
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 4:
Acceleration due to gravity is the acceleration experienced by a body during free fall.

Solution 5:
g = GM/R².

Solution 6:
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 7:
SI unit of constant of gravitation is Nm²kg^-2.

Solution 8:
we know that law of gravitation.
F = G ( m₁ x m₂)/R².

• If distance between them is halved then put R = R/2.
  F = 4 x G( m₁ x m₂)/ R².
  F₁ = 4 F.
• If distance between them is doubled then put R = 2R.
  F = G( m₁ x m₂)/ 4R².
  F₁ = F/4.
• If distance between them is made four times then put R = 4R.
  F = G( m₁ x m₂)/16 R².
  F₁ = F/16.
• If distance between them is infinite then put R = infinite.
  F = G( m₁ x m₂)/ R².
  F₁ = 0.
• If distance between them is almost zero then put R = 0.
  F = G( m₁ x m₂)/ 0.
  F₁ = infinite.

Solution 9:
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 ( m₁ x m₂)/R².
Mass of earth = 6 x 10²⁴kg.
Mass of the person = 100 kg.
G = 6.7 x 10-11 Nm²kg^-2.
Radius of earth = 6.4 x 10¹⁴.
F = (6.7 x 10^-11 x 100 x 6 x 10¹⁴ )/ (6.4 x 6.4 x 10¹²) = 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.

• 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
• We know v = u + at
  0 = 4.9 – 9.8 t
  T = 0.5 sec.
• 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 x 9.8 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 ( m₁ x m₂)/R².
Now m₁ = 2 m₁
m₂ = 2 m₂
R = 2 R
F₁ = G ( 2m₁ x 2m₂)/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 35:
Yes a body falling freely near the earth surface has a constant acceleration.

Solution 36:
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 37:

PAGE NO: 128
Solution 1:
Force is that external agency which tends to change the state of rest or the state of motion of a body.

Solution 2:
1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.

Solution 3:
Newton is the SI unit of force whereas dyne is the CGS unit of force.
1 N = 10⁵ dyne.

Solution 4:
No, force is a vector quantity.

Solution 5:
A force can produce MOTION in an objectat rest. It can ACCELERATE an object and can change its DIRECTION of motion.

Solution 6:

• force changes the shape of skin.
• force produces stretching in the rubber.
• force provides retardation to the car and finally stops the car.
• force decreases the momentum of ball and finally stops the ball.

Solution 7:
No, every force does not produce motion in every type of body.

Solution 8:
The amount of inertia of a body depends on its MASS.

Solution 9:
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:

• Weight of the book is action and normal force applied by table on book is reaction.
• Force applied by man on ground is action and force of friction is the reaction.
• Force applied by hammer on nail is action and normal force applied by nail on hammer is reaction to this force.
• Firing of bullet is the action and recoiling of gun is the reaction.
• 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:

• Objects fall on the earth due to gravitational force between the earth and object.
• Atmosphere doesnot escape because molecules of atmosphere are attracted by earth due to gravitational force of earth.
• 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.

• 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.
• ‘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 ( m₁ x m₂)/R²
Here the F is force due to attraction and this force is equal to weight of the body m2g.
So m2g = G ( m₁ x m₂)/R²
g= (G x m₂)/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 x 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:

• The force exerted by the block on is the weight of box and that is equal to 20N.
• 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 = 2s.
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.

• inertia of P/inertia of Q = m₁/m₂ = 1/2.
  So ratio of inertia of two bodies is 1:2.
• Momentum of P/momentum of Q = m₁v₁/m₂v₂ = 1
  So ratio of momentum of two bodies is 1:1.
• 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 F x 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 F x 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 F x 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 =3s.
Mass of the body = 2 kg.
Initial speed of body = 0 ms^-1
Force applied = 10 N.

• We know F x t = m (v – u)
  10 x 3 = 2 (v- 0)
  v = 15 ms^-1.
  Final velocity is 15 ms^-1.
• 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:

• We always prefer to land on sand instead of hard floor while taking a high jump because sand increases the time of contact.As F x t = m ( v – u ) and our change in momentum is constant so if time increases then force experienced would decrease.
• 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.

• We know H = ut + 1/2 gt².
  98 = 1/2 x 9.8 x t².
  t² = 98 x 2/9.8 = 20
  t= 4.47 sec.
• Final velocity when it strikes the ground
  v² – u² = 2 g H
  v² = 2 x 9.8 x 98
  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 x 9.8 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 x 1
v= 0.2 ms^-1
so velocity after 1 sec would be 0.2 ms^-1.
Velocity after 2 seconds
v= 10 – 9.8 x 2 = 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) x 19.6
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 x t².
98 = 0 + 1/2 x 9.8 x t².
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 = v x 3.47 + 1/2 x 9.8 x 3.47 x 3.47.
98 = 3.47 x v + 59
3.47 x 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 x 10^-6 kg = 2 x 10^-4 kg.
Mass of object 2 m2= 200 mg = 200 x 10^-6 kg = 2 x 10^-4 kg.
Distance between the two objects = 1 mm = 10^-3 m
We know law of gravitation is
F = G ( m1 x m2)/R²
F = (6.67 x 10^-11 x 2 x 10^-4 x 2 x 10^-4)/(10^-3 x 10^-3)
F = 6.67 x 2 x 2 x 10^-11-4-4+3+3
F = 26.68 x 10^-13 N
So these two objects would experience a force of 26.68 x 10^-13 N.

Solution 72:
Radius of earth = 6.38 x 103 km = 6.38 x 10⁶ m
G = 6.67 x 10^-11
Acceleration due to gravity = 9.8 ms^-2.
We know that
g= (G x M)/R².
9.8 = (6.67 x 10^-11 x M)/ ( 6.38 x 10⁶ x 6.38 x 10⁶)
9.8 x 6.38 x 6.38 x 10¹² = 6.67 x 10^-11 x M
398.9 x 10¹² = 6.67 x 10^-11 x M
M = 398.9 x 10¹²/6.67 x 10^-11
M = 59 x 10²³ kg
So mass of earth is 59 x 10²³ kg.

PhysicsChemistryBiologyMaths

Frank ICSE Solutions for Class 9 Biology – Interaction Between Biotic and Abiotic Factors in an Ecosystem

PAGE NO: 99

Solution 1:
Producers

Solution 2:
Consumers

Solution 3:
The climatic factors affecting ecosystem are sunlight, temperature, humidity, rainfall and wind.

Solution 4:
Low and high temperatures restrict the growth of plants and existence of animal species. Changes in temperature patterns will impact plant life which in turn will influence the animal life, since animals depend directly or indirectly on it for food. During extreme cold and hot conditions, animals either migrate to favourable places, some may hibernate or some may aestivate.

Solution 5:
Consumers which eat only plants are called consumers of the first order. Consumers that eat herbivores like deer, goats, grasshoppers, etc. are called consumers of the second order.

Solution 6:
Deserts have scanty water, either because there is little rainfall, or because the water evaporates very fast in deserts. Desert species are adapted to less amounts of water and they are capable of surviving for long periods of time in the scarcity of water. The growth of plants and animals and their vital functions are dependent on water intake. Hence water is a vital limiting factor in deserts due to the scant availability of this important resource.

Solution 7:

Solution 8:

Solution 9:
Vital atmospheric gases are oxygen, carbon dioxide and nitrogen.
Oxygen availability seldom becomes a limiting factor for land animals unless they live in soil or invade high altitudes. Plants release oxygen into the air which is used by animals for respiration. During respiration, animals release carbon dioxide which is required by plants for photosynthesis. Nitrogen is an essential gas which is vital for the growth and sustenance of organisms.

Solution 10:
In a natural ecosystem, green plants capture solar energy and convert it into chemical forms. The energy is then passed onto herbivores when they feed on green plants. From herbivores, the energy moves into carnivores that eat them. Some animals like lion and vultures are not eaten by other organisms. All the organisms ultimately pass on energy to the decomposers. Energy thus flows continuously through the ecosystem from plant to animals and from prey to predator.

Solution 11:
Energy passes through the ecosystem in a one-way path. Energy goes through each trophic level, one at a time. As it goes from one level to another, it is lost due to metabolism and in the form of heat. For example – The energy ingested by producers is used by the producers for carrying out various life activities and some amount of energy is lost as heat, so that the entire energy does not pass completely to the consumers. The energy lost as heat cannot be used anymore.

Solution 12:
The transfer of energy from autotrophs through a series of organisms that consume and are consumed is known as a food chain.

Solution 13:
The types of food chain are:

1. Grazing food chain
2. Detritus food chain
3. Auxiliary food chain
   (Write any two)

Solution 14:
Individual food chains interconnected in a complex way is called food web.

Solution 15:
An energy pyramid is a graphical representation of the flow of energy from the producers through the various consumers. It shows the amount of energy available and the loss of useful energy at each step of the food chain in an ecosystem.

Solution 16:
As the energy gets transferred from lower trophic level to the higher one, there is a loss of large amount of energy due to metabolism and as heat. As a result very little energy (i.e. 10%) gets transferred to the next level. So the trophic level at the base has maximum energy and that at the top has the least amount of energy. Hence energy pyramid is broader at the base and narrower at the top.

Solution 17:
(i)(a) unidirectional
(ii)(d) producers and consumers
(iii)(d) all the above
(iv)(a) photosynthesis
(v)(b) herbivores
(vi)(b) bacteria, fungi, etc.
(vii)(b) Communities and their physical environment
(viii)(d) solar energy
(ix)(d) biotic and abiotic
(x)(d) high temperature and high rain
BiologyChemistryPhysicsMaths

Frank ICSE Solutions for Class 9 Biology – Understanding Ecosystems

PAGE NO: 94

Solution 1:
Environment is the sum of all external conditions and influences that affect organisms. The environment may be divided into biotic i.e. living and abiotic i.e. non-living components.

Solution 2:
Ecosystems are units consisting of living things and their specific habitats in the biosphere where living things interact with each other and their environment.

Solution 3:
Man made ecosystem is formed by major human modification or alteration in the natural environment. Examples of man-made ecosystems are aquarium, park, grassland, garden etc.

Solution 4:
Two major types of ecosystems are natural ecosystem and artificial ecosystem.

Solution 5:
Two features of forests are:

1. It includes a complex assemblage of different kinds of biotic and abiotic factors.
2. The temperature and rainfall conditions of a place determine the nature and characteristics of forests.

Solution 6:
The trees in coniferous forests are found in single strands with no undergrowth. Plants predominantly found here include firs, pines, spruces and hemlock while the ground is covered with mosses, grasses, sedges and herbs that are adapted to cold.

Solution 7:
Tropical rainforests are found in regions that experience high temperature, high humidity, heavy and well-distributed rainfall all year round. These regions lack seasonal changes and there is little difference between the length of days and nights.
Tropical rainforests mainly occur inside the World’s equatorial regions. They are restricted to the small land area between the Tropic of Capricorn and the Tropic of Cancer.

Solution 8:
Forest biome refers to the naturally occurring community of flora and fauna occupying the forest.
Features of forest biomes are:

1. In tropical rainforest biomes, there is an amazing biodiversity of plants and animals. Trees are tall, with buttressed trunks and shallow roots, mostly evergreen, with large dark green leaves. Plants such as orchids, bromeliads, vines, lianas, ferns, mosses, and palms are present in tropical forests. Fauna include numerous birds, bats, amphibians, reptiles, small mammals, and a huge diversity of insects.
2. In deciduous rainforest biomes, many types of tall and short trees, shrubs, herbs, mosses and lichens are found occupying five different layers. Also a huge variety of fauna like panda, brown bear, hedgehog etc are found.
3.  Not much biodiversity occurs in coniferous forest biomes. The trees here are found in single strands with no undergrowth. Plants predominantly found here include firs, pines, spruces and hemlock while the ground is covered with mosses, grasses, sedges and herbs that are adapted to cold. Fur bearing animals are found abundantly like brown bear, fox, mink, beavers, deer and large birds of prey like red tailed hawks. Many animals migrate or hibernate during the extremely cold winters.

Solution 9:

Solution 10:
(i) (d) all plants and animal species along with environment.
(ii) (c) an artificial ecosystem
(iii) (c) communities of organisms interacting with one another
(iv) (c) Tansley
(v) (c) 100-150 cm
(vi) (b) Decomposers

BiologyChemistryPhysicsMaths

Frank ICSE Solutions for Class 9 Biology – Respiration in Plants

PAGE NO: 87

Solution 1:

Solution 2:

Solution 3:

Solution 4:
Yes, respiration is the reverse of photosynthesis.

Solution 5:

1. – (b) nutrients are oxidized without using molecular oxygen by the process of fermentation.
2. – (c) is the best organic substrate for respiration.
3.  – (a) partial breakdown of food substance.
4.  – (e) the series of change from glucose to pyruvic acid in respiration.
5.  – (d) the intermediate substance in the breakdown of glucose.

PAGE NO: 88

Solution 6:

Solution 7:
(a) Anaerobic
(b) Glycolysis
(c) Pyruvic acid
(d) Oxygen
(e) Cytoplasm

Solution 8:
(a) False
(b) False
(c) False
(d) False
(e) False

Solution 9:
(a) To show that heat is evolved during respiration
(b) In flask A, moist seeds respire and produce heat that increases the temperature.
(c) If formalin was not used, bacteria will grow on the dry seeds and respire anaerobically to produce a little heat.

PAGE NO: 89

Solution 10:
glucose ATP

Solution 11:

Solution 12:
(a) To absorb carbon dioxide produced during respiration
(b) If these are not soaked in disinfectant, the bacterial growth may be there in the tube Y and accurate result may not be obtained due to bacterial respiration.
(c) The germinating peas respire and oxygen is used which create a vacuum in the tube. So coloured water has risen in tube1.
(d) Respiration
(e) It is defined as the stepwise oxidation of glucose in the living cells to release energy.

Solution 13:
(a) mitochondria
(b) Fermentation

Solution 14:

Solution 15:

Solution 16:
(a) Respires
(b) Day and night
(c) Aerobic respiration
(d) Rises
(e) Controlled manner

PAGE NO: 90

Solution 17:
(a) Respiration
(b) ATP
(c) Aerobic respiration
(d) Anaerobic respiration
(e) Caustic potash and KOH
(f) Lime water
(g) Carbon dioxide and water
(h) Ethyl alcohol and carbon dioxide

Solution 18:

(i) (c) Glucose is converted to carbon dioxide and water, releasing energy.
(ii) (d) Glycolysis, Kreb’s cycle, electron transfer
(iii) (c) Energy is left in alcohol.
(iv) (d) To accept hydrogen and form water.
(v) (b) CO2 and alcohol
(vi) (c) mitochondria
(vii) (b) Hens Krebs
(viii) (d) fermentation
(ix) (b) ATP
(x) (c) ATP
(xi) (b) Two
(xii) (a) In cytoplasm
(xiii) (c) 673 Kcal
BiologyChemistryPhysicsMaths

Frank ICSE Solutions for Class 10 Chemistry – Analytical Chemistry

PAGE NO : 75
Solution 1:

1. Cuprous salts = Colourless
2. Cupric salts = Blue
3. Aluminium salts = Colourless
4. Ferrous salts= Light green
5. Ferric salts = Yellow
6. Calcium salts = Colourless

Solution 2:

Solution 3:

Solution 4:
K₂SO_4.

Solution 5:

Solution 6:

Solution 7:

Solution 8:

Solution 9:
Examples of amphoteric hydroxides are: Zn(OH)2, Al(OH)3.

Solution 10:

Solution 11:

PAGE NO : 76
Solution 12:

Solution 13:
The chloride of a metal which is soluble in excess of ammonium hydroxide is zinc chloride i.e. ZnCl2.

Solution 14:

Solution 15:

1. PbO
2. Al2O3
3. Na2ZnO2

Solution 16:

1. transition, Cr³⁺, Fe²⁺, MnO₄^4-.
2. Zn(OH)₂
3. NH₄Cl
4. Al₂O₃, Al
5. NH₄OH

Solution 1992-1:

1. Addition of KCN
2. Addition of excess of NaOH.
3. Addition of excess ofNH₄OH

Solution 1993-1:

PAGE NO : 77
Solution 1995-1:

1. The metal ion present in solution A is Pb²⁺
   .
2. The cation present in solution B is Cu²⁺. The probable colour of solution B is blue.

Solution 1996-1:

Solution 1996-2:
The solutions for the tests will be prepared by dissolving the given powders separately in water.

1. Solution of Calcium carbonate:
   Calcium carbonate is CaCO₃ and contains Ca²⁺ ions. Sodium hydroxide solution NaOH can be used to identify Ca²⁺ since its addition to calcium carbonate solution will give white precipitates of Ca(OH)2 which are sparingly soluble in excess of NaOH.

1. Solution of Lead carbonate:
   Lead carbonate is PbCO₃and contains Pb2+ ions. Ammonium hydroxide solution NH4OH can be used to identify Pb²⁺ since its addition to lead carbonate solution will give white precipitates of Pb(OH)2 which are insoluble in excess of NH4OH.
2. Solution of Zinc carbonate:
   Zinc carbonate is ZnCO3 and contains Zn²⁺ ions. Sodium hydroxide solution NaOH can be used to identify Zn²⁺ since its addition to zinc carbonate solution will give white gelatinous precipitates of Zn(OH)₃ which are soluble in excess of NaOH.

Solution 1996-3:

Solution 1997-1:

Solution 1998-1:

Solution 1999-1:

Solution 2000-1:

PAGE NO : 78
Solution 2001-1:

Solution 2003-1:

Solution 2003-2:

Solution 2004-1:

PAGE NO : 79
Solution 2005-1:

1. B and E (Iron (II) sulphate and Magnesium sulphate)
2. C and F (Iron (III) chloride and Zinc chloride)
3. D (Lead nitrate)
4. A (Copper nitrate)
5. F (Zinc chloride)

Solution 2006-1:

Solution 2009-1:
C ( Aluminium oxide)

Solution 2009-2:

1. P is Ferric chloride
2. Q is an ammonium salt
3. R is ferrous sulphate

Solution 2009-3:

1. When BaCl₂
2.  solution is added to the given solution ZnSO₄
3. gives a white precipitate while no precipitate is obtained with ZnCl₂ solution.
4. When NaOH solution is added to the given solution, iron (II) chloride gives dirty green precipitate while reddish brown precipitate is obtained with iron(III) chloride.

ChemistryBiologyPhysicsMaths

---

Frank ICSE Solutions for Class 9 Biology – Seeds: Structure and Germination

PAGE NO: 78

Solution 1:
(a) Seed is defined as a fertilized mature ovule which possesses an inactive embryo and reserve food for its further development.
(b) The process by which the dormant embryo of the seed resumes active growth and forms a seedling is known as germination.

Solution 2:
(a) Albuminous seed – In some dicotyledons and monocotyledons, the
food is stored mainly in the endosperm. Such seeds are called albuminous seeds. Example – Seeds of castor, cereals and grasses.
(b) Dormancy – Seed dormancy is a condition of plant seeds that prevents germination under optimal environmental conditions. Here the seed is in a state of apparent inactivity and will not grow even if favorable conditions are provided, until a definite time has elapsed.
(c) Hypogeal germination – In this germination, the seed remains inside the soil since epicotyl elongates faster than hypocotyl. Hence the cotyledons remain inside the soil. Example – Wheat, rice, pea, mango.
(d) Epigeal germination – It is a type of germination in which cotyledons are pushed above the soil into the air and light. This occurs due to rapid growth and elongation of the hypocotyl. Example – Bean, cotton, castor, papaya, onion, tamarind.

Solution 3:
This is because the seed is in a state of dormancy. In this case, even if all the favorable conditions are provided, the seed remains in a state of apparent inactivity and only germinates after a definite time has elapsed.

Solution 4:
(a) Seed coat is the outer covering of seed. It protects the inner contents of the seed.
(b) Micropyle allows entry of water into the embryo.
(c) Endosperm contains stored food mostly as starch.
(d) Cotyledons store food material for the embryo.

Solution 5:
(a) Plumule
(b) Coleorhiza
(c) Endosperm
(d) Micropyle
(e) Root and shoot
(f) Endosperm
(g) Epigeal germination
(h) Hypogeal germination
(i) Orchis seed
(j) Seed of Lodoicea moldivica

PAGE NO: 79

Solution 6:

Solution 7:
(a) castor, papaya
(b) grasses, wheat
(c) pea, mango
(d) wheat, rice

Solution 8:
The factors necessary for germination are:

1. Water – Water is essential for seed germination since protoplasm becomes active only when saturated with water. Water facilitates the necessary chemical changes in food material. Also enzymatic reaction occurs only in the water medium. Water when imbibed by the seed coat makes it soft and swollen. Then the seed coat bursts open, helping the embryo come out easily.
2. Temperature – A suitable temperature is essential for seed germination since many physiological processes occur within the seed during germination. Seeds fail to germinate below 0?C or above 45?C. Optimum temperature for seed germination is 15-30?C.
3. Oxygen – During germination, embryo resumes growth and for this energy is required. This energy comes from the oxidation of food material stored in the endosperm or cotyledons. This process requires oxygen.

Solution 9:
Apparatus required for three beans experiment are beaker, bean seeds and wooden piece.
The air-dried seeds are attached to a piece of wood, one at each end and one in the middle. This is then placed in a beaker and water is poured into it till the middle seed is half immersed in it. The beaker is then left in a warm place for a couple of days. From time to time, water is added to maintain the original level.
It is observed that after a couple of days that the bean in the middle germinates normally since it has sufficient water, oxygen and temperature. The bottom seed gets sufficient water and temperature but not oxygen hence it may develop a radicle but doesn’t grow further. The upper seed gets oxygen and temperature but not water and hence fails to germinate.
This experiment shows that water, temperature and oxygen are essential for seed germination and that germination will not occur if any one of these factors are absent.

Solution 10:
If the seeds are sown too deep in the soil, they may not get sufficient oxygen required for respiration and hence will fail to germinate.

Solution 11:

Solution 12:
(i) (d) in endosperm
(ii) (d) castor bean
(iii) (d) all the above
(iv) (d) maize
(v) (a) double coconut
(vi) (b) germination
BiologyChemistryPhysicsMaths

Frank ICSE Solutions for Class 9 Biology – Pollination and Fertilization

PAGE NO: 72

Solution 1:
Pollination is the transfer of pollen grains from the anther to the stigma of the same or another flower.
The male gametes are produced inside pollen grains located in the anthers of androecium whereas the female gametes are produced in the ovules located in the ovary of gynoecium. For forming zygote, the male gametes need to be transferred to the gynoecium for fusing with the female gametes. This is achieved through pollination. Pollination occurs through insects, wind or other agents.
There are two types of pollination – Self pollination and cross pollination.

Solution 2:
The two modes of pollination are:

(i) Self-pollination – It is the transfer of pollens produced within the anther of a flower to the stigma of the same flower or to the stigma of another flower of the same plant. In such flowers, pollination is ensured since the flowers bear similar genetic characters. Self pollination can occur in bisexual or monoecious flowers. Examples of plants showing self pollination are Mirabilis, Arachis etc.
(ii) Cross pollination – It is the transfer of pollen grains from the anthers of a flower of one plant to the stigma of a flower of another plant. Cross pollination occurs in unisexual or dioecious flowers such as papaya, maize, jasmine, rose etc.

Solution 3:

Solution 4:
Adaptations required by self pollinated plants are:

• Bisexuality – Self pollination occurs only in bisexual flowers.
• Homogamy – Both anther and stigma need to mature at the same time.
• Cleistogamy – Flowers which are bisexual and never open are called cleistogamous flowers. They are small, colourless, odourless and without nectar. The pollen grains fall on the stigma inside the closed flower. Example – Arachis

Adaptations required by cross pollinated plants are:

• Unisexuality – The stamens and carpels are found in different flowers. The male and female flowers may be borne on the same or different plants.
•  Dichogamy – In bisexual flowers, stamens and carpels mature at different times.

It is of two kinds:

1.  Protandry wherein stamens mature before carpels. E.g – jasmine
2. Protogyny wherein carpels mature before stamens. E.g. – Rose

•  Heterostyly – Here the style is either longer or shorter, thereby preventing self pollination.
•  Herkogamy – Stigma and stamen mature at the same time, but some type of barrier prevents self pollination. E.g. – In caryophyllaceous flower, the stigma projects beyond the stamens so that pollens cannot fall on it.
• Self-sterility – Pollen of one flower cannot fertilize the female gametes of the same flower.

Solution 5:

Solution 6:
Fertilisation is defined as the fusion of the male and female gametes.

Solution 7:

Solution 8:
In angiosperms, during fertilization, one male gamete fuses with the egg cell and forms diploid zygote in a process called syngamy. The other male gamete fuses with the two polar nuclei to form a triploid nucleus called primary endosperm nucleus. This process is called triple fusion. Since fertilization takes place twice here, so this process is called double fertilization.
Significance – Due to double fertilization, triploid nucleus develops into endosperm which serves as nutrition for embryo.

Solution 9:
Fruit is a ripened ovary containing one or more seeds.

Solution 10:
After fertilization, ovary undergoes two important changes:

• The ovules develop into seeds
• The ovary walls thicken and ripen into pericarp or fruit wall.

Solution 11:
Yes, fruits are important for the plant since the seeds mature inside it. Fruits are colourful and tasty and hence eaten by animals. This helps in far and wide dispersal of the seeds.

Solution 12:
(i) (c) entomophily
(ii) (a) bats
(iii) (a) ornithophily
(iv) (a) syngamy
(v) (c) pomology
(vi) (b) true fruits
BiologyChemistryPhysicsMaths

Frank ICSE Solutions for Class 9 Biology – Flowers

PAGE NO: 64

Solution 1:
A flower is the reproductive unit in angiosperms. It is a modified shoot in which internodes are shortened and leaves are modified into floral structure. Flower is meant for sexual reproduction.
A typical flower has four different kinds of whorls arranged successively on the swollen parts of a flower stalk. Flower stalk consists of the stalk called pedicel and the swollen upper part called thalamus bearing the floral leaves.
The different floral whorls are calyx, corolla, androecium and gynoecium. Calyx and corolla are accessory whorls, while androecium and gynoecium are reproductive whorls.

•  Calyx – The calyx is the outermost whorl of the flower and its members are called sepals. Generally, sepals are green, leaf like and protect the inner whorls of the flower in bud stage. They are also involved in producing food by photosynthesis. The calyx may be gamosepalous (sepals united) or polysepalous (sepals free).
• Corolla – It is the second whorl composed of floral leaves called petals. Petals are usually brightly coloured to attract insects for pollination. Petals also protect the inner whorls. Like calyx, corolla may be also free (gamopetalous) or united (polypetalous). The shape and colour of corolla vary greatly in plants.
•  Androecium – It is the third whorl and is the male reproductive whorl of a flower. Androecium is composed of one or more stamens. Each stamen consists of three parts:

1. Filament – It is the lower stalk of the stamen.
2. Anther – Filament bears a bilobed fertile structure called anther at its distal end. Each lobe contains two pollen sacs. The pollen grains are produced in pollen-sacs.
3. Connective – Filament of the stamen is extended in between the two anther lobes called connective.

• Gynoecium – It is the innermost whorl and the female reproductive part of the flower. Gynoecium is made up of one or more carpels. A carpel consists of three parts namely stigma, style and ovary.
  Ovary is the swollen basal part containing ovules. Each ovary bears one or more ovules attached to a flattened, cushion-like structure called placenta.
  Style is the elongated thread like structure attached to the apex of the ovary. It connects the ovary to the stigma.
  The stigma is situated at the tip of the style and is the receptive surface for pollen grains.
• 

Solution 2:
(a) Inflorescence – The arrangement of flowers on the floral axis is called inflorescence.
Function – Inflorescence facilitates the best arrangement and display of flowers on a branch without any sort of overcrowding. It also facilitates pollination via a prominent visual display and more efficient pollen uptake and deposition.
(b) Gynoecium – It is the innermost whorl of the flower bearing the female reproductive parts.
Function – The ovary of gynoecium produces ovules which bear the female gamete.
(c) Placentation – The manner in which placenta and ovules are arranged inside the ovary wall is known as placentation.
Function – Placentation helps in the best arrangement of ovules within the ovary. Placentation also helps in plant classification.
(d) Incomplete flower – A flower lacking one whorl out of the four whorls is said to be incomplete flower.
Function – An incomplete flower contains either male or female reproductive organs.
(e) Perianth – When the calyx and corolla are not distinct in a flower (eg. – lily), the whorl is collectively called perianth.
Function – The members of perianth, called tepals are usually brightly coloured and bear scent. This attracts insects which aids in pollination. They also protect the flower in bud condition.

Solution 3:

Solution 4:
The flower is the reproductive unit in the angiosperms and is meant for sexual reproduction. Flowers produce seeds from which new plants grow in future. So the main function of flower is to perpetuate the species.
There are six different types of flowers. These are complete, incomplete, bisexual, unisexual, actinomorphic and zygomorphic.

Solution 5:

Solution 6:
In certain flowers like tomato and brinjal, the calyx remains attached even after the formation of the fruit and does not wither away. Such calyx is called persistent calyx.

Solution 7:
Calyx is the outermost whorl of a flower which is composed of sepals. Generally these sepals are green, leaf like and protect the inner whorls of the flower in bud condition. They are also involved in producing food by photosynthesis.

Solution 8:

Solution 9:
Corolla is the second whorl composed of floral leaves called petals. Petals are usually brightly coloured to attract insects for pollination. Petals also protect the inner whorls. The shape and colour of corolla vary greatly in plants.

Solution 10:
The androecium and gynoecium are the essential parts of a flower because they are involved in sexual reproduction.
Androecium is the male reproductive organ of a flower and is involved in producing male gametes.
Gynoecium is the female reproductive part of the flower and produces the female gametes.

The non-essential or accessory parts of flowers are the calyx and corolla since they do not directly participate in the process of sexual reproduction leading to the development of seed.
Sepals of calyx are green, leaf like and protect the inner whorls of the flower in bud stage. They are also involved in producing food by photosynthesis.
Petals of corolla are usually brightly coloured to attract insects for pollination; they also protect the inner whorls.

Solution 11:

Solution 12:

Solution 13:
(a) Androecium – It is the third whorl and is the male reproductive organ of a flower. Androecium is composed of one or more stamens. Each stamen consists of three parts: Filament, Anther and Connective. The pollen grains are produced in pollen-sacs on the anthers.
(b) Gynoecium – It is the innermost whorl and is the female reproductive part of the flower. Gynoecium is made up of one or more carpels. A carpel consists of three parts namely stigma, style and ovary. Ovary is the swollen basal part containing ovules.
(c) Calyx – The calyx is the outermost whorl of the flower and its members are called sepals. Generally, sepals are green, leaf like and protect the inner whorls of the flower in bud stage. They are also involved in producing food by photosynthesis. The calyx may be gamosepalous (sepals united) or polysepalous (sepals free).
(d) Corolla – It is the second whorl composed of floral leaves called petals. Petals are usually brightly coloured to attract insects for pollination. Petals also protect the inner whorls. Like calyx, corolla may be also free (gamopetalous) or united (polypetalous).

PAGE NO: 65

Solution 14:
(a) Datura
(b) Cotton
(c) Cotton
(d) Sunflower
(e) Tomato
(f) Mulberry

Solution 15:
(i) (b) condensed stem
(ii) (b) jointed calyx
(iii) (c) thalamus
(iv) (a) reniform
(v) (c) capitulum

BiologyChemistryPhysicsMaths