
01.Matter in Our Surroundings
11
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11


02.Is Matter Around Us Pure
11
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03.Atoms and Molecules
7
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7


04.Structure of The Atom
7
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7


05.Cell  Fundamental Unit of Life
11
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06.Tissues
8
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8


07.Diversity in Living Organisms
8
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8


08.Motion
11
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11


09.Force and Newtons Laws of Motion
12
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11

Lecture9.12


10.Gravitation
9
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9


11.Work and Energy
7
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12.Sound
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Why do We Fall Ill
7
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7


14.Natural Resources
11
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11


15.Improvements in Food Resources
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Gravitation
Intext Questions
Q.1 Why is it difficult to hold a school bag having a strap made of a thin and strong string ?
Sol. It is difficult to hold a school bag having thin straps as pressure exerted by the thin straps on the shoulder will be more than broad strap.The reason being Pressure α1Area
Q.2 What do you mean by buoyancy ?
Sol. Buoyancy is the upward force experienced by an object placed in a fluid.
Q.3 Why does an object float or sink when placed on the surface of water ?
Sol. An object floats or sinks when placed on water depends on the net force. If the weight is less than buoyant force it will float, and if more than it will sink.
Page 142
Q.1 You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Sol. On the weighing machine mass is measured by comparing weights. Since our actual weight is slightly more than that measured by machine (as buoyancy reduces the weight) our mass is slightly more than that measured.
Q.2 You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighting machine. In reality, one is heavier than other. Can you say which one is heavier and why?
Sol. Actual weight = Measured weight + Buoyant Force .
Cotton bag is much larger than iron bar, so its buoyant weight is more .
Hence , actual weight of cotton is more than the iron bar.
Exercise
Q.1 How does the force of gravitation between two objects change when the distance between them is reduced to half
Sol. According to the law of gravitation , the force of attraction between any two objects of mass m_{1} and m_{2} is proportional to the product of their masses and inversely proportional to the square of the distance ‘R’ between them.
As given here F=Gm1×m2R2
Here G is the gravitational constant. When the distance (R) is reduced to half .
Then F′=Gm1×m2R2/22
Or F′=4F
Clearly, as the distance between the objects is reduced to half the force of gravitation becomes four times the original force.
Q.2 Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object ?
Sol. Weight of an object on surface of earth = mg
where ‘m’ is mass and ‘g’ is acceleration due to gravity .
Gravitational force acting on the object F = GM×mR2
Here M is mass of earth, R is radius of earth i.e. distance between the objects and G is gravitational constant
Weight of object ‘mg’ = Gravitational force F acting on it
mg = GM×mR2
g = GMR2
From above expression it is clear , ‘g’ acceleration due to gravity is independent of mass of an object and all the objects irrespective of being heavy or light experience the same acceleration due to gravity, hence fall with same speed from a given height.
Q.3 What is magnitude of gravitational force between the earth and a 1 kg object on its surface ? Take mass of earth to be 6 × 10^{24} kg and radius of the earth is 6.4 × 10^{6 }m. G = 6.67 ×10^{–11} Nm^{2} kg^{–2}.
Sol. As given in the statement :
Gravitational Constant G = 6.67 × 10–11 N m2 kg^{–2}
Mass of the object m_{1} = 1kg
Mass of the Earth m_{2}= 6 × 1024 kg ;
Radius of the earth R = 6.4 ×106 m
As we know F = Gm1×m2R2
F=6.67×10−11×1×6×1024(6.4×106)2
=6.67×6(6.4)2×10−11+24−12
⇒ F=0.977×101=9.77N
or F = 9.8 N Approximately
Q.4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth ? Why ?
Sol. The universal law of gravitation states that the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.
As given here F=GM1×M2R2
Where M_{1} = Mass of earth, M_{2} = Mass of moon, R = Distance between earth and moon. G is the gravitational constant. The above law applies to all objects anywhere in the universe. This is also true in case of force of attraction between Earth and moon. The magnitude of force (F) of attraction exerted by Earth on the moon, due to gravitation, is the same as that exerted by moon (F) on earth.These forces being equal and opposite, also in accordance with Newton’s Third Law of Motion, which states that to every action there is an equal and opposite reaction.
Q.5 If the moon attracts the earth, why does the earth not move towards the moon ?
Sol. As per Newton’s third law of motion, the mutual forces of attraction due to gravitation, between earth and moon is same. The mass of the earth is 5.97 x 1024 kg and that of moon is 7.349 x 1022 kg. Clearly the mass of earth is 82.23 times that of moon. Let Fe be the force of gravitational pull of earth on moon; Fm be the force of gravitational pull of moon on earth; me mass of earth ;mm mass of moon ; ge be the acceleration caused by earth on moon and gm be the acceleration caused by moon on earth .
Fm=mm×ge
Fe=me×gm
As per, Newton’s third law of motion,
Fm=Fe
mm×ge=me×gm
As given, me = 83.23 × mm
mm×ge=83.23×mm×gm
Therefore, ge=83.23×gm
Or gm=ge/83.23=0.012ge
Here clearly acceleration experienced by the earth due to gravitational pull of moon is very small, just .012 times ( 1.2 % ) of that experienced by the moon due to earth , which is not as effective to move the earth towards the moon.
Q.6 What happens to the force between two objects, if
(i) The mass of one object is doubled ?
(ii) The distance between the objects is doubled and tripled ?
(iii) The masses of both objects are doubled ?
Sol. According to the law of gravitation , the force of attraction between any two objects of mass m_{1} and m_{2} is proportional to the product of their masses and inversely proportional to the square of the distance ‘R’ between them.
i.e F=Gm1×m2R2
Here G is the gravitational constant. When the distance (R) is reduced to half .
(i) When the mass of one object say m1 is doubled, then
i.e. F′=G2m1×m2R2
i.e. F′=2Gm1×m2R2 = 2F
∴ As the mass of one object is doubled the force becomes 2 times.
(ii) When the distance between the bodies is doubled and tripled
when the distance is doubled
i.e. F′=2Gm1×m2(R/2)2=F/4
The force is reduced to one fourth of the original force. when the distance is tripled :
i.e. F′=2Gm1×m2(R/3)2=F/9
The force is reduced to one ninth of the original force.
(iii) When the masses of both the objects are doubled, then
F′=G2m1×2m2R2=4F
∴ When the masses of both the objects are doubled, then the force becomes four times the original force.
Q.7 What is importance of universal laws of gravitation ?
Sol. The Importance of Universal law of gravitation lies in the fact, that it was successful in explaining many phenomena such as.
(i) That, how does the different objects in this universe, affect others.
(ii) That, how gravity; the force of gravitation due to earth, is responsible for the weight of a body and keeps us on the ground. And that why force of gravity decreases with altitude.
(iii) That, how does the lunar motion around the earth occur.
(v) That, how does the planetary motion of planets in our solar system as well as that of all other celestial objects take place
(vi) That how do the tidal waves originate, due to the gravitational pull of moon and the sun.
Q.8 What is the acceleration of free fall ?
Sol. Acceleration of free fall is the acceleration experienced by body falling freely towards earth under the influence of gravitation force of earth alone. It is denoted by g and its value on the surface of earth is 9.8 m s^{–2}.
Q.9 What do we call the gravitational force between an earth and an object ?
Sol. The gravitational force between an earth and an object Weight and it is equal to product of mass(m) and acceleration due to gravity(g).
Q.10 Amit buys a few gram (force) of gold at the poles as per instruction of one of his friends. He hands over the same when he meets him at equator. Will the friend agree with the weight of gold bought ? If not, why ?
Sol. The force of gravity decreases with altitude. It also varies on the surface of the earth, decreasing from poles to the equator. Also, The weight is equal to the product of mass and acceleration due to gravity.
i.e. W = mg,
where ‘m’ is mass of the object ‘g’ is the acceleration due to gravity. As value of g is higher at polar region than the equator, the weight of an object will also be more at polar region than at equtor. Therefore, his friend will not agree with weight of the gold bought at the poles when measured at equator.
Q.11 Why will a sheet of paper fall slower than one that is crumpled into a ball ?
Sol. Surface area of a Sheet which is crumpled into a ball, is much smaller than the surface area of a plain or flat sheet. Therefore, despite both experince same force of gravity, the plain or flat sheet of paper will have to face more air resistance than the crumpled ball, so it will fall slower than the sheet crumpled into a ball.
Q.12 Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton of a 10 kg object on the moon and on the earth ?
Sol. The mass of an object always being same.
Therefore,The mass of the object on the moon will be the same, as that on earth, m = 10 kg.
We know that, Acceleration due to gravity on earth (ge) = 9.81 ms^{–2}
Acc. due to gravity on moon (gm) = 9.816ms−2
Therefore, Weight of the object on the surface of moon = mgm
= 10 × 9.816=16.35N and Weight of the object on the surface of Earth = mge
= 10 × 9.81 = 98.1 N
Q.13 A ball is thrown vertically upwards with a velocity of 49 ms^{–1}. Calculate
(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.
Sol. (i) As per the statement given
Initial velocity of the ball (u) = 49 ms^{−1}
Final velocity of the ball (v) = 0 ms^{−1}
Downward gravity ( g ) = 9.8 ms^{−2
} Upward gravity (g) = − 9.8 ms^{−2
} Max. Height attained by the ball (s ) = ?
As we know :
v^{2} − u^{2} = 2gs = > (0)^{2} − ( 49 )^{2} = 2 × (− 9.8 ) × s = > s = − 49 × 49 − 2 × 9.8
Max. Height attained by the ball (s) = 122.5 m
(ii) Also, as we know :
v = u + gt = > 0 = 49 − 9.8 × t = > t = 49 9.8 =5 s
Time for upward journey of the ball will be the same as time for downward journey i.e.
t = 5 s.
Total time taken by the ball to return to the surface of earth = 2 × t = 2 × 5 = 10 s
Q.14 A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Sol. As per given statement, initial velocity u = 0, g = – 9.8 ms^{–2}, height, s = –19.6 m
As we know that u^{2} – u^{2} = 2 gs
Therefore, u^{2} – 02 = 2 × (– 9.8) × (–19.6)
or v^{2} = (19.6)^{2}
⇒ v = –19.6 ms^{–1}
The negative sign indicates that the velocity is in the downward direction.
Q.15 A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone ?
Sol. Here,initial velocity u = 40 ms–1 and g = – 10 ms^{–2}
At the maximum height reached S , final velocity, v = 0
AS we know the v^{2} – u^{2} = 2gs
⇒ S = 40×4020=80m
Also, the total distance covered = 2 × S = 2 ×= 160 m
As the stone returns back to the original position.
Therefore, Net= S – (S)
= 80 – 80 = 0
Q.16 Calculate the force of gravitation between the earth and the sun, given that the mass of the earth = 6 × 10^{24} kg and of the sun = 2 × 10^{30} kg. The average distance between the two is 1.5 × 10^{11} m.
Sol. As given in the statement, M_{e} = 6 × 10^{24} kg, M_{s} = 2 × 10^{30
} r = 1.5 × 10^{11}m
As we know that F = G MeMsr2
Therefore, F=6.67×10−11×6×1024×2×<br/>1030(1.5×1011)2N
⇒ F=6.67×12×10211.5×1.5
Therefore, F = 3.56 × 10^{22} N
Q.17 A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet ?
Sol. Let the point at which, two stones meet after time t from the start, be at a height x from the ground. Height of the tower = 100 m
Then distance covered by the stone allowed to fall from the top of the tower,
100−x=ut+12gt2
=0×t+12gt2
=12gt2 …(1)
The distance covered by the stone thrown from the ground,
x=ut−12gt2
=25t−12gt2 …(2)
Combining eq. (1) and (2), we get
100 = 25t
⇒ t = 4s
Therefore, x = 25 × 4 – 12 × 9.8 × 4^{2
} = 100 – 78.4
= 21.6 m
Q.18 A ball thrown up vertically returns to the thrower after 6s. Find:
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches, and
(c) its position after 4s.
Sol. (a) The velocity with which ball was thrown up :
Acceleration due to gravity, g = – 9.8 ms^{–2}
As the total time taken in upward and return journey by the ball is 6 s.
Therefore, The upward journey, t = 6/2 s = 3 s
Final velocity, u = 0 ms^{–1}
Initial velocity, u = ?
As we know, by the first equation of motion,
u = u + gt
⇒ 0 = u + (9.8) × 3
⇒ 0 = 4 – 29.4
⇒ u = 29.4 ms–
Therefore, The velocity with which ball was thrown up = 29.4 ms^{–1}
T (b) Max. Height the ball reaches (h) = Distance (s) = ?
As we know, by the second equation of motion,
s=ut+12gt2
Therefore, s = 29.4 × 3 + 12 (– 9.8) × 32
⇒ = s = 88.2 m 44.1m
(c) The position of the bell after 4 seconds
= Distance, s = ?
Here as given Time, t = 4 s
As we know by equation from laws of motion,
s=ut+12gt2
Therefore, s=29.4×4+12×(−9.8)(4)2
⇒ s = 117.6 m – 78.4 m
⇒ s = 39.2 m
Q.19 In what direction does the buoyant force on an object immersed in a liquid act?
Sol. The direction of Buoyant force on an object immersed in a liquid acts is vertically upward towards the centre of buoyancy.
Q.20 Why does a block of plastic released under water come up to the surface of water?
Sol. All objects experience a force of buoyancy when they are immersed in a fluid. Objects having density less than that of the liquid in which they are immersed, float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed then it sinks in the liquid. Here the density of plastic block is less than that of water, there for, weight of water displaced by fully immersed plastic block is more than its own weight. Thus, the upward force acting on the plastic block due to buoyancy , is much more than the downward force due to the weight of the block. Due to this upward buoyant force, the block will be forced up tilll the weight of displaced liquid is equal to the weight of plastic block.
Q.21 The volume of 50 g of substance is 20 cm^{3}. If the density of water is 1 g cm^{–3}, will the substance float or sink ?
Sol. As given in the statement: The Mass of substance = 50 g
The volume of substance = 20 cm^{3
} density of water is 1 g cm^{–3
} Density of substance = Mass of substance volume of substance = 50 20 = 2.5 g cm^{−3}
Clearly As the density of substance (2.5 g cm^{−3} )is more than that of water (1 g cm^{−3}), so it will sink.
Q.22 The volume of a 500 g sealed packed is 350 cm^{3}. Will the packet float or sink in water if the density of water is 1 g cm^{–3} ? What will be the mass of water displaced by this packet ?
Sol. As per the statement of question :
Mass of sealed packed M = 500g
Volume of sealed packed V= 350 cm^{3}
Density of water = 1 g cm^{–3}
Density of sealed packet = M V
= 500 350 = 107
= 1.43 g cm^{−3} As density of packet which is 1.43 g cm^{–3} is more than that of water
(1 g cm^{–3}), it will sink in water.
The volume of water displaced is equal to the volume of packet V i.e., 350 cm³.
∴ Mass of the water displaced = Density × Volume = 1 g cm− 3 × 350 cm^{3} = 350 g
Exemplar
Multiple Choice Questions
Q.1 Two objects of different masses falling freely near the surface of moon would
(a) Have same velocities at any instant
(b) Have different accelerations
(c) Experience forces of same magnitude
(d) Undergo a change in their inertia
Sol. (a)
Q.2 The value of acceleration due to gravity
(a) Is same on equator and poles
(b) Is least on poles
(c) iIs least on equator
(d) Increases from pole to equator
Sol. (c)
Q.3 The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become
(a) F/4
(b) F/2
(c) F
(d) 2 F
Sol. (a)
Q.4 A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone
(a) Will continue to move in the circular path
(b) Will move along a straight line towards the centre of the circular path
(c) Will move along a straight line tangential to the circular path
(d) Will move along a straight line perpendicular to the circular path away from the boy
Sol. (c)
Q.5 An object is put one by one in three liquids having different densities. The object floats with 123, and 9117 parts of their volumes outside the liquid surface in liquids of densities d_{1}, d_{2} and d_{3} respectively. Which of the following statement is correct?
(a) d_{1}> d_{2}> d_{3}
(b) d_{1}> d_{2}< d_{3}
(c) d_{1}< d_{2}> d_{3}
(d) d_{1}< d_{2}< d_{3}
Sol. (d)
Q.6 In the relation F = G M m/d_{2}, the quantity G
(a) Depends on the value of g at the place of observation
(b) Is used only when the earth is one of the two masses
(c) Is greatest at the surface of the earth
(d) Is universal constant of nature
Sol. (d)
Q.7 Law of gravitation gives the gravitational force between
(a) The earth and a point mass only
(b) The earth and Sun only
(c) Any two bodies having some mass
(d) Two charged bodies only
Sol. (c)
Q.8 The value of quantity G in the law of gravitation
(a) Depends on mass of earth only
(b) Depends on radius of earth only
(c) Depends on both mass and radius of earth
(d) Is independent of mass and radius of the earth
Sol. (d)
Q.9 Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be
(a) 1/ 4 times
(b) 4 times
(c) 1/ 2 times
(d) Unchanged
Sol. (b)
Q.10 The atmosphere is held to the earth by
(a) Gravity
(b) Wind
(c) Clouds
(d) Earth’s magnetic field
Sol. (a)
Q.11 The force of attraction between two unit point masses separated by a unit distance is called
(a) Gravitational potential
(b) Acceleration due to gravity
(c) Gravitational field
(d) Universal gravitational constant
Sol. (d)
Q.12 The weight of an object at the centre of the earth of radius R is
(a) Zero
(b) Infinite
(c) R times the weight at the surface of the earth
(d) 1/R^{2} times the weight at surface of the earth
Sol. (a)
Q.13 An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N
Sol. (a)
Q.14 A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) Maximum when length and breadth form the base
(b) Maximum when breadth and width form the base
(c) Maximum when width and length form the base
(d) The same in all the above three cases
Sol. (b)
Q.15 An apple falls from a tree because of gravitational attraction between the earth and apple. If F1 is the magnitude of force exerted by the earth on the apple and F_{2} is the magnitude of force exerted by apple on earth, then
(a) F_{1} is very much greater than F_{2}
(b) F_{2} is very much greater than F_{1}
(c) F_{1} is only a little greater than F_{2}
(d) F_{1} and F_{2} are equal
Sol. (d)