Class 9 Science Chapter 9 Practice Question Answers - Gravitation

Gravitation Topic is a fundamental concept that has been studied for centuries. It is a force that exists between any two objects in the universe, and it governs the motion of planets, stars, and galaxies.  In the context of Class 9 Physics, the study of gravitation introduces students to the basic principles of the force, its effects on objects, and its application in various scenarios. Let us see some Numericals Practice Questions of this capter.

Q1. Let us find force of attraction between two block lying 1m apart. Let the mass of each block is 40 kg.
Sol. The force of attraction between two objects with mass is governed by Newton's law of universal gravitation. The formula for the force of attraction (gravitational force) between two objects is:

F = (G * m₁ * m₂) / r²

Where:

F is the gravitational force.

G is the universal gravitational constant (6.67430 x 10^-11 N·m²/kg²).

m₁ and m₂ are the masses of the two objects.

r is the distance between the centers of the two objects.

In your case, both blocks have a mass of 40 kg each, and they are lying 1 meter apart. So, m₁ = 40 kg, m₂ = 40 kg, and r = 1 m.

Now, plug these values into the formula to calculate the force of attraction: F = (6.67430 x 10^-11 N·m²/kg² * 40 kg * 40 kg) / (1 m)²

Let's calculate it step by step:

• Multiply the masses: 40 kg * 40 kg = 1600 kg²
• Square the distance: (1 m)² = 1 m²
• Now, calculate the force of attraction: F = (6.67430 x 10^-11 N·m²/kg² * 1600 kg²) / 1 m²
• Calculate the result: F = 6.67430 x 10^-11 N·m²/kg² * 1600 kg² = 1.067888 x 10^-6 N

So, the force of attraction between the two 40 kg blocks lying 1 meter apart is approximately 1.067888 x 10^-6 N.

Q2. The gravitational force between two object is 49 N. How much distance between these objects be decreased so that the force between them becomes double?

Sol. Step-1 : Here, F = 49 N
Let r = original distance between the objects.
m₁ and m₂ = masses of the objects

F = G m₁ × m₂/r² …(1)

Let new distance is R and then force is 2F,

We get, 2F = G m₁ × m₂/R²…(2)

Dividing equation (2) by (1)

2 = r²/R²

R² = r²/2

R = r/√2

Hence decrease in distance = r-r/√2 = r (√2 - 1)/√2

Q3. Two bodies A and B having masses 2 kg and 4 kg respectively are seperated by 2m. Where should a body of mass 1 kg be placed so that the gravitational force on this body due to bodies A and B is zero?

Sol. Let a body of mass 1 kg be placed at point P at a distance x from body A. Therefore, the distance of mass 1 kg from a body B = (2 x) m.
Gravitaional force on 1 kg due to body A is given by
F = N(along PA) = N (along PA)
Gravitational force on 1 kg due to body B is
F' = N (along PB).
Since gravitational force on 1 kg due to bodies A and B is zero, therefore,
F = F ' or (2 x)² 2x² or (2 x) = x = 1.414 x
or 2.414 x = 2 or x = 0.83 m. 

Q4. Calculate the force of gravitation due to a child of mass 25 kg on his fat mother of mass 75 kg if the distance between their centres is 1m from each other. Given G = (20/3) × 10^-11 Nm² kg^-2.
Sol. Here m₁ = 25 kg ; m₂ = 75 kg ; d = 1 m ; G = 20/3 × 10^-11 Nm² kg^-2
Using F = Gm₁m₂/d²
F = 20/3 x 10^-11 x 75 x 25
F = 12,500 × 10^-11
or F = 1.25 × 10^-7 N

Q5. A boy drops a stone from the edge of the roof. If passes a window 2m high in 0.1s. How far is the roof above the top of the window ?
Sol. Let a stone be dropped from the edge of the roof A. Let it passes over B with a velocity say u. Consider motion BC.
u = ?, a = 9.8 ms^_2 ; s = h = 2m ; t = 0.1 s
Using s = ut gt², we have
2 = u(0.1) × 9.8 (0.1)²
2 = 0.1u 0.049
0.1u = 2 / 0.049
or u = = 19.51 ms^_1

This initial velocity at B in motion BC is the final velocity in motion AB. Considering motion AB, we have
u = 0 ; v = 19.51 ms^_1 ; s = ? ; a = 9.8 ms^_2
Using v² - u² = 2as, we have
(19.51)² - (0)² = 2 × 9.8 s
or s = 19.4 m
Roof is 19.4 m above the window.

Q6. A ball thrown up is caught by the thrower after 4 s. With what velocity was it thrown up? How high did it go? Where was it after 3 s ? (g = 9.8 m s^_2)
Sol. Since the time of going up is the same as that of commg down, therefore time of going up = 4/2 = 2s. Let it starts upward with velocity u.
Here u = ?; a = - 9.8 m s^_2 ; t = 2s; v = 0 (at the top); s = h
Using v = u at
 or 0 = u _ 9.8 × 2
or u = 19.6 m s^_1
Again v² - u² = 2as
0 - (19.6)² = 2 (-9.8) h
h = 19.6 m
After 2s, it starts coming downwards (starting with u = 0). Considering downward motion.
u = 0; a = 9.8 m s^_2; t = 3 - 2 = 1s; s = ?
s = ut at²
or s = 0 × 9.8 (1)² = 4.9 m from top.  

Q7. Coconut is hanging on a tree at a height of 15 m from the ground. A boy launches a projectile vertically upwards with a velocity of 20 m s^_1. After what time the projectile pass by coconut? Explain the two answers in this problem.
Sol. Here u = 20 m s^_1; a = - 10 m s^_2; s = 15 m ; t = ?
Using s = ut at², we have
15 = 20t (-10) t²
Dividing throughout by 5, we have
3 = 4t - t²
or t²- 4t 3 = 0
or (t - 1) (t - 3) =t - 1 = 0 or t = 1s
or t -3 = 0 or t = 3s
After 1s, it will cross coconut while going up and after 3 s while coming down.

Q8. Fig. (a) shows how to calculate pressure exerted by a brick of mass 3 kg : (a) when standing on end; (b) when lying flat. The total force or thrust exerted is the same in both the cases.
Sol. In Fig. (a), A₁ = 5 cm × 10 cm = 50 cm² = 50 × 10^_4 m² and
F₁ = 3 kg wt = 3 kg × 10 m/s² = 30 N
Thus, pressure exerted, P₁ = = 6 × 10³ N/m² = 6 × 10³ Pa
In Fig. (b), A₂ = 10 cm × 20 cm = 200 cm² = 200 × 10^_4 m² and
F₂ = 3 kg wt = 3 kg × 10 m/s² = 30 N
Thus, pressure exerted, P₂ = = 1·5 × 10³ N/m² = 1·5 × 10³ Pa
 

Q9. What is the force of gravitation between two point masses of 1 kg and 2 kg kept 1 m apart?
Sol. 

To calculate the gravitational force (F) between two point masses (m1 and m2) separated by a distance (r), use the formula:

F = (G * m₁ * m₂) / (r²)

Where:

• F is the gravitational force.
• G is the universal gravitational constant (6.67430 x 10^-11 Nm²/kg²).
• m₁ and m₂ are the masses of the two point masses.
• r is the distance between their centers.

Here, m₁ = 1 kg, m₂ = 2 kg, and r = 1 m. G is also provided.

Now, substitute these values into the formula to calculate the gravitational force:

F = (6.67430 x 10^-11 Nm²/kg² * 1 kg * 2 kg) / (1 m²)

First, calculate the product of the masses: m₁ * m₂ = 1 kg * 2 kg = 2 kg²

Now, calculate the force: F = (6.67430 x 10^-11 Nm²/kg² * 2 kg²) / (1 m²)

Next, simplify: F = (6.67430 x 10^-11 Nm²/kg² * 2 kg²) / (1 m²)

Now, perform the calculation: F = 6.67430 10^-11 Nm²/kg² * 2 kg²

Now, multiply the constants: F = 13.34860 x 10^-11 Nm²/kg²

So, the gravitational force between the two point masses of 1 kg and 2 kg, kept 1 meter apart, is approximately 13.34860 x 10^-11 Nm²/kg²

Q10. Calculate the force of gravitation between the Earth and the sun. Mass of Earth = 6 × 10²⁴ kg, Mass of Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹m.
Sol.  To calculate the gravitational force (F) between the Earth and the Sun, we can use Newton's law of universal gravitation, which is given by the formula:

F = (G * m₁ * m₂) / (r²)

Where:

• F is the gravitational force.
• G is the universal gravitational constant (6.67430 x 10^-11 Nm²/kg²).
• m1 is the mass of the Earth (6 x 10²⁴ kg).
• m2 is the mass of the Sun (2 x 10³⁰ kg).
• r is the distance between the centers of the Earth and the Sun (1.5 x 10¹¹ m).

Now, substitute these values into the formula to calculate the gravitational force:

F = (6.67430 x 10^-11 Nm²/kg²* 6 x 10²⁴ kg * 2 x 10³⁰ kg) / (1.5 x 10¹¹ m)²

First, calculate the product of the masses: m₁ * m₂ =6 x 10²⁴ kg * 2 x 10³⁰ kg = 1.2 x 10⁵⁵ kg²

Now, calculate the force: F = (6.67430 x 10^-11 Nm²/kg² * 1.2 x 10⁵⁵ kg²) / (1.5 x 10¹¹ m)²

Next, simplify: F = (6.67430 x 10^-11 Nm²/kg² * 1.2 x 10⁵⁵ kg²) / (1.5 x 10¹¹ m)²

Now, perform the calculation: F = (8.00916 x 10⁴³ Nm²) / (2.25 x 10²² m²)

Now, divide to get the value of F: F = 3.5596296 x 10²¹ N

So, the gravitational force between the Earth and the Sun is approximately 3.5596296 x 10²¹ Newtons.

Q11. Write down the expression for acceleration experienced by a particle on the surface of the moon due to gravitational force on the moon. Find the ratio of this acceleration to that experienced by the same particle on the surface of the earth. If the acceleration due to gravity on the earth is 9.8 ms^_2, what is the acceleration of a particle on the moon's surface? Mass of moon = 7.3 × 10²² kg; Mass of Earth = 6 × 10²⁴ kg. Radius of moon= 1.74 × 10⁶ m, Radius of earth = 6.4 × 10⁶ m.
Sol.  The expression for acceleration (am) experienced by a particle on the surface of the Moon due to the gravitational force on the Moon can be calculated using Newton's law of universal gravitation:

a_m = (G * Mm) / (Rm^2)

Where:

• a_m is the acceleration on the surface of the Moon.
• G is the universal gravitational constant (6.67430 x 10^-11 Nm^2/kg^2).
• M_m is the mass of the Moon (7.3 x 10^22 kg).
• R_m is the radius of the Moon (1.74 x 10^6 m).

Now, let's calculate a_m:

a_m = (6.67430 x 10^-11 Nm^2/kg^2 * 7.3 x 10^22 kg) / (1.74 x 10^6 m)^2

First, calculate the product of the constants:

G * M_m = (6.67430 x 10^-11 Nm^2/kg^2) * (7.3 x 10^22 kg)

Now, calculate R_m^2:

(1.74 x 10^6 m)^2 = 3.0276 x 10^12 m^2

Now, calculate a_m:

a_m = (6.67430 x 10^-11 Nm^2/kg^2 * 7.3 x 10^22 kg) / (3.0276 x 10^12 m^2)

Now, perform the calculation:

a_m ≈ 1.625 m/s^2

So, the acceleration experienced by a particle on the surface of the Moon (a_m) is approximately 1.625 m/s^2.

To find the ratio of this acceleration to that experienced by the same particle on the surface of the Earth, we can use the same formula for acceleration (ae) on the Earth's surface:

a_e = (G * Me) / (Re^2)

Where:

• ae is the acceleration on the surface of the Earth.
• Me is the mass of the Earth (6 x 10^24 kg).
• Re is the radius of the Earth (6.4 x 10^6 m).

Now, let's calculate a_e:

a_e = (6.67430 x 10^-11 Nm^2/kg^2 * 6 x 10^24 kg) / (6.4 x 10^6 m)^2

First, calculate the product of the constants:

G * M_e = (6.67430 x 10^-11 Nm^2/kg^2) * (6 x 10^24 kg)

Now, calculate R_e^2:

(6.4 x 10^6 m)^2 = 4.096 x 10^13 m^2

Now, calculate ae:

a_e = (6.67430 x 10^-11 Nm^2/kg^2 * 6 x 10^24 kg) / (4.096 x 10^13 m^2)

Now, perform the calculation:

a_e ≈ 9.81 m/s^2

So, the acceleration experienced by a particle on the surface of the Earth (ae) is approximately 9.81 m/s^2.

Now, to find the ratio of am to ae:

am / ae ≈ 1.625 m/s^2 / 9.81 m/s^2 ≈ 0.1656

Therefore, the ratio of the acceleration on the surface of the Moon to that on the surface of the Earth is approximately 0.1656.

g_m =0.16 × g_E = 0.16 × 9.8 = 1.57 ms^_2 

Q12. Find the value of acceleration due to gravity at a height of (a) 6400 km, (b) 12,800 km from the surface of the earth. Radius of earth is 6400 km.
Sol. 

We know that the acceleration due to gravity at the surface of the Earth is 9.8 m/s².

To find the value of acceleration due to gravity at a height of 6400 km above the surface of the Earth, we can use the formula:

g' = (g*R^2) / (r + h)^2

where g is the acceleration due to gravity at the surface of the Earth (9.8 m/s²), R is the radius of the Earth (6400 km), r is the distance from the center of the Earth (also 6400 km), and h is the height above the surface of the Earth.

(a) When h = 6400 km:

g' = (9.8*6400^2) / (6400+6400)^2 = 2.45 m/s²

Therefore, the acceleration due to gravity at a height of 6400 km above the surface of the Earth is 2.45 m/s².

(b) When h = 12800 km:

g' = (9.8*6400^2) / (6400+12800)^2 = 0.61 m/s²

Therefore, the acceleration due to gravity at a height of 12800 km above the surface of the Earth is 0.61 m/s².

Q13. A particle is thrown up vertically with a velocity of 50 m/s. (a) What will be its velocity at the highest point of its journey? (b) How high would the particle rise? (c) What time would it take to reach the highest point.
Sol. At the highest point the velocity will be zero.
Considering activity A to B
Using v = u at
0 = 50 -9.8 × t
t = 5.1 s
Also v² - u² = 2as
0² -(50)² = 2 (-9.8) × s
5 = 127.5 m

Q14. With reference to the above sample problem, (a) Find the time the particle takes from the highest point back to the initial point (b) Find the velocity with which the particle reaches the initial point.
Sol. The data is given in the adjacent figure. Considering activity B to A
Using v² - u² = 2as
v² - 0² = 5(9.8)(127.5)
v = 50 m/s
Also v = u at
50 = 0 9.8 (t)
t = 5.1 s

Q15. A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered 20 m ? What would be its velocity when it hits the ground? Take g= 10 m/s².
Sol. Let the point B be at a height of 20 m.
Activity from A to B :
u₁ = 0, a₁ = 10 ms^_2, s₁ = 20 m, v₁ = ?
v₁² - u₁² = 2a₁s₁
v₁² - 0² = 2 (10) (20)
v₁² =20²
v₁ = 20 m/s
Activity from A to C : C is a point on the ground
u₂ = 0, a₂ = 10 ms^_2, s₂ = 40 m, v₂ = ?
v²₂ - u₂² = 2a₂s₂
v₂² - 0² = 2 (10) (40)
v₂² = 800
v₂ = 28.28 ms^_1

Q16. A body is thrown up with a speed 29.4 ms^_1.
(a) What is its speed after (i) t = 1 s, (ii) t = 2 s and (iii) t = 3 s.
(b) What is its height after (i) t = 1 s, (ii) t = 2 s and (iii) t = 3 s.
Sol. (a) (i) u = 29.4 ms^_1, a = -9.8 ms^_2, t₁ = 1 s, v₁ = ?
v₁ = u at₁
= 29.4 (-9.8) x 1 = 19.6 ms^_1

(ii) u = 29.4 ms, a = -9.8 ms^_2, t₂ = 2 s, v₂ = ?
v₂ = u at₂
= 29.4 (_9.8) × 2 = 9.8 ms^_1

(iii) u = 29.4 ms^_1, a = -9.8 ms^_2, t₃ = 3 s, v₃ = ?
v₃ = u at₃
= 29.4 (_9.8) × 3 = 0

(b) (i) u = 29.4 ms^_1, a = -9.8 ms^_2, s₁ = h₁ , t₁ = 1s
h₁ = ut₁ at₁² =29.4 × 1 (-9.8) × 1 =24.5m

(ii) u = 29.4 ms^_1, a = -9.8 ms^_2, s₂ = h₂, t₂ = 2 s
h₂ = ut₂ at₂² = 29.4 × 2 (-9.8) × 2² = 39.2 m

(iii) u = 29.4 ms^_1, a = -9.8 ms^_2, s₃ = h₃, t₃ = 3 s
h₃ = ut₃ at₃² = 29.4 x 3 (-9.8) × 3² = 44.1 m
 

Q17. What is the weight of a person whose mass is 50 kg.
Sol. The weight of the person
W = mg = 50 x 9.8 = 490 N
Note: The gravitational unit of force is kg-f (kilogram force) or kg-wt (kilogram weight) 1 kg-wt = 9.8 N = 1 kg-f
490 N = 50 kg-f

Q18. Weight of a girl is 294 N. Find her mass.
Sol.  Weight (W) = Mass (m) × Acceleration due to gravity (g)

Given that the weight of the girl is 294 N and the acceleration due to gravity on Earth is approximately 9.81 m/s², you can rearrange the formula to solve for mass (m):

m = W / g

Substitute the values:

m = 294 N / 9.81 m/s²

Now, calculate the mass:

m ≈ 29.96 kg

So, the mass of the girl is approximately 29.96 kilograms.

Q19. Weight of an object is 294 N on the surface of the earth. What is its weight at a height of 200 km from the surface of the earth. Radius of the earth= 6400 km.
Sol. Weight at a height of 200 km
W_h = mg 
Here mg = 294 N, R = 6400 km, h= 200 km
W_h = 294
= 276.45 N

Note: Weight decreases with increase of height from the surface of the earth. 

Q20. The gravitational force between two objects is F. How will this force change when
(i) distance between them is reduced to half ?
(ii) the mass of each object is quadrupled ?

Sol. (i) According to Newton's law of gravitation, gravitational force F between two objects distance r apart is
When distance between them is reduced to half, i.e., r' = r/2, the force,
Thus, F' = 4F
i.e., force becomes 4 times its previous value. "

(ii) Again, according to Newton's law of gravitation, the gravitational force F between two objects of masses m₁ and m₂ is
F µ m₁m₂
When mass of each object is quadrupled,
m'₁ = 4m₁
and m'₂ = 4m₂
The force, F ' µ m'₁m'₂
or F' = 16 F
i.e., force becomes 16 times its previous value.

Q21. A sphere of mass 40 kg is attracted by a second sphere of mass 15 kg when their centres are 20 cm apart, with a force of 0·1 milligram weight. Calculate the value of gravitational constant.
Sol. Here, m₁ = 40 kg, m₂ = 15 kg
From r = 20cm =2 × 10^_1 m
F = 0·1 milligram weight = 0·1 × 10^_3 gram weight
= 10^_4 × 10^_3 kg wt = 10^_7 × 9·8 N (1 kg wt = 9.8 N)
From G = 6·53 × 10^_11 Nm²/kg²

Q22. Calculate the force of gravity acting on your friend of mass 60 kg. Given mass of earth = 6 x 1024 kg and radius of Earth = 6·4 x 10⁶ m.
Sol. Here, m = 60 kg, M = 6 × 10²⁴ kg
R = 6·4 × 10⁶ m, F = ?
G = 6·67 × 10^_11 Nm²/kg²
Thus, F = 58·62 N

Q23. A particle is thrown up vertically with a velocity of 50 m/s. What will be its velocity at the highest point of the journey? How high would the particle rise ? What time would it take to reach the highest point? Take g = 10 m/s².
Sol. Here, initial velocity, u = 50 m/s
final velocity, v = ?
height covered, h = ?
time taken, t = ?
g = 10 m/s²
At the highest point, final velocity v = 0
From v² - u² = 2 gh, where g = - 10 m/s² for upward journey,
0 - (50)² = 2 (-10) h
h = 125 m
From v = u gt,
or 0 = 50 (-10) t
t = 50/10 = 5 s

Q24. A force of 15 N is uniformly distributed over an area of 150 m². Find the pressure in pascals.
Sol. 

Pressure (P) = Force (F) / Area (A)

Given:

• Force (F) = 15 N
• Area (A) = 150 m²

Now, simply plug these values into the formula: P = 15 N / 150 m²

Now, calculate the pressure: P = 0.1 Pa

So, the pressure is 0.1 Pascals (Pa).

Q25. How much force should be applied on an area of 1 cm² to get a pressure of 15 Pa?
Sol. Here, area, A = 1 cm² = 10^_4 m²
pressure, P = 15 Pa = 15 N/m²
As P = , F = P × A = (15 N/m²) × (10^_4 m²) = 1·5 x 10^_3 N

Q26. A block weighing 1·0 kg is in the shape of a cube of length 10 cm. It is kept on a horizontal table. Find the pressure on the portion of the table where the block is kept.
Sol. Given:

• Mass of the block (m) = 1.0 kg
• Acceleration due to gravity (g) ≈ 9.81 m/s²
• Length of one side of the cube (L) = 10 cm = 0.1 m

First, calculate the force (weight) of the block: Weight (F) = m * g

F = 1.0 kg * 9.81 m/s² ≈ 9.81 N

Now, calculate the area (A) of the bottom face of the cube:

A = Length (L) * Width (W)

A = 0.1 m * 0.1 m = 0.01 m²

Now, calculate the pressure (P):

P = F / A

P = 9.81 N / 0.01 m²

P = 981 Pascals (Pa)

Also Check at EduRev

• Newton's Universal Law of Gravitation and Gravitational Force 
• Gravitation 

Match the Column

Q1. Cloumn-I Column-II
(A) Attraction between two planets (p) gravity
(B) Attraction between a body and a planet (q) weightlessness
(C) Free fall  (r) gravitational force
(D) Weight  (s) gravitation

Q2. Cloumn-I Column-II
(A) A body falling freely (p) displacement
(B) Distance with direction (q) velocity
(C) Speed with direction (r) acceleration
(D) Rate of change of velocity (s) uniformly accelerated motion

Q3. Cloumn-I Column-II
(A) Wide straps of school bags (p) motion of planets around the sun.
(B) Archimedes principle (q) weight of an object
(C) Gravitational force  (r) submarine
(D) Upthrust  (s) balloons
(t) to reduce pressure

Q4. Cloumn-I Column-II
(A) Loss in weight  (p) gravitational force
(B) Motion of moon around the earth  (q) upthrust
(C) Airship (r) increasing pressure
(D) Sharp tip of a needle (s) archimedes principle
(t) centripetal force

True & False

1. The force of attraction between two bodies is called gravity.

2. The value of G depends upon the mass of two objects.

3. If a spring balance, holding a heavy object is released, it will read zero weight.

4. The value of G is high if the radius of the body is more and less if radius is less.

5. The centre of mass and centre of gravity for a small body lie at the same point.

6. The gravitational force between two bodies changes if a material body is placed between them.

7. The acceleration of a body thrown up is numerically the same as the acceleration of a downward falling body but opposite is sign.

8. The value of g is zero at the centre of the earth.

9. The inertia of an object depends upon its mass.

10. All objects attract each other along the line joining their centre of mass.

11. Acceleration due to gravity, g = , where symbols have their usual meanings.

12. Archimedes' principle does not apply to gases.

Fill in the Blanks

1. .................... is the force of attraction between any two bodies in the universe.

2. .................... is the force of attraction between a body and a planet.

3. Acceleration due to gravity .................... with height from the surface of the earth.

4. Acceleration due to gravity .................... with depth from the surface of the earth.

5. Acceleration due to gravity is a maximum at the ....................

6. .................... of a body changes from place to place but its .................... remains constant.

7. The upward force experienced by a body immersed partially or fully in a fluid is called ....................

8. The combined weight of the sinker and cork is .................... than the weight of the sinker alone.

9. The centre of ..........is a point, where the total upthrust, due to fluid displaced by part ...........of body acts.

10. A fish .................... by squeezing out air, from its ....................

11. When a body is partly or wholly immersed in a .................... , it experience an ....................

12. An iceberg floats with .................... of its volume below ....................

13. The unit of upthrust in SI is ....................

14. Lactometer is used to measure .................... of ....................

15. A hydrometer sinks .................... in water than in pure milk.

Answers

Match the Column

1. (A ® s), (B ® p), (C ® q), (D ® r) 2. (A ® s), (B ® p), (C ® q), (D ® r)

3. (A ® t), (B ® r, s), (C ® p, q) (D ® r, s) 4. (A ® q, s), (B ® p,t), (C ® q, s), (D ® r)

True & False

1. False 2. False 3. True 4. True 5. True 6. False 7. True

8. True 9. True 10. True 11. True 12. True 13. False 14. False

Fill in the Blanks

1. gravitation 2. gravity 3. decreases 4. decreases

5. poles 6. weight, mass 7. buoyant force 8. volume

9. 4°C 10. numerically 11. less 12. less

13. buoyancy, the immersed 14. dives, floating tube 15. fluid, upthrust 16. 11/12, water

17. Newton 18. purity, milk 19. more 20. less than