14. Natural Resources Class IX NCERT Text Book

Q.No. 1. How is our atmosphere different from the atmospheres on Venus and Mars ?

Answer:

Atmosphere of Earth has 79% nitrogen, 20% oxygen and a small amount of carbon dioxide, water vapour and other gases.
This existence of life is possible on Earth.

Whereas atmospheres of Venus and Mars have 95% - 97% carbon dioxide.
Existence of life is not possible.

Q.No. 2. How does the atmosphere act as a blanket ?

Answer:

Air is bad conductor of heat. The atmosphere keeps the average temperature of the Earth fairly constant during the day and even during the course of the whole year such as blanket.

The atmosphere prevents the sudden increase and decrease of temperature during the daylight hours and the night just like blanket.

The atmosphere slows down the escape of heat into outer space. 

Due to following behavior of atmosphere acts as blanket.

Q.No. 3. What causes winds ?

Answer:

The movement of air causes winds. When air is heated by radiation from the heated land or water, it rises. Land gets heated faster than water. A region of low pressure is created. The movement of air from one region to the other region creates winds.

The rotation of Earth and presence of mountain ranges cause winds.

Q.No. 4. How are clouds formed ?

Answer

Water vapour moves from high pressure region to low pressure region.
A large amount of water evaporates and goes into air due to heating of water bodies. The hot air rises up carrying the water vapour with it. As the air rises, it expands and cools. This cooling causes the water vapour in the air to condense in the form of droplets. 
This process forms the clouds.

Q.No. 5. List any three human activities that you think would lead to air pollution.

Answer

There are following human activities which lead to air pollution.

1. Burning of coal and petroleum.
2. Deforestation
3. Industrialization

Q.No. 6. Why do organisms need water ?

Answer

There are following reasons due to which organisms need water.

1. Usually, the reactions that take place in our body or within cells occur between substances that are dissolved in water. So, all cellular process need water.

2. Most of the substances are transported in a dissolved form, water is necessary.

Q.No. 7. What is the major source of water in the city/town/village where you live ?

Answer

River is the major source of fresh water.

Q.No. 8. Do you know any activity which may be polluting this water source ?

Answer

The discharge of waste water from homes, industries, hospitals, etc. into the river pollutes this fresh water source.

Class IX NCERT Text Book Questions Answers

Motion of a car on a banked road

               Consider about a car of mass m kg moves on banked road of inclination θ. During the motion the car experience circular motion.

Mathematical Calculation

There is no acceleration along the vertical direction. The net force along this direction must be zero.

N cosθ = mg + f sinθ ------------ (i)

The horizontal components of N and f provide the centripetal force.

N cosθ + f cosθ = mv²/r ---------- (ii)

The frictional force 

f ≤ μ_s N
For maximum velocity v

f = μ_s N
Now putting the value of f in equation no. (i)

N cosθ = mg +  μ_s N sinθ
Therefore,
N = mg/(cosθ  - μ_s sinθ)

Again putting the value of f in equation no. (ii)

N sinθ + μ_s N cosθ = mv²/r

Now substitute the value of N

mg( sinθ  + μ_s  cosθ)/(cosθ  - μ_s sinθ) = mv²/r

Therefore,

v = {rg( sinθ  + μ_s  cosθ)/(cosθ  - μ_s sinθ)}^1/2

This is maximum velocity of a car on a banked road.

Heights And Distances

Chapter 9

Class X

Exercise 9.1

Q.No. 1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30⁰

Solution:

From question it is clear that
AC = 20 m.
<ACB = 30⁰ = θ 
AB = ?

We know that

Sinθ = AB/AC

Now, putting the value of θ, AB, AC.

Sin30⁰  = AB/20
1/2 = AB/20
∴ AB = 10m Ans.

Q.No. 2. A tree breaks due to storm and broken parts bends so that the top of the tree touches the ground making an angle 30⁰ with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

From question it is clear that

Height of the tree = AB + AC
θ = 30⁰
BC = 8 m.

From figure 

Cos 30⁰ = BC/AC
√3/2 = 8/AC
∴ AC = 16/√3

Again, 

Tan 30⁰ = AB/BC
1/√3 = AB/8

∴ AB = 8/√3

Height of the tree = AB + AC 

= 8/√3 + 16/√3

= 24/√3 
= 8√3
= 8x1.732
= 13.856 m Ans.

Q.No. 3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30⁰ to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60⁰ to ground. What should be the length of the slide in each case?

Solution:

From question it is clear that
For first slide
AB = 1.5 m
α = 30⁰
Length of slide = AC = ?

For second slide
PQ = 3 m
β = 60⁰
Length of slide = PR = ?

For first slide
According to application of Trigonometry
Length of slide = AC = AB/Sin30⁰
∴ AC = 1.5x2 = 3 m

Now, for second slide
Length of slide PR = PQ/Sin60⁰
∴ PR = 3x2/√3
PR = 2√3
PR = 2x1.732
PR = 3.464 m 

Therefore length of slides are
3m and 3.464 m Ans.

Q.No. 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30⁰. Find the height of the tower.

Solution:

From question it is clear that
Let the height of the tower  = AB = ?
BC = 30 m
Angle of elevation = 30⁰

According to application of Trigonometry
AB = BCxTan30⁰
∴ AB = 30/√3
AB = 10√3
Therefore, height of the tower = 17.32 m Ans.

Q.No. 5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60⁰. Find the length of the string, assuming that there is no slack in the string.

Solution:

Let the length of the string be AC = x which represents the hypotenuse.

The inclination represents the angle of elevation = 60⁰

The height of the kite from ground = AB = 60 m.

The concerned right angle triangle is drawn below

From figure it is clear that

Sin60⁰ = AB/AC

Now, putting the value of AB and AC.

√3/2 = 60/x

∴ x = 120/√3

= 40√3 m Ans.

Q.No. 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30⁰ to 60⁰ as he walks towards the building. Find the distance he walked towards the building.

Solution:

Let the distance he walked towards the building be CD = x m.

The angle of elevation for Triangle ABC = 30⁰

The angle of elevation for Triangle ABD = 60⁰

Height of the building from the eye of the boy = AB = 30 - 1.5 = 28.5 m.

The concerned right angle triangle is drawn below

Consider in right angle triangle ABC
Tan30⁰ = AB/BC
Now, putting the value of BC
BC = 28.5.√3

Now consider in right angle triangle ABD
Tan60⁰ = AB/BD
BD = AB/Tan60⁰
∴ BD = 28.5/√3 = 9.5.√3 m.
Therefore,
CD = BC - BD = 28.5.√3  - 9.5.√3 = 19√3 m Ans.

Gravitation Class IX Chapter 10

Question - Answer

Q.No. 1. State the universal law of gravitation.

Answer:

The statement of universal law of gravitation is as follows.

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

Q.No. 2. Write the formula to find the magnitude of the gravitational force between the earth and object on the surface of the earth.

Answers:

The formula which is used to find the magnitude of the gravitational force between the earth and object on the surface of the earth is given below.

F = GMm/d²

 

Here,

F denotes gravitational force.

M and m denote mass of the earth and object respectively.

d denotes distance between the earth and the object.

G is the universal gravitational constant.

The value of G is 6.673x10^-11 Nm²/kg².

Q.No. 3. What do you mean by free fall?

Answer:

                   When an object falls towards the earth. There is no change in the direction of the motion of the object. But due to the earth’s attraction, there will be a change in the magnitude of the velocity.Therefore, whenever an object falls towards the earth, an acceleration is involved due to the earth’s gravitational force.

                   This acceleration is called acceleration due to the gravitational force of the earth or acceleration due to gravity. It is denoted by g. The SI unit of g is ms^-2.

                   According to universal law of gravitation, we know that
F = GMm/d² ................. (i)
F = mg ................. (ii)

Now, from (i) and (ii)

mg =

Therefore,

g = GM/d²

Q.No. 4. What do you mean by acceleration due to gravity?

Answer:

                   Whenever an object falls towards to the earth, acceleration is involved due to gravitational force. This is called acceleration due to gravity. This is equal to

g = GM/d²

Here,

G is the gravitational constant and equal to 6.673x10^-11 Nm²/kg².

M is the mass of the earth.

Q.No. 5. What are the difference between the mass of an object and its weight?

Answer:

                   The mass of an object is the measure of inertia. The mass of an object is constant. It does not change from place to place. Therefore, the mass of an object remains same whether on the surface of the earth, moon or outer space.

                   The weight of an object is variable. It changes from place to place because it depends on gravitational force. Therefore, weight of an object is not same whether on the surface of the earth, moon and outer space.

Q.No. 6. Why is the weight of an object on the moon 1/6^th its weight of the earth.

Answer:

                   Let the acceleration due to gravity on the surface of the earth is g and mass is m.

The weight of the object on the surface of the earth = w_e = mg

The weight of the object on the surface of the moon = w_m =mg/6

We know that acceleration due to gravity on the surface of the moon is 1/6^th of earth.

Therefore,

w_m =we/6

Therefore, the weight of an object on the moon is 1/6^th its weight of the earth.

Real Numbers

Class X

Chapter 1

NCERT Text Book Solution
Exercise 1.1

Q.No.1. Use Euclid division Algorithm to find out the HCF of the following numbers.

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

(i)

Solution:

According to Euclid division Algorithm, we know that

From question, it is clear that

a = 225

b = 135

Here, remainder is zero. Therefore, the step is over and divisor is 45.

Hence, HCF = 45 Ans.

(ii)

Solution:

According to Euclid division Algorithm, we know that

From question, it is clear that

a = 38220

b = 196

38220 = 196x195 + 0

Here, remainder is zero. Therefore, the step is over and divisor is 196.

Hence, HCF = 196 Ans.

(iii)

Solution:

According to Euclid division Algorithm, we know that

From question, it is clear that

a = 867

b = 255

867 = 255x3 + 102

255 = 102x2 + 51

102 = 51x2 + 0

Here, remainder is zero. Therefore, the step is over and divisor is 51.

Hence, HCF = 51 Ans.

Q.No. 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q +5, where q is some integer.

Solution:

According to Euclid division Algorithm, we know that

Here,

b = 6

Since

The possible remainders are 0, 1, 2, 3, 4 and 5.

That is a can be

6q, or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is quotient. However, since a is odd, a cannot be 6q, or 6q + 2, or 6q + 4 (Since they are divisible by 2)

Therefore, any odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5.

Q.No. 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

According to Euclid division Algorithm, we know that

Here,

a = 616

b = 32

616 = 32x19 + 8

32 = 8x4 + 0

Here, remainder is zero. Therefore, the step is over and divisor is 8.

Hence, HCF = 8.

Therefore, maximum number of column is 8 Ans.

Q.No. 4. Use Euclid division lemma to show that thee square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution:

According to Euclid division Algorithm, we know that

Here,

a = x

b = 3

Since

The possible remainders are 0, 1, and 2.

That is x can be

3q, or 3q + 1, or 3q + 2 where q is quotient.

Now, squaring these

9q², or 9q² + 6q + 1, or 9q² + 12q + 4

When these are divided by 3

Form of these will be

3x3q², or 3(3q² + 2q) + 1, or 3(3q² + 4q + 1) + 1

Therefore, they can be written in the form

3m or 3m + 1.

Q.No. 5. Use Euclid division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

According to Euclid division Algorithm, we know that

Here,

a = x

b = 3

Since

The possible remainders are 0, 1, 2,3 , 4, 5, 6, 7, 8.

That is x can be

9q, or 9q + 1, or 9q + 2, 9q + 3, 9q + 4, 9q + 5, 9q + 6, 9q + 7 or 9q + 8 where q is quotient.

Now, cubing these

Therefore, cube of these can be written in the form

Cube of 9q = 729q³

Form = 9x81q³ = 9m

Cube of 9q + 1= 729q³ + 243q² + 27q + 1

Form = 9(81xq³ + 27q² + 3q) + 1 = 9m + 1

and so on .....................

9m or 9m + 1.