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You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

10th Maths Exercise 5.2 Samacheer Kalvi Question 1.
What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Solution:
(i) θ = 90°
m = tan θ = tan 90° = ∝ (undefined)
(ii) m = tan θ = tan 0° = 0

Exercise 5.2 Class 10 Samacheer Kalvi Question 2.
What is the inclination of a line whose slope is (i) 0
Solution:
(i) Slope = 0
tan θ = 0
tan 0 = 0
∴ θ = 0°

(ii) Slope = 1
tan θ = 1
tan 45° = 1
∴ θ = 45°
angle of inclination is 45°

Ex 5.2 Class 10 Samacheer Question 3.
Find the slope of a line joining the points
(i) (5, \(\sqrt{5})\)) with origin
(ii) (sin θ, -cos θ) and (-sin θ, cos θ)
(i) (5, \(\sqrt{5})\)) with origin (0, 0)
Solution:

10th Maths Exercise 5.2 Question 4.
What is the slope of a line perpendicular to the line joining A(5, 1) and P where P is the mid-point of the segment joining (4, 2) and (-6,4).
Solution:
P is the mid point of the segment joining (4, 2) and (-6, 4)

10th Maths Ex 5.2 Question 5.
Show that the given points are collinear (-3, -4), (7, 2) and (12, 5)
Solution:
The verticles are A(-3, -4), B(7, 2) and C (12, 5)

Slope of AB = Slope of BC
∴ The points A, B and C lie on the same line.
∴ They are collinear.

10th Maths Exercise 5.2 Solution Question 6.
If the three points (3, -1), (a, 3), (1, -3) are collinear, find the value of a.
Solution:
Slope of AB = slope of BC.

Samacheer 10th Maths Exercise 5.2 Solutions Question 7.
The line through the points (-2, a) and (9, 3) has slope –\(\frac{1}{2}\). Find the value of a.
Solution:
A line joining the points (-2, a) and (9, 3) has slope m = –\(\frac{1}{2}\).

10th Maths Coordinate Geometry Exercise 5.2 Question 8.
The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24) . Find the value of x.
Solution:

10th Maths Exercise 5.2 In Tamil Question 9.
Show that the given points form a right angled triangle and check whether they satisfies pythagoras theorem
(i) A(1, -4), B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9, 12) and N(3, 14)
Solution:

10th Coordinate Geometry 5.2 Question 10.
Show that the given points form a parallelogram : A(2.5, 3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5)
Solution:



∴ The given points form a parallelogram.

10th Maths 5.2 Exercise Question 11.
If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Solution:
A(2, 2), B(-2, -3), C(1, -3), D(x, y)

Since ABCD forms a parallelogram, slope of opposite sides are equal and diagonals bisect each other.
Mid point of BD = Mid point of AC

10th Maths 5.2 Question 12.
Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Solution:
A (3, -4), B (9, -4), C (5, -7) and D (7, -7)

If only one pair of opposite sides of a quadrilateral are parallel, then it is said to be a trapezium.


∴ One pair of opposite sides are parallel.
∴ ABCD is a trapezium.

Samacheer Kalvi 10th Maths Exercise 5.2 Question 13.
A quadrilateral has vertices at A(- 4, -2), B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram
Solution:




In a parallelogram diagonals bisect each other. Opposite sides are parallel as their slopes are equal the mid points of the diagonals are the same.
∴ Mid points of the sides of a quadrilateral form a parallelogram.

10th Exercise 5.2 Question 14.
PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS =2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Solution:

Students who are preparing for the Science exam can download this Tamilnadu State Board Solutions for Class 10th Science Chapter 1 from here for free of cost. These Tamilnadu State Board Textbook Solutions PDF cover all 10th Science Laws of Motion Book Back Questions and Answers.

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 1 Laws of Motion

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 10th Science Chapter 1 Laws of Motion Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 1 easily after studying the Tamilnadu State Board Class 10th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 1 Laws of Motion solutions of Class 10th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Laws of Motion Textual Solved Problems

10th Science Laws Of Motion Book Back Answers Question 1.
Calculate the velocity of a moving body of mass 5 kg whose linear momentum is 2.5 kg ms^-1.
Solution:
Linear momentum = mass × velocity
Velocity = \(\frac { linear momentum }{ mass }\)
V = \(\frac { 2.5 }{ 0.5 }\) = 0.5 ms^-1.

Laws Of Motion Class 10 Questions And Answers Question 2.
A door is pushed, at a point, whose distance from the hinges is 90 cm, with a force of 40 N. Calculate the moment of the force about the hinges.
Solution:
Formula: The moment of a force M = F × d
Given: F = 40 N and d = 90 cm = 0.9 m.
Hence, moment of the force = 40 × 0.9 = 36 Nm.

Laws Of Motion Class 10 Samacheer Question 3.
At what height from the centre of the Earth the acceleration due to gravity will be \(\frac { 1 }{ 4 }\)th of its value as at the Earth.
Solution:
Data: Height from the centre of the Earth, R’ = R + h
The acceleration due to gravity at that height, g’ = \(\frac { g }{ 4 }\)
Formula:

From the centre of the Earth, the object is placed at twice the radius of the earth.

Samacheer Kalvi 10th Science Laws of Motion Textbook Evaluation

I. Choose the correct answer.

Laws Of Motion Class 10 Book Back Answers 1.
The inertia of a body depends on _____ .
(a) weight of the object
(b) acceleration due to gravity of the planet
(c) mass of the object
(d) Both a & b.
Answer:
(c) mass of the object

10th Science Unit 1 Question 2.
Impulse is equals to:
(a) rate of change of momentum
(b) rate of force and time
(c) change of momentum
(d) rate of change of mass
Answer:
(c) change of momentum

10th Science Laws Of Motion Question 3.
Newton’s III law is applicable to ______ .
(a) for a body is at rest
(b) for a body in motion
(c) both a & b
(d) only for bodies with equal masses.
Answer:
(c) both a & b

10th Law Of Motion Question 4.
Plotting a graph for momentum on the X-axis and time on Y-axis. The slope of the momentum-time graph gives:
(a) Impulsive force
(b) Acceleration
(c) Force
(d) Rate of force
Answer:
(c) Force

Laws Of Motion – Class 10 New Syllabus Question 5.
In which of the following sport the turning of the effect of force used?
(a) swimming
(b) tennis
(c) cycling
(d) hockey.
Answer:
(c) cycling

10th Physics Laws Of Motion Question 6.
The unit of ‘g’ is ms^-2. It can be also expressed as _____ .
(a) cm s^-1
(b) N kg^-1
(c) Nm² kg^-1
(d) cm² s^-2
Answer:
(b) N kg^-1

Laws Of Motion 10th Science Question 7.
One kilogram force equals to:
(a) 9.8 dyne
(b) 9.8 × 10⁴ N
(c) 98 × 10⁴ dyne
(d) 980 dyne
Answer:
(c) 98 × 10⁴ dyne

10th Standard Science Laws Of Motion Question 8.
The mass of a body is measured on planet Earth as M kg. When it is taken to a planet of radius half that of the Earth then its value will be ____ kg.
(a) 4M
(b) 2M
(c) \(\frac { M }{ 4 }\)
(d) M.
Answer:
(c) \(\frac { M }{ 4 }\)

10th Science Law Of Motion Question 9.
If the Earth shrinks to 50% of its real radius its mass remaining the same, the weight of a body on the Earth will _____ .
(a) decrease by 50%
(b) increase by 50%
(c) decrease by 25%
(d) increase by 300%.
Answer:
(c) decrease by 25%

Laws Of Motion Class 10 In Tamil Question 10.
To project the rockets which of the following
principle(s) is /(are) required?
(a) Newton’s third law of motion
(b) Newton’s law of gravitation
(c) Law of conservation of linear momentum
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

II. Fill in the blanks.

10th Science Lesson 1 Question 1.
To produce a displacement _____ is required.
Answer:
force.

10th Science Solution Samacheer Kalvi Question 2.
Passengers lean forward when the sudden brake is applied in a moving vehicle. This can be explained by ____.
Answer:
the inertia of motion.

Samacheer Kalvi Guru 10th Science Question 3.
By convention, the clockwise moments are taken as _____ and the anticlockwise moments are taken as _____.
Answer:
negative, positive.

10th Science Solutions Samacheer Kalvi Question 4.
______ is used to change the speed of the car.
Answer:
Acceleration.

10th Science Solutions Samacheer Question 5.
A man of mass 100 kg has a weight of _____ at the surface of the Earth.
Answer:
980 N.

III. State whether the following statements are true or false. Correct the statement if it is false:

Class 10 Samacheer Kalvi Science Solutions Question 1.
The linear momentum of a system of particles is always conserved.
Answer:
False.
Correct Statement: The linear momentum of a system of particles is always conserved if no external force acts.

10th Samacheer Kalvi Science Solutions Question 2.
The apparent weight of a person is always equal to his actual weight.
Answer:
True.

Question 3.
Weight of a body is greater at the equator and less at the polar region.
Answer:
False.
Correct Statement: Weight of a body is lesser at the equator and greater at the polar region.

Question 4.
Turning a nut with a spanner having a short handle is so easy than one with a long handle.
Answer:
False.
Correct Statement: Turning a nut with a spanner having a short handle is so harder than one with a long handle.

Question 5.
There is no gravity in the orbiting space station around the Earth. So the astronauts feel weightlessness.
Answer:
False.
Correct Statement: There is a gravity in the orbiting space station around the earth. Since space station and astronauts have equal acceleration. Both the astronauts and space station are in the state of weightlessness.

IV. Match the following.

Question 1.

Column I	Column II
1. Newton’s I law	(a) Propulsion of a rocket
2. Newton’s II law	(b) Stable equilibrium of a body
3. Newton’s III law	(c) Law of force
4. Law of conservation of linear momentum	(d) Flying nature of a bird

Answer:
1. (b) Stable equilibrium of a body
2. (c) Law of force
3. (d) Flying nature of a bird
4. (a) Propulsion of a rocket

V. Assertion & Reasoning

Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

Question 1.
Assertion: The sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Reason: The principle of conservation of momentum is valid if the external force on the system is zero.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 2.

1. Assertion: The value of ‘g’ decreases as height and depth increase from the surface of the Earth.
2. Reason: ‘g’ depends on the mass of the object and the Earth.

Answer:
(c) The assertion is true, but the reason is false.

VI. Answer briefly.

Question 1.
Define inertia. Give its classification.
Answer:
Inertia: The inherent property of a body to resist any change in its state of rest or the state of uniform motion, unless it is influenced upon by an external unbalanced force, is known as ‘inertia’.
Types of Inertia

• Inertia of rest
• Inertia of motion
• Inertia of direction

Question 2.
Classify the types of force based on their application.
Answer:

1. Like parallel forces
2. Unlike parallel forces

Question 3.
If a 5 N and a 15 N forces are acting opposite to one another. Find the resultant force and the direction of action of the resultant force.
Solution:
The two forces are unlike parallel forces

Let P = 5N, Q = 15N
Resultant force (R) = P – Q = 5 + (-15) = -10N
R = -10N.
The resultant force acting along the direction of “Q”.

Question 4.
Differentiate mass and weight.
Answer:

Mass	Weight
The quantity of matter contained in the body	The gravitation force exerted on it due to the Earth’s gravity alone.
Scalar quantity	Vector quantity
Unit: Kg	Unit: N
Constant at all the places	Variable with respect to gravity.

Question 5.
Define moment of a couple.
Answer:
Rotating effect of a couple is known as moment of a couple.
Moment of a couple = Force × perpendicular distance between the line of action of forces
M = F × S

Question 6.
State the principle of moments.
Answer:
When a number of like or unlike parallel forces act on a rigid body and the body is in equilibrium, then the algebraic sum of the moments in the clockwise direction is equal to the algebraic sum of the moments in the anticlockwise direction.

Question 7.
State Newton’s second law.
Answer:
“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.

Question 8.
Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?
Answer:
This is because turning effect to tighten the screws depends upon the perpendicular distance of the applied force from the axis of rotation is power arm. Larger the power armless is the force required to turn the screws. So spanner is provided with a long handle.

Question 9.
While catching a cricket ball the fielder lowers his hands backwards. Why?
Answer:
(i) When the fielder lowers his hands backwards, he increases the value of time of collision and so retardation is decreased.
(ii) Hence retarding force becomes lesser than before and the palm of the fielder is not hurt very much.

Question 10.
How does an astronaut float in a space shuttle?
Answer:
An astronaut float in a space shuttle because both are in the state of weightlessness. Both are experiencing equal acceleration towards earth as free fall bodies. Astronauts are not floating but falling freely.

VII. Solve the given problems.

Question 1.
Two bodies have a mass ratio of 3 : 4. The force applied to the bigger mass produces an acceleration of 12 ms^-2. What could be the acceleration of the other body, if the same force acts on it?
Solution:
Mass ratio of the bodies = 3 : 4 and same force is (m₁ : m₂) acting on the body and a₂ = 12 ms^-2
∴ m₁a₁ = m₂a₂
\(\frac{m_{1}}{m_{2}}=\frac{a_{2}}{a_{1}} \Rightarrow \frac{3}{4}=\frac{a_{2}}{a_{1}}\)
\(a_{1}=\frac{4}{3} \times 12=16 \mathrm{ms}^{-2}\)

Question 2.
A ball of mass 1 kg moving with a speed of 10 ms^-1 rebounds after a perfectly elastic collision with the floor. Calculate the change in linear momentum of the ball.
Solution:
Mass of a ball = 1 kg
Velocity of the bail before collision,
u = 10 m/s
Velocity of the ball after collision,
v = – u
= -10 m/s
Change in momentum,
P = m(v – u)
P = 1(-10 – 10)
= -20 kg m/s.

Question 3.
A mechanic unscrews a nut by applying a force of 140 N with a spanner of length 40 cm. What should be the length of the spanner if a force of 40 N is applied to unscrew the same nut?
Solution:
Given F₁ = 140 N, d₁ = 40 cm, F₂ = 40 N, d₂ = ?
In, both the cases, moment of forces applied are equal
F₁d₁ = F₂d₂
\(\begin{array}{l}{d_{2}=\left(\frac{F_{1}}{F_{2}}\right) d_{1}} \\ {d_{2}=40 \times \frac{140}{40}=140 \mathrm{cm}}\end{array}\)

Question 4.
The ratio of masses of two planets is 2 : 3 and the ratio of their radii are 4 : 7 Find the ratio of their accelerations due to gravity.
Solution:
The ratio of masses of two planets m₁ : m₂ = 2 : 3
The ratio of radii of two planets R₁ : R₂ = 4 : 7
Formula:

VIII. Answer in detail.

Question 1.
What are the types of inertia? Give an example for each type.
Answer:
Types of Inertia:
(i) Inertia of rest: The resistance of a body to change its state of rest is called inertia of rest. Eg: When you vigorously shake the branches of a tree, some of the leaves and fruits are detached and they fall down.

(ii) Inertia of motion: The resistance of a body to change its state of motion is called inertia of motion. Eg: An athlete runs some distance before jumping. Because, this will help him jump longer and higher.

(iii) Inertia of direction: The resistance of a body to change its direction of motion is called inertia of direction. Eg: When you make a sharp turn while driving a car, you tend to lean sideways.

Question 2.
State Newton’s laws of motion.
Answer:
Newton’s First Law: Everybody continues to be in its state of rest or the state of uniform motion along a straight line unless it is acted upon by some external force.

Newtons Second law: The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force.

Newtons Third Law: For every action, there is an equal and opposite reaction. They always act on two different bodies.

Question 3.
Deduce the equation of a force using Newton’s second law of motion.
Answer:
“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.
Let, ‘m’ be the mass of a moving body, moving along a straight line with an initial speed ‘u’ After a time interval of ‘t’, the velocity of the body changes to ‘v’ due to the impact of an unbalanced external force F.
Initial momentum of the body (P_i) = mu
Final momentum of the body (P_f) = mv
Change in momentum ∆p = P_f – P_i = mv – mu
By Newton’s second law of motion,
Force, F ∝ rate of change of momentum
F ∝ change in momentum / time
\(\begin{array}{l}{\mathrm{F} \propto \frac{m v-m u}{t}} \\ {\mathrm{F}=\frac{k m(v-u)}{t}}\end{array}\)
Here, k is the proportionality constant, k = 1 in all systems of units.
Hence, \(\mathrm{F}=\frac{m(v-u)}{t}\)
Since, acceleration = change in velocity / time,
a = (v – u)/t.
Hence, we have F = m × a
Force = mass × acceleration

• No external force is required to maintain the motion of a body moving with uniform velocity.
• When the net force acting on a body is not equal to zero, then definitely the velocity of the body will change.
• Thus, change in momentum takes place in the direction of the force. The change may take place either in magnitude or in direction or in both.

Question 4.
State and prove the law of conservation of linear momentum.
Answer:
(i) There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
(ii) Let us prove the law of conservation of linear momentum with the following illustration:

(iii) Let two bodies A and B having masses m1 and m2 move with initial velocity u₁ and u₂ in a straight line.
(iv) Let the velocity of the first body be higher than that of the second body. i.e., u₁ > u₂.
(v) During an interval of time t second, they tend to have a collision. After the impact, both of them move along the same straight line with a velocity v₁ and v₂ respectively.
Force on body B due to A,
\(\mathrm{F}_{\mathrm{B}}=\frac{m_{2}\left[v_{2}-u_{2}\right]}{t}\)
Force on body A due to B,
\(\mathrm{F}_{\mathrm{A}}=\frac{m_{1}\left[v_{1}-u_{1}\right]}{t}\)
By Newton’s III law of motion, Action force = Reaction force
F_A = -F_B
\(\begin{aligned} \frac{m_{1}\left[v_{1}-u_{1}\right]}{t} &=\frac{m_{2}\left[v_{2}-u_{2}\right]}{t} \\ m_{1} v_{1}+m_{2} v_{2} &=m_{1} u_{1}+m_{2} u_{2} \end{aligned}\)
The above equation confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
Hence the law of conservation linear momentum is proved.

Question 5.
Describe rocket propulsion.
Answer:
Propulsion of rockets is based on the law of conservation of linear momentum as well as Newton’s III law of motion. Rockets are filled with a fuel (either liquid or solid) in the propellant tank. When the rocket is fired, this fuel is burnt and a hot gas is ejected with a high speed from the nozzle of the rocket, producing a huge momentum. To balance this momentum, an equal and opposite reaction force is produced in the combustion chamber, which makes the rocket project forward.

While in motion, the mass of the rocket gradually decreases, until the fuel is completely burnt out. Since, there is no net external force acting on it, the linear momentum of the system is conserved. The mass of the rocket decreases with altitude, which results in the gradual increase in velocity of the rocket. At one stage, it reaches a velocity, which is sufficient to just escape from the gravitational pull of the Earth. This velocity is called escape velocity.

Question 6.
State the universal law of gravitation and derive its mathematical expression.
Answer:
This law states that every particle of matter in this universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between the centres of these masses. The direction of the force acts along the line joining the masses.

The force between the masses is always attractive and it does not depend on the medium where they are placed.
Let, m₁ and m₂ be the masses of two bodies A and B placed r metre apart in space.
Force F ∝ m₁ × m₂
F ∝ \(1 / r^{2}\)
On combining the above two expressions
\(\begin{array}{l}{\mathrm{F} \propto \frac{m_{1} \times m_{2}}{r^{2}}} \\ {\mathrm{F}=\frac{\mathrm{G} m_{1} m_{2}}{r^{2}}}\end{array}\)
Where G is the universal gravitational constant.
Its value in SI unit is 6.674 × 10^-11 Nm² kg^-2.

Question 7.
Give the applications of the universal law of gravitation.
Answer:
Application of Newton’s law of gravitation are:
(i) Dimensions of the heavenly bodies can be measured using the gravitation law. Mass of the Earth, radius of the Earth, acceleration due to gravity, etc., can be calculated with a higher accuracy.
(ii) Helps in discovering new stars and planets.
(Hi) Helps to explain germination of roots is due to the property of geotropism which is the property of a root responding to the gravity.
(iv) One of the irregularities in the motion of stars is called ‘Wobble’ lead to the disturbance in the motion of a planet nearby. In this condition the mass of the star can be calculated using the law of gravitation.
(v) Helps to predict the path of the astronomical bodies.

IX. HOT Questions.

Question 1.
Two blocks of masses 8 kg and 2 kg respectively, lie on a smooth horizontal surface in contact with one other. They are pushed by a horizontally applied force of 15 N. Calculate the force exerted on the 2 kg mass.
Solution:

Given: m₁ = 8 kg, m₂ = 2 kg, F = 15 N
F = mtotal, F = (m₁ + m₂) a = (8 + 2) a = 10 a
15 = 10 a
⇒ a = \(\frac{15}{10}=\frac{3}{2}\) ms^-2
Force exerted by mass of 8 kg
F = m₁ a = \(8 \times \frac{3}{2}\) = 12 N.

Question 2.
A heavy truck and bike are moving with the same kinetic energy. If the mass of the truck is four times that of the bike, then calculate the ratio of their momenta. (Ratio of momenta = 1 : 2)
Solution:
Given: Let m₁, m₂ are the masses of truck and bike.
m₁ = 4m₂
Here kinetic energies of both truck and bike are same
\(\begin{aligned} m_{1} v_{1}^{2} &=m_{2} v_{2}^{2} \\ 4 m_{2} v_{1}^{2} &=m_{2} v_{2}^{2} \\ \frac{v_{1}}{v_{2}} &=\frac{1}{2} \\ v_{2} &=2 v_{1} \end{aligned}\)
Ratio of momenta: \(\frac{p_{1}}{p_{2}}=\frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{4 m_{2}}{m_{2}} \cdot \frac{v_{1}}{2 v_{1}}\) = 2
P₁ : P₂ = 2 : 1.

Question 3.
“Wearing a helmet and fastening the seat belt is highly recommended for the safe journey”.Justify your answer using Newton’s laws of motion.
Answer:
During the motion of car and two wheelers, when the brakes are applied, the vehicles slow down but our body tends to continue in the same state of motion due to inertia. So this may cause injury to passengers. Hence they are advised to wear a helmet and seat belt.

Samacheer Kalvi 10th Science Laws of Motion Additional Questions

I. Choose the correct answer.

Question 1.
A cricketer catches a ball of mass 150 gm in 0.1s and which is moving with a speed of 20 ms^-1, then he experiences the force of ____ _.
(a) 300 N
(b) 30 N
(c) 3 N
(d) 0.3 N.
Answer:
(b) 30 N
Hint: Impulse = change in momentum
F.∆t = mv – mu
\(\mathrm{F}=\frac{m v-m u}{\Delta t}\)
\(=\frac{150 \times 10^{-3} \times 20}{0.1}=30 \mathrm{N}\)

Question 2.
SI unit of force is:
(a) Dyne
(b) newton
(c) kgms^-1
(d) Joule
Answer:
(b) newton

Question 3.
A coin is dropped in a lift. It takes time t₁ to reach the floor, when the lift is stationary, it takes time t₂, when the lift is moving up with constant acceleration, then ____ _.
(a) t₁ > t₂
(b) t₁ < t₂
(c) t₁ = t₂
(d) None.
Answer:
(a) t₁ > t₂

Question 4.
An unbalanced force acts on a body, the body:
(a) must remain at rest
(b) must be accelerated
(c) must move with uniform velocity
(d) move with uniform motion
Answer:
(b) must be accelerated

Question 5.
A satellite in its orbit around the earth is weightless on account of its _____.
(a) velocity
(b) momentum
(c) angular momentum
(d) acceleration.
Answer:
(c) angular momentum

Question 6.
When two or more forces acting on a body and the body does not change its position, then the forces are:
(a) imbalanced
(b) mechanical force
(c) balanced forces
(d) none
Answer:
(c) balanced forces

Question 7.
What would be the acceleration due to gravity at another planet, whose mass and radius core twice that of earth?
(a) g
(b) \(\frac{g}{2}\)
(c) \(\frac{g}{4}\)
Answer:
(b) \(\frac{g}{2}\)
Hint: We know that \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
\(\frac{g_{1}}{g_{2}}=\frac{\left(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)}{\left(\frac{\mathrm{G} .2 \mathrm{M}}{4 \mathrm{R}^{2}}\right)}\)
\(\frac{g_{1}}{g_{2}}=2 \quad \Rightarrow \quad g_{2}=\frac{1}{2} g_{1}\)

Question 8.
At sea level, the value of “g” is maximum at _____.
(a) the poles
(b) the equator
(c) 45° south latitude
(d) 45° north of longitude.
Answer:
(a) the poles

Question 9.
An object cannot change the state of rest or motion, until a force is applied. This inability of the object is called:
(a) inertia
(b) mass
(c) weight
(d) acceleration
Answer:
(a) inertia

Question 10.
The ability of a body to maintain its state of rest or motion is called ______.
(a) mechanics
(b) kinematics
(c) kinetics
(d) Inertia.
Answer:
(d) Inertia.

Question 11.
_____ deals with the bodies, which are at rest under the action of forces.
(a) Statics
(b) Dynamics
(c) Kinematics
(d) Kinetics.
Answer:
(a) Statics

Question 12.
A motor car starts from rest and moves after 5 seconds. If its velocity is 200 m/s then its acceleration is:
(a) 100 m/s²
(b) 40 m/s²
(c) 20 m/s²
(d) 80 m/s²
Answer:
(b) 40 m/s²

Question 13.
_____ deals with the motion of bodies considering the cause of motion.
(a) Force
(b) Dynamics
(c) Statics
(d) Kinetics.
Answer:
(d) Kinetics

Question 14.
Linear momentum = _____
(a) mass × velocity
(b) mass × distance
(c) distance × time
(d) \(\frac{\text { mass }}{\text { velocity }}\).
Answer:
(a) mass × velocity

Question 15.
The inability of the body to change its state is:
(a) force
(b) momentum
(c) acceleration
(d) inertia
Answer:
(d) inertia

Question 16.
Two or more forces of equal or unequal magnitude acting along the same direction, parallel to each other are called _____.
(a) like parallel forces
(b) unlike parallel forces
(c) resultant force
(d) balanced force.
Answer:
(a) like parallel forces

Question 17.
The axis of the fixed edge about which the thing is rotated is called as _____ .
(a) axis of rotation
(b) fixed axis rotation
(c) point of rotation
(d) Fixed point.
Answer:
(a) axis of rotation

Question 18.
When a net force acts on an object, the object will be accelerated in the direction of force with an acceleration proportional to:
(a) force on the object
(b) velocity
(c) mass
(d) inertia
Answer:
(a) force on the object

Question 19.
Rotating effect of a couple is known as ______ .
(a) product of forces
(b) the momentum of a couple
(c) mass
(d) momentum.
Answer:
(b) momentum of a couple

Question 20.
The amount of force required to produce an acceleration of 1 ms^-2 in a body of mass _____ is called unit force.
(a) 10 kg
(b) 100 kg
(c) 1 kg
(d) 0 kg.
Answer:
(c) 1 kg

Question 21.
The acceleration of a body is due to:
(a) balance force
(b) electrostatic force
(c) unbalanced force
(d) conservative force
Answer:
(c) unbalanced force

Question 22.
Universal gravitational constant ______ .
(a) G = 6.684 × 10^-10 Nm² kg^-1
(b) G = 7.4 × 10¹⁰ Nm²
(c) G = 6.623 × 10¹¹ Nm² kg^-1
(d) G = 6.674 × 10^-11 Nm² kg^-2
Answer:
(d) G = 6.674 × 10^-11 Nm² kg^-2

Question 23.
Mean value of the acceleration due to gravity is ______ .
(a) 10.1 ms^-2
(b) 8.8 ms^-2
(c) 9.8 ms^-2
(d) 9.8 ms.
Answer:
(c) 9.8 ms^-2

Question 24.
The unit of weight is:
(a) kg
(b) g
(c) Newton
(d) ms^-1
Answer:
(c) Newton

Question 25.
The value of accelaration due to gravity on the surface of the moon is _____ .
(a) 1.75 ms^-1
(b) 3.8 ms^-2
(c) 1.625 ms^-2
(d) 1.625 ms^-1
Answer:
(c) 1.625 ms^-2

Question 26.
The unit of weight is _____ .
(a) kg m
(b) kg
(c) newton
(d) kg m^-1
Answer:
(c) newton

Question 27.
The weight of a body is _____ poles than at the equatorial region.
(a) more
(b) less
(c) zero
(d) one.
Answer:
(a) more

Question 28.
In a collision between a heavier body and a lighter body, which body experiences greater force?
(a) heavier body
(b) lighter body
(c) both the body experience same force
(d) both body exchange acceleration
Answer:
(c) both the body experience same force

II. Fill in the blanks.

Question 1.
Turning a tap is an example of ____
Answer:
couple.

Question 2.
Torque is a _______ quantity.
Answer:
vector.

Question 3.
1 gf is equal to _____ dyne.
Answer:
980.

Question 4.
The resultant force acting on a body is equal to zero then the body will be in ______
Answer:
equilibrium.

Question 5.
The force equal to resultant but opposite in direction is ______
Answer:
equilibrate.

Question 6.
The product of force and time is ______
Answer:
impulse.

Question 7.
The force between the masses is always ______
Answer:
attractive.

Question 8.
The quantity of matter contained in the object is known as _____
Answer:
mass.

Question 9.
The magnitude of the universal gravitational constant is _____.
Answer:
6.674 × 10^-11 Nm² kg^-2

Question 10.
Propulsion of rockets is based on the ____ and ____
Answer:
Law of conservation of linear momentum & Newton’s third law.

Question 11.
Parallel unequal forces are acting in ______ directions.
Answer:
Opposite.

Question 12.
Torque and force are the ____ quantities.
Answer:
vector.

Question 13.
The unit of moment of a couple is _____ .
Answer:
newton metre (Nm).

Question 14.
A _____ enables you to manoeuvre a car easily by transferring a _______ to the wheels with less effort.
Answer:
steering wheel, torque.

Question 15.
_____ is required to produce the acceleration of a body.
Answer:
Force.

Question 16.
The acceleration is produced along the radius is called ______
Answer:
centripetal acceleration.

Question 17.
______ is equal to the magnitude of change in momentum.
Answer:
impulse.

Question 18.
A large force acting for a very short interval of time is called as ______ .
Answer:
impulse Force.

Question 19.
Mass of the earth _____
Answer:
5.972 × 10²⁴ kg.

Question 20.
The relation between acceleration due to gravity (g) and the universal gravitational constant (G) is _____ .
Answer:
\(g=\frac{G M}{R^{2}}\).

III. State whether the following statements are true or false, correct the statement if it is false.

Question 1.
Rest and motion are interrelated terms.
Answer:
True.

Question 2.
In the C.G.S. system, the unit of linear momentum is kg ms^-1.
Answer:
False.
Correct Statement: In the C.G.S. system, the unit of linear momentum is g cms^-1.

Question 3.
An external force is required to maintain the motion of a body moving with uniform velocity.
Answer:
False.
Correct Statement: No external force is required to maintain the motion of a body moving with uniform velocity.

Question 4.
The amount of force required for a body of mass 1 gram produces an acceleration of 1 cm s^-2
Answer:
True.

Question 5.
By Newton’s III – law of motion, the action force is not equal to the reaction force.
Answer:
False.
Correct Statement: By Newton’s III – law of motion, the action force is equal to the reaction force.

Question 6.
The value of acceleration due to gravity (g) is not the same at all the points on the surface of the earth.
Answer:
True.

Question 7.
The value of acceleration due to gravity on the surface of the moon is 1.625 ms^-2.
Answer:
True.

Question 8.
The regularities in the motion of stars are called ‘wobble’.
Answer:
False.
Correct Statement: The irregularities in the motion of stars is called ‘wobble’.

Question 9.
Mechanics is divided into kinematics and kinetics.
Answer:
False.
Correct Statement: Mechanics is divided into statics and dynamics.

Question 10.
Application of Newton’s law of gravitation helps to predict the path of the astronomical bodies.
Answer:
True.

IV. Match the following.

Question 1.

1. Linear momentum	(a) Mass and acceleration
2. Force	(b) Change in momentum
3. Moment of force	(c) GM/R2
4. Impulse	(d) Mass and velocity
5. Acceleration due to gravity	(e) Force and perpendicular distance

Answer:
1. (d) Mass and velocity
2. (a) Mass and acceleration
3. (e) Force and perpendicular distance
4. (b) Change in momentum
5. (c) GM/R²

Question 2.

1. Kinetics	(a) Causes the motion
2. Kinematics	(b) In equilibrium
3. Balanced force	(c) The motion of bodies without cause
4. Unbalanced force	(d) The motion of bodies with cause

Answer:
1. (d) The Motion of bodies with cause
2. (c) The Motion of bodies without cause
3. (b) In equilibrium
4. (a) Causes the motion

V. Assertion & Reasoning

Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

Question 1.
Assertion: At poles value of acceleration due to gravity (g) is greater than that of the equator.
Reason: Earth rotates on its axis in addition to revolving around the sun.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 2.
Assertion: The force exerted by the earth on an apple is more than that exerted by apple on the earth.
Reason: The force on apple exerts on the earth is determined by the mass of the apple only.
Answer:
(d) The assertion is false but the reason is true

Question 3.
Assertion: A freely falling body is in the state of weightlessness
Reason: A body becomes conscious of its weight only when it is opposed
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 4.
Assertion: Newton’s third law of motion is applicable only when bodies are in motion.
Reason: Newton’s third law applies to all types of forces, e.g. gravitational, electric or magnetic forces.
Answer:
(d) The assertion is false but the reason is true.

Question 5.
Assertion: The apparent weight of the person is zero, in which condition or state is known as weightless.
Reason: When the person in a lift moves down with an acceleration (a) is equal to the acceleration due to gravity (g)
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 6.
Assertion: A gear is a circular wheel with teeth around its rim.
Reason: It helps to change the speed of rotation of a wheel by changing the force and helps to transmit power.
Answer:
(c) The assertion is true, but the reason is false.

Question 7.
Assertion: Mass of a body is defined as the gravitational force exerted on it due to earth’s gravity alone
Reason: Weight = mass × acceleration due to gravity.
Answer:
(d) The assertion is false, but the reason is true.

Question 8.
Assertion: Weight is a vector quantity.
Reason: Direction of weight is always towards the centre of the earth.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 9.
Assertion: Resultant force is equal to the vector sum of all the forces.
Reason: A system cannot be brought to equilibrium by applying another force.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

VI. Answer briefly

Question 1.
Define Linear momentum.
Answer:
The product of mass and velocity of a moving body gives the magnitude of linear momentum. It acts in the direction of the velocity of the object. Linear momentum is a vector quantity.
Linear Momentum = mass × velocity.
Unit of momentum in SI system is Kg ms^-1 and in C.G.S system its unit is g cm s^-1

Question 2.
What is the resultant force?
Answer:
When several forces act simultaneously on the same body, then the combined effect of the multiple forces can be represented by a single force, which is termed as ‘resultant force’.

Question 3.
What is meant by equilibrant?
Answer:
A system can be brought to equilibrium by applying another force, which is equal to the resultant force in magnitude, but opposite in direction. Such force is called as ‘Equilibrant’.

Question 4.
Explain the Newton third law of motion with examples.
Answer:
‘For every action, there is an equal and opposite reaction. They always act in two different bodies’.
Example: When you fire a bullet, the gun recoils backwards and the bullet is moving forward (Action) and the gun equalises this forward action by moving backwards (Reaction).

Question 5.
How did the change in momentum achieve?
Answer:
Change in momentum can be achieved in two ways. They are:

• A large force acting for a short period of time and
• A smaller force acting for a longer period of time.

Question 6.
Define impulse.
Answer:
A large force acting for a very short interval of time is called as ‘Impulsive force’.When a force F acts on a body for a period of time t, then the product of force and time is known as ‘impulse’ represented by ‘J’
Impulse, J = F × t …….. (1)
By Newton’s second law
F = ∆p / t (A refers to change)
∆p = F × t ………. (2)
From (1) and (2)
J = ∆p
Impulse is also equal to the magnitude of change in momentum.
Its unit is kg ms^-1 or Ns.

Question 7.
What is meant by free fall?
Answer:

1. When the person in a lift moves down with an acceleration (a) equal to the acceleration due to gravity (g), i.e., when a = g, this motion is called as ‘free fall’.
2. The apparent weight (R = m (g – g) = 0) of the person is zero. This condition or state refers to the state of weightlessness.

Question 8.
Define weightlessness.
Answer:
Whenever a body or a person falls freely under the action of Earth’s gravitational force alone, it appears to have zero weight. This state is referred to as ‘weightlessness’.

Question 9.
Explain the various causes of the apparent weight of a person in a moving lift.
Answer:

Case 1: Lift is moving upward with an acceleration ‘a’	Case 2: Lift is moving downward with an acceleration ‘a’	Case 3: Lift is at rest.	Case 4: Lift is falling down freely
R – W = Fnet = ma ⇒ R = W + ma ⇒ R = mg + ma ⇒ R = m(g + a)	W – R = Fnet = ma ⇒ R = W – ma ⇒ R = mg – ma ⇒ R = m(g – a)	Here,the acceleration is zero a = 0 R = W R = mg	Here,the acceleration is equal to g a = g R = m(g – g) = 0
R > W	R < W	R = W	R = 0
Apparent weight is greater than the actual weight.	Apparent weight is lesser than the actual weight.	Apparent weight is equal to the actual weight.	Apparent weight is equal to zero.

Question 10.
Explain the variation of acceleration due to gravity.
Answer:
Variation of acceleration due to gravity (g):

1. Since, g depends on the geometric radius of the Earth, (g ∝ 1 / R²), its value changes from one place to another on the surface of the Earth.
2. The geometric radius of the Earth is maximum in the equatorial region and minimum in the polar region, the value of g is maximum in the polar region and minimum at the equatorial region.
3. When you move to a higher altitude from the surface of the Earth, the value of g reduces.
4. when you move deep below the surface of the Earth, the value of g reduces. Value of g is zero at the centre of the Earth.

Question 11.
Define one newton and one dyne.
Answer:
Definition of 1 newton (N): The amount of force required for a body of mass 1 kg produces an acceleration of 1 ms^-2, 1 N = 1 kg ms^-2
Definition of 1 dyne: The amount of force required for a body of mass 1 gram produces an acceleration of 1 cms^-2, 1 dyne = 1 g cms^-2; also 1 N = 10⁵ dyne.

Question 12.
How can you measure the moment of the couple?
Answer:
(i) Rotating effect of a couple is known as the moment of a couple.
(ii) Moment of a couple is measured by the product of any one of the forces and the perpendicular distance between the line of action of two forces. The turning effect of a couple is measured by the magnitude of its moment.
(iii) Moment of a couple = Force × perpendicular distance between the line of action of forces
M = F × S
(iv) The unit of moment of a couple is newton metre (N m) in SI system and dyne cm in the CGS system.
(v) By convention, the direction of moment of a force or couple is taken as positive if the body is rotated in the anti-clockwise direction and negative if it is rotating in the clockwise direction.
They are shown in Figures.

Question 13.
Define Torque.
Answer:
(i) The rotating or turning effect of a force about a fixed point or fixed axis is called the moment of the force about that point or torque (τ).
(ii) τ = F × d
(iii) Torque is a vector quantity.
(iv) Its SI unit is Nm.

VII. Answ er in detail.

Question 1.
Explain any three application of Torque.
Answer:
Application of Torque:
(i) Gears: A gear is a circular wheel with teeth around its rim. It helps to change the speed of rotation of a wheel by changing the torque and helps to transmit power.

(ii) Seasaw: Most of you have played on the seesaw. Since there is a difference in the weight of the persons sitting on it, the heavier person lifts the lighter person. When the heavier person comes closer to the pivot point (fulcrum) the distance of the line of action of the force decreases. It causes less amount of torque to act on it. This enables the lighter person to lift the heavier person.

(iii) Steering Wheel: A small steering wheel enables you to manoeuvre a car easily by transferring torque to the wheels with less effort.

Question 2.
State Newton’s third law. Explain it with three examples.
Answer:
Newton’s third law of motion: Newton’s third law states that ‘for every action, there is an equal and opposite reaction. They always act in two different bodies’.
If a body A applies a force F_A on a body B, then the body B reacts with force F_B on the body A, which is equal to F_A in magnitude, but opposite in direction.
F_B = -F_A
Examples:

• When birds fly they push the air downwards with their wings (Action) and the air pushes the bird upwards (Reaction).
• When a person swims he pushes the water using the hands backwards (Action), and the water pushes the swimmer in the forward direction (Reaction).
• When you fire a bullet, the gun recoils backwards and the bullet is moving forward (Action) and the gun equalises this forward action by moving backwards (Reaction).

Question 3.
Derive the relation between ‘g’ and G. Explain how to determine the mass of earth.
Answer:
(i) Let us compute the magnitude of this force in two ways. Let, M be the mass of the Earth and m be the mass of the body.
(ii) The entire mass of the Earth is assumed to be concentrated at its centre.
(iii) The radius of the Earth is R = 6378 km (= 6400 km approximately). By Newton’s law of gravitation, the force acting on the body is given by
\(\mathrm{F}=\frac{\mathrm{GM} m}{\mathrm{R}^{2}}\) ….(1)

(iv) The radius of the body considered is negligible when compared with the Earth’s radius. Now, the same force can be obtained from Newton’s second law of motion.
(v) According to this law, the force acting on the body is given by the product of its mass and acceleration (called weight). Here, acceleration of the body is under the action of gravity hence a = g
F = ma = mg
F = weight = mg ……. (2)
Comparing equations J = F × t and ΔP = F × t, we get
\(m g=\frac{G M m}{R^{2}}\) …….. (3)
Acceleration due to gravity
\(g=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) ……. (4)
Mass of the Earth (M):
Rearranging the equation (4), the mass of the Earth is obtained as follows:
Mass of the Earth M = g R² / G
Substituting the known values of g, R and G, you can calculate the mass of the Earth as M = 5.972 × 10²⁴ kg.

VIII. Problems.

Question 1.
A cricket ball of mass 0.5 kg strikes a bat normally with a velocity of 30 ms^-1 and rebounds with a velocity of 20 ms^-1 in the opposite direction, calculate the impulse of the force exerted by the ball on the bat.
Solution:
Impulse = change in momentum = mu – (-mv)
= m (u + v)
= 0.5 (30 + 20)
= 25 Ns

Question 2.
A force exerted on a body of mass 100 g changes its speed by 0.2 ms^-1 in each second. Calculate the magnitude of the force.
Given, mass m = 100 g = 0.1 kg and acceleration a = 200 cms^-2 = 0.2 ms^-2.
Solution:
F = ma = 0.1 × 0.2 = 0.02 N.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.3

10th Maths Exercise 8.3 Samacheer Kalvi Question 1.
Write the sample space for tossing three coins using tree diagram.
Solution:

Ex 8.3 Class 10 Samacheer Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Solution:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Exercise 8.3 Class 10 Samacheer Question 3.
If A is an event of a random experiment such that P(A) : P(\(\overline{\mathbf{A}}\)) =17 : 15 and n(S) = 640 then find
(i) P(\(\overline{\mathbf{A}}\))
(ii) n(A).
Solution:
P(A): P(\(\overline{\mathbf{A}}\)) = 17 : 15

10th Maths 8.3 Solutions Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution:
Outcomes {O}: {(HHH), (THH), (HTH), (HHT), (HTT), (THT), (TTH), (TTT)}
Two consecutive tails {F} : {(HTT), (TTH), (TTT)}
n{F} = 3
n{O} = 8

10th Maths Exercise 8.3 Question 5.
At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Answer:
Sample space = {1, 2, 3,… ,1000}
n(S) = 1000
(i) Let A be the event of setting square number greater than 500
A = {529, 576, 625, 676, 729, 784, 841, 900, 961}
n(A) = 9
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{9}{1000}\)
The probability that the first player wins prize = \(\frac{9}{1000}\)
(ii) If the first player wins, the number is excluded for the second player.
n(A) = 8 and n(S) = 999
P(A) = \(\frac{n(A)}{n(S)}=\frac{8}{999}\)
Probability the second player wins a prize = \(\frac{8}{999}\)

10th Maths Probability Exercise 8.3 Question 6.
A bag contains 12 blue balls and x red balls. If one ball is drawn at random
(i) what is the probability that it will be a red ball?
(ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Solution:
12 ➝ blue balls
x ➝ red balls
(i) P (red ball) = \(\frac{x}{x+12}\)
(ii) 8 red balls are added to the bag.
∴ 12 ➝ blue balls
x + 8 ➝ red balls

Given that P(ii) = 2 × P(i)

⇒ (x + 8)(x + 12) = 2x(x + 20)
⇒ (x² + 20x + 96) = 2x² + 40x
⇒ x² + 20x – 96 = 0
⇒ x² + 24x – 4x – 96 = 0
⇒ x(x + 24) – 4(x + 24) = 0
⇒ (x – 4)(x + 24) = 0
∴ x = 4 (or) x = -24
x cannot be negative ⇒ x = 4
Substituting x = 4 in (i),

10th Maths Exercise 8.3samacheer Kalvi Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Solution:
Doublet = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6,6)}
Total number of outcomes = 6 × 6
n(S) = 36
Number of favourable outcomes = 6

(ii) Number of favourable outcomes = 6
as favourable outcomes = (1, 2), (2, 1), (1, 3), (3, 1),(1, 5),and (5, 1)

(iii) Sum as prime numbers = {(1, 1), (1, 2), (2, 3), (1, 4), (1, 6), (4, 3), (5, 6)}
Number of favourable outcomes = 7
⇒ Probability = \(\frac{7}{36}\)

(iv) With two dice, minimum sum possible = 2
∴ Prob (sum as 1) = 0 [Impossible event]

10th Maths 8.3 Question 8.
Three fair coins are tossed together. Find the probability of getting
(i) all heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Answer:
Three fair coins are tossed together
Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting all heads
A = {HHH}
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting atleast one tail.
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}\)
(iii) Let C be the event of getting atmost one head
C = {HTT, THT, TTH, TTT}
n(C) = 4
\(P(C)=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)
(iv) Let D be the event of getting atmost two tails.
D = {HTT, TTT, TTH, THT, THH, HHT, HTH}
n(D) = 7
\(P(D)=\frac{n(D)}{n(S)}=\frac{7}{8}\)

10 Maths Exercise 8.3 Question 9.
Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution:
Dice 1
S = {1,2, 3, 4, 5, 6}
Dice 2
S = {1,1,2, 2, 3, 3}
Total possible outcomes when they are rolled

n(S) = 36
Event of sum (2) = A = {(1,1), (1,1)},
n(A) = 2,P(A) = \(\frac{2}{36}\)
Event of sum 3 is B = {(1, 2), (1, 2), (2, 1), (2, 1)}

Event of sum 4 is C= {(1, 3), (1, 3), (2, 2), (2, 2), (3, 1) (3, 1)}
n(C) = 6

Event of getting the sum 5 is
D = {(2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1)}
n(D) = 6, P(D) = \(\frac{6}{36}\) .
Event of getting the sum 6 is
E = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}
n(E) = 6, P(E) = \(\frac{6}{36}\)
Event of getting the sum 7 is
F = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}
n(F) = 6
P(F) = \(\frac{6}{36}\)
Event of getting the sum 8 is
G = {(5, 3), (5, 3), (6, 2), (6, 2)}

Event of getting the sum 9 is
H = {(6, 3), (6, 3), n(H) = 2

10th Standard Maths Exercise 8.3 Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball is drawn
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Solution:
5 red 6 white 7 green 8 black total no. of balls = 5 + 6 + 7 +8= 26

Exercise 8.3 Class 10 Samacheer Kalvi Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is 3/8 then, find the number of defective bulbs.
Solution:
Let number of defective bulbs be ‘x’
Total number of bulbs = x + 20

⇒ 8x = 3x + 60
⇒ 5x = 60
⇒ x = 12
∴ No.of defective bulbs are = 12.

10th Maths Exercise 8.4 Samacheer Kalvi Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Solution:

(i.e) remaining number of cards = 52 – 6 = 46 13
(i) P(a clavor) = \(\frac{13}{46}\)
(ii) P(queen of red card) = 0 as both Queen of diamond and heart have been removed.
(iii) only K of clavor is in the deck
⇒ P(king of black card) = \(\frac{1}{46}\)

10th Maths Ex 8.3 Question 13.
Some boys are playing a game, in which the stone was thrown by them landing in a circular region (given in the figure) is considered as a win and landing other than the circular region is considered as a loss. What’is the probability to win the game?

Solution:

Samacheer Kalvi 10th Maths Exercise 8.3 Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?
Solution:

10th Maths Exercise 8.3 4th Sum Question 15.
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Solution:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.1

Exercise 3.1 Class 10 Maths Samacheer Question 1.
Solve the following system of linear equations in three variables
(i) x + y + z = 5; 2x – y + z = 9; x – 2y + 3z = 16
(ii) \(\frac { 1 }{ x } \) – \(\frac { 2 }{ y } \) + 4 = 0; \(\frac { 1 }{ y } \) – \(\frac { 1 }{ z } \) + 1 = 0; \(\frac { 2 }{ z } \) + \(\frac { 3 }{ x } \) = 14
(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)
Solutions:
(i) x + y + z = 5 ………….. (1)
2x – y + z = 9 …………. (2)
x – 2y + 3z = 16 …………. (3)

Substitute z = 4 in (4)
3x + 2(4) = 14
3x + 8 = 14
3x = 6
x = 2
Substitute x = 2, z = 4 in (1)
2 + y + 4 = 5 ⇒ y = -1
x = 2, y = -1, z = 4

You can Download Samacheer Kalvi 10th Maths Solution Book Pdf Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

\(\frac{1}{y}\) = b
\(\frac{1}{z}\) = c in (1), (2) & (3)
a – 2b + 4 = 0 ⇒ a – 2b = -4 …………. (1)
b – c + 1 = 0 ⇒ b – c = -1 ……….. (2)
2c + 3a = 14 ⇒ 2c + 3a = 14 …………. (3)

(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)
x = \(\frac { 3y }{ 2 } \) – 10 …………. (1)
2z + 5 = 110 – (y + z)
2z = 105 – y – z
y = 105 – 3z ………….. (2)
Substitute (2) in (1), x = \(\frac { 315 }{ 2 } \) – \(\frac { 9z }{ 2 } \) – 10
= 2z + 5 – 20
∴ 315 – 9z – 20 = 4z – 30
13 z = 315 – 20 + 30
= 325
z = \(\frac { 325 }{ 13 } \) = 25
x + 20 = 2z + 5
x + 20 = 50 + 5
x = 35
Substitute z = 25 in (2)
y = 105 – 3z = 105 – 75 = 30
y = 30
x = 35, y = 30, z = 25
The system has unique solutions.

10th Maths Exercise 3.1 Question 2.
Discuss the nature of solutions of the following system of equations
(i) x + 2y – z = 6 ; -3x – 2y + 5z = -12 ; x – 2z = 3
(ii) 2y + z = 3 (-x + 1); -x + 3y -z = -4 3x + 2y + z = – \(\frac { 1 }{ 2 } \)
(iii) \(\frac { y+z }{ 4 } \) = \(\frac { z+x }{ 3 } \) = \(\frac { x+y }{ 2 } \); x + y + z = 27
Solution:
(i) x + 2y – z = 6 …………. (1)
-3x – 2y + 5z = -12 ……… (2)
x – 2z = 3 …………… (3)

We see that the system has an infinite number of solutions.
(ii) 2y + z = 3(-x + 1);
-x + 3y – z = -4;
3x + 2y + z = –\(\frac { 1 }{ 2 } \)
2y + z + 3x = 3 ⇒ 3x + 2y + z = 3 ………….. (1)
-x + 3y – z = -4 …………. (2)
3x + 2y + z = –\(\frac { 1 }{ 2 } \) ………………. (3)


This is a contradiction. This means the system is inconsistent and has no solutions.

Sub. x = 3 in (4) ⇒ 5(3) – z = 0
15 – z = 0
-z = -15
z = 15
Sub, x = 3, z = 15 in (3)
x + y + z = 27
3 + y + 15 = 27
y = 27 – 18 = 9
x = 3, y = 9, z = 15
∴ The system has unique solutions.

Ex 3.1 Class 10 Samacheer Question 3.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. Four years ago if Vani’s grandfather was four times as old as Vani then how old are they all now?
Solution:
Let Vani’s age be x
Let Vani’s father’s age be y
Let Vani’s grand father’s age be z.

Sub, z = 84 in (3), we get
4x – 84 = 12
4x = 96
x = 24
Sub, x = 24, z = 84 in (1) we get
24 + y + 84 = 159
y = 159 – 108
= 51
∴ Vani’s age = 24 years
Her father’s age =51 years
Her grand father’s age = 84 years.

10th Maths Exercise 3.1 Samacheer Kalvi Question 4.
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number?
Solution:
Let the number be 100x + 10y + z.
Reversed number be 100z + 10y + x.
x + y + z = 11 …………… (1)
100z + 10y + x = 5(100x + 10y + z) + 46
100z + 10y + x = 500x + 50y + 5z + 46
499x + 40y – 95z -46 ………….. (2)
x + 2y = z
x + 2y – z = 0 ……………. (3)

10th Maths Algebra Exercise 3.1 Solutions Question 5.
There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Solution:
Let x, y and z be number of currency pieces of 5,10,20 rupees
x + y + z = 12 ………. (1)
5x + 10y + 20z = 105 ………… (2)
10x + 5y + 20z = 125 …………. (3)

Sub, z = 2 in (5), we get
15y + 20 × 2 = 85
15y = 45
y = 3
Sub; y = 3, z = 2 in (1)
x + y + z = 12
x = 7
∴ The solutions are
the number of ₹ 5 are 7
the number of ₹ 10 are 3
the number of ₹ 20 are 2

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science History Solutions Chapter 2 Ancient Civilisations

Ancient Civilisations Textual Exercise

I. Choose the correct answer.

9th Social Guide Question 1.
The earliest signs to denote words through pictures ……………
(a) Logographic
(b) Pictographic
(c) Ideographic
(d) Stratigraphic
Answer:
(b) Pictographic

Ancient Civilizations 9th Class Question 2.
The preservation process of dead body in ancient Egypt ……………
(a) Sarcophagus
(b) Hyksos
(c) Mummification
(d) Polytheism
Answer:
(c) Mummification

Ancient Civilizations 9th Class Question 3.
The Sumerian system of writing : ……………
(a) Pictographic
(b) Hieroglyphic
(c) Sonogram
(d) Cuneiform
Answer:
(b) Hieroglyphic

9 Social Guide Question 4.
The Harappans did not have the knowledge of ……………
(a) Gold and Elephant
(b) Horse and Iron
(c) Sheep and Silver
(d) Ox and Platinum
Answer:
(d) Ox and Platinum

9th Social Science Guide Question 5.
The Bronze image suggestive of the use of lost-wax process known to the Indus people.
(a) Jar
(b) Priest king
(c) Dancing girl
(d) Bird
Answer:
(c) Dancing girl

Ancient Civilizations Questions And Answers Question 6.
(i) The oldest civilisation in Mesopotamia belonged to the Akkadians.
(ii) The Chinese developed the Hieroglyphic system.
(iii) The Euphrates and Tigris drain into the Mannar Gulf.
(iv) Hammurabi, the king of Babylon was a great law maker. .
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(d) (iv) is correct

Samacheer Kalvi 9th Social Book Answers Question 7.
(i) Yangtze River is known as Sorrow of China.
(ii) Wu-Ti constructed the Great Wall of China.
(iii) Chinese invented gun powder.
(iv) According to traditions Mfencius was the founder of Taoism.
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iii) and (iv) are correct
Answer:
(c) (iii) is correct

Samacheer Kalvi Guru 9th Social Science Question 8.
What is the correct chronological order of four civilisations of Mesopotamia?
(a) Sumerians – Assyrians – Akkadians – Babylonians
(b) Babylonians – Sumerians – Assyrians – Akkadians
(c) Sumerians – Akkadians – Babylonians – Assyrians
(d) Babylonians – Assyrians – Akkadians – Sumerians
Answer:
(c) Sumerians – Akkadians – Babylonians – Assyrians

9th Social Science Guide Pdf Download Question 9.
Assertion (A): Assyrians of Mesopotamian civilisation were contemporaries of Indus civilisation.
Reason (R): The Documents of an Assyrian ruler refer to the ships from Meluha.
(a) A and R are correct and A explains R
(b) A and R are correct but A doesn’t explain R
(c) A is incorrect but R is correct
(d) Both A and R are incorrect
Answer:
(d) Both A and R are incorrect

II. Fill in the blanks.

1. ……………. is a massive lime stone image of a lion with a human head.
2. The early form of writing of the Egyptians is known as ……………
3. ……………. specifies the Laws related to various crimes in ancient Babylonia.
4. …………… was the master archive keeper of Chou state, according to traditions.
5. The …………… figurines and paintings on the pottery from the sites suggest the artistic skills of the Harappans.
Answers:
1. The Great Sphinx of Ghiza
2. Hieroglyphic
3. Hammurabi code
4. Lao Tze
5. terracotta

III. Find out the correct statement.

Social Guide For Class 9 Question 1.
(a) The Great Bath at Harappa is well-built with several adjacent rooms.
(b) The cuneiform inscriptions relate to the epic of Gilgamesh.
(c) The terracotta figurines and dancing girl made of copper suggest the artistic skills of Egyptians.
(d) The Mesopotamians devised a solar calendar system.
Answer:
(a) correct
(b) correct
(c) incorrect
(d) incorrect

9th Standard Social Guide Question 2.
(a) Amon was considered the king of god in ancient Egypt.
(b) The fortified Harappan city had the temples.
(c) The great sphinx is a pyramid-shaped monument found in ancient Mesopotamia.
(d) The invention of the potter’s wheel is Credited to the Egyptians.
Answer:
(a) correct
(b) incorrect
(c) incorrect
(d) incorrect

IV. Match the following.

Answer:
1. (e)
2. (a)
3. (d)
4. (b)
5. (c)

V. Answer the following briefly.

9th Std Social Science Guide Pdf Question 1.
The Egyptians excelled in art and architecture. Illustrate.
Answer:

1. The Egyptians excelled in art and architecture. Their writing is also a form of art.
2. Numerous sculptures, painting and carvings attest to the artistic skills of Egyptians.
3. The Pyramids are massive monuments, built as tombs of mourning to the Pharaohs.
4. The Great Sphinx of Giza is a massive limestone image of a lion with a human head.

9th Standard Social Science Guide Question 2.
State the salient features of the Ziggurats.
Answer:

1. Ziggurats were pyramid – shaped monuments found in the ancient Mesopotamia (Modem Iraq).
2. One of the most famous Ziggurats of the time is the one in the city of Ur.
3. The Ziggurats were at the city centre on a platform and appeared like steep pyramids with staircases leading to the top.

Samacheer Kalvi 9th Social Science Guide Free Download Pdf Question 3.
Hammurabi Code is an important legal document. Explain.
Answer:

1. Hammurabi Code is an important legal document that specifies the laws related to various crimes.
2. It has 282 provisions specifying cases related to family rights, trade, slavery, taxes and wages. ‘
3. It is carved on a stone, which portrays Hammurabi as receiving the code from the Sun god Shamash.
4. It was a compilation of old laws based on retributive principles. .
5. The ‘eye for eye’ and ‘tooth for tooth’ form of justice is used in the Hammurabi Code.

Indus Valley Civilization Samacheer Kalvi Question 4.
Write a note on the Great Wall of China.
Answer:

• The Great Wall of China, one of the wonders of the world, was a massive effort undertaken for the protection of China from the Mongols.
• In 220 BCE, under Qin Shi Huang, earlier fortifications were connected by walls as a form of defence against invasions.
• It was built from third century BCE until 17th century CE.
• It ran for over 20,000 kilometres covering the hills and plains, from the border of Korea in the east to the Ordos Desert in the west.

VI. Answer all the questions given under each caption.

1. Early Civilizations:
Social Science Class 9 Guide Question 1.
What is meant by civilization?
Answer:
(a) Civilisation is seen as an advanced, organised way of life.
(b) It instilled a way of life that could be considered as an adaptation to particular environmental and cultural contents.

Social 9th Guide Question 2.
Name the important early Civilizations.
Answer:
The Egyptian, Mesopotamian, the Chinese and the Indus were the important Civilisations.

Samacheer Kalvi 9th Social Science Book Pdf Question 3.
What supported the livelihood of a large number of people?
Answer:
The surplus food production by the farmers in the fertile regions supported the livelihood of. a large number of people.

Samacheer Kalvi 9th Social Science Book Answers Question 4.
What happened when civilization began to take shape?
Answer:
As civilizations began to take shape, huge buildings were built, the art of writing developed and science and technology contributed to the betterment of society.

2. Features of Egyptian Civilization:
Question 1.
Who built the pyramids and why?
Answer:
The pyramids are massive monuments built by Egyptians as tombs of mourning to the Pharaohs. The Great Pyramids near Cairo are known as the Gizza Pyramids.

Question 2.
What is the process of mummification?
Answer:
(a) The Egyptians had the tradition of preserving the dead bodies using Natron salt;, a combination of sodium carbonate and sodium bicarbonate.
(b) The preservation process is called mummification.
(c) After 40 days when the salt absorbed all the moisture, the body was filled with sawdust and wrapped in strips of linen clothes and covered with fabric. The body was stored in a stone coffin called sarcophagus.

Question 3.
What is the belief system of ancient Egyptians?
Answer:
(a) The Egyptians believed in life after death.
(b) Egyptian practiced polytheism. They worshipped many Gods.

Question 4.
What is the importance of great sphinx?
Answer:
The Great sphinx of Giza is a massive limestone image of a lion with a human head.
It is dated to the time of Pharaoh Khaffe. It is one of the largest sculptures of the world and measures seventy three metres in length and twenty metres in height.

VII. Answer the following in detail.

Question 1.
Define the terms Hieroglyphics and Cuneiform with their main features.
Answer:
Hieroglyphics: Egyptians are well known for their writing system. Their form of writing is known as hieroglyphic. Hieroglyphic was used in the inscriptions on seals and other objects. The heretic, an another form of writing, was used for common purposes. This form of writing used a pictogram-based system. It was developed around 3000 BCE and many texts and books were written using this script.

The Egyptian writing system was deciphered by the French scholar, Francois Champollion (1822 CE). He used the Rosetta stone, a trilingual inscription, for deciphering the script. This inscription, which was written in Hieroglyphic, Demotic and Greek,- was taken to France by Napoleon and from there it was taken to England. Now this inscription is on display in the British Museum London.

Cuneiform: Cuneiform is the Sumerian writing system. The shape of the letter is in the form of wedge and hence it is called cuneiform. Evolving around 3000 BCE, it is one of the earliest scripts of the world. The epic of Gilgamesh was written in this script. They used this script for commercial transactions and writing letters and stories. The clay tablets contain loads of information on the Sumerian civilization.

Question 2.
To what extent is the Chinese influence reflected in the fields of philosophy and literature.
Answer:
(i) Chinese poets and philosophers such as Lao Tze, Confucius, Mencius, Mo Ti (Mot Zu) and Tao Chien (365-427 CE) contributed to the development of Chinese civilization. Sun-Tzu, a military strategist, wrote the work called Art of War.

(ii) The Spring and Autumn Annals is the official chronicle of the state at the time. The Yellow Emperor’s Canon of Medicine is considered China’s earliest written book on medicine. It was codified during the time of Han Dynasty.

(iii) Lao Tze (c. 604-521 BCE) was the master archive keeper of Chou state. He was the founder of Taoism. He argued that desire is the root cause of all evils.

(iv) Confucius (551 —497 BCE) was famous among the Chinese philosophers. He was a political reformer. His name means Kung the master.

(v) He insisted on cultivation of one’s own personal life. He said, “If personal life is cultivated, family life is regulated; and once family life is regulated, national life is regulated.

(vi) Mencius (372-289 BCE) was another well-known Chinese philosopher. He travelled throughout China and offered his counsel to the rulers.

Student Activities

Question 1.
Mark the areas of Bronze Age civilization on the world map.
Answer:
The Bronze Age period: 3300 – 1200 B.C.
The location: China
(i) The Congshan people of China lived from around 300 B.C. to 2400 B.C. in China
(ii) Around 2300 B.C. in Europe.
The teacher can help the students to locate the place on the world map.

Question 2.
Prepare a chart on the pyramids and the mummies.
Answer:
Prepare a chart on the Pyramids and the Mummies with guidelines from the Textbook and the Internet. Students can be divided into groups & prepare the charts on the Pyramids and Mummies.

Question 3.
Collect the pictures of the seals and the pottery of Indus people.
Answer:
Download the pictures & collect them.
Assignment with teacher’s guidance
(i) Prepare a hand out comparing the ancient world civilizations.
Reference: Textbook & Internet
(ii) Prepare a scrap book collecting pictures on Indus civilization from website.
Students can collect pictures on Indus Civilization and paste it in their scrap book.

Ancient Civilisations Additional Questions

I. Choose the correct answer.

Question 1.
……………… times were mostly egalitarian in nature.
(a) Mesolithic
(b) Neolithic
(c) Palaeolithic
(d) Iron Age
Answer:
(a) Mesolithic

Question 2.
……………… became intimately connected with the Sangam Age Tamitagam by the Sea route.
(a) Romans
(b) Persians
(c) Egyptians
(d) Greeks
Answer:
(c) Egyptians

Question 3.
…………… were the contemporaries of the people of Indus and Egyptian civilisation.
(a) The Egyptians
(b) The Sumerians
(c) The Romans
(d) The Indus people
Answer:
(b) The Sumerians

Question 4.
……………. did not give much attention to the life after death.
(a) The Sumerians
(b) The Babylonians
(c) The Egyptians
(d) The Chinese
Answer:
(a) The Sumerians

Question 5.
The yellow river is known as the sorrow of …………
(a) Egypt
(b) Rome
(c) China
(d) Iraq
Answer:
(c) China

Question 6.
(i) The Egyptian king was known as the Pharaoh.
(ii) The preserved dead body is called the mummy..
(iii) The Egyptians have no belief in life after death.
(iv) Paprus was used for making paper.
(a) (i) is incorrect
(b) (ii) is incorrect
(c) (iii) is correct
(d) (iv) is incorrect
Answer:
(c) (iii) is correct

Question 7.
(i) Egypt was called as the Gift of Nile
(ii) Egyptians practiced polytheism
(iii) Thoth is the God of death
(iv) Cuneiform is Egyptian way of writing
(a) (i) & (ii) are incorrect
(b) (iii) & (iv) are correct
(c) (iii) is correct
(d) (i) & (ii) are correct
Answer:
(d) (i) & (ii) are correct

Question 8.
Confucius was famous among the Chinese Philosophers.
(a) Lao Tze
(b) Confucius
(c) Mencius
(d) Sun Tzu
Answer:
(b) Confucius

II. Fill in the blanks.

1. The river Nile originates in ……………
2. The preserved dead body is called …………….
3. …………. was the God of writing and learning.
4. The city of Akkad later became the city of ……………. a commercial and cultural centre of largest Asia.
5. …………. is perhaps the oldest written epic on earth.
6. ………….. was the popular ruler of ten late (or) neo Assyrian Empire.
7. ………….. was the first military power in History.
8. Agriculture was the main occupation of the …………..
9. The ……………. refers to the large collection of terrcotta warrior images found in China.
10. The Indus Valley civilization is also known as ………….. civilisation.
11. The Indus people worshipped …………… trees.
12. The Harappans had close trade link with the ………………
Answers:
1. Lake Victoria
2. the mummy
3. Thoth
4. Babylon
5. The Epic of Gilgamesh
6. Ashurbanipal
7. Assyrian Empire
8. Mesopotamians
9. The Terracotta Army
10. Harappan
11. Pipal
12. Mesopotamians

III. Find out the correct statement.

Question 1.
(a) The oldest civilisation in Mesopotamia belonged to the Sumerians.
(b) The Sumerians believed to have originated from Central Asia.
(c) Sargon and his descendants ruled Mesopotamia for more than 200 years.
(d) The Akkadians dominated Sumeria briefly from 2500 BCE to 2450 BCE.
Answer:
(a) correct
(b) correct
(c) incorrect
(d) incorrect

Question 2.
(a) The Harappans used painted Pottery.
(b) The Harappans have no knowledge about weights and measures.
(c) The Indus people buried the dead.
(d) The Indus civilization completely disappeared.
Answer:
(a) correct
(b) incorrect
(c) correct
(d) incorrect

IV. Match the following.

Answer:
1. (d)
2. (a)
3. (e)
4. (b)
5. (c)

V. Answer the following briefly.

Question 1.
The term civilization is used to distinguish the Urban society from early forms of societies. Why?
Answer:

• The Urban society practised crafts, engaged in trade and exchange, adopted science and technology and formed political organisation.
• Hence the term civilisation is used to distinguish them from the early forms of societies.

Question 2.
What is Egyptian civilization known for?
As one of the oldest civilizations, the Egyptian civilisation is known for its monumental architecture, art, sciences and crafts at a very early age.

Question 3.
Mention the contributions of the Mesopotamian civilization.
Answer:

1. The invention of the potter’s wheel is credited to the Sumerians.
2. They developed the calender system of 360 days and divided a circle into 360 units.
3. The Cuneiform system of writing was their contribution.
4. The Hammurabi’s law code was another legacy of the Mesopotamians.

Question 4.
Write a short note on “Silk Road” in China.
Answer:

1. The greatest of the Han emperors Wu Ti (Han Wu the Great, 141 to 87 BCE) sent Zhang Qian as emissary to the West in 138 BCE.
2. It paved the way for the opening of the Silk Road in 130 BCE to encourage trade activities.
3. Because of the Silk Road and the resultant trade connections, China benefitted immensely during the rule of Emperor Zhang.

Question 5.
The Indus civilisation is known as Harappan civilisation rather than Indus. Why?
Answer:

• The Indus valley civiliSation is also known as the Harappan civilisation since Harappa was the first site to be discovered.
• This civilisation is known as Harappan civilisation rather than Indus valley civilisation, since it extended beyond the Indus river valley.

VI. Answer all the questions given under each caption.

Question 1.
The Babylonians.
(i) Who were the Babylonians?
Answer:
The Semitic people called Amorites who moved from the Arabian desert into Mesopotamia, were known as Babylonians

(ii) Name the oldest written epic on earth. .
Answer:
The Epic of Gilgamesh is the oldest written epic on earth.

(iii) Who was a great law-maker?
Answer:
Hammurabi the sixth king of Babylon was a great law-maker.

(iv) What was the previous name of Babylon?
Answer:
The previous name of Babylon was the city of Akkad.

Question 2.
Indus civilisation.
(i) What is the other name of Indus civilisation?
Answer:
The Harappan civilisation.

(ii) What did the Harappans use for construction?
Answer:
The Harappans used baked and unbaked bricks and stones for construction.

(iii) What kind of potterys were used by the Harappans?
Answer:
The Harappans used painted potteries. Their potteries have a deep red slip and black paintings.

(iv) What was their belief for the dead?
Answer:
The Indus people buried the dead. Burials were done elaborately.

VII. Answer the following in detail.

Question 1.
How did the Egyptians excel in art and architecture? Explain.
Answer:

1. The Egyptians excelled in art and architecture. Their writing is also, a form of art.
2. Numerous sculptures, painting and carvings attest to the artistic skills of Egyptians.
3. The pyramids are massive monuments built as tombs of mourning to the Pharaohs.
4. The great pyramids near Cairo are known as the Giza Pyramids.
5. Pyramids are considered to be one of the wonders of the world, and they were built between 2575 and 2465 BCE.
6. These monuments display the engineering, architectural and human resource management skills of the Egyptians.
7. The Great Sphinx of Giza is a massive limestone image of a lion with a human head.
8. It is dated to the time of Pharaoh Khaffe. It is one of the largest sculptures of the world
   and measures seventy three metres in length and twenty metres in height.

Question 2.
Compare the Indus Civilization with Tamil Civilization.
Answer:
(i) The similarity of the graffiti found on the megalithic burial pots of South India with the Indus script and the identical place names of Tamil Nadu and Indus region of Pakistan are presented as arguments to establish the relationship between the Indus civilisation and Tamil culture.

(ii) Researchers like Father Henry Heras, Asko Parpola and Iravatham Mahadevan find similarity between the Indus script and the Dravidian/Tamil language.

(iii) Archaeological evidence points out that several groups of people have been living in Tamil Nadu and South India continuously from the Mesolithic period.

(iv) A few groups from the Indus region might have migrated into southern India.

(v) Some of the ideas and technologies of the Indus civilisations had reached South India in the Iron Age.

(vi) The camelian beads, shell bangles and bronze mirrors found in the Megalithic/Early Historic sites of Tami Nadu were first introduced by the people of the Indus civilisation. More research is needed to arrive at any definite conclusion in this matter.

(vii) The towns of ancient Tamilagam such as Arikkamedu, Uraiyur and Keezhadi that flourished are part of the second urbanisation of India and these towns are much different from the Indus cities. These towns emerged approximately 1,200 years after the decline of the Indus civilisation.

Students can Download Tamil Chapter 1.3 தமிழரின் கப்பற்கலை Questions and Answers, Summary, Notes Pdf, Samacheer Kalvi 7th Tamil Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 2 Chapter 1.3 தமிழரின் கப்பற்கலை

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
தமிழர்கள் சிறிய நீர்நிலைகளைக் கடக்கப் பயன்படுத்தியது ……..
அ) கலம்
ஆ) வங்கம்
இ) நாவாய்
ஈ) ஓடம்
Answer:
ஈ) ஓடம்

Question 2.
தொல்காப்பியம் கடற்பயணத்தை …………………….. வழக்கம் என்று கூறுகிறது.
அ) நன்னீ ர்
ஆ) தண்ணீ ர்
இ) முந்நீர்
ஈ) கண்ணீ ர்
Answer:
இ)
முந்நீர்)

Question 3.
கப்பலை உரிய திசையில் திருப்புவதற்குப் பயன்படும் கருவி ……………….
அ) சுக்கான்
ஆ) நங்கூரம்
இ) கண்ண டை
ஈ) சமுக்கு
Answer:
அ) சுக்கான்

கோடிட்ட இடங்களை நிரப்புக

Question 1.
கப்பல் கட்டுவதற்குப் பயன்படும் மர ஆணிகள் ……….. …….. என அழைக்கப்படும்.
Answer:
தொகுதி

Question 2.
கப்பல் ஓரிடத்தில் நிலையாக நிற்க உதவுவது …
Answer:
நங்கூரம்

Question 3.
இழைத்த மரத்தில் காணப்படும் உருவங்கள் எனக் குறிப்பிடப்படும்.
Answer:
கண்ணடை

பொருத்துக –

1. எரா – திசைகாட்டும் கருவி
2. பருமல் – அடிமரம்
3. மீகாமன் – குறுக்கு மரம்
4. காந்த ஊசி – கப்பலைச் செலுத்துபவர்
Answer:
1. எரா – அடிமரம்
2. பருமல் – குறுக்கு மரம்
3. மீகாமன் – கப்பலைச் செலுத்துபவர்
4. காந்த ஊசி – திசைகாட்டும் கருவி

தொடர்களில் அமைத்து எழுதுக

1. நீரோட்டம்
கடல் நீரோட்டங்களின் திசையை தமிழர்கள் தம் பட்டறிவால் நன்கு அறிந்திருந்தனர்.

2. காற்றின் திசை
காற்றின் திசையை அறிந்து கப்பல்களைச் செலுத்தும் முறையைத் தமிழர்கள் நன்கு அறிந்திருந்தனர்.

3. வானியல் அறிவு
கப்பல் ஓட்டும் மாலுமிகள் சிறந்த வானியல் அறிவைப் பெற்றிருந்தனர்.

4. ஏற்றுமதி
பூம்புகார் துறைமுகத்திலிருந்து கப்பல்கள் மூலம் பொருள்கள் ஏற்றுமதியும் இறக்குமதியும் செய்யப்பட்டன.

குறுவினா

Question 1.
தோணி என்னும் சொல்லின் பெயர்க்காரணத்தைக் கூறுக.
Answer:

• எடைக் குறைந்த பெரிய மரங்களின் உட்பகுதியைக் குடைந்து எடுத்துவிட்டுத்த தோணியாகப் பயன்படுத்தினர் தமிழர்கள்.
• உட்பகுதி தோண்டப்பட்டவை என்பதால் அவை தோணிகள் எனப்பட்டன.
• சிறிய நீர்நிலைகளைக் கடக்கத் தமிழர்கள் தோணியைப் பயன்படுத்தினர்.

Question 2.
கப்பல் கட்டும்போது மரப்பலகைகளுக்கு இடையே தேங்காய் நார் அல்லது பஞ்சு வைப்பதன் நோக்கம் என்ன?
Answer:

• மரங்களையும் பலகைகளையும் ஒன்றோடு ஒன்று இணைக்கும் போது அவற்றுக்கு இடையே தேங்காய் நார், பஞ்சு ஆகியவற்றில் ஒன்றை வைத்து நன்றாக இறுக்கி ஆணிகளை அறைந்தனர்.
• சுண்ணாம்பையும் சணலையும் கலந்து அரைத்து அதில் எண்ணெய் கலந்து கப்பலின் அடிப்பகுதியில் பூசினர்.
• இதனால் கப்பல்கள் பழுதடையாமல் நெடுங்காலம் உழைத்தன.

Question 3.
கப்பல் உறுப்புகள் சிலவற்றின் பெயர்களைக் கூறுக.
Answer:
கப்பலின் உறுப்புகள் :

• கப்பல் பல்வேறு வகையான உறுப்புகளை உடையது.
• எரா, பருமல், வங்கு , கூம்பு, பாய்மரம், சுக்கான், நங்கூரம் போன்றவை கப்பலின் உறுப்புகளுள் சிலவாகும்.

சிறுவினா

Question 1.
சிறிய நீர்நிலைகளையும் கடல்களையும் கடக்கத் தமிழர்கள் பயன்படுத்திய ஊர்திகளின் பெயர்களை எழுதுக.
Answer:

• தமிழர்கள் தோணி, ஓடம், படகு, புணை, மிதவை, தெப்பம் போன்றவற்றைச் சிறிய நீர்நிலைகளைக் கடக்கப் பயன்படுத்தினர்.
• கலம், வங்கம், நாவாய் முதலியவை அளவில் பெரியவை.
• இவற்றைக் கொண்டு தமிழர்கள் கடல் பயணம் மேற்கொண்டனர்.

Question 2.
பண்டைத் தமிழரின் கப்பல் செலுத்தும் முறை பற்றி எழுதுக.
Answer:
கப்பலைச் செலுத்தும் முறை :

• காற்றின் திசையை அறிந்து கப்பல்களைச் செலுத்தும் முறையைத் தமிழர்கள் நன்கு அறிந்திருந்தனர்.
• இவ்வுண்மையை வெண்ணிக்குயித்தியார் தம் புறப்பாடலில் குறிப்பிடுகிறார். “நளியிரு முந்நீர் நாவாய் ஓட்டி வளி தொழில் ஆண்ட உரவோன் மருக – வெண்ணிக் குயத்தியார்
• கடலில் காற்று வீசும் திசை, கடல் நீரோட்டங்களின் திசை ஆகியவற்றைத் தமிழர்கள் தம் பட்டறிவால் நன்கு அறிந்து அவற்றுக்கு ஏற்ப உரிய காலத்தில் சரியான திசையில் கப்பலைச் செலுத்தினர்.

Question 3.
கப்பல் பாதுகாப்பானதாக அமையத் தமிழர்கள் கையாண்ட வழிமுறைகள் யாவை?
Answer:

• பெருந்திரளான மக்களையும் பொருள்களையும் ஏற்றிச் செல்லும் வகையில் பெரிய கப்பல்களைத் தமிழர் உருவாக்கினர்.
• நீண்ட தூரம் கடலிலேயே செல்ல வேண்டி இருந்ததால் கப்பல்களைப் பாதுகாப்பனவையாகவும் வலிமைமிக்கவையாகவும் உருவாக்கினர்.
• கப்பல் கட்டுவதற்கு உரிய மரங்களைத் தேர்ந்தெடுப்பதில் தமிழர்கள் மிகுந்த கவனம் செலுத்தினர். தண்ணீரால் பாதிப்பு அடையாத மரங்களையே கப்பல் கட்டப் பயன்படுத்தினர். நீர்மட்ட வைப்பிற்கு வேம்பு, இலுப்பை, புன்னை, நாவல் போன்ற மரங்களைப் பயன்படுத்தினர்.
• பக்கங்களுக்குத் தேக்கு, வெண்தேக்கு போன்ற மரங்களைப் பயன்படுத்தினர். நீளம், அகலம், உயரம் ஆகியவற்றைச் சரியான முறையில் கணக்கிட்டுக் கப்பலை உருவாக்கினர்.
• மரங்களையும் பலகைகளையும் இணைக்கும் போது அவற்றுக்கு இடையே தேங்காய் நார் அல்லது பஞ்சு இவற்றில் ஒன்றை வைத்து இறுக்கி ஆணிகளை அறைந்தனர். மரத்தால் ஆன ஆணிகளையே பயன்படுத்தினர்.

 

சிந்தனை வினா

Question 1.
இக்காலத்தில் மக்கள் வெளிநாடுகளுக்குச் செல்வதற்கு கடற்பயணத்தைப் பெரிதும் மேற்கொள்ளாதது ஏன் எனச் சிந்தித்து எழுதுக.
Answer:
தற்போது விமானம் கண்டுபிடிக்கப்பட்ட பிறகு, கடற்வழி பயணம் விரும்பத்தகாத ஒன்றாக மாறிவிட்டது. காலதாமதம் காரணமாக தற்பொழுது கடல் வழி பயணத்தைத் தவிர்த்து வான்வெளிப் பயணத்தை ஏற்றனர். அதன் காரணமாகவே கடற்வழிப் பயணத்தை பெரிதும் மேற்கொள்ளுதலைத் தவிர்த்தனர்.

கற்பவை கற்றபின்

Question 1.
பலவகையான கப்பல்களின் படங்களைச் சேகரித்துப் படத்தொகுப்பு ஒன்று உருவாக்குக.
Answer:
பாய்மரக்கப்பல்கள் :

• காற்றின் உதவியால் செலுத்தப்படும் கப்பல்கள் பாய்மரக்கப்பல்கள் எனப்பட்டன.
• பெரிய பாய்மரம், திருக்கைத்திப் பாய்மரம், காணப் பாய்மரம், கோசுப் பாய்மரம் போன்ற பலவகையான பாய்மரங்களைத் தமிழர் பயன்படுத்தினர்.

நீர்மூழ்கிக் கப்பல்கள் :

• இவை நீரில் மூழ்கியபடியே வெகு தொலைவு செல்லக்கூடிய நீர் ஊர்தி ஆகும்.
• நீர்மூழ்கிக் கப்பல்கள் போரில், பகைவர் கப்பல்களைத் தாக்குதல், முற்றுகையை முறியடித்தல், நீரில் இருந்தபடியே நிலப்பகுதியைத் தாக்குதல், இரகசியமாகச் சிறப்புப் E படைகளை முக்கியப் பகுதிகளில் இறக்கி வியூகம் அமைத்தல் ஆகிய பல பணிகளைச் ஓ செய்ய வல்லவை.

பயணிகள் கப்பல்கள் :

• பயணிகள் கப்பல்களின் அளவு சிறிய ஆற்று படகுகளில் இருந்து மிக பெரிய கப்பல்கள் வரை இருக்கும்.
• ஓர் இடத்திலிருந்து மற்றோர் இடத்திற்குப் பயணிகளைக் கொண்டு செல்லும் கப்பல்கள்.
• குறுகிய பயணங்களுக்குப் பயணிகள் மற்றும் வாகனங்களை ஏற்றிச் செல்லும் கப்பல்கள், இன்பப் பயணங்கள் மேற்கொள்ளப்படும் கப்பல்கள் இந்த வகையில் அடங்கும்.

எண்ணெய்க் கப்பல்கள் :

திரவ பெட்ரோலிய வாயு , திரவ இயற்கை எரிவாயு , வேதிப்பொருட்கள் ஆகியவற்றைச் சுமந்து செல்லும் கப்பல்கள்.

சரக்குக் கப்பல்கள் :

• சரக்குக் கப்பல் என்பது ஒரு துறைமுகத்தில் இருந்து இன்னொரு துறைமுகத்துக்குச் சரக்கு மற்றும் பல்வேறு  பொருட்களை எடுத்துச் செல்லும் கப்பலாகும்.
• சரக்குக் கப்பல்கள் பொதுவாக அவற்றின் தேவைக்கு ஏற்ப வடிவமைப்புச் செய்யப்படுகின்றன.
• பொருட்களை ஏற்றி இறக்கும் வசதிக்காக அவற்றில் பாரந்தூக்கிகளும் பொருத்தப்படுவதுண்டு.

Question 2.
தரைவழிப்பயணம், கடல்வழிப்பயணம், வான்வழிப்பயணம் ஆகியவை குறித்து வகுப்பறையில் கலந்துரையாடுக.
Answer:
இடம் : அரசு உயர்நிலைப்பள்ளி.

நடிகர்கள் : கபிலன், மணிவண்ண ன், கயல்விழி, தேன்மொழி, தமிழாசிரியர்,

விளக்கம் : ஏழாம் வகுப்பு மாணவர்கள் வகுப்பறையில் கலந்துரையாடுகின்றனர். உடன் தமிழாசிரியர் தமிழ்வேந்தன் அவர்கள் உரையாடலில் உடன் பங்கேற்கின்றார்.

தமிழ்வேந்தன் : மாணவச் செல்வங்களே! இன்று நாம் எதைப் பற்றி உரையாடப் போகிறோம் தெரியுமா?

கயல்விழி : சொல்லுங்கள் ஐயா! தெரிந்து கொள்கிறோம்.

தமிழ்வேந்தன் : கோடை விடுமுறையில் நீங்கள் ஊர்ப்பயணம் போவீர்களா?

மணிவண்ணன் : நிச்சயமாக

தமிழ்வேந்தன் : எங்கு போவீர்கள்?

தேன்மொழி : எங்கள் பாட்டி வீட்டிற்குப் போவேன் ஐயா!

தமிழ்வேந்தன் : எப்படி போவீர்கள்?

தேன்மொழி : பேருந்தில் பயணம் செய்வேன் ஐயா!

தமிழ்வேந்தன் : அதாவது தரைவழிப் பயணம் அப்படித்தானே!

தேன்மொழி : ஆமாம் ஐயா!

தமிழ்வேந்தன் : பேருந்து, சிற்றுந்து, மகிழுந்து, இரு சக்கரவாகனம் இவையாவும் தரைவழிப்பயணம் தானே!

கயல்விழி : ஆமாம் ஐயா!

தமிழ்வேந்தன் : தரைவழிப்பயணம் செல்வதில் சிக்கல்கள் ஏதாவது உண்டா?

கபிலன் : ஒன்றா இரண்டா எடுத்துச் சொல்ல ! பெருகி வரும் மக்கள்தொகையில் நகர்புறம், கிராமப்புறம் எங்கும்போக்குவரத்து இடையூறுகள் அதிகம். வாகனங்கள் மக்கள் தொகைப் போல பன்மடங்கு பெருகி உள்ளன ஐயா!

தமிழ்வேந்தன் : இதனைக் கட்டுப்படுத்த என்ன வழி?

கபிலன் : சாலைகளை விரிவாக்கம் செய்ய வேண்டும் ஐயா!

தமிழ்வேந்தன் : நல்ல கருத்து இது வரவேற்கத்தக்கது. இதைத்தவிர வேறு பயணம் ஏதாவது உண்டா ?

தேன்மொழி : உண்டு ஐயா ! கடல்வழிப் பயணம்

தமிழ்வேந்தன் : ஆதியில் பெருந்திரளான மக்கள் தங்களையும் தங்கள் உடைமைகளையும் ஏற்றிச் செல்ல கப்பலைமலைபோல நம்பி இருந்தனர்.

தேன்மொழி : அது மட்டுமல்ல ஐயா, நீண்ட தூரம் கடலிலேயே செல்ல வேண்டி இருந்தால் கப்பல் பயணம் பாதுகாப்பானது மற்றும் மனதிற்கு மகிழ்ச்சி தரக்கூடியது.

கயல்விழி : போக்குவரத்து இடையூறு கப்பல் பயணத்தில் இல்லை
ஐயா ! கடல் வழிப்பயணத்தில் துரிதமாக போக வேண்டிய இடத்திற்கு போய்ச் சேரலாம். மனதிற்கும் உடலுக்கும் புது தெம்பு தருவது இந்த ..
கடல் வழிப் பயணம்தான்.

தமிழ்வேந்தன் : நல்லது. இதைத் தவிர வேறு பயணம் ஏதாவது உண்டா ?

மணிவண்ண ன் : தரைவழிப் பயணம், கடல் வழிப்பயணம் இரண்டையும் பின்னுக்குத்
தள்ளும் பயணம் வான் வழிப் பயணம்தான்.

தமிழ்வேந்தன் : எப்படி?

மணிவண்ணன் : கட்டணச் செலவு கூடுதலாக இருந்தாலும் நாம் போய் சேர வேண்டிய இடத்திற்குச் சில மணிநேரங்களில் அலுங்காமல் குலுங்காமல் போய் சேரலாம். இதனால் நேரம் மிச்சம். சுகமான பயணம் ஐயா!

தமிழ்வேந்தன் : உண்மைதான்! இருப்பினும் நடுத்தர மக்களுக்கும், ஏழை எளிய மக்களுக்கும் சாதகமாக இருக்கும் பயணம் தரைவழிப் பயணம் மட்டும் தான். காத்திருப்பதிலும் ஒரு சுகம்தான்! கூட்ட நெரிசலில் பயணிப்பதும் ஒரு தனி சுகம்தான் !

(பள்ளி மணி ஒலிக்கிறது)

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
முந்நீர் வழக்கம் என்று தொல்காப்பியம் எதைக் குறிப்பிடுகிறது?
அ) கடற்பயணம்
ஆ) வான்வழிப்பயணம்
இ) தரைவழிப்பயணம்
ஈ) இவை ஏதுமில்லை
Answer:
அ) கடற்பயணம்

Question 2.
பூம்புகார் துறைமுகத்தில் இருந்து கப்பல்கள் மூலம் பொருள்களை ஏற்றுமதி இறக்குமதி செய்யப்பட்டன. இக்கூற்றை மெய்ப்பித்த நூல் எது?
அ) அகநானூறு
ஆ) பட்டினப்பாலை
இ) கலித்தொகை
ஈ) குறுந்தொகை
Answer:
ஆ) பட்டினப்பாலை

Question 3.
“உலகு கிளர்ந்தன்ன உருகெழு வங்கம்” – இத்தொடர் இடம் பெற்ற நூல் எது?
அ) அகநானூறு
ஆ) புறநானூறு
இ) பட்டினப்பாலை
ஈ) மதுரைக்காஞ்சி
Answer:
அ) அகநானூறு

Question 4.
பலவகையான கப்பல்களின் பெயர்கள் குறிப்பிடப்பட்டுள்ளன என்பதனை மெய்ப்பிக்கும் நூல் எது?
அ) சேந்தன் திவாகரம்
ஆ) ஏலாதி
இ) திருமந்திரம்
ஈ) சீவகசிந்தாமணி
Answer:
அ) சேந்தன் திவாகரம்

Question 5.
தமிழர்கள் பயன்பாட்டில் உருவான கப்பல்கள் பழுதடையாமல் நெடுங்காலம் உழைத்தன. இம்முறையைக் கண்டு வியந்தவர்.
அ) மார்க்கோபோலோ
ஆ) ஆம்ஸ்ட்ராங்
இ) கலிலியோ
ஈ) இவர்களில் யாருமில்லை
Answer:
அ) மார்க்கோபோலோ

Question 6.
கப்பலின் முதன்மையான உறுப்பாகிய அடிமரம் ……………. எனப்படும்
அ) எரா
ஆ) பருமல்
இ) சுக்கான்
ஈ) வங்கு
Answer:
அ) எரா

Question 7.
குறுக்கு மரத்தை ………….. என்பர்
அ) நங்கூரம்
ஆ) சமுக்கு
இ) கூம்பு
ஈ) பருமல்
Answer:
ஈ) பருமல்

Question 8.
கப்பல் செலுத்துபவரை ….. …… என அழைப்பர்.
அ) மாலுமி
ஆ) கம்மியர்
இ) கனரக வாகனம் ஓட்டுநர்
ஈ) இவர்களில் யாருமில்லை
Answer:
அ) மாலுமி

கோடிட்ட இடத்தை நிரப்புக

Question 1.
கப்பலை அழைக்கும் விளக்கு என்னும் பொருளுக்கு ………………… என்று பெயர்.
Answer:
கலங்கரை விளக்கம்

Question 2.
கடலில் செல்லும் கப்பல்களுக்குத் துறைமுகம் இருக்கும் இடத்தைக் காட்டுவதற்காக அமைக்கப்படுவது ………………
விடை:
கலங்கரை விளக்கம்

Question 3.
கலம் என்பதன் பொருள் …………
Answer:
கப்பல்

Question 4.
கரைதல் என்பதன் பொருள் ……………
Answer:
அழைத்தல்

Question 5.
கப்பலைச் செலுத்துவதற்கும், உரிய திசையில் திருப்புவதற்கும் பயன்படும் முதன்மையான கருவி ……………………. எனப்ப டும்.
Answer:
சுக்கான்

Question 6.
கப்பலை நிலையாக ஓரிடத்தில் நிறுத்தி வைக்க உதவும் உறுப்பு …………………… ஆகும்.
Answer:
நங்கூரம்]

Question 7.
மரத்தின் வெட்டப்பட்ட பகுதியை ……………. என்பர்.
Answer:
வெட்டுவாய்

Question 8.
……… என்பது இழைத்த மரத்தில் காணப்படும் உருவங்கள் ஆகும்.
Answer:
கண்ணடை

Question 9.
நீளம், அகலம், உயரம் ஆகியவற்றை சரியான முறையில் கணக்கிட்டுக் கப்பலை உருவாக்கினர். இவற்றை ……………………என்னும் நீட்டலளவையால் கணக்கிட்டனர்.
Answer:
தச்சு முழம்

Question 10.
பெரிய படகுகளில் முன் பக்கம் ………………………. என அழைக்கப்பட்டன.
Answer:
கரிமுக அம்பி, பரிமுக அம்பி

Question 11.
கப்பல்கள் பழுதடையாமல் நெடுங்காலம் உழைக்க மரத்தால் ஆன ஆணிகளைப் பயன்படுத்தினர் இந்த ஆணிகளை……………….. என்பர்.
Answer:
தொகுதி

Question 12.
பாய்மரங்களைத் ……………….. பயன்படுத்தினர்.
Answer:
தமிழர்]

Question 13.
தமிழர் கட்டிய கப்பல்களை ஐம்பது ஆண்டுகள் வரை பழுது பார்க்க வேண்டிய அவசியமில்லை எனக் கூறியவர் …………………
Answer:
வாக்கர் என்னும் ஆங்கிலேயர்

விடையளி :

Question 1.
பாய்மரக் கப்பல்கள் என்றால் என்ன?
Answer:
காற்றின் உதவியால் செலுத்தப்படும் கப்பல்கள் பாய்மரக் கப்பல்கள் எனப்பட்டன.

Question 2.
பாய்மரங்களின் பல வகைகளைக் குறிப்பிடுக.
Answer:
பெரிய பாய்மரம், திருக்கைத்திப் பாய்மரம், காணப் பாய்மரம், கோசுப் பாய்மரம் போன்ற ல பலவகையான பாய்மரங்களைத் தமிழர் பயன்படுத்தினர்.

Question 3.
பாய்மரங்களைக் கட்டும் கயிறுகளின் வகைகளை வரிசைப்படுத்துக.
Answer:

1. ஆஞ்சான் கயிறு
2. தாம்பாங்கயிறு
3. வேடாங்கயிறு
4. பளிங்கைக் கயிறு
5. மூட்டங்கயிறு
6. இளங்கயிறு
7. கோடிப்பாய்க்கயிறு.

Question 4.
கப்பல் செலுத்துபவரை எவ்வாறு அழைப்பர்?
Answer:
கப்பல் செலுத்துபவரை மாலுமி, மீகாமன், நீகான், கப்பலோட்டி முதலிய பல பெயர்களால் அழைப்பர்.

Question 5.
எரா என்றால் என்ன?
Answer:
கப்பலின் முதன்மையான உறுப்பாகிய அடிமரம் எரா எனப்படும்.

Question 6.
சுக்கான் என்றால் என்ன?
Answer:
கப்பலைச் செலுத்துவதற்கும், உரிய திசையில் திருப்புவதற்கும் பயன்படும் முதன்மையான கருவி சுக்கான் எனப்படும்.

Question 7.
நங்கூரம் எதற்கு பயன்படுகிறது?
Answer:
கப்பலை நிலையாக ஓரிடத்தில் வைக்க உதவும் உறுப்பு நங்கூரம் ஆகும். இது கப்பலை ஓரிடத்தில் நிறுத்துவதற்குப் பயன்படுகிறது.

Question 8.
சமுக்கு என்றால் என்ன?
Answer:

• சமுக்கு என்னும் ஒரு கருவியையும் கப்பலில் பயன்படுத்தினர் என்று கப்பல் சாத்திரம் என்னும் நூல் குறிப்பிடுகிறது.
• இது காந்த ஊசி பொருத்தப்பட்ட திசை காட்டும் கருவியாக இருக்கலாம் என ஆய்வாளர்கள் கருதுகின்றனர்.

Question 9.
மீகாமன் – குறிப்பு வரைக.
Answer:

• மீகாமன் என்பதற்குக் கப்பலைச் செலுத்துபவர் என்பது பொருளாகும்.
• கப்பல் ஓட்டும் மாலுமிகள் சிறந்த வானியல் அறிவையும் பெற்றிருந்தனர்.
• கோள்களின் நிலையை வைத்துப் புயல், மழை போன்றவை தோன்றும் காலங்களையும் கடல் நீர் பொங்கும் காலத்தையும் அறிந்து தகுந்த காலத்தில் கப்பல்களைச் செலுத்தினர்.
• கடலில் காற்று வீசும் திசை, கடல் நீரோட்டங்களின் திசை ஆகியவற்றைத் தமிழர்கள் அறிந்திருந்தனர்.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

10th Maths Exercise 5.1 Samacheer Kalvi Question 1.
Find the area of the triangle formed by the points
(i) (1, -1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:

(ii) (-10, -4), (-8, -1) and (-3, -5)

Exercise 5.1 Class 10 Samacheer Kalvi Question 2.
Detemine whether the sets of points are collinear ?

Solution:
(i)


∴ The given points are collinear

10th Maths Exercise 5.1 Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’.

Solution:
Area 20 sq. units.

8p = 104
p = 13

Ex 5.1 Class 10 Samacheer Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2, 3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and(-4 – a,6 – 2a)
Solution:

10th Maths Coordinate Geometry Exercise 5.1 Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8, 6), (-1, -2) and (-6, -3)
Solution:
(i) (-9, -2), (-8, -4), (2, 2), and (1, -3)

(ii) (-9, 0), (-8, 6), (-1, -2) and (-6, -3)

10th Maths Ex 5.1 Question 6.
Find the value of k, if the area ofa quadrilateral is 28 sq.units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:

10th Maths 5.1 Question 7.
If the points A(-3, 9) , B(a, b) and C(4,-5) are collinear and if a + b = 1, then find a and b.
Solution:

Maths Ex 5.1 Class 10 Samacheer Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the mid-points of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
p (11, 7), Q (13.5, 4), and R (9.5, 4) are the mid points of the sides of a ∆ABC.

Samacheer Kalvi 10th Maths Exercise 5.1Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:
Area of the patio = Area of the quadrilateral ABCD – Area of the swimming pool EFGFI.
Area of the quadrilateral ABCD

10th Standard Maths Exercise 5.1 Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(1, 6) and C(7, -4) has to be painted.
If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:

10th Maths Exercise 5.1 Answers Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG
Solution:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.15

10th Maths Exercise 3.15 Solutions Question 1.
Graph the following quadratic equations and state their nature of solutions,
(i) x² – 9x + 20 = 0
Solution:

Step 1:
Points to be plotted : (-4, 72), (-3, 56), (-2, 42), (-1, 30), (0, 20), (1, 12), (2, 6), (3, 2), (4, 0)
Step 2:
The point of intersection of the curve with x axis is (4, 0)
Step 3:

The roots are real & unequal
∴ Solution {4, 5}

(ii) x² – 4x + 4 = 0

Step 1: Points to be plotted : (-4, 36), (-3, 25), (-2, 16), (-1, 9), (0, 4), (1, 1), (2, 0), (3, 1), (4, 4)
Step 2: The point of intersection of the curve with x axis is (2, 0)
Step 3:

Since there is only one point of intersection with x axis, the quadratic equation x² – 4x + 4 = 0 has real and equal roots.
∴ Solution{2, 2}

(iii) x² + x + 7 = 0
Let y = x² + x + 7
Step 1:

Step 2:
Points to be plotted: (-4, 19), (-3, 13), (-2, 9), (-1, 7), (0, 7), (1, 9), (2, 13), (3, 19), (4, 27)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect with the x-axis.

Step 4:
The roots of the equation are the points of intersection of the parabola with the x axis. Here the parabola does not intersect the x axis at any point.
So, we conclude that there is no real roots for the given quadratic equation,

(iv) x² – 9 = 0
Let y = x² – 9
Step 1:

Step 2:
The points to be plotted: (-4, 7), (-3, 0), (-2, -5), (-1, -8), (0, -9), (1,-8), (2, -5), (3, 0), (4, 7)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect the x-axis.

Step 4:
The roots of the equation are the co-ordinates of the intersecting points (-3, 0) and (3, 0) of the parabola with the x-axis which are -3 and 3 respectively.
Step 5:
Since there are two points of intersection with the x axis, the quadratic equation has real and unequal roots.
∴ Solution{-3, 3}

(v) x² – 6x + 9 = 0
Let y = x² – 6x + 9
Step 1:

Step 2:
Points to be plotted: (-4, 49), (-3, 36), (-2, 25), (-1, 16), (0, 9), (1, 4), (2, 1), (3, 0), (4, 1)
Step 3:
Draw the parabola and mark the co-ordinates of the intersecting points.

Step 4:
The point of intersection of the parabola with x axis is (3, 0)
Since there is only one point of intersection with the x-axis, the quadratic equation has real and equal roots. .
∴ Solution (3, 3)

(vi) (2x – 3)(x + 2) = 0
2x² – 3x + 4x – 6 = 0
2x² + 1x – 6 = 0
Let y = 2x² + x – 6 = 0
Step 1:

Step 2:
The points to be plotted: (-4, 22), (-3, 9), (-2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30)
Step 3:
Draw the parabola and mark the co-ordinates of the intersecting point of the parabola with the x-axis.

Step 4:
The points of intersection of the parabola with the x-axis are (-2, 0) and (1.5, 0).
Since the parabola intersects the x-axis at two points, the, equation has real and unequal roots.
∴ Solution {-2, 1.5}

Question 2.
Draw the graph of y = x² – 4 and hence solve x² – x – 12 = 0
Solution:



Point of intersection (-3, 5), (4, 12) solution of x² – x – 12 = 0 is -3, 4

10th Maths Graph 3.15 Answers Question 3.
Draw the graph of y = x² + x and hence solve x² + 1 = 0.
Solution:

Draw the parabola by the plotting the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12), (4, 20), (5, 30)

To solve: x² + 1 = 0, subtract x² + 1 = 0 from y = x² + x.
x² + 1 = 0 from y = x² + x

Plotting the points (-2, -3), (0, -1), (2, 1) we get a straight line. This line does not intersect the parabola. Therefore there is no real roots for the equation x² + 1 = 0.

10th Maths Guide Graph Question 4.
Draw the graph of y = x² + 3x + 2 and use it to solve x² + 2x + 1 = 0.
Solution:

Draw the parabola by plotting the point (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20), (4, 30).

To solve x² + 2x + 1 = 0, subtract x² + 2x + 1 = 0 from y = x² + 3x + 2

Draw the straight line by plotting the points (-2, -1), (0, 1), (2, 3)
The straight line touches the parabola at the point (-1,0)
Therefore the x coordinate -1 is the only solution of the given equation

10th Graph Exercise 3.15 Solutions Question 5.
Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0. y = x² + 3x – 4
Solution:

Draw the parabola using the points (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14), (4, 24).

To solve: x² + 3x – 4 = 0 subtract x² + 3x – 4 = 0 from y = x² + 3x – 4 ,

The points of intersection of the parabola with the x axis are the points (-4, 0) and (1, 0), whose x – co-ordinates (-4, 1) is the solution, set for the equation x² + 3x – 4 = 0.

10th Maths Graph Answers Question 6.
Draw the graph of y = x² – 5x – 6 and hence solve x² – 5x – 14 = 0.
Solution:

Draw the parabola using the points (-5, 44), (-4, 30), (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10)

To solve the equation x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The co-ordinates of the points of intersection of the line and the parabola forms the solution set for the equation x2 – 5x – 14 = 0.
∴ Solution {-2, 7}

10th Class Maths Graph Pdf Question 7.
Draw the graph of y = 2x² – 3x – 5 and hence solve 2x² – 4x – 6 = 0. y = 2x² – 3x – 5
Solution:

Draw the parabola using the points (-4, 39), (-3, 22), (-2, 9), (-1, 0), (0, -5), (1, -6), (2, -3), (3, 4), (4, 15).

To solve 2x² – 4x – 6 = 0, subtract it from y = 2x² – 3x – 5

Draw a straight line using the points (-2, -1), (0, 1), (2, 3). The points of intersection of the parabola and the straight line forms the roots of the equation.
The x-coordinates of the points of intersection forms the solution set.
∴ Solution {-1, 3}

10th Maths Graph Question 8.
Draw the graph of y = (x – 1)(x + 3) and hence solve x² – x – 6 = 0.
Solution:
y = (x – 1)(x + 3) = x² – x + 3x – 3 = 0
y = x² + 2x – 3

Draw the parabola using the points (-4, 5), (-3, 0), (-2, -3), (-1,-4), (0, -3), (1, 0), (2, 5), (3, 12), (4, 21)

To solve the equation x² – x – 6 = 0, subtract x² – x – 6 = 0 from y = x² – 2x – 3.

Plotting the points (-2, -3), (-1, 0), (0, 3), (2, 9), we get a straight line.
The points of intersection of the parabola with the straight line gives the roots of the equation. The co¬ordinates of the points of intersection forms the solution set.
∴ Solution {-2, 3}