Category: Class 10

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.

All these concepts of Chapter 1 Laws of Motion are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 1 Laws of Motion State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 1 Laws of Motion.

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 English Solutions Poem Chapter 3 I am Every Woman

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I am Every Woman Textual Questions

A. Read the lines and answer the questions.

(i) The summer of life she’s ready to see in spring.
She says, “Spring will come again, my dear Let me care for the ones who ’re near.”
(a) What does the word summer mean here?
(b) How does she take life?
(c) What does she mean by “spring will come again”?
Answer:
(a) Summer here means development.
(b) She takes life optimistically.
(c) By the phrase ‘spring will come again’, the poet means that the future will be better.

(ii) Strong is she in her faith and belief.
“Persistence is the key to everything, ” says she.
(a) What is she strong about?
(b) How does she deal with the adversities in life?
Answer:
(a) She is strong about her faith and belief.
(b) She is strong in her faith and belief and is determined while dealing with the adversities in life.

(iii) Despite the sighs and groans and moans,
She’s strong in her faith, firm in her belief!.
(a) Is she complaining about the problems of life?
(b) Pick out the words that show her grit.
Answer:
(a) No she is not complaining about the problems of life.
(b) The words that show her grit are strong and firm.

(iv) Don’t ever try to saw her pride, her self-respect.
She knows how to thaw you, saw you – so beware!
(a) What do the words thaw and saw mean here?
(b) What is the tone of the author?
Answer:
(a) Thaw and saw means that she will reduce you to nothing.
(b) The tone of the poetess is a caution about careful intervention.

(v) She’s today’s woman. Today’s woman dear.
Love her, respect her, keep her near…
(a) Describe today’s woman according to the poet.
(b) How should a woman be treated?
Answer:
(a) Today’s woman is a woman born with determination, ready to take risks in life and is strong in her faith and beliefs.
(b) A woman should be treated with love and respect.

Additional Questions

(i) “A woman is beauty innate,
A symbol of power and strength.
She puts her life at stake,
She’s real, she’s not fake!”
(a) What is the rhyme scheme of the above stanza?
(b) Pick out the rhyming words.
(c) What is meant by ‘innate’?
(d) Explain the line, ‘She puts her life at stake’.
(e) What do you understand by the term, ‘not a fake’?
Answer:
(a) ‘abcc’ is the rhyme scheme of the above stanza.
(b) The rhyming words are stake and fake.
(c) ‘Innate’ means inborn.
(d) The line is an indication that a woman is willing to take risks in life.
(e) By ‘not a fake’ one can understand that a woman is genuine and real.

(ii) “Let me care for the ones who ’re near.
She’s The Woman – she has no fear!”
(a) What kind of a woman do you see in the above lines?
(b) Pick out the rhyming words.
(c) What is her immediate duty that she wants to concentrate on?
Answer:
(a) I see a fearless woman in the above line,
(b) The rhyming words are near and dear.
(c) She wants to concentrate on the thought of caring for the near and dear ones.

(iii) “She’s a lioness; don’t mess with her.
She ’ll not spare you if you ’re a prankster.
Don’t ever try to saw her pride, her self-respect.
She knows how to thaw you, saw you – so beware!”
(a) Who is a prankster?
(b) What is the rhyme scheme?
(c) To whom is the poem addressed?
(d) What is the warning given by the poetess?
(e) What is the woman capable of?
Answer:
(a) A prankster is one who pretends to be mischievous.
(b) ‘aaba’ is the rhyme scheme in the above lines.
(c) The poem is addressed to the male readers who generally underestimate a woman.
(d) The poetess warns the male readers to refrain from sawing a woman’s pride and self-respect.
(e) The woman is capable of thawing and slicing down a person’s underestimated thoughts.

B. Read the lines and identify the figure of speech.

Answers:
1. (a) The rhyming words are stake and fake.
(b) innate-checkmate, strength-length, stake-spake, fake-make
(c) The rhyme scheme for the above lines is ‘abcc’.

2. (a) She’s a lioness.
(b) She is a dynamite. She is a rock.
She is a bubbly personality. She is a roller coaster of emotions.

3. (a) Faith, firm is the alliterated words in the given line.
(b) The other alliterated words from the poem are symbol/strength; summer/see and saw/self-respect.

C. Fill in with a word in each blanks to complete the summary of the poem. Use the help box given below.

Every woman is beautiful …………….(1) …………….. She is the……………… (2) ………….. of power and ………….. (3) …………….. She is prone to put her ………….(4) ………….. at risk. Every woman is true in expressing her love and she is never ……………… (5) …………….. She is very ……….. (6) ………………. in her approach even at times of ………… (7) …………… she finds a ray of ……………. (8) ………….. and she continues to …………. (9) ………….or her ……….. (10) …………….. ones. She is the ………………. (11) ………….. and she has no …………… (12) ……………. She is forceful in her ……………. (13) ………… and …………….(14) …………….. She is never a…………… (15) ………………..and she is ……………. (16) ……………. She is ferocious like a …………….(17)…………………..It’s better for the …………… (18) ……………… to stay away from her. Never should one try to bring ……………….(19) …………………to her pride and ……………. (20) ………………for she knows how to ………………… (21) …………… and …………….. (22) ………….. them. She is …………….. (23) ………………. woman. It is ……………….. (24) ………………….. to love her …………… (25) ……………. her and to keep her ………………. (26) …………… .
Answers:

1. innate
2. symbol
3. strength
4. life
5. fake
6. optimistic
7. adversity
8. hope
9. care
10. dignified
11. woman
12. fear
13. faith
14. belief
15. quitter
16. persistent
17. lioness
18. prankster
19. disgrace
20. self-respect
21. thaw
22. saw
23. today’s
24. healthier
25. respect
26. near

D. Answer the following in a paragraph of about 80 to 100 words.

I Am Every Woman Poem Question 1.
How are today’s women portrayed by the poet?
Answer:
Rakhi Nariani Shirke is an academician with a passion for writing poems as a medium of self-expression. This poem talks about the multifaceted nature of women. A woman is born beautiful and beauty is an inborn trait. She is a symbol of supremacy and strength. She takes risks in life for she is real and always genuine. Today’s women are empowered, brave, strong and resolute.

They are always ready to take up new ventures. They are persistent and work tirelessly to prove what they are capable of. Women have to be treated respectfully for the growth of a nation. As stated earlier, she is the symbol of strength as she is strong in her faith and beliefs. In spite of all the outbreaks and cries and laments, she is strong in her hope and firm in her trust. She is a lioness, brave and daring. She is a disciplinarian and will not tolerate your pranks with her. She is a woman of today smart and brave.

“Each time a woman stands up for herself, she stands up for all women.”

I Am Every Woman Poem Questions And Answers Question 2.
What qualities have made women powerful?
Answer:
The woman is made powerful due to her multifaceted personality. Today’s women are empowered, brave, strong and resolute. They love to explore and venture into new horizons. – They are persistent and work tirelessly to prove what they are capable of. Her inner beauty, her supremacy and strength makes her all-powerful. She exhibits her power by showcasing her fearless nature. She is the symbol of strength as she is strong in her faith and beliefs. She is seen as all-powerful because of her determined nature.

She may sigh, cry and moan but is strong in her hope and firm in her trust. She is a lioness, brave and daring and one dare not muddle with her. If anyone is a mischief maker, she will not spare them. The woman should be feared cautiously since she would thaw or saw you for your deeds against her.

“A Woman believes in being strong, even when things seem to go wrong.”

I am Every Woman (Rakhi Nariani Shirke)
Literary Devices At A Glance (Figures of Speech)

I am Every Woman by Rakhi Nariani Shirke About the Poet:
Rakhi Nariani Shirke is an academician with a passion for writing poems as a medium of self-expression. She is a post graduate, with a Bachelor’s degree in Education. She graduated from Jai Hind College, Mumbai and lives in Navi Mumbai. She is a teacher at -Ryan International School at Raigarh, Maharashtra, India.

I am Every Woman Summary:

Introduction:
This poem talks about the multifaceted nature of women. Today’s women are empowered, brave, strong and resolute. They are always ready to take up new ventures. They are persistent and work tirelessly to prove what they are capable of. Women have to be treated respectfully for the growth of a nation.

Inborn trait:
A woman is born beautiful and beauty is an inborn trait. She is a symbol of supremacy and strength. She takes risks in life for she is real and always genuine.

Fearless by Nature:
The development of life a woman is ready to see in spring. She says that there will always be spring after summer and so she cares for the ones who are near for she is the fearless woman.

Women of Strength:
As stated earlier, she is the symbol of strength as she is strong in her faith and beliefs. The poetess who is also a woman insists that persistence is the key to everything. In spite of all the out breaths, cries and laments, she is strong in her hope and firm in her trust. She is a lioness, brave and daring and one should not muddle with her.

Dare to be a Mischief Maker?
If anyone is a mischief maker, trying to play with her, the woman will not spare her. One should be careful and not saw her conceit or her self-respect. One should beware of such a woman for she would not only thaw you but also saw you for she is the modem day woman, the woman of the Present. Therefore, she should be loved, respected and kept at close quarters.

Conclusion:
The poet Rakhi Nariani Shirke dares at those who make a mockery of women. Being today’s woman she warns those who are ruthless towards women. She states that today’s women are bold and smart, ready to take up new ventures and achieve the goals.

I am Every Woman Glossary:

Textual:

Additional:

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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:

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Tamilnadu Samacheer Kalvi 10th English Solutions Poem Chapter 1 Life

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Life Textual Questions

A. Read the following lines from the poem and answer the questions that follow:

1. Let me but live my life from year to year,
With forward face and unreluctant soul;
(a) Whom does the word ‘me’ refer to?
(b) What kind of life does the poet want to lead?
Answer:
(a) ‘Me’ refers to the poet, Henry Van Dyke.
(b) The poet wants to lead a life facing every year with courage and willingness to do anything.

Additional:
(a) Pick out the alliterated words in the first line.
(b) What is the figure of speech used in the second line?
(c) What do you understand by the words, ‘forward face’?
(d) With what kind of attitude does the speaker wish to lead his life in the poem, ‘Life’ by Henry Van Dyke?
Answer:
(a) ‘Let, live, life’ are the alliterated words in the first line.
(b) The figure of speech is alliteration, (e.g. forward face)
(c) The words, ‘forward face’ indicates the courage with which one is ready to face anything in life.
(d) ’In the poem, ‘Life’ the speaker wishes to lead his life with a forward face and unreluctant soul. Hence the speaker’s attitude is optimistic.

2. Not hurrying to, nor turning from the goal;
Not mourning for the things that disappear
(a) Why do you think the poet is not in a hurry?
(b) What should one not mourn for?
Answer:
(a) The poet does not want to do anything in haste as he is determined to work towards reaching his goal.
(b) One should not mourn for the things that disappear.

Additional:
(a) Give the synonym for ‘mourn’.
(b) What is ‘goal’?
(c) Why do things disappear?
Answer:
(a) ‘Lament’ is the synonym for ‘mourn’.
(b) Goal is nothing but the aim or objective in this context.
(c) Things disappear because they are inconsistent and keep changing.

3. In the dim past, nor holding back in fear
From what the future veils; but with a whole
And happy heart, that pays its toll
To Youth and Age, and travels on with cheer.
(а) What does the poet mean by the phrase ‘in the dim past’?
(b) Is the poet afraid of future?
(c) How can one travel on with cheer?
Answer:
(a) The poet means that the past was very dull and glum.
(b) No, the poet isn’t afraid of the future.
(c) One can travel cheerfully with a happy heart.

Additional:
(a) What is meant by ‘veil’?
(b) Mention the figure of speech in the third line.
(c) Pick out the contrasting words in the above lines.
(d) Pick out the rhyming words.
(e) What is the rhyme scheme of the above lines?
Answer:
(a) A veil is a cover that is used to partially hide something.
(b) Alliteration – happy heart
(c) The contrasting words are ‘youth’ and ‘age’.
(d) The rhyming words are fear and cheer; whole and toll.
(e) The rhyme scheme is ‘abba’.

4. So let the way wind up the hill or down,
O ’er rough or smooth, the journey will be joy:
Still seeking what I sought when but a boy,
New friendship, high adventure, and a crown,
(a) How is the way of life?
(b) How should be the journey of life?
(c) What did the poet seek as a boy?
Answer:
(a) The way of life is rough or smooth.
(b) The journey of life should be joyful.
(c) The poet as a boy sought new friendships, high adventure and a crown.

Additional:
(a) Give the rhyme scheme.
(b) What is the rhyming word for down and boy?
(c) Pick out the alliterated words in the third line.
(d) When is new friendship needed?
(e) What is rough and smooth?
Answer:
(a) ‘abba’ is the rhyme scheme.
(b) The rhyming word for down is ‘crown’ and the rhyming word for boy is ‘joy’.
(c) The alliterated words are ‘still seeking’.
(d) New friendship is needed during boyhood.
(e) The way of life is rough and smooth.

5. My heart will keep the courage of the quest,
And hope the road’s last turn will be the best.
(a) What kind of quest does the poet seek here?
(b) What is the poet’s hope?
Answer:
(a) The quest sought here by the poet is a courageous heart.
(b) The poet hopes that the end journey in life will be the best to cherish.

Additional:
(a) Pick out the rhyming words.
(b) What is meant by ‘road’s last turn’?
(c) Which word in the above lines indicate ‘life’?
Answer:
(a) ‘Quest and best’ are the rhyming words.
(b) Life’s last phase is the road’s last turn.
(c) ‘Road’ indicates life.

(vi) In the dim past, nor holding back in fear
From what the future veils; but with a whole
And happy heart, that pays its toll – To Youth and Age, and travels on with cheer.
(a) Identify the rhyming words of the given lines.
Answer:
(a) Fear and cheer; whole and toll are the rhyming words.

Additional:
(a) Mention the two main stages of life.
(b) Pick out the alliterated words in the third line.
Answer:
(a) The two main stages of life are youth and old age.
(b) ‘Happy, Heart’ is the alliterated words in the third line.

(vii) Let me but live my life from year to year,
With forward face and unreluctant soul;
Not hurrying to, nor turning from the goal;
Not mourning for the things that disappear

(a) Identify the rhyme scheme of the given lines.
(b) What kind of life does the poet want to lead?
Answer:
(a) ‘abba’ is the rhyme scheme.
(b) The poet wants to lead a life without haste and steadfast in its target.

Additional:
(a) Pick out the couplet in the above lines.
With forward face and unreluctant soul;
Not hurrying to, nor turning from the goal;

B. Answer the following question in about 80-100 words:

10th Life Poem Paragraph Question 1.
Describe the journey of life as depicted in the poem by Henry Van Dyke.
Answer:
Henry Van Dyke, one of the greatest American short story writers and poets, has surpassed the act of writing skillfully. ‘Life’ is no doubt one of his priced literary pieces. This poem is a pinnacle of expressive embarkment on the quest of self-revival from the glum beats of monotony. It has a very deep and farsighted meaning held within it and this is evident from the very beginning of the poem. The poem is the poet’s own reflection on his life and tells his point of view on the more important things in life.

The poet advises the readers from his life experiences. We feel that he is now an older man reflecting on his younger days. Through his words he is explaining to us what he is taking away as most important to live is the best life possible. Life is too short to get caught up in the moment or worry about the past. On the other hand, it suggests that we look forward to what the future holds. We sometimes find ourselves brooding on the bad times and we forget about how much good there is in the future. The poet is making us understand this concept and be more aware of reality.

“Life is short and if we enjoy every moment of everyday,
We will be happy no matter what happens or changes our way! ”

C. Based on your understanding of the poem, complete the following passage by the using the phrases given in the box.

The poet wants to live his life …………… (a) ……………., willing to do something. He neither wants ………….. (b) …………. from his goal. He does not want to …………… (c) ………….. the things he has lost, not hold back for fear of the future. He instead prefers to live his life with a whole and happy heart which cheerfully travels from ………….. (d) …………. Therefore, it does not matter to him whether the path goes ……………. (e) ……………., rough or smooth, the journey will be …………… (f) …………. He will continue to seek what he wanted as a boy – new friendship, ……………. (g) ……………. and a crown (prize). His heart will remain courageous and pursue his desires. He hopes that every turn in his life’s journey will be the best.
Answers :
(a) looking ahead
(b) to hurry nor move away
(c) mourn
(d) youth to old age
(e) up or down the hill
(f) joyful
(g) high adventure

Life by Henry Van Dyke About the Poet :
Henry Van Dyke (1852 – 1933) was born in Pennsylvania. A nature lover and avid reader, he earned degrees from Princeton and served as a Presbyterian minister for more than 20 years. He eventually returned to Princeton and nearly for 20 years worked as a professor of English with a bit of service as the U.S. ambassador to Luxembourg and the Netherlands in between. Henry is well known for his works, “ Joyful, Joyful, We Adore Thee” and the Christmas stories,
“The Story of the Other Wise Man” and “The First Christmas Tree.”

Life Summary: 

Introduction :
Life is an experience. No matter whatever it brings, it should be approached with courage, dedication and a cool mind. One should learn to be neutral in success or failure. One must stand together in facing challenges. To be optimistic is the solution to set; the end result of success.

Will power the most needed trait In this poem, life is described not as an object but as an experience, one that should be lived with courage, dedication, without urgency and with a clear sense of purpose that drives the mind and soul. For mankind, failure or disaster is something that often makes our path incomprehensible and morose. But all it needs is a strong willpower and dedicated commitment to make that path easier appeasing the dubious notes.

Cheerful efforts :
So here the central character is cheerful about his efforts that will see him through with an eagerly pace in search of his goal. Standing at the edge of dilemma, it isn’t easy to challenge the odds and negativism all alone. So here the poet has urged his supporters to stand by his unanimity among all disparages.

Optimistic determination He knows that a persistent effort is the need of the hour to see every obstacle removed. So no matter what happens he should be prepared to go along the journey to his goal while facing many adventures. No matter what happens optimism is the ultimate solution that makes him stand up after every fall. Upon staying erect he hopes the end result will bring him all the fortune that he has ever looked for.

Conclusion:
One should not feel saddened by what life unfolds before him/her. One should stand up to fight and achieve his/her utmost goal.

Life Glossary:

Textual :

Additional :

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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

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\(\frac{1}{y}\) = 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

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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:

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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}

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Tamilnadu Samacheer Kalvi 10th Social Science Civics Solutions Chapter 5 India’s International Relations

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India’s International Relations Textual Exercise

I. Choose the correct answer.

Question 1.
Me Mahon Line is a border between ………………
(a) Burma and India
(b) India and Nepal
(c) India and China
(d) India and Bhutan
Answer:
(c) India and China

Question 2.
India is not a member of which of the following
(1) G20
(2) ASEAN
(3) SAARC
(4) BRICS
Select the correct option:
(a) 4 only
(b) 2 and 4
(c) 2, 4 and 1
(d) 1,2 and 3
Answer:
(b) 2 and 4

Question 3.
OPEC is ………..
(a) An international insurance Co.
(b) An international sports club
(c) An Organisation of Oil Exporting Countries
(d) An international company
Answer:
(c) An Organisation of Oil Exporting Countries

Question 4.
With which country does India share its longest land border?
(a) Bangladesh
(b) Myanmar
(c) Afghanistan
(d) China
Answer:
(a) Bangladesh

Question 5.
Match the following and choose the correct answer form the codes given below.

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

Question 6.
How many countries share their border with India?
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 7.
Which two island countries are India’s neighbours?
(a) Sri Lanka and Andaman island
(b) Maldieves and Lakshadweep island
(c) Maldieves and Nicobar island
(d) Sri Lanka and Maldieves
Answer:
(d) Sri Lanka and Maldieves

Question 8.
Which Indian state is surrounded by three countries?
(a) Arunachal Pradesh
(b) Meghalaya
(c) Mizoram
(d) Sikkim
Answer:
(d) Sikkim

Question 9.
How many Indian states have their boundary with Nepal?
(a) Five
(b) Four
(c) Three
(d) Two
Answer:
(a) Five

Question 10.
Who drew up the borders for newly independent Pakistan?
(a) Lord Mountbatten
(b) Sir Cyril Radcliffe
(c) Clement Atlee
(d) None of the above
Answer:
(b) Sir Cyril Radcliffe

II. Fill in the blanks.

1. …………… is a small Himalayan kingdom.
2. India’s gateway to South East Asia is ………….
3. …………… is a buffer country between India and China.
4. A strip of land …………. belongs to India on West Bengal and Bangladesh border.
5. …………… is known as the Land of a thunderbolt.
6. India and Sri Lanka are separated by …………
Answers:
1. Bhutan
2. Myanmar
3. Nepal
4. Teen Bigha Corridor
5. Bhutan
6. Palk strait

III. Consider the following statement and tick the appropriate answer

Question 1.
The Kaladan transport project by India and Myanmar consists of which of the following modes of transport?
1. Roads
2. Railways
3. Shipping
4. Inland water transport
Select the correct answer using the codes given below
(a) 1, 2 and 3 only
(b) 1, 3 and 4 only
(c) 2, 3 and 4 only
(d) 1, 2, 3 and 4
Answer:
(b) 1, 3 and 4 only

Question 2.
Assertion (A): India and France launched International Solar Alliance.
Reason (R): It was done to bring together countries between Tropic of Cancer and Tropic of Capricorn for co-operation of solar energy.
(a) A is correct and R is the correct explanation of A
(b) A is correct and R is not the correct explanation of A
(c) A is wrong and R is correct
(d) Both are wrong
Answer:
(a) A is correct and R is the correct explanation of A

Question 3.
Which of the following statements are true?
Statement 1. ICCR has initiated a Tagore Chair in University of Dhaka.
Statement 2. Mayanmar is India’s gateway to western countries.
Statement 3. Nepal and Bhutan are land locked nations.
Statement 4. Sri Lanka is one of the partner in Nalanda University Project of India.
(a) 1, 2 and 3
(b) 2, 3 and 4
(c) 1, 3 and 4
(d) 1, 2 and 4
Answer:
(c) 1, 3 and 4

Question 4.
Assertion (A): OPEC has vested interest in India’s economic growth.
Reason (R): Devoid of necessary oil resources India strongly focuses on agriculture and industrial production.
(a) A is correct and R explains A
(b) A is wrong and R is correct
(c) Both are correct
(d) Both are wrong
Answer:
(c) Both are correct

IV. Match the following.

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

V. Give Short Answers.

Question 1.
Name the neighbouring countries of India.
Answer:
Afghanistan, Pakistan, Bhutan, China, Nepal, Bangladesh, Myanmar, Sri Lanka, and the Maldives.

Question 2.
Write a short note on Strategic partnetship Agreement (SPA).
Answer:
Indo-Afghan relation was strengthened by the Strategic Partnership Agreement (SPA). SPA provides assistance to re-build Afghan’s infrastructure, institutions, agriculture, water, education, health and providing duty-free access to the Indian market.

Question 3.
Mention the member countries of BRICS.
Answer:
Brazil,. Russia, India, China and South Africa.

Question 4.
What do you know about Kaladan Multi – Model Transit Transport?
Answer:
India is building the Kaladan Multi-Model Transit Transport, a road-river-port cargo transport project to link Kolkata to Sittwe in Myanmar. A project aiming to connect Kolkata with Ho Chi Minh City on the South Sea for the formation of an economic zone will have a road pass through Myanmar, Cambodia and Vietnam and work on the first phase connecting Guwahati with Mandalay is currently undemay.

Question 5.
How do you assess the importance of Chabahar agreement?
Answer:

1. A trilateral agreement called the Chabahar Agreement was signed between India, Afghanistan and Iran, which has led to the establishment to transit and transport corridor among three countries using Chabahar port.
2. This port is seen as golden gateway for India to access land locked markets of Afghanistan and central Asia by passing Pakistan.

Question 6.
List out any five global groupings in which India is a member.
Answer:
India is a member of formal groupings like UNO, NAM, SAARC, G20 and the Commonwealth.

Question 7.
What is the role of Japan India Institute of Manufacturing (JIM)?
Answer:
In the manufacturing sector Japan announced its co-operation of training 30,000 Indian people in the Japan India Institute of Manufacturing (JIM) providing Japanese style manufacturing skills to enhance India’s manufacturing Industry base and contribute to “Make in India” and “Skill India” initiatives.

VI. Answer in detail.

Question 1.
Highlight India and International organisation with special reference to any three India’s global groupings.
Answer:
India is potential superpower and has a growing international influence all around the world. Being a newly industralised county, India has great history of collaboration with several countries. It has acted as prominent member of several international organizations and has been a founding member of some. India is a member of formal grouping like UNO, NAM, SAARC, G20 and the Common Wealth.

India has been extending a helping hand to the UNO, in all its efforts in ending military conflicts, and in promoting peace and progress among the nations.

Name of the Global Grouping	Name of the Member Countries	Objectives
IBSA	India, Brazil, South Africa	To focus on agriculture, education, energy, trade, culture and defence among others
BCIM	Bangladesh, China, India,Myanmar	To respond to threats such as natural disasters and data breaches and protect business interests
BBIN	Bangladesh, Bhutan, India, Nepal	For energy development

Question 2.
Trace the reason for the formation of BRICS and write its objectives. BRICS:
Answer:

1. Brazil, Russia, India, China, and South Africa are leading emerging economies and political powers at the regional and international level.
2. The BRICS organisations headquarters is in Shangai, China.
3. BRICS opened up a possibility for countries of Global South to challenge the Global North.

Reason for the formation of BRICS:

1. To be an alternative to world bank and IMF to challenge U.S supremacy.
2. To provide self owned and self – managed organisations to carry out developmental and economical plans in its member nations.

Objectives of BRICS

1. To achieve regional development.
2. It act as a bridge between developed and developing countries.
3. To contribute extensively to development of humanity.
4. To establish a more equitable and fair world.
5. Boost intra BRICS trade in their local currencies to increase trade co-operation and cope with the current international financial crisis.
6. To promote the technological information exchange among the member states.
7. To enhance inclusive economic growth that will lead to an increase in the creation of jobs, fight against poverty and accelerate the economic transformation of members.

Question 3.
Mention OPEC missions and how does it help other countries?
Answer:
OPEC’s mission:

1. To coordinate oil policies in its member countries
2. Help stabilise oil markets
3. To secure fair and stable income to petroleum producers
4. An efficient, economic and regular supply of oil to consuming nations
5. A fair return on capital to those investing in the petroleum industry

How does OPEC help other countries:
The OPEC Fund for International Development (OPID) is an institution that helps finance projects with low interest loans. It also provides grants to social and humanitarian projects. OPEC has an Information Centre with over 20,000 volumes including books, reports, maps and conference proceedings related to petroleum, energy and the oil market. The Information Centre is open to the public and is often used by researchers and students.

VII. Project and activity

Question 1.
Students can be asked to collect information form newspapers about India’s relatio with world countries.
Answer:
Do it yourself.

Question 2
Group project involving students to prepare an album with pictures on India’s latest projects with its neighboring countries.
Answer:
Do it yourself.

India’s International Relations Additional Questions

I. Choose the correct answer.

Question 1.
India was a dependent country till …………
(a) August 13, 1947
(b) August 15, 1947
(c) January 26, 1980
Answer:
(b) August 15, 1947

Question 2.
In which year Farakka accord on sharing of Ganga water signed?
(a) 1970
(b) 1973
(c) 1975
(d) 1977
Answer:
(d) 1977

Question 3.
In spite of past conflicts both ……………. are trying to come closer.
(a) India and Bangladesh
(b) India and Afghanistan
(c) India and Pakistan
Answer:
(c) India and Pakistan

Question 4.
…………….. is a landlocked nation.
(a) Bhutan
(b) Nepal
(c) Both (a) and (b)
(d) Pakistan
Answer:
(c) Both (a) and (b)

Question 5.
…………. was the first country to recoginse the Republic of china.
(a) England
(b) Russia
(c) India
Answer:
(c) India

Question 6.
Bangladesh got freedom from …………….. in …………
(a) India, 1951
(b) Nepal, 1961
(c) Pakistan 1971
Answer:
(c) Pakistan 1971

Question 7.
Which country is an important partner in our energy needs for petroleum and natural gas?
(a) Bangladesh
(b) Sri Lanka
(c) Myanmar
(d) All the above
Answer:
(c) Myanmar

II. Fill in the blanks :

1. ……………… relation was strengthened by the strategic-Partnership Agreement.
2. India was the first nation to acknowledge ……………. as an independent country.
3. By the Farakka Barrage issue, the distribution of …………. water was settled amicably.
4. India and Bangladesh share ……………. common rivers.
5. ………….. known as a land of thunder bolt.
6. India declared the bilateral trade relation known as …………. to …………..
7. ………….. being the manufacturing hub of the world.
8. Maldives is located south of Lakshadweep islands in the ……………
9. India’s second largest border is shared with …………….
10. …………… is a small land locked country.
Answers:
1. Indo-Afghan
2. Bangladesh
3. Ganga
4. 54
5. Bhutan
6. Bharat, Bhutan
7. China
8. Indian Ocean
9. Myanmar
10. Nepal

III. Match the following.

a.

1.	Pasupati and Janakpu	(a)	India
2.	Maldives	(b)	The borderline between India and China
3.	Bhutan	(c)	Nepal
4.	Varanasi and four Dhaams	(d)	Himalayan kingdom
5.	Me Mahoon Line	(e)	Indian Ocean

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

IV. Short Answer

Question 1.
Explain about LOC.
Answer:

1. The ceasefire line determined in 1949 was called the LOC after 1972.
2. This is the boundary that came to be agreed between India and Pakistan under the Shimla agreement of 1972.
3. It was called Radcliffe line at the time of partition 1947. (Radcliffe was the chairman of the border commission) This is now called LOC.

Question 2.
Which is the fundamental factor of India’s foreign policy?
Answer:
The recognition of sovereign equality of all people living in various parts of the world is the fundamental factor in India’s foreign policy.

Question 3.
Give a short note on BRICS payment system.
Answer:
At the 2015 BRICS summit, ministers from the BRICS nations initiated consultation a payment system that would be an alternative to the society for world wide Inter Bank Financial Telecommunication system (SWIFT).

Question 4.
Explain India’s role towards China.
Answer:
When China became republic in 1949. India was the first country to recognize it. Both the countries have successfully attempted to restore the economic lines. China has formally declared that she will back India’s claim for becoming a permanent member of united Nation’s Security Council.

Question 5.
List out the important Pilgrimage destination in India.
Answer:
Pashupati and Janakpur are traditional centres in Nepal where as Varanasi and the four Dhaans (Badrinathpuri, Dwareka and Rameshwaram) are important pilgrimage destination in India.

Question 6.
Prove that India is a very good friend of Bangladesh.
Answer:
It is due to the effort and support of Smt.Indira Ghandhi, the Prime Minister of India, Bangladesh got freedom from Pakistan in 1971. In 1972, a 25 years treaty of friendship co-operation and peace was sighed in Dacca by India and Bangladesh. The Farakka Barrage issue regarding the distribution of Ganga water was settled amicably. Thus India is a very good friend of Bangladesh. Our friendship with Bangladesh will go on forever.

V. Detail.

Question 1.
Discuss India and its Neighbours?
Answer:

1. India has always been known as a peace – loving country. India is surrounded by many neighbouring countries with whom she has traditionally tried to maintain friendly and good neighbourly relations.
2. India’sposition is unique in its neighbourhood.
3. India’s neighbours had been a part of a homogeneous culture prevailing in the Indian subcontinent for last five thousand years.
4. India is a vast country with Pakistan and Afghanistan to the north – west.
5. China, Nepal, Bhutan to the North.
6. Bangladesh to the east.
7. Myanmar to the Far East.

Sri Lanka (from South – East) and Maldives (from South – West) are two countries that lie close to India separated by the Indian ocean. India has cordial historical, religious, economic, ethnic and linguistic relationship will all of these countries.

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In the Samacheer Kalvi 10th English Guide for textbook solutions, subject experts covered all types of questions and answers related to the topics, quick notes, summary, solved & unsolved exercises, etc. If you are planning to prepare Chapter via textbook, then you’re suggested to go with this Samacheer Kalvi 10th English Book Solutions Questions and Answers PDF for better understanding and preparation.

Tamilnadu Samacheer Kalvi 10th English Vocabulary Compound Words

English Subject experts who are having max years of experience prepared this Tamilnadu State Board Solutions for 10th English Vocabulary Compound Words Questions and Answers. They have explained all the topics covered in the board prescribed latest syllabus in a simple way to understand easily. So, students can prepare English from this Samacheer Kalvi 10th English Book Questions and Answers PDF. Download the Tamilnadu State Board 10th English Workbook Answers PDF by accessing the below links and learn properly for the final exams to score well.

Question 1.
Choose the word from the options given to form a compound word with “flash”.
(a) table (b) post (c) back (d) pen
Answer:
(c) back

Question 2.
Choose the word from the options given to form a compound word with “make”.
(a) late (b) plaza (c) life (d) over
Answer:
(d) over

Question 3.
Choose the word from the options given to form a compound word with “soft”.
(a) hand (b) ware (c) head (d) back
Answer:
(b) ware

Question 4.
Choose the word from the options given to form a compound word with “deep”.
(a) road (b) fry (c) wind (d) thick
Answer:
(b) fry

Question 5.
Choose the word from the options given to form a compound word with “down”.
(a) Low (b) slope (c) safe (d) cast
Answer:
(d) cast

Question 6.
Choose the w&d from the options given to form a compound word with “light”.
(a) glare (b) low (c) sensitive (d) land
Answer:
(c) sensitive

Question 7.
Choose the word from the options given to form a compound word with “fast”.
(a) food (b) slow (c) mode (d) pace
Answer:
(a) food

Question 8.
Choose the word from the options given to form a compound word with “good”.
(a) person (b) will (c) rid (d) call
Answer:
(b) will

Question 9.
Choose the word from the options given to form a compound word with “sea”.
(a) food (b) space (c) snare (d) cast
Answer:
(a) food

Question 10.
Choose the word from the options given to form a compound word with ‘radio”.
(a) street (b) active (c) park (d) top
Answer:
(b) active

Question 11.
Choose the word from the options given to form a compound word with ‘out’.
(a) field (b) sourcing (c) ground (d) area
Answer:
(b) sourcing

Question 12.
Choose the word from the options given to form a compound word with “welt”.
(a) done (b) light (c) defined (d) fine
Answer:
(c) defined

Question 13.
Choose the word from the options given to form a compound word with “walking”.
(a) stick (b) post (c) park (d) fast
Answer:
(a) stick

Question 14.
Choose the word from the options given to form a compound word with “river”.
(a) drive (b) flowing (c) bed (d) nice
Answer:
(c) bed

Question 15.
Choose the word from the options given to form a compound word with “safe”.
(a) role (b) guard’ (c) mat (d) mate
Answer:
(b) guard

Question 16.
Choose the word from the options given to form a compound word with “hand”.
(a) machine (b) beaten (c) written (d) will
Answer:
(c) written

Question 17.
Choose the word from the options given to form a compound word with “gazing”.
(a) goat (b) boy (c) land (d) star
Answer:
(d) star

Question 18.
Choose the word from the options given to form a compound word with “head”.
(a) leader (b) master (c) manager (d) up
Answer:
(b) master

Question 19.
Choose the word from the options given to form a compound word with “walk”.
(a) man (b) sat (c) milk (d) load
Answer:
(a) man

Question 20.
Choose the word from the options given to form a compound word with “back”.
(a) box (b) rest (c) tin (d) on
Answer:
(b) rest

Question 21.
Choose the word from the options given to form a compound word with ‘Grass’.
(a) rest (b) green (c) garden (d) hopper
Answer:
(d) hopper

Question 22.
Choose the word from the options given to form a compound word with “skate”.
(a) stake (b) blank (c) board (d) road
Answer:
(c) board

Question 23.
Choose the word from the options given to form a compound word with ‘Sun”.
(a) flower (b) drive (c) gaze (d) glare
Answer:
(a) flower

Question 24.
Choose the word from the options given to form a compound word with “grand”.
(a) gear (b) rose (c) mother (d) car
Answer:
(c) mother

Question 25.
Choose the word from the options given to form a compound word with “base”.
(a) post (b) boast (c) ball (d) long
Answer:
(c) ball

Question 26.
Choose the word from the options given to form a compound word with “night”.
(a) right (b) fame (c) mare (d) mire
Answer:
(c) mare

Question 27.
Choose the word from the options given to form a compound word with “doer”.
(a) step (b) more (c) look (d) up
Answer:
(a) step

Question 28.
Choose the word from the options given to form a compound word with “cycle”.
(a) river (b) more (c) motor (d) bike
Answer:
(c) motor

Question 29.
Choose the word from the options given to form a compound word with “band”.
(a) rag (b) width (c) bog (d) up
Answer:
(b) width

Question 30.
Choose the word from the options given to form a compound word with “match”.
(a) right (b) box (c) hatch (d) mixing
Answer:
(b) box

We hope the data given here will benefit you to the fullest extent at the time of preparation. For better understanding of English subject this Samacheer Kalvi 10th English Answers for Class 10th English Vocabulary Compound Words PDF is the best resource. Download & ace up your preparation. Keep in touch with us and get the latest information on Tamilnadu State board Textbook Answers PDF.

Looking to improve English skills and gain more subject knowledge then the best resources that you can use here is Samacheer Kalvi 10th English Solutions for Poem Chapter 7 The House on Elm Street Questions and Answers.

In the Samacheer Kalvi 10th English Guide for Chapter 7 The House on Elm Street textbook solutions, subject experts covered all types of questions and answers related to the topics, quick notes, summary, solved & unsolved exercises, etc. If you are planning to prepare Chapter 7 The House on Elm Street via textbook, then you’re suggested to go with this Samacheer Kalvi 10th English Book Solutions Questions and Answers PDF for better understanding and preparation.

Tamilnadu Samacheer Kalvi 10th English Solutions Poem Chapter 7 The House on Elm Street

English Subject experts who are having max years of experience prepared this Tamilnadu State Board Solutions for 10th English Poem Chapter 7 The House on Elm Street Questions and Answers. They have explained all the topics covered in the board prescribed latest syllabus in a simple way to understand easily. So, students can prepare Chapter 7 English from this Samacheer Kalvi 10th English Book Questions and Answers PDF. Download the Tamilnadu State Board 10th English Chapter 7 The House on Elm Street Workbook Solutions PDF by accessing the below links and learn properly for the final exams to score well.

The House on Elm Street Textual Questions

A. Read the given lines and answer the questions given below.

(i) It sat alone.
What happened there is still today unknown.
It is a very mysterious place,
And inside you can tell it has a ton of space,
But at the same time it is bare to the bone.
(a) What does ‘It’ refer to?
(b) Pick out the line that indicates the size of the house?
Answer:
(a) ‘It’ refers to the mysterious house.
(b) And inside you can tell it has a ton of space, – this line indicates the size of the house.

The House On Elm Street Figure Of Speech Additional:
(a) Do you know what happened inside that house?
(b) What is meant by ‘bare to the bone’?
(c) Which adjective is used to describe the house?
(d) Mention the figure of speech in the last line.
(e) Give the rhyming word for place and bone.
It Sat Alone What Is The Figure Of Speech Answer:
(a) No, I don’t know what happened inside the house.
(b) ‘Bare to the bone’ means there isn’t anything in the house.
(c) The adjective ‘mysterious’ is used to describe the house.
(d) The figure of speech is alliteration in the last line. (eg. bare bone)
(e) The rhyming word for place is space and for bone, it is unknown and alone,

(ii) I drive past the house almost every day.
The house seems to be a bit brighter.
On this warm summer day in May.
It plays with your mind.
(a) To whom does ‘I’ refer to?
(b) Pick out the alliterated words in the 2nd line.
Answer:
(a) ‘I’ refers to the poetess, Nadia Bush.
(b) The alliterated words in the second line are bit brighter.

I Sat Alone Figure Of SpeechAdditional:
(a) How does the house appear in May?
(b) How is the day in Summer?
(c) Pick out the rhyming words in the above stanza.
(d) What is the rhyme scheme followed in this stanza?
(e) What plays with your mind?
(f) How often does the poetess see this house?
Answer:
(a) The house appears a bit brighter in May.
(b) The day is warm and bright in summer.
(c) The rhyming words are day and May in the above stanza.
(d) The rhyme scheme followed in this stanza is ‘abac’.
(e) The thought about the house plays with your mind.
(f) The poetess sees this house almost every day.

(iii) It never grows leaves,
Not in the winter, spring, summer or fall.
It just sits there never getting small or ever growing tall
(a) What does ‘it’ refer to?
(b) In what way the tree is a mystery?
Answer:
(a) ‘It’ refers to the tree that is beside the house.
(b) The tree is a mystery because no leaves grow in it during any one of the season. It neither grow tall nor does it become short.

Additional:
(a) Why is the poet using the word ‘fall’?
(b) What do you understand by the word, ‘fall’?
(c) Which are the four seasons mentioned here?
(d) What sits there without getting small nor growing tall?
(e) What is the rhyming word for ‘fall’?
(f) Give the American English word for autumn.
Answer:
(a) The poet is using the word ‘fall’ because the poetess is an American.
(b) Autumn is the season which represents the season ‘fall’.
(c) The four seasons mentioned here are winter, spring, summer and fall.
(d) The tree which never grows leaves beside the mysterious house sits there without getting small nor growing tall.
(e) Tall is the rhyming word for fall.
(f) The American English word for autumn is fall.

(iv) Rumors are constantly being made,
And each day the house just begins to fade.
What happened inside that house?
(a) Does the house remain the same every day?
(b) How does the poet consider the house to be a mystery?
Answer:
(a) No the house doesn’t remain the same as it has started to fade away.
(b) There are many rumours about the house and hence the poetess considers the house to be a mystery.

Words That Rhyme With Fade Additional:
(a) Give the rhyming word for ‘fade’.
(b) Give the British English spelling for ‘Rumor’.
(c) What do you understand by the word, ‘constantly’?
(d) Which line indicates the mysterious nature?
Answer:
(a) The rhyming word for fade is made.
(b) The British English spelling for Rumor is Rumour.
(c) Here ‘constantly’ means regularly.
(d) ‘What happened inside that house?’ is the line that indicates that the house is mysterious,

(v) What happened inside that house?
I really don’t know I guess it will always be a mystery
(a) Does the poet know what happened in the house?
(b) What is the mystery about the house?
Answer:
(a) No, the poetess doesn’t know what happened in the house.
(b) The mystery about the house is nothing but the fact that no one will know what is happening inside the house.

Additional:
(a) Who does ‘I’ refer to?
(b) What is the guess made by the poetess, Nadia Bush?
Answer:
(a) ‘I’ refers to the poetess, Nadia Bush.
(b) The guess made by the poetess, Nadia Bush is that the house will be a mystery forever.

Additional Questions

(i) At night the house seems to be alive,
Lights flicker on and off.
I am often tempted to go to the house,
To just take a look and see what it is really about,
But fear takes over me.
(а) When is the house filled with life?
(b) How does one know that there is life in the house?
(c) What is the temptation of the poetess?
(d) Does the poetess go inside the house?
(e) Why doesn’t the poetess go inside the house?
(f) What is ‘flicker’?
Answer:
(a) The house is filled with life at night.
(b) One knows that there is life in the house when you see the lights switched on and switched off.
(c) The temptation of the poetess is to enter the house and see what is inside.
(d) The poetess doesn’t go inside the house.
(e) The poetess doesn’t go inside the house because she is frightened because of the mystery behind the house.
(f) ‘Flicker’ is flash or glimmer.

The House On The Elm Street Textual Questions

B. Answer the following in a paragraph.

Question 1.
Where is the house located? Why is it a mysterious place?
Answer:
The house is located on Elm Street. There aren’t any house around it. It stood all alone in an isolated place on Elm Street. Next to the house, is a tree. The tree too is mysterious like the house since no leaves sprout in any of the seasons. It is said to be a mysterious place ’ since no one knows who lives there or what is inside. No one knows what happens inside that house and hence it is very mysterious. It is for sure a big house with vast space inside the house. Generally, at night, the house looks like it is alive with people in it.

Lights are switched on and off. Every day the poetess, drives past the house. The house seems to look a bit brighter on a fine warm summer day in the month of May. The very thought of this mysterious house plays with your mind since it is just one house of this kind in the areas known around. Likewise, the tree too is barren during winter, spring, summer and autumn. The poetess says that the tree just stays there and never grows tall nor becomes short. She wonders how a tree could survive without any leaves or without any growth and hence feels it is a mysterious place.
‘The mysteiy of existence will always remain a mystery and secret.”

Question 2.
How is the mystery depicted in the poem?
Answer:
Nadia Bush the poetess talks about a house that stood alone in an isolated place on Elm Street. No one knows what happens inside that house and it is very mysterious. Even by looking from outside, one can easily say that it has a vast space but at the same time it is bare to the bone meaning there is no one living inside the mysterious house or no basic necessities inside the house on Elm Street. Generally at night, the house looks like it is alive with people in it. Lights are switched on and off.

It gives a feel that someone is inside the house. When Nadia sees such a situation, she is tempted to go inside the house to just take a look and see what is actually happening inside the house. However, Nadia the poetess is frightened and never dares to do so. Every day the poetess, drives past the house. The house seems to look a bit brighter. The very thought of this mysterious house plays with your mind since it is just one house of this kind in the area around.

Next to the house, is also a tree with no leaves the entire year. It doesn’t grow tall nor does it shrink in size. The tree too is mysterious like the house since it has no leaves in any of the seasons to make one wonder how a tree could survive without any leaves or without any growth. Every day, the house also begins to fade making it all the more mysterious.
‘ The best secrets are the most twisted. ’

The House On Elm Street Additional Questions

Question 1.
Describe the tree which adds to the mystery of the house.
Answer:
The poem, ‘The House on Elm Street’ by Nadia Bush is a dark poem. The house and the tree depict darkness and also a mysterious feel. Trees generally grow green. It is a representation of fertility and robust nature. Unlike a tree’s characteristic features the tree beside the mysterious house is also mysterious. The tree stands erect with no leaves in them. There is no fall of leaves during autumn and there is no leaves sprouting during spring. There aren’t any leaves on the tree for the leaves to dry during summer nor for the snow to cover the leaves with snow during winter. This tree doesn’t grow tall nor does it get small. It is quite a mystery to see a living tree appearing lifeless!
‘Mystery creates wonder and wonder is the basis of man s desire to understand.’

The House on Elm Street Poem Figure Of Speech Textual Questions

C. Read the poem and write the rhyming words and rhyme scheme for the given stanzas.

Answers:

1. unknown
2. bone
3. face
4. day
5. kind
6. be
7. fall

D. Identify the poetic lines where the following figures of speech are employed and complete the tabular column.

The House on Elm Street (.Nadia Bush)
Literary Devices At A Glance (Figures of Speech)

The House on Elm Street by Nadia Bush About the Poet and Biography
The House on Elm Street was published by Nadia Bush a budding poetess in April 2017. Born on‘September 24th, she lives in Somerset, Pennsylvania. She loves the movie Paranormal Activity. TRVL channel’s Ghost Adventures is one of her favorite programmes. She has a pet cat and perhaps her life style makes her pen such a mysterious poem about the dark when it was given as an assignment.

The House On Elm Street Poem Summary By Nadia Bush

Introduction:
Nadia Bush wasn’t really sure what to write at first. She wrote this poem for her English class because they were told to write a “dark” poem. She decided to write about a house that no one knows what happened inside, but they did know something is definitely not right about it.

A strange house:
Nadia Bush the poetess talks about a house that stood alone in an isolated place on Elm street. No one knows what happens inside that house and it is very mysterious. Even by looking from outside, one can easily say that it has a vast space but at the same time it is bare to the bone meaning there is no one living inside the mysterious house or no basic necessities inside the house on Elm Street.

Nadia’s curiosit:
Generally at night, the house looks like it is alive with people in it. Lights are switched on and off. When Nadia sees such a situation, she is tempted to go inside the house to just take a look and see what is actually happening inside the house. However, Nadia the poetess is frightened and never dares to do so.

Thoughts on the mysterious house
Every day the poetess, drives past the house. The house seemed to look a bit brighter on one fine warm summer day in the month of May. The very thought of this mysterious house plays with your mind since it is just one house of this kind in the areas known to her.

Strange barren tree:
Next’to the house, is a tree. The tree too is mysterious like the house since it has no leaves in any of the seasons. The tree is barren during winter, spring, summer and autumn. The poetess says that the tree just stays there and never grows tall nor becomes short. She wonders how a tree could survive without any leaves or without any growth.

Rumours spread:
There are rumours always spreading regularly about this house on Elm Street. Every day, the house begins to fade. No one knows what happens inside the house just as she is unaware of the same. The poetess guesses that it will always be a mystery about this house on Elm street.

Conclusion:
Though it is fictional, the poet creates an impression that such a house exists and stimulates a crave in us to visit the house. This is the highlight of the poem.

The House on Elm Street Glossary:
Textual:

We hope the data given here will benefit you to the fullest extent at the time of preparation. For better understanding of English subject this Samacheer Kalvi 10th English Solutions for Class 10th English Poem Chapter 7 The House on Elm Street PDF is the best resource. Download & ace up your preparation. Keep in touch with us and get the latest information on Tamilnadu State board Textbook Solutions PDF.