Class 9

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. below). If AD is extended to intersect BC at P, show that

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. AP bisects ∠A as well as ∠D.
4. AP is the perpendicular bisector of BC.

Solution:
Given
AB = AC
DB = DC
To prove:

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. ∠1 = ∠2 and ∠5 = ∠6
4. ∠3 = ∠4 = 90° and BP = PC

Proof
1. In ∆ABD and ∆ACD
AB = AC (given)
BD = CD (given)
AD =AD (common)
∆ABD = ∆ACD (by SSS)
and so ∠1 = ∠2 (by CPCT)

2. In ∆ABP and ∆ACP
AB = AC (given)
∠1 = ∠2 (proved)
AP = AP (common)
∆ABP ≅ ∆ACP (by SAS)
and so ∠3 = ∠4
and BP = CP (by CPCT)

∠1 = ∠2 AP bisects ∠A

3. In ∆DBP and ∆DCP
BP = CP (proved)
∠3 = ∠4 (proved)
DP = DP (common)

∆DBP ≅ ∆DCP (by SAS)
and so ∠5 = ∠6 (by CPCT)
∴ AP bisects ∠D.

4. BP = CP (proved)
∠3 = ∠4 (proved)
∠3 + ∠4 = 180° (LPA’s)
∠3 + ∠3 = 180° (∠3 = ∠4)
2∠3 = 180°
∠3 = 90°
∠3 = ∠4 (each 90°)
and therefore, AP is the perpendicular bisector of BC

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

1. AD bisects BC
2. AD bisects ∠A.

Solution:
Given
AB = AC and AD ⊥ BC
To prove
∠1 = ∠2 and
BD = CD
Proof:
In ∆ABD and ∆ACD
AB = AC
AD = AD
∠3 = ∠4
∆ABD = ∆ADC
and so BD = CD and ∠1 = ∠2

Question 3.
Two sides AS and BC and median AM on one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see-Fig. below). Show that:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Solution:
Given
AB = PQ
BC = QR
AM = PN
To prove:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Proof:
BC = QR (given)
\(\frac{1}{2}\) BC = \(\frac{1}{2}\)QR
BM = QN

1. In ∆ABM and ∆PQN
AB = PQ (given)
BM = QN (proved)
AM = PN (given)
∆ABM ≅ ∆PQN (by SSS)
and so ∠ABC = ∠PQR (by CPCT)

2. In ∆ABC and ∆PQR
AB = PQ (given)
BC = QR (given)
∠B = ∠Q (proved)
∆ABC ≅ ∆PQR by (SAS)

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that triangle ABC is isosceles.

Solution:
Given
∠E = ∠F
BE = CF
To prove
AB = AC
Proof:
In ∆FBC and ∆ECB
BE = CF (given)
∠F = ∠E (each 90°)
BC = CB (common)
∴ ∆FBC = ∆ECB (by RHS)
and so ∠B = ∠C (by CPCT)
In ∆ABC, ∠B = ∠C
AB = AC
(sides opposite to equal angles of a A are equal)
and so ABC is isosceles.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
Given
AB = AC
∠APB = ∠APC = 90°
To prove:
∠B = ∠C

Proof:
In ∆ABP and ∆ACP
AP = AP (common)
∠APB = ∠APC (each 90°)
AB = AC (given)
∴ ∆ABP = ∆ACP (by RHS)
and ∠B = ∠C (by CPCT)

Theorem 7.6.
If two angles of a triangle are equal, then the sides opposite to them are also equal.
Given
In ∆ABC, ∠C = ∠B

To prove: AB = AC
Construction:
Draw the bisector of ∠A and let it meet BC an D
Proof:
In ∆s ABD and ACD, we have ∠B = ∠C (Given)
∠BAD = ∠CAD (Construction)
AD = AD (Common)
∴ ∆ABD ≅ ∆ACD (AAS Cong. Criterion)
Hence, AB = AC (CPCT)

Theorem 7.7.
If two sides of a triangle are unequal, the longest side has greater angle opposite to it.
Given:
In ∆ABC; AC > AB
To prove: ∠ABC > ∠ACB.

Construction:
Mark point D on AC such that AB = AD. Join BD.
Proof:
In ∆ABD,
AB = AD
∴ ∠1 = ∠2 (Const. As opp. equal sides) ….(1)
But ∠2 is an exterior angle of ABCD.
∠2 > ∠ACB (Exterior Angle Theorem) …(2)
From (1) and (2), we have
∠1 > ∠ACB (Const.)
But ∠ABC > ∠1
∴ ∠ABC > ∠ACB

Theorem 7.8. (Converse of Theorem 7.7)
In a triangle the greater angle has the longer side opposite to it.
Given:
In ∆ABC, ∠ABC > ∠ACB
To prove: AC > AB

Proof:
For ∆ABC, there are only three possibilities of which exactly one must be true.

1. AC = AB
2. AC < AB (iii) AC > AB.

Case 1:
If AC = AB, then ∠ABC = ∠ACB, which is contrary to what is given.
AB ≠ AC

Case 2:
If AC < AB, the longer side has the greater angle opposite to it.
∴ ∠ACB > ∠ABC.
This is also contrary to what is given.

Case 3:
We are left with the only possibility, namely AC > AB which is true.
AC > AB.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 9 Force and Laws of Motion

Force and Laws of Motion Intext Questions

Force and Laws of Motion Intext Questions Page No. 118

Question 1.
Which of the following has more inertia:

1. a rubber ball and a stone of the same size?
2. a bicycle and a train?
3. a five – rupees coin and a one – rupee coin?

Answer:
As we know, inertia is the calculated value for the mass of the body. It is proportional to mass of the body:

1. Inertia of the stone is greater than that of a rubber ball as mass of a stone is more than the mass of a rubber ball for the same size.
2. Inertia of the train is greater than that of the bicycle. As mass of a train is more than the mass of a bicycle.
3. Mass of a five rupee coin is more than that of a one – rupee coin. Hence, inertia of the five – rupee coin is greater than that of the one – rupee coin.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kick it towards a player of his own team”. Also, identify the agent supplying the force in each case.
Answer:
Four times:

• First, when a football player kicks to another player. Agent supplying the force: First case – First player. Second when that player kicks the football to the goalkeeper. Agent supplying the force. Second case – Second player.
• Third when the goalkeeper stops the football. Agent supplying the force: Third case – Goalkeeper.
• Fourth when the goalkeeper kicks the football towards a player of his own team. Agent supplying the force: Fourth case – Goalkeeper.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
When we shake any tree’s branches vigorously some leaves of that tree get detached because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
Due to inertia of motion, we fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest.

1. Case I: Since the driver applies brakes and bus comes to rest. But, the passenger tries to maintain its inertia of motion. As a result, a forward force is exerted on him.
2. Case II: The passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus.

Force and Laws of Motion Intext Questions Page No. 126

Question 1.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
According to Newton’s third law of motion, a force is exerted by the Earth on the horse in the forward direction while horse pushes the ground in the backward direction. As a result, the cart moves forward.

Question 2.
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
According to Newton’s third law of motion, a reaction force is exerted over fireman by the ejecting water in the backward direction when a fireman holds a hose, which is ejecting large amounts of water at a high velocity. As a result of the backward force, the stability of the fireman get affected. Hence, it is difficult for him to remain stable while holding the hose.

Question 3.
From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35 ms^-1. Calculate the initial recoil velocity of the rifle.
Answer:
Given,
Mass of the rifle, m₁ = 4kg
Mass of the bullet, m₂ = 50g = 0.05kg
Recoil velocity of the rifle = v₁
Bullet is fired with an initial velocity, v₂= 35 m/s
Condition:
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the rifle and bullet system = (m₁+ m₂)v = 0
Total momentum of the rifle and bullet system after firing = m₁v₁ + m₂v₂
= 0.05 × 35 = 4v₁ + 1.75
As per law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing
⇒ 4v₁ + 1.75 = 0
v₁ = –\(\frac { 1.75 }{ 4 }\) = -0.4375 m/s
So, the rifle recoils backwards with a velocity of 0.4375 m/s because value is negative.

Question 4.
Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2 ms^-1 and 1  ms^-1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^-1. Determine the velocity of the second object.
Answer:
Given,
m₁ = 100g = 0.1kg
m₂ = 200g = 0.2kg
Velocity of m₁ before collision, v₁ = 2 m/s
Velocity of m₂ before collision, v₂ = 1 m/s
Velocity of m₁ after collision, v₃ = 1.67 m/s
Velocity of m₂ after collision = v₄
As per the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
Hence,
m₁v₁ + m₂v₂ = m₁v₃+ m₂v₄
Putting values
2(0.1) + 1(0.2) = 1.67(0.1) + v₄(0.2)
0.4 = 0.167 + 0.2v₄
v₄ = 1.165 m/s
Velocity of the second object = 1.165 m/s.

Force and Laws of Motion NCERT Textbook Exercises

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non –  zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Yes, an object may travel with a non – zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the upthrust and the viscosity of air. The net force on the drop is zero.

Question 2.
When a carpet is beaten with a stick, dust comes out of it Explain.
Answer:
When we beat the carpet with a stick, it comes into motion. But the dust particles continue to be at rest due to inertia and get detached from the carpet.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
Due to sudden jerks or due to the bus taking sharp turn on the road, the luggage may fall down from the roof because of its tendency to continue moving in the original direction. To avoid this, the luggage is tied with a rope on the roof.

Question 4.
A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would come to rest.
Answer:
(c) there is a force on the ball opposing the motion.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)
Answer:
Here, u = 0, s = 400m, t = 20 s, a = ?, F = ?.
m = 7 tonnes
= 7 × 1000kg
= 7000kg
⇒ s = ut + \(\frac { 1 }{ 2 }\)at²
400 = (0 + 20) + \(\frac { 1 }{ 2 }\)a(20)²
a = \(\frac { 400\times 2 }{ { 20 }^{ 2 } } \)
∴ a = 2 m/s²
Force,
F = ma
= 7000 × 2 = 14,000 N.

Question 6.
A stone of 1kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice?
Answer:
m = 1kg, u = 20 m/s, s = 50m, v = 0, F = ? a = ?.
⇒ v² – u² = 2as
(0)² – (20)² = 2a (50)
∴ – 400 = 100a
= \(\frac { 400 }{ 100 }\) – 4 m/s²
Force of friction, F = m × a
= 1kg × -4 m/s² = -4 N

Question 7.
A 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer:
m = 8,000 + 5 × 2,000 = 18,000kg
(a) The net accelerating force,
F = Engine force – friction force
= 40,000 – 5,000 = 35,000 N.

(b) The acceleration of the train,
a = \(\frac { F }{ m }\) = \(\frac { 35,000 }{ 18,000 }\) = \(\frac { 35 }{ 18 }\) = 1.94 ms^-2

(c) The force of wagon 1 on wagon 2
= The net accelerating force – mass of wagon × acceleration
= 35,000 – 2,000 × \(\frac { 35 }{ 18 }\)
= 35,000 – 3888.8 = 31,111.2 N.

Question 8.
An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is stopped with a negative acceleration of 1.7 ms^-2?
Answer:
Here, mass = 1500kg
a = -1.7 ms²
F = m × a
= 1500 × (-1.7)
= -2550 N
The force between the vehicle and the road is 2,550 is, m a direction opposite to the direction of the vehicle.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v? Choose correct option.
(a) (mv)²
(b) mv²
(c) \(\frac { 1 }{ 2 }\) × mv²
(d) mv.
Answer:
(d) mv.

Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
The cabinet will move with constant velocity only when the net force on it is zero.
Force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.

Question 11.
Two objects, each of mass 1.5kg, ate moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms^-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Here, m₁ = m₂ = 1.5kg,
u₁ = 2.5 ms^-1, u₂ = – 2.5 ms^-1
Let v be the velocity of the combined object after the collision.
By conservation of momentum,
Total momenta after collision = Total momenta before collision
= (m₁ + m₂) v = m₁u₁ + m₂u₂
= (1.5 + 1.5) v = 1.5 × 2.5 + 1.5 × (-2.5)
= 3.0 v = 0
or
⇒ v = 0 ms^-1

Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
Action and reaction always act on different bodies, so they cannot cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so, the truck does not move.

Question 13.
A hockey ball of mass 200g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Here, m = 200g = 0.2kg,
u = 10 ms^-1,
v = -5 ms^-1
change in momentum = m (v – u)
= 0.2 (- 5 – 10) = -3kg ms^-1.

Question 14.
A bullet of mass 10g travelling horizontally with a velocity of 150 ms^-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Here m = 10g = 0.01kg,
u = 150 ms^-1,
v = 0, t = 0.03 s
a = \(\frac { v-u\quad }{ t } \) = \(\frac { 0-150\quad }{ 0.03 } \) = -5,000 ms^-1
The distance of penetration of the bullet into the block,
s = ut + \(\frac { 1 }{ 2 }\)at²
150 × 0.03 + \(\frac { 1 }{ 2 }\) × (-5,000) × (0.03)
= 4.5 – 2.25 = 2.25
The magnitude of the force exerted by the wooden block on the bullet
= ma = 0.01 × 5,000 = 50 N.

Question 15.
An object of mass 1kg travelling in a straight line with a velocity of 10 ms^-1 collides with, and sticks to, a stationary wooden block of mass 5kg. Then, they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
Here, m₁ = 1kg, u₁ = 10 ms^-1, m₂ = 5kg, u₂ = 0
Let v be the velocity of the combined object after the collision.
Total momentum just before the impact
= m₁u₁ + m₂u₂  = 1 × 10 + 5 × 0 = 10kg ms^-1
Total momentum just after the impact = (m₁ + m₂)v = (1 + 5)v
= 6v kg ms^-1 by conservation of momentum,
6v = 10
= v = \(\frac { 10 }{ 6 }\) = \(\frac { 5 }{ 3 }\) ms^-1
∴ Total momentum just after the impact
= 6 × \(\frac { 5 }{ 3 }\) = 10 ms^-1

Question 16.
An object of mass 100kg is accelerated uniformly from a velocity of 5 ms^-1 to 8 ms^-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Here, m = 100kg,
u = 5 ms^-1,
v = 8 ms^-1,
t = 6 s
Initial momentum, P₁ = mu = 100 × 5 = 500kg ms^-1
Final momentum, P₂ = mu = 100 × 8 = 800kg ms^-1
The magnitude of the force exerted on the object.
F = \(\frac { { P }_{ 2 }-{ P }_{ 1 } }{ t } \) = \(\frac { 800-500 }{ 6 }\) = \(\frac { 300 }{ 6 }\) = 50 N.

Question 17.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
Both, the motorcar and insect experience the equal force and hence, a same change in their momentum. So, we agree with Rahul. But due to smaller mass or inertia, the insect dies.

Question 18.
How much momentum will a dumb – bell of mass 10kg transfer to the floor if it falls from a height of 80cm? Take its downward acceleration to be 10 ms^-2.
Answer:
Here, m = 10kg, u = 0,
s = 80cm = 0.80m,
a = 10 m/s^-2
Let v be the velocity gained by the dumb-bell as it reaches the floor.
As v² – u² = 2as
v² – 0² = 2 × 10 × 0.80 = 16
or
v = 4 ms^-1
Momentum transferred by the dumb-bell to the floor
p = mv = 10 × 4 = 40kg ms^-1

Force and Laws of Motion Additional Questions

Force and Laws of Motion Multiple Choice Questions

Question 1.
Which of the following statements is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing.
(b) Its velocity always changes.
(c) It always goes away from the earth.
(d) A force is always acting on it.
Answer:
(c) It always goes away from the earth.

Question 2.
According to the third law of motion, action and reaction ________ .
(a) Always act on the same body.
(b) Always act on different bodies in opposite directions.
(c) Have same magnitude and directions.
(d) Act on either body at normal to each other.
Answer:
(b) Always act on different bodies in opposite directions.

Question 3.
A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to ________ .
(a) Exert larger force on the ball.
(b) Reduce the force exerted by the ball on hands.
(c) Increase the rate of change of momentum.
(d) Decrease the rate of change of momentum.
Answer:
(b) Reduce the force exerted by the ball on hands.

Question 4.
The inertia of an object tends to cause the object ________ .
(a) To increase its speed.
(b) To decrease its speed.
(c) To resist any change in its state of motion.
(d) To decelerate due to friction.
Answer:
(c) To resist any change in its state of motion.

Question 5.
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is ________ .
(a) Accelerated
(b) Uniform
(c) Retarded
(d) Along circular tracks.
Answer:
(a) Accelerated

Question 6.
An object of mass 2kg is sliding with a constant velocity of 4 ms^-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is ________ .
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N.
Answer:
(b) 0 N

Question 7.
Rocket works on the principle of conservation of ________ .
(a) Mass
(b) Energy
(c) Momentum
(d) Velocity.
Answer:
(c) Momentum

Question 8.
A water tanker filled up to \(\frac { 2 }{ 3 }\) of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would ________ .
(a) Move backward
(b) Move forward
(c) Be unaffected
(d) Rise upwards.
Answer:
(b) Move forward

Question 9.
Which of the following represents example(s) of potential energy?
(a) A moving car
(b) A moving fan
(c) A book resting on the table
(d) Both (a) and (c).
Answer:
(d) Both (a) and (c).

Question 10.
Unit of force is ________ .
(a) Ampere
(b) Volt
(c) Joule
(d) Hertz.
Answer:
(c) Joule

Question 11.
Product of mass and acceleration of a body is called ________ .
(a) Acceleration
(b) Work
(c) Power
(d) Energy.
Answer:
(b) Work

Question 12.
Which of the following is correct about energy?
(a) Energy is not required to do work.
(b) Work can be expressed as Force × Displacement.
(c) Unit of power is joule.
(d) Power is the amount of work done per unit time.
Answer:
(c) Unit of power is joule.

Question 13.
An object of mass 3kg is falling from the height of 1m. The kinetic energy of the body will be when it touches the ground ________ .
(a) 29.4 N
(b) 29.4 J
(c) 30 N
(d) 15 J
Answer:
(b) 29.4 J

Question 14.
Two objects with masses 1kg and 9kg, and equal momentum. Calculate the ratios of their kinetic energies ________ .
(a) 3 : 1
(b) 9 : 1
(c) 1 : 1
(d) 1 : 2
Answer:
(b) 9 : 1

Question 15.
Considering air resistance negligible, the sum of potential and kinetic energies of the free falling body would be ________ .
(a) zero
(b) infinite
(c) would decrease
(d) remains fixed.
Answer:
(d) remains fixed.

Force and Laws of Motion Very Short Answer Type Questions

Question 1.
What do we call to the product of mass and velocity of an object?
Answer:
Momentum.

Question 2.
Define inertia.
Answer:
The property by which an object tends to remain in the state of rest or of uniform motion unless acted upon by some force is called inertia.

Question 3.
Which property has S.I. unit kilogram metres per second i.e., 1kg m/s?
Answer:
Momentum.

Question 4.
Give an example of scalar quantity.
Answer:
Mass.

Question 5.
Give an example of vector quantity.
Answer:
Momentum.

Question 6.
Calculate the total momentum of the bullet and the gun before firing.
Answer:
For both, it would be zero because both of them are at rest.

Question 7.
Which force slows down a moving bicycle when we try to stop?
Answer:
The force of friction.

Question 8.
Which kind of force of gravity work when an object is under free fall?
Answer:
Unbalanced force.

Question 9.
Which property of an object resist a change in their state of rest or motion?
Answer:
Inertia.

Question 10.
Which law of Newton is also known as Galileo’s law of inertia?
Answer:
First law.

Question 11.
Is force a vector quantity?
Answer:
Yes.

Question 12.
Which force of motion opposes motion of an object?
Answer:
Force of friction.

Question 3.
When action and reaction forces act on two different bodies, what kind of magnitude they have?
Answer:
Action and reaction forces act on two different bodies but they are equal in magnitude.

Question 14.
When gun moves in the backward direction, which kind of velocity is generated?
Answer:
Recoil.

Question 15.
Which factor of body is dependent on its mass?
Answer:
Inertia of a body depends on its mass.

Force and Laws of Motion Short Answer Type Questions

Question 1.
There are three solids made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have highest inertia?
Answer:
Steel has highest inertia because it has greatest density and greatest mass, therefore, it has highest inertia.

Question 2.
Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer.
Answer:
If the breaks are applied suddenly then, the balls will start rolling in the direction in which the train was moving. Due to the application of the brakes, the train comes to rest but due to inertia the balls try to remain in motion, therefore, they begin to roll. Direction and speed of all balls will not be same because the masses of the balls are not the same, therefore, the inertial forces are not same on both the balls. Thus, the balls will move with different speeds.

Question 3.
Two identical bullets are fired, one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?
Answer:
According to law of conservation of momentum or explanation by Newton’s laws of motion, light rifle will hurt the shoulder more.

Question 4.
A horse continues to apply a force in order to move a cart with a constant speed. Explain why?
Answer:
The force applied by the horse balances the force of friction

Question 5.
Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain.
Answer:
Law of conservation of momentum is applicable to isolated system (no external force is applied). In this case, the change in velocity is due to the gravitational force of earth.

Question 6.
In which of the following conditions work done will be equal to zero?
Answer:
In the absence of any one of the two conditions given below, work done will be equal to zero, that is work is not considered to be executed:

1. Force should act on the object.
2. Object must be displaced.

Question 7.
Define energy and explain its forms.
Answer:

1. Energy: Energy is the capacity of doing work. More the power, more will be energy and vice – versa. For example, a motorcycle has more energy than a bicycle.
2. Forms of energy: There are many forms of energy, such as kinetic energy, potential energy, mechanical energy, chemical energy, electrical energy etc.

Force and Laws of Motion Long Answer Type Questions

Question 1.
Give the formulation of work. In which conditions work can occur?
Answer:
Work = Force × Displacement
or W = F × s
where, W is work ‘F’ is force and ‘s’ is displacement.
If force, F = 0
Therefore, work done, W = 0, s = 0
If displacement, s = 0
Therefore, Work done, W = F × 0 = 0
It proves that, there are two conditions for work to occur or be done:

1. Force should act on the object.
2. Object must be displaced.

Question 2.
Give the conditions when work done become positive and negative.
Answer:
When force is applied in the direction of displacement, the work done is considered as positive.
i.e., W = F × s
When force is applied in opposite direction of displacement, the work done is considered as negative.
i.e., W = -F × s = -Fs.

Question 3.
Explain positive and negative work.
Answer:
1. Positive work:
If a force displaces the object in its direction, then the work done is positive.
Here,
W = Fd
Example:
Motion of ball falling towards ground where displacement of ball is in the direction of force of gravity.

2. Negative work. If the force and the displacement are in opposite directions, then the work is said to be negative.
Here,
W = -Fd.
Example:
If a ball is thrown in upward direction but the force due to earth’s gravity is in the downward direction.

Question 4.
A cyclist moving along a circular path of radius 63m completes three rounds in 3minutes.
1. The total distance covered by him during this time.
2. Net displacement of cyclist.
3. The speed of the cyclist
Answer:
1. Total distance covered
s = 2πr × t
s = 2πr × 3
= 2 × \(\frac { 22 }{ 7 }\) × 63 × 3 = 1188m

2. Displacement = Zero

3. Speed = \(\frac { Distance }{ Time }\)
= \(\frac { 1188 }{ 180 }\)
= 6.6 m/s.

Force and Laws of Motion Higher Order Thinking Skills (HOTS)

Question 1.
As per Newton’s third law, every force is accompanied by equal and opposite force. How then can anything move?
Answer:
According to the Newton’s third law, action and reaction are two equal and opposite forces but they act on different bodies. This make the motion of a body possible.

Question 2.
The passengers travelling in a bus fall ahead when a speeding bus stops suddenly. Why?
Answer:
When the speeding bus stops suddenly lower part of the body, a long with the bus comes to rest while the upper part tends to remain in motion due to inertia of motion. That is why passengers fall ahead.

Question 3.
A player always runs some distance before taking a jump. Why?
Answer:
A player always runs for some distance before taking a jump because inertia of motion helps him to take a longer jump.

Force and Laws of Motion Value Based Question

Question 1.
Sushil saw his karate expert breaking a slate. He tried to break the slate but Sushil’s friend stopped him from doing so and told him that it would hurt, one needs lot of practice in doing such activity.

1. How can a karate expert break the slate without any injury to his hand?
2. What is Newton’s third law of motion?
3. What value of Sushil’s friend is seen in the above case?

Answer:

1. A karate expert Sushil applies the blow with large velocity in a very short interval of time on the slate, therefore large force is exerted on the slate and it breaks.
2. To every action, there is an equal and opposite reaction, both act on different bodies.
3. Sushil’s friend showed the value of being responsible and caring friend.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3

Question 1.
In Fig. given below, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
Given
∠SPR = 135°
∠PQT = 110°
To find. ∠PRQ
Calculation:
∠Q = 110°
110° + ∠PQR = 180° (LPA’s)
∠1 = 70°
∠P = ∠1 + ∠PRQ
135° = 70° + ∠PRQ
135° – 70° = ∠PRQ
65° = ∠PRQ

Question 2.
In Fig. given below, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∠XYZ, find ∠OZY and ∠YOZ.

Solution:
Given
∠X = 62°, ∠XYZ = 54°
∠OYZ = \(\frac{1}{2}\)∠XYZ
∠OZY = \(\frac{1}{2}\)∠XZY
To find ∠OZY and ∠YOZ
Calculation:
In ∠XYZ
∠X + ∠Y + ∠Z = 180° (ASP)
62° + 54° + ∠Z = 180°
116° + ∠Z = 180°
∴ ∠Z = 64°
In ∠OYZ
∠OYZ + ∠OZY + ∠O = 180° (ASP)
⇒ \(\frac{1}{2}\)∠XYZ + \(\frac{1}{2}\)∠XZY + ∠O = 180°
32° + 27° + ∠O = 180°
59° + ∠O = 180°
∠O = 121°

Question 3.
In Fig. given below, if AS ∥ DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:
Given
AS ∥ DE,
∠BAC = 35°
∠CDE = 53°
To find. ∠DCE
Calculation:
AB ∥ DE and AE is the transversal.
∠BAE = ∠AED = 35°
In ∆DEC
∠D + ∠E + ∠C = 180°
53° + 35° + ∠C = 180°
88° + ∠C = 180°
∠C = 180° – 88° = 92°

Question 4.
In Fig. given below, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:
Given
∠PRT = 40°
∠RPT = 95°
∠TSQ = 75°
To find. ∠SQT
Calculation:
In ∆PRT
∠P + ∠R + ∠T= 180° (ASP)
95° + 40 ° + ∠T= 180°
∠T = 180° – 135° = 45°
∠T = ∠STQ = 45°
In ∆TSQ
∠STQ + ∠S + ∠Q = 180° (V.O.A’S)
45° + 75° + ∠Q = 180° (ASP)
∠Q = 180° – 120° = 60°

Question 5.
In Fig. given below, if PQ ⊥ PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:
Given
PQ ⊥ PS
PQ ∥ SR
∠SQR = 28°
∠QRT = 65°
To find, x and y
Calculation:
PQ ∥ SR and QR is the transversal
x + 28° = 65° (A.I.A’s)
x = 65° – 28° = 37°
In ∆PQS
x + y + 90° = 180° (ASP)
37° + y + 90° = 180°
∴ y = 180° – 127° = 53°

Question 6.
In Fig. given below, the side QR to ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.

Solution:
Given
∠TQR = \(\frac{1}{2}\)∠PQR
∠TRS = \(\frac{1}{2}\)∠PRS
To prove. ∠QTR = \(\frac{1}{2}\)∠QPR
Proof:
In ∆PQR
∠PRS = ∠P + ∠Q (EAP)
In ∆TQR, ∠TRS = ∠T + ∠TQR (EAP)
\(\frac{1}{2}\)∠PRS = ∠T + \(\frac{1}{2}\)∠PQR
Multiplying both sides by 2
∠PRS = 2∠T+ ∠PQR
From (1) and (2) we get,
∠P + ∠Q = 2∠T + ∠PQR
∠P=2∠T
∠T= \(\frac{1}{2}\)∠P
∠QTR = \(\frac{1}{2}\)∠QPR
Proved.

MP Board Class 9th Maths Solutions

MP Board Class 9th Special English Short Composition with Guidance

1. A reporter made the following notes on a theft case news item. Complete the news using the given notes. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes Place — dace — time Market — Raghav Colony — 12-2-20xx — 3 A.M. Items stolen — jewelry — worth 3 lacs — cash 50 thousand — electronic goods — broken window — no clue.

“Theft At Raghav Colony”

It was reported that a theft (a) ……………………. on 12-2-20xx at 3-00 A.M. near the (b) ……………………. place. The burglars (c) ……………………. Some (d) ……………………. were reported to be missing from the house. So far (e) ……………………. found about the thieves.
Answer:
(a) took place in Raghav Colony
(b) market
(c) gained access into the house through the broken window.
(d) electronic goods jewelry worth 3 lacs and 50 thousand cash
(e) no clue has been.

2. A reporter made the following notes for a news item in the newspaper. complete the news using the given notes. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes Train Accident — Level crossing — 2 P.M. — 1.3.20xx 12 KM — Nagpur — collides with a truck — a high speed — 4 dead — several hurts.

I Accident At Level Crossing

I At a level crossing near Nagpur (a) ……………………… away a rail accident (b) ……………………… The train was running at (c) ……………………… It was (d) ……………………… on (e) ……………………… The train driver saw the truck and gave (f) The truck was also (g) ……………………… Its driver did not (h) ……………………… and tried to (i) ……………………… but failed. The driver of the truck was (j) ……………………… of those killed. The injured were
Answer:
(a) 12 KM (kilometers)
(b) occurred
(c) a high speed
(d) a collision with a truck
(e) the level crossing
(f) a signal
(g) at high speed
(h) pay heed to the signal
(i) the level crossing
(j) among one.

3. Salim made the following notes for the report regarding ‘Onion procession’ held in the town. Making use of the notes, complete the report by filling in the blanks. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes Procession — Cars — Jeeps — People — Delhi Gate — Connaught Place — Leader — Open jeep Balloon — Flower garlands — Onion garlands — Onions of different sizes — Huge balloon — shape of onion.

Onions Draw Crowds

In the form of huge a (a) ………………… threaded in (b) ………………… adorning every neck which made the long (c) ………………… proceeding from Delhi Gate to Connaught atop a car, onions appeared Place, it was a great attraction. The cavalcade was led by an (d) ………………… jeeP on which Mr. Dutta stood with (e) ………………… around his neck addressing the crowd against the rising prices.
Answer:
(a) balloon in the shape of an onion
(b) the garlands
(c) procession
(d) open
(e) a garland of onions.

4. Here are a few details about Gandhiji. Use this information to make a paragraph. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Birth — 2-10-1869 Residence — Rajkot — Gujarat. Parents — Father — Diwan, Mother — religious woman Education — India and England Profession — Lawyer — South Africa

Gandhiji was (a) ………………… They lived in (b) ………………… His father (c) ………………… and his mother! He got his education (d) ………………… He became (e) ………………… and
Answer:
(a) born on 2nd October 1869.
(b) Rajkot in Gujarat.
(c) was a Diwan and his mother was a religious woman.
(d) in India and England.
(e) a lawyer and went to South Africa.

5. Karishma got the grades given below. She asked her teacher to give her a written report on her abilities and achievements. Given below is a part of the report her teacher wrote. Based on the grades complete the sentences. Do not add any new information. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Karishma is a very bright girl who shows great application and diligence in class. Her special (a) ………………… and Computer Science where (b) ………………… this-year. She should, however, (c) ………………… in the Languages, (d) ………………… and is able to work as a team. (e) ………………… in harmony. Always neatly turned out, Karishma is a pleasure to have in class.
Answer:
(a) interests lie in Mathematics, Art
(b) she has scored A+ grades
(c) improve upon her grades
(d) In her music lessons and physical education classes she can improve
(e) Her leadership qualities, application and team work are

6. Prabha is a reporter for a paper. She made these notes for a news item. Read them and complete the news item given below. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes local train — traveling at full speed — Canning to Talki — Gauranga — 7 yr. old — tending cattle nearby — saw a break in track — took off and waved red shorts like a flag — driver noticed, braked, stopped — avoided the accident.

Seven Year Old Averts Train Accident

New Delhi, April 18
A local train was traveling at full speed from Canning to Talki in the Sealdah South section. Gauranga (a) ………………… near the railway line, saw that (b) ………………… at one place. He immediately (c) ………………… and waved it (d) ………………… rushing towards the train. The driver noticed the boy (e) ………………… just before the point of danger. Gauranga’s alertness saved hundreds of lives.
Answer:
(a) a 7-year-old boy who was tending cattle
(b) the track had broken
(c) took off his red shorts
(d) like a flag
(e) braked and stopped the train.

7. Meera, the editor of her school magazine, has made the sched-ule given below for the school magazine. She has written a note to her friends on the editorial board based on it. Complete the gaps left in the note. Do not add any new information. Write the answers in your answer sheet against the correct blank numbers. Do not copy the whole sentence. (5 marks)

Magazine Schedule For 2007-08 ”

Collection of material : through competitions in April-Aug.’ 07 (short story, poetry, cartoons) collect articles, cartoons comics in Aug. Sept’ 07 Reporters write on school activities as they occur. Selection of material : Jul.-Sept.’ 07 Editing and Illustrations: Aug.-Nov.’ 07 – Printing : Hand material to printer Nov.’ 07 Final proofs Feb.’ 08 delivery of magazine : March 08

This yeir (a) ……………………….. printed by March 2008, so we have to start our work immediately. Different (b) ……………………….. between April and August 2007 so that we can collect enough good irìterial for inclusion in the magazine. All material to he printed in the Åfl agazine should (c) ……………………….. by September 2007. Of course, the reporters on the editorial board will (d) ……………………….. school, activities as they occur. The magazine should be rediy and printed (e) ………………………..
Answer:
(a) we intend to have the school magazine
(b) competitions will be organised
(c) be selected
(ii) give in their write-ups on
(e) by February 1997.

8. Read the information given in the columns below and then fill in gaps in the paragraph appropriately. Do not use any extra information. Write the answers against the correct blank numbers in your answer sheet. (5 Marks)

Amrita and Salma are good friends though they are very different ‘from each other. Amrita is quiet and reserved though she likes the company of friends, (a) ……………………… and likes to talk a lot Salma also likes (b) ……………………… Amrita on the other hand, prefers Indian food. Amrita (c) ……………………… very well and reads a lot while Salma is a champion (d) ……………………… to relax. Even in television programmes their tastes are quite different. However, (e) ……………………… in classical music and old film songs and often attend concerts together.
Answer:
(a) Salma, on the other hand, is cheerful and outgoing
(b) all kinds of food, especially junk food
(c) can play the sitar
(d) swimmer and loves to do painting in order
(e) their choice of music is similar and both are interested.

9. Here is a diary extract of a tourist about his tour to Rajasthan. Use the information to complete the account of his visit to Rajasthan. Do not add any new information. Write the answers in your answer sheet against the correct blank number. (5 marks)
5th March :
Arrival at Jaipur
Visit to Hawa Mahal
Man Singh Art Gallery
Amer Fort: Examples of advanced architecture

7th March :
Arrival at Ajmer
Visit to Pushkar
Dargah

8th March :
Arrival at Udaipur
Visit to Palace, Temples (incl. Jain Temples)
Lakes, Boating, Chittorgarh, .
fort mines, Pratap Memorial — all very fascinating

10th March :
To Jodhpur forts, palaces

12th March :
Visit to the Jaisalmer desert.
Evidence of bravery, generosity of people, rulers

People: Cooperative

Visit To Land Of Forts And Palaces
My visit to Jaipur, Ajmer, Udaipur, Jodhpur and Jaisalmer made me feel that Rajasthan ,(a) …………………… During my visit to Jaipur I (b) …………………… and I concluded that (c) …………………… I was extremely thrilled during my visit to Udaipur. I enjoyed (d) …………………… The story of Rana Pratap and Rani Padmavati of Chittorgarh made me feel that (e) ……………………
Answer:
(a) is a place where the generosity and co-operative attitude of the people is most evident.
(b) visited the Hawa Mahal, the Man Singh Art Gallery and Amer Fort.
(c) these were excellent examples of advanced architecture.
(d) the fascinating sights of the palaces/ the Jain Temples, Chittorgarh and Pratap Memorial.
(e) the rulers of the place were indeed brave.

10. Aparijate has been asked to write an article for the ‘Nature’ section of the School magazine. She has referred to an encyclopedia and made the following notes about the ‘golden monkey’. Use the information in the notes to complete her article. Do not add any new information. Write the answers against the correct blank number in your answer sheet. (5 Marks)
Animal : Golden Monkey
Habitat : mountain forests — China
Size : Adult male 83 cm Adult feftiale 74 cm
Weight : Male 20 kg Female 10 kg
Movement : Winter grcfcips — 70 Summer groups — 300 (food plentiful)
Endangered : hunting, golden hair — make coats
Protection : reseryes, sanctuaries

The golden monkey is mainly found (a) ……………………. An adult male is 83 cm long and weighs. 20 kg. (b) ……………………. than the male beihg 74 cm long and weighing 10 kg. In winter (c) ……………………. while in summer (d) ……………………. they form groups of 300. In the past the monkeys were hunted for their gplden hair. Now-a-days (e) …………………….
Answer:
(a) in the mountain forests of China.
(b) An adult female is comparatively smaller
(c) they form groups of 70
(d) when food is in plentiful
(e) they are protected in reserves and sanctuaries

MP Board Class 9th English Solutions

MP Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Question 1.
In Fig. given below, find the values of x and y and then showthat AB ∥ CD.

Solution:
∠y = 130° (VOA’s)
x + 50° = 180° (LPA’)
∴ x = 180° – 50° = 130°
∠x and ∠y are alternate interior angles and are equal AB ∥ CD

Question 2.
In Fig. given below, if AB ∥ CD, CD ∥ EF and y : z = 3 : 7, find x.

Solution:
Given
AB ∥ CD
CD ∥ EF
y : z = 3 : 7
∴ y = 3k
and z = 7k
To find, x
Solution:
AB ∥ EF and PQ is the transversal
x = z (A. I.A’s)
AB ∥ CD and PQ is the transversal
∴ x + y = 180° (C.I.A.’s)
z + y = 180° (∴ x = z)
7k + 3k = 180°
∴ k = 18°
y = 3 x 18° = 54°
2 = 7 x 18° = 126°
x = z = 126°

Question 3.
In Fig. given below, if AS ∥ CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:
Given
AB ∥ CD
EF ⊥ CD
To find
∠AGE, ∠GEF, ∠FGE
AB ∥ CD and GE is the transversal
∴ ∠AGE = ∠GED = 126°
∠AGE + ∠FGE = 180°
126° + ∠FGE = 180°
∠FGE = 54°
∠GED = ∠GEF + ∠FED
126° = ∠GEF + 90°
126° – 90° = ∠GEF
∠GEF = 36°

Question 4.
In Fig. given below, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Solution:
Given
PQ ∥ ST
∠PQR = 110°; ∠TSR = 130°
To find
∠QRS Construction.
Draw a line EF ∥ PQ passing through point R.
Calculation. PQ ∥ EF (by construction)
PQ ∥ ST (Given)
EF ∥ ST (lines ∥ to the same line are parallel to each other)
PQ ∥ EF and QR is the transversal
110° + ∠1 = 180° (CIA’s)
∴ ∠1 = 70°
ST ∥ EF and SR is the transversal
130° + ∠3 = 180° (CIA’s)
∠3 = 50°
∠1 + ∠2 + ∠3 = 180° (angles on the same line)
70° + ∠2 + 50° = 180°
∠2 = 180° – 120°
∴ ∠QRS = 60°

Question 5.
In Fig. given below, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Given
AB ∥ CD
∠APQ = 50°; ∠PRD = 127°
To find, x and y
Calculation:
AB ∥ CD and PQ is the transversal
∠APQ = x = 50° (AIA’s)
y + 50° = 127°
∴ y = 77°

Question 6.
In Fig. given below, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB ∥ CD.

Solution:
Given
PQ ∥ RS
To prove:
AB ∥ CD
Construction:
Draw MB and NC normals at point B and C respectively.
Proof:
∠1 = ∠2
(Angle of incident = Angle of reflection)
and ∠3 = ∠4
∠ABC = 2∠2 and ∠BCD = 2∠3

MB ⊥ PQ and NC ⊥ RS
MB ∥ CN (Lines perpendicular to two parallel lines are parallel to each other)
MB ∥ CN and BC is the transversal
∠2 = ∠3 (AIA’s)
⇒ 2∠2 = 2∠3
(on multiplying both sides by 2)
⇒ ∠ABC = ∠BCD
∠ABC and ∠BCD are alternate interior angles and are equal AB ∥ CD

Angle Sum Property of a Triangle:

1. Triangle:
A plane figure bounded by three line segments in a plane is called a triangle. A triangle has six parts –

• Three sides – AB, BC and AC
• Three angles – ∠A, ∠B and ∠C

2. Median:
The median of a ∆ corresponding to a particular side is the line segment joining the midpoint of that side to its opposite vertex. In Fig., D, E and F are the midpoints of sides BC, AC and AB respectively. AD, BE and CF are the medians.

3. Incentre of a triangle:
The incentre of a ∆ is the point of intersection of angle bisectors of the triangle. In Fig., OA, OB and OC are the angle bisectors of ∠A, ∠B and ∠C respectively. These angle bisectors intersect at point O. Therefore, point O is the incentre of ∆ABC.

4. Circumcenter of a triangle:
The circumcenter of a triangle is the point of intersection of the perpendicular bisectors of the sides of the triangle. In Fig., O is the circumcenter of ∆ABC where the perpendicular bisectors of AB, BC and AC intersect.

Theorem 1.
The sum of the angles of a triangle is 180°.
Given
ABC is a ∆ in which ∠1, ∠2 and ∠3 are angles.
To prove:
∠1 + ∠2 + ∠3 = 180°

Construction:
Draw a line PQ passing through point A parallel to BC.
Proof:
PQ is a line.
∠4 + ∠1 + ∠5 = 180° (angles on the same line) …..(i)
PQ ∥ BC and AB is the transversal
∴ ∠4 = ∠2 (AIA’s) …..(ii)
PQ ∥ BC and AC is the transversal
∴ ∠5 = ∠3 (AIA’s) …..(iii)
Putting ∠4 = ∠2 and ∠5 = ∠3 in (i), we get
∠2 + ∠1 + ∠3 = 180°
∴ ∠1 + ∠2 + ∠3 = 180°. Proved

Exterior Angle Property:

Theorem 2.
If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two opposite interior angles. This is also known as exterior angle property (EAP).
Given
ABC is a ∆ in which side BC is produced to point D.
To prove. ∠4 = ∠1 + ∠2

Proof:
In ∆ABC,
∠1 + ∠2 + ∠3 = 180° (ASP)
∠3 + ∠4 = 180° (LPA)
From (i) and (ii), we get
∠1 + ∠2 + ∠3 = ∠3,+ ∠4
∠1 + ∠2 = ∠4. Proved

Corollary

Theorem 3.
An exterior angle of a triangle is greater than either of the opposite interior angles.

Given
In ∆ABC
∠4 = ∠1 + ∠2 (EAP)
∠4 > ∠1 and ∠4 > ∠2
Hence an exterior angle of A is greater than either of the opposite interior angle.

Lines And Angles:

Example 1.
In the Fig., AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Find the value of y.
Solution:
Let ∠DAB = x and ∠BAC = 3x
In ∆ABD
AB = BD
∠ADB = ∠DAB = x
(∴ In a ∆ angles opposite to equal sides are equal)
∠ABC = x + x = 2x
(In a ∆ Exterior angle is equal to sum of two opp. interior angles) In ∆ABC
2x + 3x + y = 180° (ASP)
5x + y = 180° …..(i)
In ∆ADC
x + y = 180° (ASP)
y = 180° – X …..(ii)

Putting y = 180° – x in (i), we get
5x + 108° – x = 180°
4x + 108° = 180°
4x = 180° – 108°
= 72°
∴ x = \(\frac{72}{4}\) = 18°
Putting x = 18° in (ii), we get
y = 108° – 18° = 90°.

Example 2.
In the Fig., given below, find the value of x.
Solution:
Given
∠ABD = 100°
∠ACE = 115°

To find. x°
∠ABC + ∠ABD = 180° (LPA’s)
∠ABC + 100° = 180°
∠ABC = 180° – 100° = 80°
∠ACB + ∠ACE = 180°
∠ACB + 115° = 180°
∠ACB = 180° – 115° = 65°
Now, ∠A + ∠ABC + ∠ACE = 180° (ASP)
⇒ x° + 80° + 65° = 180°
⇒ x° + 145° = 180°
x° = 180° – 145° = 35°

Example 3.
In Fig., PQ ⊥ PS, PQ ∥ SR, ∠SQR = 25° and ∠QRT = 65°.
Solution:
Given
∠SQR = 25°
∠QRT = 65°

To find x and y
In ∆SQR
∠QRT = ∠SQR + ∠QSR (FAP)
⇒ 65° = 25° + ∠QSR
⇒ 65° – 25° = ∠QSR
∠QSR = 40°
PQ ∥ SR and SQ is the transversal.
∴ ∠PQS = ∠QSR (AIA’s)
∴ x = 40°

Example 4.
In Fig., AM ⊥ L BC and AN is the bisector of ∠A. Find the measure of ∠MAN.

Solution:
In ∆AMB
∠ABM + ∠AMB + ∠MAB = 180°
65° + 90° + ∠MAB = 180°
155° + ∠MAB = 180°
∠MAB = 180° – 155° = 25°
In ∆BAC
∠A + ∠B + ∠C = 180° (ASP)
∠A + 65° + 30° = 180°
∠A + 95° = 180°
∠A = 180° – 95° = 85°
∠MAN = ∠BAN – ∠MAB
= \(\frac{1}{2}\) ∠A – 25°
= \(\frac{85^{\circ}}{2}\) – 25 = \(\frac{85^{\circ}-50^{\circ}}{2}\)
∠MAN = \(\frac{35^{\circ}}{2}\) = 17.5°

Example 5.
If the sides of a triangle are produced in order, prove that sum of the exterior angles so formed is equal to four right angles.
Solution:

∠EAB = ∠2 + ∠3 (EAP) …..(i)
∠FBC = ∠1 + ∠3 (EAP) …..(ii)
∠ACD = ∠1 + ∠2 (EAP) …(iii)
Adding (i), (ii) and (iii), we get
∠EAB + ∠FBC + ∠ACD = ∠2 + ∠3 + ∠1 + ∠3 + ∠1 + ∠2
= 2(∠1 + ∠2 + ∠3) = 2 x 180° = 360°. Proved

Example 6.
Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 90° + \(\frac{1}{2}\) ∠A.

Solution:
Given
∠1 = \(\frac{1}{2}\) ∠ABC
⇒ ∠ABC = 2∠1
∠2 = \(\frac{1}{2}\) ∠ACB
⇒ ∠ACB = 2∠2
To prove:
∠BOC = 90° + \(\frac{1}{2}\) ∠A
Proof:
In ∆BOC ∠O + ∠1 + ∠2 = 180°
∠1 + ∠2 = 180° – ∠O …..(i)
In ∆ABC
∠A + ∠ABC + ∠ACB = 180°
∠A + 2∠1 + 2∠2 = 180°
2∠1 + 2∠2 = 180° – ∠A
2(∠1 + ∠2) = 180° – ∠A
∠1 + ∠2 = \(\frac{180^{\circ}-\angle A}{2}\) = 90° – \(\frac{\angle A}{2}\) …..(ii)
From (i) and (ii) we get
180° – ∠O = 90° – \(\frac{\angle A}{2}\)
180° – 90° + \(\frac{\angle A}{2}\) = ∠O
90° + \(\frac{\angle A}{2}\) = ∠O
∴ ∠BOC = 90° + \(\frac{\angle A}{2}\)

Example 7.
In Fig. show that ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°
Solution:
In ∆ACE
∠A + ∠C + ∠E = 180° (ASP) …..(i)

In ABDF
∠B + ∠D + ∠F = 180° (ASP) …..(ii)
Adding (i) and (ii), we get
∠A + ∠C + ∠E + ∠B + ∠D + ∠F = 180° + 180°
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°

Example 8.
In Fig., ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = \(\frac{1}{2}\) (∠Q – ∠R)

Solution:
Given
∠Q > ∠R
∠QPA = ∠RPA
PM ⊥ QR
To prove:
∠APM = \(\frac{1}{2}\) (∠Q – ∠R)
Proof:
In ∆PMQ
∠Q + ∠QPM + ∠M= 180° (ASP)
∠Q + ∠QPA – ∠APM + 90° = 180°
(∴ ∠QPM = ∠QPA – ∠APM)
∠Q + ∠QPA – ∠APM = 180° – 90°
∠Q + ∠QPA – ∠APM = 90°
∠Q + ∠QPA – ∠APM = 90° …..(i)
In ∆PMR
∠R + ∠RPM + ∠PMR = 180°
∠R + ∠RPA + ∠APM + 90° = 180°
(∴ ∠RPM = ∠RPA + ∠APM)
∠R + ∠QPA + ∠APM = 180° – 90°
(∴ ∠RPA = ∠QPA)
∠R + ∠QPA + ∠APM = 90° …..(ii)
From (i) and (ii), we get
∠Q + ∠QPA – ∠APM = ∠R + ∠QPA + ∠APM
∠Q – ∠R = ∠APM – ∠QPA + ∠QPA + ∠APM
∠Q – ∠R = 2∠APM
∠APM = \(\frac{1}{2}\) (∠Q – ∠R) Proved

Example 9.
In fig., AB ∥ DC, ∠BDC = 30° and ∠BAD = 80°, find x, y and z.
Solution:
AB ∥ DC and BD is the transversal.
x = 30° (AIA’s)

In ∆ABD
x + y + 80° = 180° (ASP)
30° + y + 80° = 180°
110° + y = 180°
∴ y = 180° – 110° = 70°
y – 30° = 70 – 30° = 40°
In ∆BCD
30° + 40° + z = 180° (ASP)
[∴ y – 30° = 40°]
70° + z = 180°
∴ z = 180° – 70° = 110°

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

1. OB = OC
2. AO bisect ∠A.

Solution:
Given
AB = AC
∠1 = ∠2, ∠3 = ∠4
To prove:

1. OB = OC
2. ∠5 = ∠6

Proof:
In ∆ABC,
AB = AC
∠B = ∠C
\(\frac{1}{2}\) ∠B = \(\frac{1}{2}\) ∠C
∠1 = ∠3 or ∠2 = ∠4
In ∆OBC
∠2 = ∠4
and so OB = OC
(In a A, sides opposite to equal angles are equal)
In ∆ABO and ∆ACO
BO = CO (proved)
∠1 = ∠3 (proved)
AB = AC (given)
∆ABO = ∆ACO (by SAS)
and so ∠5 = ∠6 (by CPCT)

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. below). Show that AABC is an isosceles triangle in which AB =AC.

Solution:
Given
∠ABC = ∠ADC
To prove:
AB = AC
Proof:
In A ABD and A ACD
BD = CD (given)
∠ADB = ∠ADC (given each 90°)
AD = AD (common)
∴ ∆ABD = ∆ACD (BySAS)
and so AB = AC (by CPCT)

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. below). Show that these altitudes are equal.

Solution:
Given
AB = AC
∠E = ∠F (each 90°)
To prove: BE = CF
Proof:
In ∆ABE and ∆ACE
∠A = ∠A (common)
∠E = ∠F (each 90°)
AB = AC (given)
∆ABE = ∆ACE (byAAS)
and so BE = CF (by CPCT)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. below). Show that:

1. ∆ABE ≅ ∆ACF
2. AB = AC,

i. e., ABC is an isosceles triangle.

Solution:
Given
BE = CF
∠E = ∠F (each 90°)
To prove:

1. ∆ABE = ∆ACE
2. AB = AC

Proof:
In ∆ABE and ∆ACF
∠A = ∠A (common)
BE = CF (given)
∠E = ∠F (each 90°)
∆ABE = ∆ACF (by AAS)
and so AB = AC (by CPCT)

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. below). Show that ∠ABD = ∠ACD.

Solution:
Given
AB = AC
BD = CD
To prove
∠ABD = ∠ACD
Construction: Join AD
Proof:
In ∆ABD and ∆ACD
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD = ∆ACD (by SSS)
and so ∠ABD = ∠ACD (by CPCT)

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. below). Show that ∠BCD is a right angle.
Solution:
Given: AB = AC
AD = AB
To show: ∠BCD = 90°
i.e., ∠2 + ∠3 = 90°

Proof:
AB = AC …..(1)
AB = AD …..(2)
From (1) and (2), we get
AC = AD
In ∆ABC
AB = AC
∠1 = ∠2
(In a A, angles opposite to equal sides are always equal) …p) …(3)
In ∆ACD
AC = AD
∠3 = ∠4
(In a A, angles opposite to equal sides are always equal) …(4)
In ∆BCD
∠1 + ∠2 + ∠3 + ∠4 = 180° (ASP)
∠2 + ∠2 + ∠3 + ∠3 = 180°
(∴ ∠1 = ∠2, ∠3 = ∠4)
2 (∠2 + ∠3) = 180°
(∠2 + ∠3) = 90°
∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A – 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆BAC
AB =AC
∠B = ∠C = x
∠A + ∠B + ∠C= 180°
∠B + ∠C = 180° – 90°
∠B + ∠C = 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\)

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given
ABC is an equilateral ∆
i. e., AB = BC = AC
To prove
∠A = ∠B = ∠C = 60°
Proof:
In ∆BAC
AB = AC
∠B = ∠C
(In a A, angles opposite to equal sides are always equal) ……(1)
AC = BC
∠A = ∠B
(In a A, angles opposite to equal sides are always equal) …..(2)
From (1) and (2), we get,

∠A = ∠B = ∠C = x (say)
∠A + ∠B + ∠C = 180° (ASP)
⇒ x + x + x = 180°
⇒ 3x = 180°
⇒ x = \(\frac{180^{\circ}}{3}\) = 60°
∴ ∠A = ∠B = ∠C = 60

Theorem 7.4.
SSS (Side-Side-Side) Congruence Theorem:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given
In ∆s ABC and DEF we have,

AB =DE
BC = EE
and AC = DF
To prove:
∆ABC = ∆DEF
Construction:
Suppose BC is the longest side.
Draw EF such that EE = AB and FEG = ∠CBA.
Join GF and DG.

Proof:
In ∆s ABC and GEE, we have
AB = GE (Const.)
∠ABC = ∠GEF (Const.)
and BC = EF (Given)
∴ ∆ABC = ∆GEF (SAS Cong. Axiom)
∠A = ∠G (CPCT) …..(1)
and AC = GF (CPCT) …..(2)
Now AB = EG (Const.)
AB = DE (Given)
∴ DE = EG ……(3)
Similarly, DF = GF ……(4)
In ∆EDG
DE = EG (Proved above)
∴ ∠A = ∠2 (∠s opp. equal side) …..(5)
In ∆DFG
FD = FG (Proved above)
∴ ∠3 = ∠4 (∠s ppp. equal side) …..(6)
∴ ∠1 + ∠3 – ∠2 + ∠4 [From (5) and (6)]
i. e. ∠D = ∠G …..(7)
But ∠G = ∠A [From (1)]
∴ ∠A = ∠D …..(8)
In ∆s ABC and DEF,
AB – DE (Given)
AC = DF (Given)
∠A = ∠D [From (8)]
∆ABC ≅ ∆DEF (SAS Cong. Axiom)

Theorem 7.5.
RHS (Right Angle Hypotenuse Side) Congruence Theorem:
Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.
Given
In ∆s ABC and DEF,
∠B = ∠E = 90°
AC = DF
BC = EF.

∆ABC ≅ ∆DEF
Construction:
Produce DE to M so that
EM = AB, Join ME.
Proof:
In ∆s ABC and MEF
AB = ME (Const.)
BC = EE (Given)
∠B = ∠MEF (each 90°)
∴ ∆ABC = ∆MEF (SAS Cong. Axiom)
Hence ∠A = ∠M (CPCT) …(1)
AC = MF (CPCT) …(2)
Also AC =DF (Given)
∴ DF = MF
∴ ∠D = ∠M (∠s opp. equal side of ADFM) …(3)
From (1) and (3), we have
∠A = ∠D …..(4)
Now, in ∆s ABC and DEF, we have
∠A = ∠D [From (4)]
∠B = ∠E (Given)
∴ ∠C = ∠F …..(5)
Again, in ∆s ABC and DEF, we have
BC = EF (Given)
AC = DF (Given)
∠C = ∠F [From (5)]
∴ ∆ABC = ∆DEF (SAS Cong. Axiom)

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 7 Diversity in Living Organisms

Diversity in Living Organisms Intext Questions

Diversity in Living Organisms Intext Questions Page No. 80

Question 1.
Why do we classify organisms?
Answer:
Classification of organism make it easy to study the millions of organisms on this earth. Similarities among them is the basis to classify them into different classes. Classification makes study easier.

Question 2.
Give three examples of the range of variations that you see in life – forms around you.
Answer:
Variations observed in life are:

1. Size: Organisms vary greatly in size – from microscopic bacteria to elephants, whales and large trees.
2. Appearance: The colour of various animals is quite different. Number of pigments are found in plants. Their body – built also varies.
3. Life time: The life span of different organisms is varied.

Diversity in Living Organisms Intext Questions Page No. 82

Question 1.
Which do you think is a more basic characteristic for classifying organisms?
(a) the place where they live.
(b) the kind of cells they are made of. Why?
Answer:
(a) Different organisms may share same habitat but may have entirely different form and structure. So, the place where they live cannot be a basis of classification.

(b) The kind of cells they are. made of. Because placement of organism to other destination can create a easy confusion.

Question 2.
What is the primary characteristic on which the first division of organisms is made?
Answer:
The primary characteristic on which the first division of organisms is made is the nature of the cell – prokaryotic or eukaryotic cell.

Question 3.
On what basis are plants and animals put into different categories?
Answer:
Plants and animals are very different from each other but main basis to differentiate is “Mode of nutrition’’. Plants are autotrophs. They can make their food own while animals are heterotrophs which are dependent on others for food. Locomotion, absence of chloroplasts etc. also make them different.

Diversity in Living Organisms Intext Questions Page No. 83

Question 1.
Which organisms are called primitive and how are they different from the so-called advanced organisms?
Answer:
A primitive organism is the one which has a simple body structure and ancient body design or features that have not changed much over a period of time. As per the body design, the primitive organisms which have simple structures are different from those so – called advanced organisms which have complex body structure and organization.

Question 2.
Will advanced organisms be the same as complex organisms? Why?
Answer:
Yes, they are developed from same ancestor once. They have relatively acquired their complexity recently. There is a possibility that these advanced or ‘younger’ organisms acquire more complex structures during evolutionary time to compete and survive in the changing environment.

Diversity in Living Organisms Intext Questions Page No. 85

Question 1.
What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?
Answer:
The organisms belonging to kingdom Monera are unicellular and prokaryotic whereas the organisms belonging to Kingdom Protista are unicellular and eukaryotic. This is the main criterion of their classification.

Question 2.
In which kingdom will you place an organism which is single – celled, eukaryotic and photosynthetic?
Answer:
Kingdom Protista.

Question 3.
In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?
Answer:
In the hierarchy of classification, “species” will have the smallest number of organisms with a maximum of characteristics in common whereas “the kingdom” will have the largest number of organisms a Arthropoda.

Diversity in Living Organisms Intext Questions Page No. 88

Question 1.
Which division among plants has the simplest organisms?
Answer:
Division thallophyta.

Question 2.
How are pteridophytes different from the phanerogams?
Answer:
1. Pteridophyta: They have inconspicuous or less differentiated reproductive organs. They produce naked embryos called spores.
Examples:

• Ferns
• marsilea
• equisetum, etc.

2. Phanerogams: They have well developed reproductive organs. They produce seeds.
Example:

• Pinus
• cycas
• fir etc.

Question 3.
How do gymnosperms and angiosperms differ from each other?
Answer:
Gymnosperm:

1. They are non – flowering plants.
2. Naked seeds not enclosed inside fruits are produced.
3. Examples:
   • Pinus
   • Cedar
   • Fir
   • Cycas etc.

Angiosperm:

1. They are flowering plants.
2. Seeds are enclosed inside fruits.
3. Examples:
   • Coconut
   • Palm
   • Mango etc.

Diversity in Living Organisms Intext Questions Page No. 94

Question 1.
How do poriferan animals differ from coelenterate animals?
Answer:

Poriferan	Coelenterate
1. Mostly marine, non – motile.	1. Motile marine animals that either live in colonies or have a solitary life – span.
2. Cellular level of organisation.	2. Tissue level of organisation.
3. Spongilla, Euplectella etc.	3. Hydra, sea anemone.

Question 2.
How do annelid animals differ from arthropods?
Answer:

Annelids	Arthropods
1. Closed circulatory system	1. An open circulatory system
2. The body is divided into several identical segments	2. The body is divided into few specialized segments

Question 3.
What are the differences between amphibians and reptiles?
Answer:

Amphibian	Reptiles
1. They live at land and water both.	1. They are completely terrestrial.
2. Scales are absent.	2. Skin is covered with scales.
3. They lay eggs in water.	3. They lay eggs on land.
4. Example: frogs, toads and salamanders.	4. Example: lizards, snakes, turtles, chameleons etc.

Question 4.
What are the differences between animals belonging to the Aves group and those in the mammalia group?
Answer:
Most birds have feathers and they possess a beak.Mammals do not have feathers and the beak is also absent. Birds lay eggs. Hence, they are oviparous. Some mammals lay eggs and some give birth to young ones. Hence, they are both oviparous and viviparous.

Diversity in Living Organisms NCERT Textbook Exercises

Question 1.
What are the advantages of classifying organisms?
Answer:
Advantages of classification:

1. Better categorization of living beings based on common characters.
2. Easier study for scientific research.
3. Better understanding of human’s relation and dependency on other organisms.
4. Helps in cross breeding and genetic engineering for commercial purposes.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
Gross character will form the basis of start of the hierarchy and fine character will form the basis of further steps of single hierarchy.
Examples:

• Presence of vertebral column in human beings can be taken under vertebrata.
• Presence of four limbs makes them members of Tetrapoda.
• Presence of mammary glands keeps them under mammalia.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
Basis of classification:

1. Number of cells: unicellular or multicellular.
2. Complexity of cell structure: Prokaryote and Eukaryote.
3. Presence or absence of cell wall.
4. Mode of nutrition.
5. Level of organization.

Question 4.
What are the major divisions in the Plantae? What is the basis for these divisions?
Answer:
Major divisions of Kingdom Plantae:

Division	Basis for classification
1. Thallophyta or Algae	1. Thallus like body, plant body is not differentiated into roots, stems etc.
2. Bryophyta	2. Body is divided into leaf and stem, lack vascular tissue.
3. Pteridophyta	3. Body is divided into root, stem and leaf, lack seeds.
4. Gymnosperm	4. Seed bearing, naked seeds, lack flowers.
5. Angiosperm	5. Seed bearing covered seeds, produce flowers.

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
In plants, basic structure of their body is a major criteria based on which thallophytes are different from bryophytes. Apart from this, absence or presence of seeds is another important criteria. Gymnosperms and angiosperms are further segregated based on if seeds are covered or not. It is clear that it is the morphological character which makes the basis for classification of plants.

In animal, classification is based on more minute structural variations. So in place of morphology, cytology forms the basis. Animals are classified based on layers of cells, presence or absence of coelom. Further, higher hierarchy animals are classified based on the presence or absence of smaller features, like presence or absence of four legs.

Question 6.
Explain how animals in Vertebrata are classified into further subgroups.
Answer:
Vertebrata is divided into two super classes, viz. Pisces and Tetrapoda. Animals of pisces have streamlined body with fins and tails to assist in swimming. Animals of tetrapoda have four limbs for locomotion.
Tetrapoda is further classified into following classes:

1. Amphibia: Amphibians are adapted to live in water and on land. They can breathe oxygen through kin when under water.
2. Reptilia: These are crawling animals. Skin is hard to withstand extreme temperatures.
3. Aves: Forelimbs are modified into wings to assist in flying. Beaks are present. Body is covered with feathers.
4. Mammalia: Mammary glands are present to nurture young ones. Skin is covered with hair. Most of the animals are viviparous.

Diversity in Living Organisms Additional Questions

Diversity in Living Organisms Tissues Multiple Choice Questions

Question 1.
Find out incorrect sentence.
(a) Protista includes unicellular eukaryotic organisms.
(b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms.
(c) Both Monera and Protista may be autotrophic and heterotrophic.
(d) Monerans have well defined nucleus.
Answer:
(d) Monerans have well defined nucleus.

Question 2.
Which among the following has specialised tissue for conduction of water?
(i) Thallophyta
(ii) Bryophyta
(iii) Pteridophyta
(iv) Gymnosperms
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv).
Answer:
(c) (iii) and (iv)

Question 3.
Which among the following produce seeds?
(a) Thallophyta
(b) Bryophyta
(c) Pteridophyta
(d) Gymnosperms.
Answer:
(d) Gymnosperms.

Question 4.
Which one is a true fish?
(a) Jellyfish
(b) Starfish
(c) Dogfish
(d) Silverfish.
Answer:
(c) Dogfish

Question 5.
Which among the following is exclusively marine?
(a) Porifera
(b) Echinodermata
(c) Mollusca
(d) Pisces.
Answer:
(b) Echinodermata

Question 6.
Which among the following animals have pores all over their body?
(a) Porifera
(b) Aves
(c) Mollusca
(d) Pisces.
Answer:
(a) Porifera

Question 7.
Which among the following have chi tin as cell wall?
(a) Sycon
(b) Yeast
(c) Jelly fish
(d) Euplectella.
Answer:
(c) Jelly fish

Question 8.
Which among the following is not a Monocotyledonous plant?
(a) Wheat
(b) Rice
(c) Maize
(d) Gram.
Answer:
(d) Gram.

Question 9.
Which among the following is not a dicotyledonous plant?
(a) Wheat
(b) Sunflower
(c) Mango
(d) Gram.
Answer:
(a) Wheat

Question 10.
An organism with a single cell is called _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(c) Unicellular

Question 11.
The amphibians of the plant is _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(b) Bryophyta

Question 12.
Plant bearing naked seeds are _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Gymnosperm.
Answer:
(d) Gymnosperm.

Diversity in Living Organisms Very Short Answer Type Questions

Question 1.
Name a saprophyte and also tell, why are they called so.
Answer:
Aspergillus: They are called so because they obtain their nutrition from dead and decaying matter.

Question 2.
Why are lichens called dual organisms?
Answer:
Lichens are permanent symbiotic association between algae and fungi. Therefore, they are called dual organisms.

Question 3.
State the phylum to which centipede and prawn belong.
Answer:
Arthropoda.

Question 4.
Name one reptile with four – chambered heart.
Answer:
Crocodile.

Question 5.
Identify kingdom in which organisms do not have well defined nucleus and do not show multicellular body designs.
Answer:
Monera.

Diversity in Living Organisms Short Answer Type Questions

Question 1.
Why do we differentiate organism, give two main basis?
Answer:
Due to variation in various characteristics, we differentiate organism. Two main basis are mode of nutrition and habitat.

Question 2.
Which kingdom generate food on earth and initiate food chain?
Answer:
Plantae.

Question 3.
Which kingdom do not have cell wall to their cell?
Answer:
Animalia.

Question 4.
What do you understand by biodiversity?
Answer:
Biodiversity: The variety of living beings found in a particular geographical area is called biodiversity of that area. Amazon rainforests is the largest biodiversity hotspot in the world.

Question 5.
Why classification is required?
Answer:
Classification is necessary for the study of living beings in easy way. Without proper classification, it would be impossible to study millions of organisms which exist on this earth.

Question 6.
What was the basis of classification of Ancient Greek philosopher Aristotle?
Answer:
Aristotle classified living beings on the basis of their habitat. He classified them into two groups, i.e. those living in water and those living on land.

Question 7.
How can we divide organism on the basis of mode of nutrition ?
Answer:
On this basis, organisms can be divided into two broad groups, i.e. autotrophs and heterotrophs.

Question 8.
Define Monocotyledonous plants. Give examples.
Answer:
Monocotyledonous: There is single seed leaf in a seed. A seed leaf is a baby plant.
Examples:

• wheat
• rice
• maize, etc.

Question 9.
Give example of Dicotyledonous plants.
Answer:
Dicotyledonous plants: Mustard, gram, mango etc.

Question 10.
Give one difference between prokaryotes and eukaryotes
Answer:

1. Prokaryotes: When nucleus is not organized, i.e., nuclear materials are not membrane bound; the organism is called prokaryote.
2. Eukaryotes: When nucleus is organized, i.e., nuclear materials are membrane bound; the organism is called eukaryote.

Question 11.
What is the difference between unicellular and multicellular organism?
Answer:

1. Unicellular organism: An organism with a single cell is called unicellular organism.
2. Multicellular organism: An organism with more than one cell is called multicellular organism.

Question 12.
Write short notes on the following:
(a) Thallophyta
(b) Bryophyta
Answer:
(a) Thallophyta: The plant body is thallus type. The plant body is not differentiated into root, stem and leaves. They are known as algae also.
Examples:

• Spirogyra
• chara
• volvox
• ulothtrix etc.

(b) Bryophyta: Plant body is differentiated into stem and leaf like structure. Vascular system is absent, which means there is no specialized tissue for transportation of water, minerals and food. Bryophytes are known as the amphibians of the plant kingdom, because they need water to complete a part of their life cycle.
Examples:

• Moss
• marchantia.

Question 13.
What are cryptogams and phanerogams?
Answer:
Plant body is differentiated into root, stem and leaf. Vascular system is present. They do not bear seeds and hence are called cryptogams. Plants of rest of the divisions bear seeds and hence are called phanerogams.
Examples:

• Marsilear
• ferns
• horse tails etc.

Question 14.
How gymnosperms are different from angiosperms?
Answer:

1. Gymnosperms: They bear seeds. Seeds are naked i.e., are not covered. The word ‘gymnos’ means naked and ‘sperma’ means seed.
2. Angiosperms: The seeds are covered. The word ‘angios’ means covered. There is great diversity in species of angiosperm.

Question 15.
What is porifera?
Answer:
Porifera: These animals have pores all over their body. The pores lead into the canal system. They are marine animals. Examples:

• Sycon
• Spongilla
• Euplectella, etc.

Question 16.
What is coelenterata?
Answer:
Coelenterata: The body is made up of a coelom (cavity) with a single opening. The body wall is made up of two layers of cells (diploblastic).
Examples:

• Hydra
• jelly fish
• sea anemone, etc.

Question 17.
What is Platyhelminthes?
Answer:
The body is flattened from top to bottom and hence the name platyhelminthes. These are commonly known as flatworms. The body wall is composed of three layers of cells (triploblastic).
Example:

• Planaria
• liver fluke
• tapeworm etc.

Question 18.
What is Nematohelminthes and Annelida?
Answer:
Nematohelminthes: Animals are cylindrical in shape and the body is bilaterally symmetric and there are three layers in the body wall.
Example:

• Roundworms
• pinworms
• filarial parasite (Wuchereria) etc.

Annelida: True body cavity is present in these animals. The body is divided into segments and hence the name annelida.
Example:

• Earthworm
• leech etc.

Question 19.
Explain the followings:
(a) Arthropoda
(b) Mollusca
(c) Echinodermata
(d) Protochordata
(e) Chordata.
Answer:
(a) Arthropoda: Animals have jointed appendages which gives the name arthropoda. Exoskeleton is present which is made of chitin. This is the largest group of animals; in terms of number of species.
Examples:

• cockroach
• housefly
• spider
• prawn
• scorpion etc.

(b) Mollusca: The animal has soft body; which is enclosed in a hard shell. The shell is made of calcium carbonate.
Examples:

• Snail
• mussels
• octopus etc.

(c) Echinodermata: The body is covered with spines, which gives the name echinodermata. Body is radially symmetrical. The animals have well developed water canal system, which is used for locomotion.
Examples:

• Starfish
• sea urchins etc.

(d) Protochordata: Animals are bilaterally symmetrical, triploblastic and ceolomate. Notochord is present at least at some stages of life.
Examples:

• Balanoglossus
• herdmania
• amphioxus etc.

(e) Chordata: Animals have notochord, pharyngeal gill slits and post anal tail; for at least some stages of life. Phylum chordata is divided into many sub – phyla; out of which we shall focus on vertebrata.

Diversity in Living Organisms Long Answer Type Questions

Question 1.
What is the different levels of organizations in case multicellular organism?
Answer:
Level of organization: Even in case of multicellular organisms, there can be different levels of organization:
(a) Cellular level of organization: When a cell is responsible for all the life processes, it is called cellular level of organization.

(b) Tissue level of organization: When some cells group together to perform specific function, it is called tissue level of organization.

(c) Organ level of organization: When tissues group together to form some organs, it is called organ level of organization. Similarly organ system level of organization is seen in complex organisms.

Question 2.
“Classification of living organism is based on evolution.” Explain.
Answer:
It is a well – established fact that all the life forms have evolved . from a common ancestor. Scientists have proved that the life begun on the earth in the form of simple life forms. During the course of time, complex organism evolved from them. So, classification is also based on evolution. A simple organism is considered to be primitive while a complex organism is considered to be advanced.

Question 3.
Explain five kingdom classification by Robert Whittaker (1959).
Answer:
Five Kingdom Classification by Robert Whittaker (1959):
This is the most accepted system of classification. The five kingdoms and their key characteristics are given below:

1. Monera: These are prokaryotes; which means nuclear materials are not membrane bound in them. They may or may not have cell wall. They can be autotrophic or heterotrophic. All organisms of this kingdom are unicellular. Examples: bacteria, blue green algae (cyanobacteria) and mycoplasma.

2. Protista: These are eukaryotes and unicellular. Some organisms use cilia or flagella for locomotion. They can be autotrophic or heterotrophic. Examples: unicellular algae, diatoms and protozoans.

3. Fungi: These are heterotrohic and have cell wall. The cell wall is made of chitin. Most of the fungi are unicellular. Many of them have the capacity to become multicellular at certain stage in saprophytic. Some fungi live in symbiotic relationship with other organisms, while some are parasites as well.
Examples:

• Yeast
• penicillum
• aspergillus
• mucor etc.

4. Plantae: These are multicellular and autotrophs. The presence of chlorophyll is a distinct characteristic of plants, because of which they are capable of doing photosynthesis. Cell wall is present.

5. Animalia: These are multicellular and heterotophs. Cell wall is absent. They feed on decaying organic materials.

Diversity in Living Organisms Higher Order Thinking Skills (HOTS)

Question 1.
What are the differences between Platyhelminthes and Nematohelminthes?
Answer:

Platyhelminthes	Nematohelminthes
1. Form: They are flat in shape and are called flat worms.	1. They are cylindrical in form and are called round worms.
2. Sexuality: Animals are hermaphrodite.	2. Animals are uni – sexual.
3. Coelom: Platyhelminthes are acoelomate.	3. Nematohelminthes are pseudocoelomate.
4. Digestive Tract: It is incomplete.	4. It is complete

Question 2.
Differentiate between animals belonging to the Mammalia group and those in the Aves group.
Answer:
Differences between mammals and aves.

Mammals	Aves
1. Give birth to young ones except platypus and the echidna.	1. Lay eggs.
2. Mammary glands are present.	2. Mammary glands are absent.
3. Body covered with hair.	3. Body covered with feathers.
4. Sweat and sebaceous glands are present in the skin.	4. Sweat and sebaceous glands are not present in the skin.

Diversity in Living Organisms Value Based Questions

Question 1.
Ashish, a IX class student, was studying chapter, ‘Diversity in Living Organisms’. He thought that all the fungi are harmful as these spoil food and cause various diseases. However, his elder sister Dimple told him that not all fungi are harmful as these are also used in making bread, vitamins and medicines.

1. Name any fungus which is the source of some medicine.
2. Name any fungus which is used in bread making.
3. What value are displayed by Ashish’s sister?

Answer:

1. Pencillium.
2. Yeast.
3. Dimple acted as elder sister and enhanced his younger brother’s scientific knowledge about fungi and their functions.

Question 2.
Coral is getting diminished in all the oceans due to global warming. People in Goa island protects their coral by not allowing people / tourist to take it away.

1. What is the phylum of coral?
2. What is coral made up of?
3. What value of people in Goa island is reflected here?

Answer:

1. Coelenterates is the phylum of coral.
2. It is made up of calcium carbonate.
3. They reflect the value of being responsible citizen, respecting environment and nature.

MP Board Class 9th Science Solutions

MP Board Class 9th Science Solutions Chapter 4 Structure of the Atom

Structure of the Atom Intext Questions

Structure of the Atom Intext Questions Page No. 47

Question 1.
What are canal rays?
Answer:
Canal rays are positively charged radiations which led to the discovery of positively charged sub-atomic particle called proton. These rays were discovered by E. Goldstein.

Question 2.
If an atom contains one electron and one proton, will it carry any charge or not?
Answer:
The atom will not contain any charge and will be electrically neutral because both electron and proton will balance each other.

Structure of the Atom Intext Questions Page No. 49

Question 1.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Answer:
According to Thomson’s model, an atom consist of a positively charged sphere and electrons are embedded in it. So, both charges are equal which makes the atom electrically neutral.

Question 2.
On the basis of Rutherford’s model of an atom, which sub – atomic particle is present in the nucleus of an atom?
Answer:
Proton is the sub – atomic particle which is present in the nucleus of an atom.

Question 3.
Draw a sketch of Bohr’s model of an atom with three shells.
Answer:

Question 4.
What do you think would be the observation if the a-particle scattering experiment is carried out using a foil of a metal other than gold?
Answer:
The observations would be same as that of gold foil.

Structure of the Atom Intext Questions Page No. 49

Question 1.
Name the three sub – atomic particles of an atom.
Answer:

1. Positively charged – Protons
2. Negatively charged – Electrons
3. No charged – Neutrons.

Question 2.
Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
Answer:
Atomic mass = Number of protons + Number of neutrons
∴ 4 = 2 + Number of neutrons
∴ Number of neutrons = 4 – 2 = 2.

Structure of the Atom Intext Questions Page No. 50

Question 1.
Write the distribution of electrons in carbon and sodium atoms.
Answer:
Atomic number of Carbon = 6

Question 2.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Answer:
K shell is the 1 shell
So, n = 1
Then maximum electron’s = 2n² = 2 × (1)²
= 2 × 1 = 2
and L shell is the second shell.
So, n = 2
Then maximum electrons = 2(n)²
= 2 × (2)² = 8
∴ Total number of electrons = 2 + 8 = 10.

Structure of the Atom Intext Questions Page No. 52

Question 1.
How will you find the valency of chlorine, sulphur and magnesium?
Answer:
We know that valency is the number of electrons lost, gained or shared by atom to become stable or to complete 8 electrons in the shell.
Now, Chlorine,
Atomic number = 17

Then, it will take 8 – 7 = 1 electron to complete its shell.
∴ Its valency is ‘I’
Sulphur, Atomic number = 16

It will take 8-6 = 2 electrons to complete its shell.
∴ Its valency is ‘2’.
Magnesium, Atomic number = 12

It will lose 2 electrons from its outermost shell to become stable.
∴ Its valency will be ‘2’.

Structure of the Atom Intext Questions Page No. 52

Question 1.
If number of electrons in an atom is 8 and number of protons is also 8 then,
(i) What is the atomic number of the atom? and
(ii) What is the charge on the atom?
Answer:
(i) Number of electrons = 8 and,
Number of protons =8
Then, Atomic number = Number of protons = 8

(ii) Now, total electrons (-) = Total protons (+)
So, atom will be electrically neutral.

Question 2.
With the help of table 4.1 of Textbook, find out the mass number of oxygen and sulphur atom.
Answer:
From the table, we have,
Oxygen,
Mass Number = Number of protons + Number of neutrons
= 8 + 8 = 16
Sulphur,
Mass number = Number of protons + Number of neutrons
= 16 + 16 = 32.

Structure of the Atom Intext Questions Page No. 53

Question 1.
For the symbol H, D, and T tabulate three sub – atomic particles found in each of them.
Answer:
H, D, and T stand for protium, deuterium and tritium as isotopes of hydrogen atom.
Table:

Question 2.
Write the electronic configuration of any one pair of isotopes and isobars.
Answer:
Pair of isotopes: \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)
Electronic configuration:

Structure of the Atom NCERT Textbook Exercises

Question 1.
Compare the properties of electrons, protons and neutrons.
Answer:

Question 2.
What are the limitations of J.J. Thomson’s model of the atom?
Answer:
J.J. Thomson’s model explained the existence of positive charge in the form of sphere and electrons embedded in it. But, he was unable to explain the Rutherford’s gold foil experiment in which most of positive α – particles passed straight, existence of electrons in the circular path and protons at the centre of the atom.

Question 3.
What are the limitations of Rutherford’s model of the atom?

Answer:
Rutherford explained that electrons revolve in a circular path which is found contradictory in terms of stability of atom. Because electrons are negatively charged and when they move continuously in circular paths then they should lose their energies and finally, fall into the positively charged nucleus making atoms unstable and collapse.

Question 4.
Describe Bohr’s Model of the atom.
Answer:
Neils Bohr proposed the theory for model of the atom. It is explained as:

1. Atom is made up of three sub – atomic particles as electrons, protons and neutrons.
   
2. The electrons move round the nucleus in fixed circular paths called orbits or shells.
3. The orbits are represented by the letters K, L, M, N, or the number n = 1, 2, 3, 4.
4. Centre of the atom is called the nucleus.
5. Electrons do not radiate energies while revolving in the orbits.
6. Electrons gain energy when they jump from lower shell to higher shell and lose energy when they return down from higher energy level to lower energy level.

Question 5.
Compare all the proposed models of an atom given in this chapter.
Answer:

Question 6.
Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer:
Rules:

(a) Maximum electrons present in a shell is given by 2n² whereas n is the number of that shell.
Like,

• K Shell, n = 1 → 2n² = 2 × (1)² = 2
• L Shell, n = 2 → 2n² = 2 × (2)² = 8
• M Shell, n = 3 → 2n² = 2 × (3)² = 18
• N Shell, n = 4 → 2n² = 2 × (4)² = 32.

(b) The outermost shell can have maximum of 8 electrons.
(c) Electrons cannot be occupied in a shell till its inner shells or orbits are completely filled.

Question 7.
Define valency by taking examples of silicon and oxygen.
Answer:
Valency is the combining capacity of an atom to become electrically stable. Or It means how many electrons are lost or gained by an atom to become stable.
In Silicon,

It has 4 valence electrons.
So, it will lose 4 electrons to become stable.
∴ Its valency is 4.
In Oxygen,

It has 6 valence electrons.
So, it will gain 8 – 6 = 2 electrons to become stable.
∴ Its valency is 2.

Question 8.
Explain with examples:
(i) Atomic number
(ii) Mass number
(iii) Isotopes
(iv) Isobars
Give any two uses of isotopes.
Answer:
(i) Atomic number: It is equal to the total number of
protons in the nucleus of its atom.
E.g.,

• Carbon has 6 protons. So, its atomic number is 6.

(ii) Mass number: It is equal to the sum of total number of protons and neutrons in the nucleus.
E.g.,

• Sodium has 11 protons and 12 neutrons. So, its mass number is 11 + 12 = 23

(iii) Isotopes: These are atoms of the same element having same atomic number, but different mass number.
E.g.

• \(_{ 35 }^{ 79 }{ Br }\), \(_{ 35 }^{ 81 }{ Br }\), \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)

(iv) Isobars: These are the atoms of different elements having different atomic number but same mass number.
E.g.

• \(_{ 18 }^{ 40 }{ Ar }\), \(_{ 20 }^{ 40 }{ Ca }\), \(_{ 11 }^{ 24 }{ Na }\), \(_{ 12 }^{ 24 }{ Mg }\)

Use of Isotopes:

• Uranium isotope is used as a fuel in nuclear reactor for generating electricity.
• Sodium isotope is used to detect the blood clots.

Question 9.
Na+ has completely filled K and L shells. Explain.
Answer:
Atomic number of sodium (Na) is 11.

Now, if Na loses 1 electron then it will become Na⁺.

Now K shell can have maximum of 2 electrons and L shell can have maximum of 8 electrons.
Then, Na⁺ has completely filled K and L shell.

Question 10.
If bromine atom is available in the form of, say, two isotopes \(_{ 35 }^{ 79 }{ Br }\) (49.7%) and \(_{ 35 }^{ 81 }{ Br }\)Br (50.3%), calculate the average atomic mass of bromine atom.
Answer:
Average atomic mass of bromine atom
= 49.7% of atomic mass of \(_{ 35 }^{ 79 }{ Br }\) + 50.3% of atomic mass of \(_{ 35 }^{ 81 }{ Br }\)
= 49.7% of 79 + 50.3% of 81
= \(\frac { 49.7 }{ 100 }\) × 49 + \(\frac { 450.3 }{ 100 }\) × 81
= (39.263 + 40.743)u = 80.006u

Question 11.
The average atomic mass of a sample of an element X is 16.2u. What are the percentages of isotopes If \(_{ 8 }^{ 16 }{ X }\) and \(_{ 8 }^{ 18 }{ X }\) in the sample?
Answer:
Let the percentage of \(_{ 8 }^{ 16 }{ X }\) in sample be x% and percentage of \(_{ 8 }^{ 18 }{ X }\) in sample be (100 – x)%.
Now,
x% of 16 + (100 – x)% of 18 = 16.2


-2x + 1800 = 16.2 × 100 – 2x + 1800 = 1620
∴ -2x = 1620- 1800 = -180
x = \(\frac {180}{2}\) = 90.
∴ Percentage of  \(_{ 8 }^{ 16 }{ X }\) is 90% and percentage of \(_{ 8 }^{ 18 }{ X }\) is (100 – 90)% = 10%.

Question 12.
If Z = 3, what would be the valency of the element? Also, name the element.
Answer:
Z = 3
So, atomic number = 3 (∵ Z = atomic number)
∴ Electronic configuration = 2, 1
Valency = 1
The name of the element is lithium (Li).

Question 13.
Composition of the nuclei of two atomic species X and Y are given as under

Give the mass number of X and Y. What is the relation between the two species?
Answer:
Mass number of X = Protons + Neutrons = 6 + 6 = 12
And,
Mass number of Y = Protons + Neutrons = 6 + 8 = 14
Both species have same atomic number.
So, they are isotopes of the same element.

Question 14.
For the following statements, write T for True and F for False.

1. J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
2. A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
3. The mass of an electron is \(\frac {1}{2000}\) times that of proton.
4. An isotope of iodine is used for making tincture iodine which is used as a medicine.

Answer:

1. False
2. False
3. True
4. False.

Put tick (✓) against correct choice and cross (✗) against wrong choice in questions 15, 16 and 17.

Question 15.
Rutherford’s alpha – particle scattering experiment was responsible for the discovery of.
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron.
Answer:
(a) Atomic nucleus

Question 16.
Isotopes of an element have.
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.
Answer:
(c) different number of neutrons

Question 17.
Number of valence electrons in Cl^– ion are:
(a) 16
(b) 8
(c) 17
(d) 18
Answer:
(b) 8

Question 18.
Which one of the following is a correct electronic configuration of sodium?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1.
Answer:
(d) 2, 8, 1.

Question 19.
Complete the following table:

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Name of the Atomic Species
9	–	10	–	–	–
16	32	–	–	–	Sulphur
–	24	–	12	–	–
–	2	–	1	–	–
–	1	0	1	0	–

Answer:

Atomic Number	Mass Number	Number of Neutrons	Number of Pro-­tons	Number of Elec­trons	Name of the Atomic Species
9	19	10	9	9	Fluorine
16	32	16	6	6	Sulphur
12	24	12	12	12	Magnesium
1	2	1	1	1	Hydrogen
1	1	0	1	0	Deuterium

Structure of the Atom Additional Questions

Structure of the Atom Multiple Choice Questions

Question 1.
Which is a positive sub – atomic particle?
(a) Proton
(b) Neutron
(c) Electron
(d) None of these.
Answer:
(a) Proton

Question 2.
Electron is discovered by _____ .
(a) J.Chadwick
(b) Neils Bohr
(c) J.J Thomson
(d) Rutherford.
Answer:
(c) J.J Thomson

Question 3.
Proton is discovered by _____ .
(a) Rutherford
(b) J. Chadwick
(c) J J. Thomson
(d) E. Goldstein.
Answer:
(d) E. Goldstein.

Question 4.
Neutron is discovered by _____ .
(a) J.J. Thomson
(b) J. Chadwick
(c) Neils Bohr
(d) Rutherford.
Answer:
(b) J. Chadwick

Question 5.
Nucleus is discovered by _____ .
(a) Rutherford
(b) J. Chadwick
(c) J.J. Thomson
(d) Neils Bohr.
Answer:
(a) Rutherford

Question 6.
Mass of electron is _____ .
(a) 9 × 10^-25g
(b) 6 × 10^-28g
(c) 8 × 10^-24g
(d) 9 × 10^-28g.
Answer:
(d) 9 × 10^-28g.

Question 7.
Mass of Neutron is _____ .
(a) 1.6 × 10^-22g
(b) 1.6 × 10^-23g
(c) 1.6 × 10^-25g
(d) 1.6 × 10^-24g.
Answer:
(d) 1.6 × 10^-24g.

Question 8.
Charge on an electron is _____ .
(a) -1.8 × 10^-18C
(b) -1.7 × 10^-20C
(c) -1.6 × 10^-19C
(d) -1.5 × 10^-21C.
Answer:
(c) -1.6 × 10^-19C

Question 9.
The energy paths in an atom in which electrons revolve are called _____ .
(a) Rings
(b) Cycles
(c) Orbits
(d) Circles.
Answer:
(c) Orbits

Question 10.
ass number is the sum of _____ .
(a) Protons and Electrons
(b) Protons and Neutrons
(c) Electrons, Protons and Neutrons
(d) None of these.
Answer:
(c) Electrons, Protons and Neutrons

Question 11.
Atomic number is equal to _____ .
(a) Number of protons
(b) Number of neutrons
(c) Number of electrons
(d) Both (a) and (c).
Answer:
(d) Both (a) and (c)

Question 12.
Maximum number of electrons that can be filled in ‘M’ shell are _____ .
(a) 17
(b) 19
(c) 18
(d) 20.
Answer:
(c) 18

Question 13.
An atom has atomic number ‘17’, then its valency will be _____ .
(a) 7
(b) 2
(c) 1
(d) 8.
Answer:
(c) 1

Question 14.
Isotopes of an element have same number of _____ .
(a) Neutrons
(b) Protons
(c) Electrons
(d) Both (b) and (c).
Answer:
(c) Electrons

Question 15.
Isobars of different elements have same _____ .
(a) Atomic number
(b) Electrons
(c) Mass number
(d) Neutrons.
Answer:
(d) Neutrons

Structure of the Atom Very Short Answer Type Questions

Question 1.
Who discovered canal rays?
Answer:
E. Goldstein.

Question 2.
Name the fruit which resembles J.J. Thomson model of atom.
Answer:
Watermelon.

Question 3.
Who discovered nucleus?
Answer:
Ernest Rutherford.

Question 4.
Who discovered neutrons?
Answer:
James Chadwick.

Question 5.
Name the central part of an atom where protons and neutrons are held together.
Answer:
Nucleus.

Question 6.
What is Alpha Particle?
Answer:
It is a Helium ion (He²⁺) which has 2 units of positive charge and 4 units of mass.

Question 7.
What are cathode rays?
Answer:
Cathode rays are a beam of fast moving electrons.

Question 8.
What was the main drawback of Rutherford’s model of the atom?
Answer:
Inability to explain the stability of atom.

Question 9.
Write the symbolic representation of an element A with atomic number 10 and mass number 20.
Answer:
\(_{ 10 }^{ 20 }{ A }\)

Question 10.
Name three Isotopes of Hydrogen.
Answer:

1. Protium (\(_{ 1 }^{ 1 }{ H }\))
2. Deuterium (\(_{ 1 }^{ 2 }{ H }\))
3. Tritium (\(_{ 1 }^{ 3 }{ H }\))

Question 11.
Write the electronic configuration of potassium (K).
Answer:
Atomic number of potassium (K) =19

Question 12.
Define valency.
Answer:
It is the combining capacity of an atom to become electrically stable.

Question 13.
What is the charge of a proton?
Answer:
1.6 × 10^-19C.

Question 14.
Write the year of discoveries of these sub – atomic particles – electron, proton, neutron and neucleus.
Answer:

1. Electron – 1897
2. Proton – 1866
3. Neutron – 1932
4. Nucleus – 1911.

Question 15.
Which radioactive Isotope is used in treatment of goitre?
Answer:
Iodine – 131.

Structure of the Atom Short Answer Type Questions

Question 1.
Define:
(a) Canal rays
(b) Cathode rays
(c) Atomic number
(d) Mass number
(e) Energy shells
(f) Valency
(g) Octet
(h) Isotopes
(t) Isobars.
Answer:
(a) Canal rays: These are positively charged radiations which led to the discovery of sub – atomic positively charged particles called protons through an experiment conducted by J.J. Thomson in 1897.

(b) Cathode rays: These are negatively charged radiations which led to the discovery of sub – atomic negatively charged particles called electrons during an experiment conducted by E. Goldstein in 1866.

(c) Atomic number: It is the number of protons present in the nucleus of an atom. It is represented by the letter ‘Z’.

(d) Mass number: It is the total number of protons and neutrons present in an atom of an element.
So, Mass number = Number of protons + Number of neutrons.

(e) Energy Shells: These are fixed circular paths around the nucleus of an atom in which electrons revolve continuously with high speed. These are also called orbits. They are represented by the alphabets K, L, M, N.

(f) Valency: It is the combining capacity of an atom to become electrically stable, or it also means the number of valency electrons lost or gained by an atom to complete the eight electrons in the valence shell.

(g) Octet: The completely filled outermost shell like L, M or N with 8 electrons is called an octet. When an atom completes its octet, then it become stable.

(h) Isotopes: These are atoms of same element having same atomic number, but different mass number.

E.g.

• (\(_{ 1 }^{ 1}{ H }\)) , (\(_{ 1 }^{ 2 }{ H }\)), (\(_{ 1 }^{ 3 }{ H }\)) and \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)are the isotopes of hydrogen and carbon respectively.

(i) Isobars: These are atoms of different elements having different atomic number but same mass number.
E.g.

• \(_{ 18 }^{ 40 }{ Ar }\), \(_{ 20 }^{ 40 }{ Ca }\) and \(_{ 11 }^{ 24 }{ Na }\), \(_{ 12 }^{ 24 }{ Mg }\).

Question 2.
Differentiate between:
(a) Electrons and protons.
(b) Atomic number and mass number.
(c) Isotopes and isobars.
(d) Valence electrons and valency.
Answer:
(a)

Electrons	Protons
(i) This is negatively charged sub – atomic particle.	(i) This is positively charged sub – atomic particle.
(ii) Its mass is 9 × 10–28gms.	(ii) Its mass is 1.6 × 10-24gms.
(iii) Its symbol is “e–”.	(iii) Its symbol is “P+”.

(b)

Atomic number	Atomic mass
(i) It is the total number of protons present in the atom.	(i) It is the sum of protons and neutrons present in the atom.
(ii) It is represented by ‘Z’.	(ii) It is represented by ‘A’.
(iii) It is written on the bottom left as a subscript with the symbol the of element.	(iii) It is written on top left as a subscript with the symbol the of element.

(c)

Isotopes	Isobars
(i) These are atoms of the same element.	(i) These are atoms of the different elements.
(ii) They have same atomic number.	(ii) They have different atomic number.
(iii) They have different mass number.	(iii) They have same mass number.
(iv) They have same chemical properties.	(iv) They have different chemical properties.

(d)

Valence Electrons	Valency
(i) These are electrons present in the outermost shell of an atom.	(i) These are electrons lost or gained through valence shell of an atom to become stable.
(ii) Valence electrons can be 1, 2, 3 …….. 8 or more.	(ii) Valency can be 0, 1, 2, 3, 4 only.

Question 3.
Draw the diagrams of:
(a) J.J. Thomson’s model of atom.
(b) Rutherford’s model of atom.
(c) Neils Bohr’s model of atom.
Answer:
(a) J.J. Thomson’s Model:

(b) Rutherford’s Model:

(c) Neils Bohr’s model of atom.

Question 4.
Write the postulates of J.J. Thomson’s model of the atom.
Answer:
J.J. Thomson’s postulates for model of the atom are as follows:

1. An atom is a positively charged sphere or ball and negatively charged electrons are embedded in it.
2. The atom is electrically neutral because negative and positive charges are equal in magnitude.

Question 5.
Write the main points of the theory given by Rutherford for model of atom.
Answer:
Main points of theory of Rutherford regarding model of atom are:

1. There is an existence of positively charged centre in the atom called as nucleus which contains all the mass of the atom.
2. The electrons revolve round the nucleus in circular paths called orbits at high speeds.
3. The size of nucleus (centre of the atom) is very small as compared to size of the atom.
4. Most of the sphere in an atom is empty.

Question 6.
Write the rules given by Bohr – Bury for arrangement of electrons in different orbits in an atom.
Answer:
Rules given by Bohr – Bury are as follows:

1. Maximum electrons present in a shell is given by 2n² where
   • n is the number of that shell.
   • Like, for first shell K, n = 1
   • For second shell L, n = 2
   • third shell M, n = 3
   • fourth shell N, n = 4 called as nucleus which contains all the mass of the atom.
2. The outermost shell can have maximum of 8 electrons.
3. Electrons cannot occupy a shell till its inner shells or orbits are completely filled.

Question 7.
An atom ‘X’ has a mass number ‘23’ and atomic number ‘11’. Find its electrons, protons and neutrons. Also, name the element
Answer:
We know,
Atomic number = Number of protons.
∴ 11 = Number of protons
And, Number of protons = Number of electrons.
∴ Number of electrons = 11
Now, Mass number = Protons + Neutrons.
23 = 11 + Neutrons
∴ Neutrons = 23 – 11 = 12
∴ Atom ‘X’ has 11 electrons, 11 protons and 12 neutrons.
The element is Sodium (Na).

Question 8.
Write the electronic configuration of neon, aluminium, sulphur, argon. Also, find valencies.
Answer:

Question 9.
What are radioactive isotopes? Write down the type of isotopes used in:
(a) Tracing blood clots and tumours in human body
(b) Treatment of cancer
(c) Treatment of Goitre
(d) Nuclear reactor as a fuel.
Answer:
Radioactive isotopes: These are unstable isotopes due to extra neutrons in their nucleus and emits different types of radiations.
Examples:

• Uranium – 235
• Cobalt – 60
• Carbon – 14.

Types of Isotopes used in:
(a) Sodium – 24 to detect blood clots and Arsenic – 72 to detect tumours.
(b) Cobalt – 60
(c) Iodine-131
(d) Uranium – 235.

Question 10.
Draw the atomic structure of:
(a) Fluorine atom (F)
(b) Sodium atom (Na)
(c) Potassium atom (K)
Answer:
(a) Fluorine atom (F):
Atomic number: 9

(b) Sodium atom (Na)
Atomic number: 11

(c) Potassium atom (K)
Atomic number: 19

Question 11.
Write the atomic number, mass number, electrons, protons and neutrons of following atoms:
(a) \(_{ 14 }^{ 24 }{ X }\)
(b) \(_{ 13 }^{ 27 }{ X }\)
Answer:
(a) \(_{ 14 }^{ 24 }{ X }\)
Atomic number = 14
Mass number = 24
Electrons = 14
Protons = 14
Neutrons = 24 – 14 = 10

(b) \(_{ 13 }^{ 27 }{ X }\)
Atomic number = 13
Mass number = 27
Electrons = 13
Protons = 13
Neutrons = 27 – 13 = 14

Question 12.
Pick out the Isotopes and Isobars from the following atoms:
\(_{ 17 }^{ 37 }{ A }\), \(_{ 18 }^{ 40 }{ A }\), \(_{ 17 }^{ 33 }{ A }\), \(_{ 20 }^{ 40 }{ A }\).
Answer:

1. Isotopes: \(_{ 17 }^{ 37 }{ A }\), \(_{ 17 }^{ 33 }{ A }\)
2. Isobars: \(_{ 18 }^{ 40 }{ A }\), \(_{ 20 }^{ 40 }{ A }\).

Question 13.
What are noble gases? Why they are stable? Give three examples.
Answer:
Noble gases are the elements which are stable and do not take part in chemical reaction.
They are stable because they have completely filled outer- most shell with 8 electrons
example:

Helium is the only noble gas which has 2 electrons in outermost shell.

Question 14.
Write all the Isotopes of:

1. Hydrogen
2. Oxygen
3. Chlorine
4. Bromine
5. Carbon
6. Neon.

Answer:

1. Hydrogen (H) – \(_{ 1 }^{ 1 }{ H }\), \(_{ 1 }^{ 2 }{ H }\)
2. Oxygen (O) – \(_{ 8 }^{ 16 }{ O }\), \(_{ 8 }^{ 17 }{ O }\), \(_{ 8 }^{ 18 }{ O }\)
3. Chlorine (Cl) – \(_{ 17 }^{ 35 }{ Cl }\), \(_{ 17 }^{ 37 }{ Cl }\)
4. Bromine (Br) – \(_{ 35 }^{ 79 }{ Br }\), \(_{ 35 }^{ 81 }{ Br }\)
5. Carbon (C) – \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)
6. Neon (Ne) –  \(_{ 10 }^{ 20 }{ Ne }\), \(_{ 10 }^{ 21 }{ Ne }\), \(_{ 10 }^{ 22 }{ Ne }\)

Question 15.
Why is it wrong to say that atomic number of an atom is equal to its number of electrons?
Answer:
We know that in an atom number of electrons is equal to the number of protons. But, we cannot say that atomic number is equal to number of electrons because number of electrons can be changed after losing or gaining by an atom during chemical reaction. But, number of protons remain constant.

Question 16.
What explanation did Neils Bohr gave on stability of atoms?
Answer:
Neils Bohr explained the stability of atom through following points:

1. The electrons revolve around the nucleus in fixed orbits or energy levels or shells and each orbit has its fixed radius.
2. While revolving electrons do not radiate their energies, so they do not fall into the nucleus and make the atom stable.

Question 17.
What are nucleons? What is the name given to the atoms having same number of nucleons?
Answer:
Protons and neutrons together in the nucleus are called nucleons. It means number of nucleons is equal to the sum of protons and neutrons. Atoms having same number of nucleons are called isobars.

Structure of the Atom Long Answer Types Questions

Question 1.
The average atomic mass of a sample of an element X is 13u. What are the percentages of isotopes \(_{ 6 }^{ 12 }{ X }\) and \(_{ 6 }^{ 14 }{ X }\) in the sample?
Answer:
Let, the percentage of isotope \(_{ 6 }^{ 12 }{ X }\) be x%.
So, percentage of isotope \(_{ 6 }^{ 14 }{ X }\) is (100- x) %.
Now, Average atomic mass = Mass of \(_{ 6 }^{ 12 }{ X }\) + Mass of \(_{ 6 }^{ 14 }{ X }\) According to percentages,
∴ 13 = x% of 12 + (100 – x)% of 14

∴ 13 × 100 = 1400 – 2x
1300 = 1400 – 2x
1300 = 1400 – 2x
-100 = -2x
x = \(\frac { 100 }{2 }\) = 50
So, percentage of \(_{ 6 }^{ 12 }{ X }\) is 50% and percentage of \(_{ 6 }^{ 14 }{ X }\) is
= (100 – x)%
= (100 – 50)%
= 50%

Question 2.
Explain Rutherford’s Gold Foil Experiment. Also, explain its observations conclusion, theory proposed and drawback of his model.
Answer:
Ernest Rutherford performed an alpha-particles scattering experiment in which he passed a-particles on the gold foil.
Observation:

1. Most of α – particles passed straight without any deflection.
2. Some of the α – particles get deflected from their path.
3. Very few α – particles get completely bounced back.

Conclusions:

1. Maximum space in an atom is vacant as most of α – particles passed straight without any deflection.
2. Some α – particles get deflected from their paths show the existence of positive charge in the atom.
3. Very few α – particles get completely bounced back indicating the concentration of all mass with positive charge in a small volume at the centre.

Theory proposed:

1. There is an existence of positively charged centre in the atom called nucleus which contains all the mass of the atom.
2. The electrons revolve around the nucleus in circular paths called orbits at high speeds.
3. The size of nucleus (centre of the atom) is very small as compared to size of the atom.
4. Most of the space in an atom is empty.

Drawback: Rutherford’s model did not explain the stability of the atom. He proposed that electron revolves around the nucleus in circular paths. So, electrons should radiate their energies as they are continuously in circular motion. Then, they should fall into the positively charged nucleus making the atom unstable and collapse.

Question 3.
Draw the electronic structure of sodium and calcium with atomic number 11 and 20 respectively.
Answer:
Sodium has electronic distribution as 2, 8, 1
Calcium has electronic distribution as 2, 8, 8, 2
Electronic structures of sodium and calcium are given:

Question 4.
Both helium (He) and beryllium (Be) have two valence electrons. Whereas ‘He’ represents a noble gas element, ‘Be’ does not. Assign reason.
Answer:
The element He (Z = 2) has two electrons present in the only shell
i.e., K – shell. Since, this shell can have a maximum of two electrons only therefore,
‘He’ is a noble gas element.
The element ‘Be’ (Z = 4) has the electronic configuration as: 2,2.
Although, the second shell has also two electrons but it do not represent a noble gas element.

Structure of the Atom Higher Order Thinking Skills (HOTS)

Question 1.
Which isotope of hydrogen contain same number of electrons, protons and neutrons?
Answer:
Deuterium (\(_{ 1 }^{ 2 }{ D }\))
Number of electron (1) = Number of proton (1)
= Number of neutron (2 – 1 = 1)

Question 2.
Which element of these two would be chemically more reactive: element A with atomic number 18 or element B with atomic number 16 and why?
Answer:
Electric configuration of

• A – 2,8,8
• B – 2, 8, 6

Since, the outermost shell of A is complete, it would be inert and will not react. Whereas element B require two atoms to complete its octet. Therefore, B would be more reactive.

Structure of the Atom Value Based Question

Question 1.
Shivek could not solve the following question in the group. His group – mate explained him and solved his difficulty.
The question was as follows:

What information do you get from the given figure about the atomic number, mass number and valency of the given atom ‘X’:

1. What is the atomic number, the mass number and valency of the atom?
2. Name the element ‘X’.
3. What value of Shivek’s friend are reflected in this behaviour?

Answer:

1. The atomic number is 5, The mass number is 11, The valency is 3.
2. The element ‘X’ is boron.
3. Shivek’s friend showed the values of helping and caring nature.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 5 Introduction to Euclid’s Geometry Ex  5.2

Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution:
If a line p intersects two lines l and m such that (∠1 + ∠2) is less than 180°, then lines l and m will meet at O, as shown in Fig. below.

Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution:
Yes, Euclid’s fifth postulate is important to express parallel lines. Two lines will never meet if they are not according to Euclid’s fifth postulate.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 5 Introduction to Euclid’s Geometry Ex  5.1

Question 1.
Which of the following statements are true and which are false? Give reasons for your answers.

1. Only one line can pass through a single point.
2. There are infinite number of lines which pass through two distinct points.
3. A terminated line can be produced indefinitely on both the sides.
4. If two circles are equal, then their radii are equal.
5. In Fig. below, if AB = PQ and PQ = XY, then AB = XY.

Solution:

1. False, infinitely many lines can pass through a given point.
2. False, only one line can pass through two distinct points.
3. True, by postulate 2 i.e., a terminated line can be produced indefinitely.
4. True, equal circles coincide each other. Therefore their radii will be equal.
5. True, by Euclid’s axiom 1, i.e., things which are equal to the same thing are equal to one another.

Question 2.
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?

1. parallel lines
2. perpendicular lines
3. line segment
4. radius of a circle
5. square

Solution:
1. Parallel lines:
Two distinct lines in a plane are called parallel lines if they do not have a common point. Here the undefined terms are lines and plane.

2. Perpendicular lines:
Two lines are perpendicular to each other if they intersect each other at right angle. Here the undefined term is right angle.

3. Line segment:
A part of a line between two points on a line is called a line segment. Here the undefined term is part of a line.

4. Radius of a circle:
Radius of a circle is the distance of a point on the circle from the center of the circle.

5. Square:
A square is a rectangle having all sides equal. Here undefined term is rectangle.

Question 3.
Consider two ‘postulates’ given below:

1. Given any two distinct points A and B, there exist a third point C which is in between A and B.
2. There exist at least three points that are not on the same line.

Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Solution:
Yes, these postulates contain undefined terms such as point, line, distinct points. They are consistent because they deal with two different situations:

1. Point C is lying between two distinct points A and B on a line.
2. Point C is not lying on the line through A and B.

These postulates do not follow from Euclid’s postulates. However they follow from axiom “given two distinct points, there is a unique line that passes through them.

Question 4.
If a point C lies between two points A and B such that AC = BC, then prove that AC = \(\frac{1}{2}\)AB. Explain by drawing the figure.
Solution:
Given: AC = BC
To prove: AC = \(\frac{1}{2}\)AB
Proof:
AC = BC
Adding AC on both sides
AC + AC = BC + AC
2AC = AB
AC = \(\frac{1}{2}\)AB

Question 5.
In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one midpoint.
Solution:
If possible, Let us assume that a line segment AB has two mid points C and D when C is the mid point of AB

AC = 1/2 AB …(i)
where D is the mid point of AB
AD = 1/2 AB …(ii)
From (i) and (ii), we get
AC = AD
(By Euclid’s axiom, things which are half of the same thing are equal to one another.). This is possible only if C and D coincides.
∴ Our assumption that C and D are two mid points of AB are wrong and hence a line segment has one and only one mid point.

Question 6.
In Fig. below, if AC = BD, then prove that AB = CD.

Solution:
Given: AC = BD
To prove: AB = CD
Proof:
AC = BD
Subtracting BC on both sides, we get
AC – BC = BD – BC (By Euclid’s axiom-3)
∴ AB = CD

Question 7.
Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate).
Solution:
We know that whole is always greater than its part.

MP Board Class 9th Maths Solutions