Class 9

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 2 Equations of Motion Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 2 Notes Solutions Equations of Motion

SCERT Class 9 Physics Chapter 2 Notes Solutions Kerala Syllabus Equations of Motion Questions and Answers

Class 9 Physics Chapter 2 Let Us Assess Answers Equations of Motion

Question 1.
A car starts from rest and moves with uniform acceleration. Calculate the acceleration of the car if it covers a distance of 200 m in 20 s.
Answer:
Initial speed u = 0
distance travelled s = 200 m
time taken t = 20 sec
acceleration a = ?
Using the second equation
s = ut + \(\frac{1}{2}\) at²
⇒ 200 = (0)t + \(\frac{1}{2}\) a(20)²
a = 200/200 = 1 m s^-2

Question 2.
If an object starts from rest and moves with an acceleration of 2 m/s², what will be the velocity of the object after 10 s?
Answer:
Given, a = 2 m/s², time = 10 s
Initial velocity u=0
According to the equation of motion:
v = u + at
=0 + 2 × 10
= 20 m/s

Question 3.
Three different graphs related to the motion of a vehicle are given below. Analyse the graphs and find the characteristics of the motion.

Answer:

• Graph 1 shows a vehicle moving with uniform acceleration.
• Graph 2 shows a vehicle moving with uniform deceleration.
• Graph 3 shows a vehicle moving with non-uniform acceleration.

Question 4.
Graph related to the motion of Car A and Car B is given.
a) Which car has more acceleration? Why?
b) Redraw the graph by changing the scale and compare the graphs.

Answer:
a) Car A

When the scale of a graph changes, it affects the appearance and readability of the graph, but the underlying data and relationships remain the same.

Question 5.
Observe the figure.

A child runs along a circular path of circumference 440 m at a constant speed. The radius of the circular path is 70 m. The time taken to run from A and reach A in clockwise direction through B is 80 s. Find the distance, displacement, speed and velocity in each case in the table.
Answer:

Question 6.
A train starts from rest and attains a speed of 72 km/h in 5 minutes. Find the acceleration and displacement of the train.
Answer:
Given,
v = 72 km/h = 20 m/s
(I km/hr = 1000/3600 (m/s))
Time (t) = 5 minutes = 300 s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{20-0}{300}\) = \(\frac{1}{15}\) = m/s
Displacement s= ut + \(\frac{1}{2}\) at²
= 0 × 300 + \(\frac{1}{2}\) × \(\frac{1}{15}\) 300²
= 3000 m

Question 7.
Analyse Table and draw the velocity-time graph.

X Time (s)	0	2	4	6	8	10
Y Velocity(m/s)	10	15	20	20	20	15

a) From the graph, find the time interval during which there is no acceleration.
b) Find the time interval during which deceleration occurs.
c) Find the displacement between the fourth second and the eighth second.
Answer:

a) 4 s to 8 s
b) 8 s to 10 s
c) Time t = 4s
Velocity = 20 m/s
Displacement = Velocity × Time = 20 × 4 = 80 m

Question 8.
Observe the graphs.

In which graph,
a) does the object have uniform acceleration?
b) does the object have uniform velocity?
c) does the object have acceleration and deceleration?
Answer:
a) Graph 2
b) Graph 1
c) Graph 3

Question 9.
The velocity-time graph of an object in straight line motion is given.

Identify the regions of the graph where the object is moving with:
a) acceleration.
b) uniform velocity.
c) deceleration.
Answer:
a) Region OA
b) Region AB
c) Region BC

Question 10.
It is not safe for pedestrians to wear dark coloured clothes at night and in conditions with dim light. The school authorities decided to choose a dark coloured uniform for your school. Record your response to the decision. Justify the answer with respect to road safety.
Answer:
Choosing a dark coloured uniform for our school raises significant safety concerns, particularly regarding pedestrian safety at night and in low- light conditions. This decision is problematic because: visibility issues, road safety. For the safety of all students, it would be better for the school to choose a uniform that includes lighter or reflective elements. This change would help ensure that students are more visible to drivers, particularly in low-light conditions, thereby reducing the risk of accidents and enhancing overall road safety.

Question 11.
In the figure given below, a child travels from P to S through Q and R and comes back straight to P.

Analyse the figure and complete the table given below.
Answer:

Starting from P	Speed	Velocity
On reaching Q	\(\frac{60}{10}\) = 6 m/s	\(\frac{60}{10}\) = 6 m/s
On reaching R	\(\frac{150}{25}\) = 6 m/s	\(\frac{125}{25}\) = 5 m/s
On reaching S	\(\frac{204}{34}\) = 6 m/s	\(\frac{136}{16}\) = 8.5 m/s
While returning to P	\(\frac{340}{50}\) = 6.8 m/s	\(\frac{0}{50}\) = 0 m/s

Question 12.
A stone is thrown vertically upwards with a velocity of 20 m/s. (a = -10 m/s²)
a) What is the maximum height that the stone has reached?
b) How far will this stone travel in 3 s after it is thrown?
c) How high will the stone be from the ground, 3 s after it is released?
Answer:
Given,
v = 0m/s; u 20 m/s; a = -10 m/s²
a) Using equation of motion, v² = u² + 2as
0 = 20² + 2(-10)h
0 = 400 – 20 h
20 h = 400
h = 400/20 = 20 m

b) s = ut + 1⁄2 at²
20 = (20×t) + 1⁄2 (-10 t²)
20 = 20 t – 5 t²
t² – 4t + 4 = 0
(t-2)² = 0
t = 2s
During the first 2 seconds, it reaches the maximum height of 20 m. Then it falls for 1 second. Now the stone falls down, then u=0.
s = 1⁄2 × 10 × (1)² = 5 m
Total distance travelled by the stone in 3 s = 20 + 5 = 25 m.
c) During the first 2 seconds, it reaches the maximum height of 20 m. Then it falls for 1 second. Now the stone falls down, then u=0.
s = 1⁄2 × 10 × (1)² = 5 m
Therefore, the height from the ground after 3 seconds: 20 m – 5m = 15 m.

Question 13.
An object is moving with a speed of 40 m/s.
If it is given a deceleration of 8 m/s²,
a) how long will it take to come to rest?
b) what is the displacement of the object in this time?
Answer:
a) Here, v = 0 m/s
u = 40 m/s
a = -8 m/s
By using the equation, v = u + at
0 = 40 + (-8) t
= 40 – 8t
8t = 40
t = 40/8 = 5 s
So, it will take 5 seconds for the object to come to rest.

b)Displacement s = ut + \(\frac{1}{2}\) at²
= (40 × 5 ) + \(\frac{1}{2}\) (-8)5²
= 200 + \(\frac{1}{2}\) (-200)
=200 – 100
= 100 m

Question 14.
An object is moving with a velocity of 20 m/s. This object is given an acceleration of 5 m/s². What is the velocity when the displacement is 120 m?
Answer:
Given,
u = 20 m/s
a = 5 m/s²
s = 120 m
By using equation, v² = u² + 2as
v² = 20² + 2 × 5 × 120
v² = 400 + 1200
v² = 1600
v = √1600 = 40 m/s

Question 15.
A bullet travelling with a velocity of 60 m/s comes to rest after penetrating 2 cm into a wooden block. What is the acceleration of this bullet? How much is its deceleration?
Answer: Given,
Initial velocity (u) = 60 m/s
Final velocity (v) = 0m/s
Distance (s) = 2 cm = 0.02 m.
We know, v² = u² + 2as
0² = 60² + 2 × a × 0.02
0 = 3600 + 0.04 a
a = \(\frac{-3600}{0.04}\) = 90000 m/s²
The deceleration (since it is negative acceleration) is 90000 m/s.

Question 16.
Observe the table illustrating the motion of an object, select an appropriate scale and draw the graph. Interpret the graph and write down the answers to the questions given below.

Time (s)	0	5	10	15	20	25	30
Velocity (m/s)	20	25	30	30	30	25	20

a) Which is the time interval with no acceleration?
b) Which is the time interval with deceleration?
c) Calculate the displacement of this object in 30 s.
Answer:

a) 10s to 20s
b) 20s to 30s
c) Time t = 30 s
Velocity v = 20 m/s
Displacement-Area under velocity-time graph.
= \(\frac{1}{2}\) × (20+30) × 10 + (10×30) + \(\frac{1}{2}\) ×(20+30) × 10
= \(\frac{500}{2}\) + 300 + \(\frac{500}{2}\)
= 250 + 300 + 250
= 800 m

Question 17.
The velocity-time graph of the motion of Car A and Car B is given below.

a) Which car started first?
b) How much time did each car take to attain the same speed?
c) Which car possesses more acceleration?
d) Which car has more displacement?
Answer:
a) Car A
b) 14 s
c) Car B
d) Car A

Class 9 Physics Chapter 2 Extended Activities Answers Equations of Motion

Question 1.
The Science Club organises an awareness class on ‘Road accidents due to over speeding. Prepare the necessary slides for the presentation. (Hints: traffic rules, signboards, traffic rules to be followed by pedestrians, etc.)
Answer:
Hint:
Slide 1:
Title: Road Accidents Due to Over speeding Subtitle: Understanding the Risks and Prevention Measures
Introduction: Brief overview of the importance of road safety and the focus on over speeding.

Slide 2: Introduction
Road accidents are a major cause of injury and death.
Over speeding is a leading factor in many accidents.
Today, we’ll learn about how to stay safe on the road.

Slide 3: Importance of Traffic Rules
Traffic rules are designed to keep everyone safe.
Following these rules helps prevent accidents.

Slide 4: Common Traffic Signs
Stop Sign: Stop your vehicle completely. Speed Limit Sign: Do not exceed the speed indicated.
Pedestrian Crossing Sign: Slow down and be ready to stop for people crossing.

Slide 5: Traffic Rules for Pedestrians
Use pedestrian crossings.
Look both ways before crossing the road. Avoid crossing between parked cars. Walk on sidewalks whenever possible.

Slide 6: Tips for Safe Driving
Always stay within speed limits.
Keep a safe distance from the vehicle in front of you.
Stay alert and focused on the road.

Slide 7: Conclusion
Following traffic rules saves lives.
Over speeding is dangerous for everyone.
Stay safe and be responsible on the road.

Question 2.
An object is thrown upwards with a velocity of 30 m/s. It returns to the same position after some time. Draw the velocity-time graph of this object and display it in the class (Consider the deceleration as 10 m/s²).
Hint:
Initial velocity (u): 30 m/s (upwards).
Final velocity (v) when it comes back to the same
position: -30 m/s (downwards).
Deceleration (a): 10 m/s².
Time to reach the highest point (t): Calculate using V = u + at.
After finding the value, then take the correct scale and draw the graph similar to this.

Prepare a project on whether the safety measures implemented in your area are adequate to reduce road accidents.
Project planning can be done with the help of your teacher.
Bring relevant findings to the attention of the Road Safety Authority

(For more information, the services of the National Transportation Planning and Research Conor (NATPAC) and the Motor Vehicles Department can be availed.

The report should include:

• Introduction
• Hypothesis
• Objectives
• Methodology
• Analysis
• Results
• Conclusion
• Suggestions

Hint:
Introduction
Road accidents are a serious issue affecting our community. This project aims to evaluate whether the current safety measures in our area are sufficient to reduce road accidents.

Hypothesis
The existing safety measures in our area are adequate to significantly reduce road accidents.

Objectives

• To identify the current safety measures implemented in our area.
• To assess the effectiveness of these safety measures.
• To gather opinions from local residents and authorities.
• To make suggestions for improvement based on findings.

Methodology
Planning: Work with our teacher to create a project plan.

Data Collection:

• Conduct surveys with local residents.
• Interview local traffic authorities.
• Observe road conditions and safety measures (e.g., traffic signs, speed limits, pedestrian crossings).

Research: Use resources from NATPAC and the Motor Vehicles Department.
Analysis: Compare collected data with road safety standards.

Analysis

• Survey Results: Summarize the responses from local residents about their perception of road safety.
• Interviews: Key points from interviews with traffic authorities.
• Observations: Note the presence and condition of traffic signs, speed bumps, pedestrian crossings, etc.

Results

• Positive Findings: List of effective safety measures currently in place (e.g., well-marked pedestrian crossings, functional traffic lights).
• Areas of Concern: Identify any shortcomings (e.g., lack of speed bumps in critical areas, poor visibility of traffic signs).

Conclusion
Summarize whether the safety measures are adequate. If they are not, highlight the main issues identified.

Equations of Motion Class 9 Notes Questions and Answers Kerala Syllabus

Activity
Observe figure.

A,B,C and Dare four electric poles erected on the side of the road. The distance between any two adjacent poles is 40 m. A child starts walking from pole B, passes C and reaches D. After that, the child returns from D and reaches the pole C.

Question 1.
What is the total distance travelled by the child?
Answer:
120 m

Question 2.
What is the distance between the current position C and the initial position B of the child?
Answer:
40 m

Question 3.
If the child travels 40 m from B, which are the possible positions of the child?
Answer:
Child can reach at A or C.

Question 4.
In which direction should the child travel 40 m to reach C from B? (towards the east /towards the west)
Answer:
Towards the east
Here 40 m eastward from B to C is the displacement of the child. So, Displacement is the shortest distance.

Question 5.
Is displacement a vector or a scalar?
Answer:
Vector

Question 6.
If the child travels 40 m in the westward direction from B, the current position of the child is …………
Answer:
A
Note: If the displacement from B in the forward direction (towards the east) is considered as positive, the displacement in the backward direction (towards the west) should be considered as negative. (These can also be considered in the reverse order). Once the direction is determined, the positive and negative directions should not be changed thereafter. Here, B is the initial position and A is the final position. Hence the displacement is negative.

Question 7.
Complete table based on the child’s travel given above.
Answer:

Stages of path covered by the child	Distance covered	Displacement
Directly from B to C	40 m	40 m
Starts from B, reaches D and returns to C	120 m	40 m (from B to C)
Starts from B and reaches D	80 m	80 m (from B to D)
From B to A	40 m	40 m (from B to A)
Starts from B, reaches A and returns to B	80 m	Zero

Question 8.
The child moved from A to D and returned to A. What is the distance covered? What is the displacement? Aren’t the initial position and final position the same?
Answer:
The distance covered is 240 m and the displacement is zero because of initial position and final position are same.

Question 9.
Write down the situations in which the distance covered and displacement are equal.
Answer:

• When you travel directly from the start point to the end point.
• When there are no turns or curves in your path.
• When you move in a straight line without changing direction.

Question 10.
Two different paths taken by a child to move from position P through Q are depicted.

a) What is the distance covered in figure (a)? What is the displacement?
b) What is the distance covered as per the motion in figure (b)? What is the displacement?
c) In which situation is the magnitude of the distance and displacement equal?
Answer:
a) Distance is 100 m. Displacement is 80 m. (From P to R)
b) From P to Q, the distance is 100 m and displacement is also 100 m. (From P to S)
c) In the second case; i.e., Displacement and distance are equal only when an object moves in a straight line without changing direction.

Question 11.
Tabulate the differences between distance and displacement related to the path traversed by a person in Table.
Answer:

Distance	Displacement
Length of the path covered	Straight-lone distance
Can’t be zero	Can be zero
Scalar	Vector

Question 12.
The classrooms and some other locations of a school are depicted.

During the interval, a student from Class 9B went to the staff room. The child then proceeded to the statue of the Father of the Nation in the school garden and returned to the class via the school office. Complete Table based on the path followed by the child.
Answer:

Child’s path	Distance (m)	Displacement (m)
When the child reaches the corridor in front of the staff room from Class 9B	15 m	15 m
When the child reaches the garden near the statue of the Father of the Nation from 9B via the staff room.	85 m	60 m
When the child returns to Class 9B	180 m	0 m

A child travels from P to R via Q in 18 seconds, as shown in the figure.

Question 13.
What is the total distance covered by the child to reach R from P through Q?
Answer:
90 m

Question 14.
What is the speed of the child when the child travels from P to R through Q?
Answer:
Speed = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{90}{18}\)
= 5 m/s

Question 15.
What is the displacement of the child?
Answer:
72 m

Question 16.
Isn’t the child’s displacement of 72 m taking place in 18 s?
Answer:
Yes
Let’s find out the displacement in one second.
Displacement in one second = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{72 m}{18 s}\) = 4 m/s
The displacement in unit time is the velocity.

Question 17.
What is its direction?
(P → R/R → P/P → Q → R)
Answer:
P → R
So the direction of displacement and velocity are same.

Question 18.
Now, let’s consider the doubt raised by the child at the beginning of this lesson. A train of length 200 m travels with a velocity of 20 m/s. What is the time taken by this train to cross a straight bridge of length 800 m?
Answer:

Displacement = Length of the bridge + length of the rain
s = 800 m + 200 m = 1000 m
Velocity (v) = 20 m/s
Time (t) = ?
t = \(\frac{s}{v}\) = \(\frac{1000 m}{20 m/ s}\)
Time = 50 s

Question 19.
Calculate the speed and velocity of the child from P to Q in figure c. What inferences can you draw from this?
Answer:
Speed = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{40}{8}\) = 5 m/s.
Velocity = \(\frac{\text { Displacement }}{\text { Time}}\)
= \(\frac{40}{8}\) = 5 m/s.
This shows that the speed and velocity is same when it travels through a straight line.

Question 20.
A vehicle travels along a straight line with a velocity of 25 m/s and covers a distance of 400 m. Calculate the time taken for this.
Answer:
Given,
Velocity = 25 m/s
Distance = 400m
Here Speed and Velocity is same (along a straight line)
So, Speed = \(\frac{\text { Distance }}{\text { Time}}\)
Time = \(\frac{\text { Distance }}{\text { Speed}}\)
= \(\frac{400}{25}\) = 16 s.

Question 21.
What is the displacement of an object moving with a velocity of 36 m/s in one minute?
Answer:
Given,
Velocity = 36 m/s
Time = 1 minute = 60 s
Displacement = Velocity × Time
=36 × 60 = 2160 m

Question 22.
Is the velocity of CarA always the same? Why?
Answer:
Yes. Because it travels equal distance in equal interval of time.

Question 23.
What about the velocity of Car B? Why?
Answer:
Not same. Because it travels unequal distance in equal interval of time.

Question 24.
Haven’t you noticed the mud sticking to the tyres of vehicles being thrown off when they rotate? Does the mud splash in the same direction every time?
Answer:
No, the mud does not splash in the same direction every time. It is thrown off tangentially due to the rotation of the tires.

Question 25.
Isn’t the direction of motion of an object moving along a circular path always changing?
Answer:
Yes, the direction of motion of an object moving along a circular path is always changing.

Question 26.
Is the velocity of Car C the same every second? Does the velocity change?
Answer:
Even though the magnitude of the speed does not change, the velocity changes because the direction changes.

Question 27.
Complete the table based on the information in figures (1), (2), (3).

Answer:

Question 28.
Classify the situations given below as uniform velocity and non-uniform velocity. Record it in the science diary.

• Motion of a stone dropped from a height
• When light travels through vacuum
• A bus starts from a bus stop and is moving forward
• A train travelling at a uniform speed in the same direction
• Swinging on a swing

Answer:

Uniform velocity	Non-uniform velocity
When light travels through vacuum A train travelling at a uniform speed in the same direction	Motion of a stone dropped from a height A bus starts from a bus stop and is moving forward Swinging on a swing

Question 29.
The velocity of a bus starting from a bus stop keeps changing. Will the change in velocity be the same in each second?
Answer:
No.

The data related to the straight line motion of the bus is given below. Analyse the information and answer the questions.

Question 30.
Imagine that you are sifting in a bus. When the bus starts and moves forward in a straight line, doesn’t the velocity change?
Answer:
Yes

Question 31.
When the bus travels from A to B, the velocity at A is……………
(initial velocity / final velocity)
Answer:
Initial velocity.

Question 32.
The velocity at B is……………(initial velocity / final velocity)
Answer:
Final velocity.

Question 33.
While considering the motion from B to C, velocity at B is……………………
Answer:
Initial velocity.

Question 34.
Complete the table using the data of the motion of the bus.

Answer:

Question 35.
A car is moving along a straight road with a velocity of 10 m/s. It is given an acceleration of 5 m/s². Calculate the velocity of the car after 2s.
Answer:
Initial velocity u = 10 m/s
Acceleration a = 5 m/s²
time t = 2s
Final velocity v = ?
a = \(\frac{\text { v-u }}{\text { t}}\)
V – u = at
V = u + at = 10 + 5 × 2 = 20 m/s
To calculate the final velocity, we can use the equation v = u + at

Question 36.
The velocity of an object changes from 4 m/s to 28 m/s in 4 s. Calculate the acceleration.
Answer: Given,
Initial velocity (u) = 4 m/s
Final velocity (y) = 28 m/s
Time (t) = 4s
Acceleration = \(\frac{\text { change in velocity }}{\text { t}}\)
a = \(\frac{\text { v-u }}{\text { t}}\)
a = \(\frac{28-4}{4}\) = 6m/s²

Question 37.
Have a look at the scene in an amusement park. List the instances in which acceleration occurs.
Answer:
Motion of a giant wheel
Roller Coasters
Swing Rides: When the swings begin to move, accelerate outward, decelerate back inward.

Question 38.
Find instances of acceleration in your daily life and record them in the science diary.
Answer:

• Motion of a coconut falling from a coconut tree
• A car speeding up
• A bicycle starting from rest
• An airplane taking off
• An elevator starting to ascend or descend

Question 39.
Can you think of some instances ¡n everyday life where the velocity decreases? Expand the list by giving more examples.
Answer:

• Train arrivingata station
• The upward motion of a stone thrown upwards
• A car approaching a red traffic light
• A ball rolling up hill
• A parachutist landing

Question 40.
Observe the figure, and complete the table.


Answer:

Question 41.
Should negative sign be given while writing the value of retardation?
Answer:
Yes, a negative sign should be given while writing the value of retardation, as it indicates a decrease in velocity.

Question 42.
An object starts from rest and attains a velocity of 10 m/s in 5 s.
a) What is its acceleration?
b) What is the acceleration if it comes to rest in 5 s? What is the retardation?
Answer:
Given,
a) Initial velocity (u)=0 m/s
Final velocity (v) = 10 m/s
Time (t) = 5 s
Acceleration (a) = \(\frac{\text { v-u }}{\text { t}}\)
= \(\frac{\text { 10-0 }}{\text { 5}}\) = 2 m/s²

b) Here,
initial velocity (u) = 10 m/s
Final velocity (v) = 0 m/s
Time (t) = 5 s
Acceleration (a) = \(\frac{\text { v-u }}{\text { t}}\) = \(\frac{\text { 0-10 }}{\text { 5}}\) = -2 m/s²
Retardation is simply the magnitude ofthe negative acceleration.
Thus, retardation = 2 m/s²

Question 43.
A vehicle travelling at a speed of 5 m/s is brought to. rest ¡n 2 s by applying brakes. Calculate the retardation of the vehicle.
Answer:
Initial velocity u = 5 m/s
Final velocity v = 0
Time taken = 2 s
Acceleration a = \(\frac{\text { v-u }}{\text { t}}\)
= \(\frac{\text { 0-5 }}{\text {2}}\) = -2.5 m/s²
Thus, retardation = 2.5 m/s²

Question 44.
If the velocity of an object in the 2nd second is 40 m/s and 30 m/s in the 4th second, what is its acceleration? What is its retardation? What is its velocity at the 8th second?
Answer:
Velocity at the 2^nd second u = 40 m/s
Velocity at the 4^th second v = 30 m/s
Time (t) = 2s
Acceleration a = \(\frac{\text { v-u }}{\text { t}}\) = \(\frac{\text { 30 – 40 }}{\text {2}}\) = -5 m/s²
So retardation is 5 m/s².
To find velocity at 8^th second.
Initiai velocity (u) = 40 m/s
Time (t) = 6 s
So, a = \(\frac{\text { v-u }}{\text { t}}\)
V = u + at
= 40 + (-5 × 6)
=40 – 30 = 10 m/s.

Question 45.
Was the acceleration obtained as per Table 1 the same on each occasion?
Answer:
Yes

Question 46.
What about the acceleration obtained as per Table 2?
Answer:
Not same and also have negative sign.
You have learned about speed, velocity, and acceleration. Over speeding of vehicles can cause accidents. We must follow traffic rules to reduce accidents. However, accidents aren’t caused only by over speeding. Pedestrians should also follow traffic rules.

Question 47.
Which are the traffic rules for pedestrians to follow?
Answer:

• Pedestrians should walk along the right side of the road.
• Cross the road only at the zebra crossing, obeying the traffic signal.
• Use sidewalks
• Avoid crossing between parked cars

Question 48.
In addition to the signs mentioned above, collect more symbols for each category, prepare separate posters and display them on the school bulletin board.
Answer:
Hint: An example shown below.

Question 49.

What information can be gathered from the above graph?
Answer:
From 2020 to 2022 there is a significant spike in accidents. The highest number of accidents happened in 2022. From 2019 to 2020 the number of accidents decreased.

Question 50.
List down your findings.
Answer:
The number of road accidents is generally increasing, this might indicate issues with traffic management, increasing vehicle numbers, or lack of road safety measures. Significant drops in 2019 to 2020 could be attributed to the COVID-19 pandemic, which resulted in lockdowns and reduced road traffic.

Question 51.
In which year is the number of accidents the least?
Answer:
2020

Question 52.
How many accidents occurred in 2019?
Answer:
43000

Question 53.
What is the nature of the graph obtained?
Answer:
An inclined straight line

Question 54.
By what name is this graph known?
Answer:
Position-Time graph

Question 55.
From the shape of the graph obtained, what is the nature of the velocity of the object?
Answer:
Uniform velocity

Question 56.
What is the displacement of the object in 5 s?
Answer:
2.5 m

Question 57.
What is the time taken to travel 1.5 m?
Answer:
3 s
Let’s consider another situation.
Position-Time graph of a moving car.

Question 58.
How can we find the velocity of the car from A to B from above graph?
Answer:
Here,
Displacement=30 m, Time=6 s
Velocity = Displacement/Time = \(\frac{30}{6}\) = 5 m/s

Question 59.
What is the displacement of the car from A to B in the graph?
Answer:
30 m

Question 60.
What is the time taken by the car to travel from A to B?
Answer:
6 s

Question 61.
What is the position of the car at the sixth second in the graph?
Answer:
30 m

Question 62.
Which type of velocity does this car have?
Answer:
Uniform velocity

Question 63.
Find out the velocity of the car between 6 s and 8 s from the graph.
Answer:
Here,
Displacement = 10m
Time = 2 m
Velocity = Displacement/ Time
= \(\frac{10}{2}\) = 5 m/s

A velocity-time graph is a graph that plots velocity on the Y-axis and time on the X-axis.

Question 64.
From the graph, find the displacement of the vehicle between the fourth and tenth second.
Answer:
Velocity = Displacement/Time
Then, displacement = velocity × time
On the graph, it will be equal to AB × AD (Equal to the area of the rectangle ABCD)
Displacement = 40 × 6 = 240 m

Question 65.
Isn’t this equal to the area of the portion below BC on the graph?
Answer:
Yes.
Area = 40 × 6 = 240 m

Question 66.
Find the displacement during the first 4 s from graph.
Answer:
displacement Area of triangle AOB
= \(\frac{1}{2}\) × 40 × 4 = \(\frac{160}{2}\) = 80 m

Question 67.
What is the change in velocity in the first 4 s? What is the acceleration?
Answer:
Initial velocity (u) = 0 m/s
Final velocity (v) = 40 m/s
Change in velocity = v – u – 0 – 40 = -40 m/s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{-40}{4}\) = -10 m/s²

Question 68.
What is the acceleration of this vehicle between 4 s and 10 s?
Answer:
Initial velocity (u) = 40 m/s
Final velocity (v) = 40 m/s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{40-40}{6}\) = 0 m/s²

Question 69.
A body starts from rest and acquires a velocity of 20 m/s in 2 s and 40 m/s in 6 s. What is the displacement of the object during this time interval?
Answer:
Initial velocity, u = 20 m/s
Time t = t₂ – t₁ = 6s – 2s = 4 s
Final velocity, v = 40 m/s
Acceleration, a = \(\frac{v-u}{t}\)
= \(\frac{40 \mathrm{~m} / \mathrm{s}-20 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}\)
= 5 m/s²
Displacement s = ut + \(\frac{1}{2}\) at²
= (20 m/s × 4 s) + [\(\frac{1}{2}\) × 5 m/s² (4 s)²]
= 80 m + 40 m
=120 m

Question 70.
If the velocity of a car increases from 6 m/s to 16 m/s in 10 s.
a) calculate the acceleration of the car.
b) what is the displacement of the car during this time?
Answer:
a) u = 6 m/s
v = 16 m/s
t = 10 s
Acceleration a = \(\frac{v-u}{t}\)
= \(\frac{16 \mathrm{~m} / \mathrm{s}-6 \mathrm{~m} / \mathrm{s}}{10 \mathrm{~s}}\)
= \(\frac{10 \mathrm{~m} / \mathrm{s}}{10 \mathrm{~s}}\)
= 1 m/s²

b) s = ut + \(\frac{1}{2}\) at²
=(6 m/s × 10 s) + [\(\frac{1}{2}\) × 1 m/s² (10 s)²]
=60 m + 50 m
= 110 m

Question 71.
The velocity of a train that started from a railway station becomes 90 km/h in 10 minutes. Calculate the acceleration of the train.
Answer:

Question 72.
An object falls down from rest and moves with an acceleration 10 m/s, hits the ground with a velocity 20 m/s. From what height does the object fall?
Answer:
u = 0
a = 10 m/s²
v = 20 m/s
v² = u² + 2as
(20 m/s)² = 0² + 2 × 10 × s
400 = 20 × s
s = \(\frac{400}{20}\) = 20 m

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 7 Non Metals Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 7 Solutions Non Metals

Kerala Syllabus Std 9 Chemistry Chapter 7 Non Metals Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 7 Let Us Assess Answers Non Metals

Question 1.
Some gases are given in the following box. Choose suitable answers from the box and answer
the questions below.
Hydrogen, Chlorine, Oxygen, Nitrogen
a. Which gas is formed in the thermal decomposition of KMn04?
b. Which gas is used for the purification of water?
c. Which element is essential for the growth of plants?
d. Which is the inflammable gas obtained during the electrolysis of water?
Answer:
a. Oxygen.
b. Chlorine.
c. Nitrogen.
d. Hydrogen.

Question 2.
Find examples of allotropes of carbon and complete the table.

Crystalline allotropes	Amorphous allotropes

Answer:

Crystalline allotropes	Amorphous allotropes
Diamond, Graphite, Fullerenes, Graphene	Charcoal, Carbon Black, Soot, Lampblack

Question 3.
An experiment is given to identify chloride salt. Write down the observations and inferences.

Experiment	Observation	Inference
i. Add AgNO3 solution to the given salt solution.
ii. Add NH4OH solution to the precipitate formed

Answer:

Experiment	Observation	Inference
i. Add AgNO3 solution to the given salt solution.	A white precipitate forms.	The formation of this white precipitate indicates the presence of Cl– ions in the original salt solution.
ii. Add NH4OH solution to the precipitate formed	The white precipitate of AgCl dissolves in the NH4 OH solution, forming a colouriess solution.	This solubility of AgCl in NH4OH is a characteristic property of AgCl. It confirms the presence of Cl– ions in the original salt solution.

Question 4.
Find the suitable gas for each of the following situations. (Chlorine, nitrogen, CFC, oxygen)
• As an oxidising agent in rocket fuels.
• Depletion of ozone.
• To inflate the tyres of vehicles.
Answer:
As an oxidising agent in rocket fuels – Oxygen
Depletion of ozone – CFC, Chlorine
To inflate the tyres of vehicles – Nitrogen

Question 5.
Potassium permanganate is heated in a boiling tube.
a. A burning matchstick is inserted into the mouth of the boiling tube. Write down your observation.
b. Which gas is produced?
Answer:
a. Observation: When a burning matchstick is inserted into the mouth of the boiling tube containing heated potassium permanganate, the matchstick will reignite with a bright flame.

b. The gas produced during the decomposition of potassium permanganate is oxygen. The oxygen gas supports combustion, hence the reignite of the matchstick.

Question 6.
Write the answers to the following questions on the preparation of chlorine gas in the laboratory.
a. What are the chemicals required to prepare chlorine gas?
b. Chlorine gas is collected by passing it through water. Why?
c. Chlorine gas is passed through concentrated sulphuric acid. Why?
Answer:
a. Potassium permanganate (KMnO₄)

Concentrated hydrochloric acid (HCl)

b. To remove the HCl vapours.

c. Chlorine gas is passed through concentrated sulphuric acid to remove any water vapour that may have been carried over from the reaction mixture. Concentrated sulphuric acid is a dehydrating agent, which means it can absorb water from other substances.

Question 7.
The bleaching action of chlorine requires moisture. Give reason
Answer:
Chlorine gas reacts with water to form hypochlorous acid.

• Hypochlorous acid is the active bleaching agent.
• The reaction between chlorine gas and water requires moisture.
• Without moisture, chlorine gas will not have any bleaching action.

Question 8.
“Chemical fertilisers must be banned completely, and organic fertilisers must be promoted.” What is your opinion about this argument? Justify your answer.
Answer:
Reasons to Ban:

• Environmental Health: Chemical fertilisers can harm soil and water, so reducing them can help ecosystems.
• Sustainability: Organic fertilisers are often better for long-term soil health
• Human Health: Less chemical use means fewer harmful residues in food and water.

Reasons Against Ban:

• Food Production: Chemical fertilisers help increase crop yields, which is vital for feeding a growing population.
• Farmer Struggles: Farmers who rely on chemicals may have a hard time switching to organic
  methods without support.

Question 9.
a. What are the chemicals required for preparing hydrogen in the laboratory? b. How will you identify that the gas obtained is hydrogen?
Answer:
a. Zinc granules: A reactive metal that reacts with acids to produce hydrogen.
Dilute hydrochloric acid: The acid reacts with the zinc to produce hydrogen gas
Zn + 2HCl → ZnCl₂ + H₂

b. Hydrogen gas can be identified by a simple test known as the “pop” test.
Collect the gas produced in a test tube.
Bring a lighted splint near the mouth of the test tube.
Observe the reaction: If the gas is hydrogen, it will ignite with a “pop” sound

Question 10.
a. What are the merits of hydrogen when it is used as a fuel?
b. What are the isotopes of hydrogen?
c. Which among these is an isotope without a neutron?
Answer:
a. High calorific value.
Clean combustion.
Abundant availability.
Possibility of pollution is low.

b. Hydrogen has three isotopes:
Protium: The most common isotope, with one proton and no neutrons in its nucleus.
Deuterium: An, isotope with one proton and one neutron in its nucleus.
Tritium: An isotope with one proton and two neutrons in its nucleus.

c. Protium

Question 11.
a. What is meant by calorific value?
b. Some fuel is given in the box. Which has the highest calorific value?
Petrol, Coal, Ethanol, Hydrogen, Methanol
c. Hydrogen is not used as a domestic fuel. What are the reasons for this limitation?
Answer:
a. The calorific value of a fuel is the heat energy released from unit mass of the fuel on complete combustion.

b. Hydrogen.

c. Hydrogen is a gas that bums with an explosion.
It is difficult to store and distribute hydrogen.
Safety concerns.
High Cost.
Technological limitations.

Question 12.
One of the crystalline forms of carbon is diamond. Diamond does not conduct electricjty. Why?
Answer:
In diamond, each carbon atom is covalently bonded to four other carbon atoms surrounding it. This
strong bond is responsible for the hardness of the diamond. Due to the absence of free electrons in this crystalline structure, diamond does not conduct electricity.

Question 13.
a. Which allotrope of carbon is a conductor of electricity?
b. Give a reason for this.
c. Write down any two uses of this allotrope.
Answer:
a. Graphite

b. The reason for graphite’s electrical conductivity lies in its unique structure:
Layered Structure: Graphite has a layered structure, where each layer consists of carbon atoms arranged in hexagonal rings.

Free Electrons: Within each layer, one of the carbon atom’s valence electrons is delocalised and can move freely between the hexagonal rings. These delocalised electrons act as charge carriers, allowing graphite to conduct electricity.

c) Graphite is used as the “lead” in pencils.
Graphite is used as an electrode in various types of batteries, such as lithium-ion batteries.

Question 14.
Match the following

A	B
Diamond	Nanotechnology
Graphite	Making of ornaments
Graphene	Lubricants

Answer:

A	B
Diamond	Making of ornaments
Graphite	Lubricants
Graphene	Nanotechnology

Question 15.
a. Dilute hydrochloric acid and calcium carbonate are taken in a test tube. Which gas is produced? b. Complete the chemical equation of the reaction.
CaCO₃ + 2HCl → CaCl₂ + …………… + ……………
c. The gas obtained is passed through clear lime water. Write down the observation.
Answer:
a. Carbon dioxide (CO₂) gas is produced.

b. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

c. When the gas obtained r(carbon dioxide) is passed through clear lime water (calcium hydroxide solution), it turns the lime water milky.

Extended Activities

Question 1.
Write a short note on the bleaching property of chlorine and present it.
Answer:
Chlorine is a powerful oxidising agent, which means it can remove electrons from other substanCes. This property allows chlorine to act as a bleaching agent, breaking down coloured compounds and making them colourless.
When chlorine comes into contact with a coloured compound, it oxidises the coloured molecules, breaking down their chemical structure.
The oxidation process results in the loss of colour, as the coloured molecules are transformed into colourless products.

Applications of Chlorine Bleaching:
Textile Industry: Chlorine is used to bleach cotton, linen, and other fabrics.
Paper Industry: Chlorine is used in the pulp and paper industry to bleach wood.
Water Treatment: Chlorine is added to drinking water and swimming pools to disinfect and remove impurities.

Question 2.
Prepare a table showing the use of oxygen, nitrogen and hydrogen in daily life.
Answer:

Gas	Common use
Oxygen	Breathing, medical treatments, industrial processes (combustion, steelmaking), rocket fuel, oxyacetylene flames is used for welding.
Nitrogen	Fertilizer production, food packaging (to prevent spoilage), industrial gases (creating inert atmospheres), automotive airbags, In the production of ammonia.
Hydrogen	Fuel cells (generating electricity), rocket fuel, production of ammonia (fertilizers), hydrogenation of fats and oils, used as an reducing agent in metal production.

Question 3.
Visit any chemical fertiliser factory, prepare a project report and present it
Answer:
Guidelines for Preparing Your Chemical Fertiliser Factory Project Report
1. Introduction:

• Clearly state the purpose of your visit and the overall goal of the report.
• Provide a brief overview of the factory, including its name, location, and any relevant background information.

2. Factory Tour:

• Describe the physical layout and facilities of the factory.
• Discuss the production process, step-by-step, including any diagrams or flowcharts to illustrate the process

3. Raw Materials and Production:

• Identify the primary raw materials used in fertiliser production.
• Explain the chemical reactions involved in the manufacturing process.
• Discuss any specific equipment or machinery used.

4. Environmental Impact:

• Assess the potential environmental impacts of fertiliser production (Eg: water pollution, airpollution, soil contamination).
• Discuss any sustainability initiatives or measures taken by the factory to minimise
  environmental harm.

5. Safety Measures:

• Describe the safety precautions and regulations implemented at the factory.
• Discuss any potential hazards or risks associated with fertiliser production.

6. Technology and Innovation:

• Highlight any advanced technologies or innovative practices used by the factory.
• Discuss the potential benefits of these technologies.

7. Challenges and Future Outlook:

• Identify any challenges or obstacles faced by the factory.
• Discuss the factory’s future plans and goals.

8. Conclusion:

• Summarize the key findings and observations from your visit.
• Offer your personal reflections or insights.
• Provide any recommendations or suggestions for improvement.

Non Metals Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Examine the periodic table, find out some examples of metals, non metals and metalloids, and

Metals	Non metals	Metalloids

Answer:

Metals	Non metals	Metalloids
Sodium Potassium Magnesium Calcium Aluminium Iron Copper Gold	Carbon Nitrogen Oxygen Hydrogen Chlorine Sulphur Fluorine	Metalloids Boron Silicon Germanium Arsenic Antimony Tellurium

Hydrogen, carbon, oxygen, nitrogen and chlorine are some important non-metals. Let us examine the methods of their preparation and some of their physical and chemical properties.

Hydrogen is the most abundant element in the universe. It is the major component of the sun and other stars. Hydrogen is seen in the atmosphere in a very small quantity in free state.

Question 2.
Complete the given table and write a short note on hydrogen.
Answer:

Atomic Number	1
Mass number	1
No. of electrons	1
No. of protons	1
No. of neutrons	No neutrons
Solubility in water	Insoluble in water
Isotopes	Protium, Deuterium, Tritium

Question 3.
Take 5 mL of dilute hydrochloric acid in a test tube. Add some zinc granules to it. Record your observation.
a) Is the gas produced hydrogen? How do you confirm it? Insert a burning splint to the mouth of the test tube. What do you observe?
Answer:
Yes, the hydrogen gas bums with a pop sound. Hydrogen is a flammable gas.

b) Write the balanced chemical equation of the above reaction.
Answer:
Zn + 2HCl → ZnCl₂ + H₂
Hydrogen is industrially prepared by the electrolysis of water.

Question 4.
To which category do the above reactions belong?
(Combination reaction, displacement reaction, decomposition reaction)
Answer:
Combination reaction.

Question 5.
The graph given below shows some fuels and their calorific values.

a. Which fuel has the lowest calorific value?
Answer:
Carbon monoxide.

b. Which fuel has the highest calorific value?
Answer:
Hydrogen.

c. What is the product obtained when hydrogen burns in air?
Answer:
Water.
The calorific value of hydrogen is higher than that of other fuels.

The merits of hydrogen as a fuel:

• When hydrogen is used as a fuel, the possibility of pollution is very low since water is the only product formed.
• High calorific value

In spite of these merits, hydrogen is not used as a domestic fuel due to the below-given limitations:

• Hydrogen is a gas that bums with an explosion.
• It is difficult to store and distribute hydrogen.

If these limitations are overcome, hydrogen will become a universal fuel. The problems like the scarcity of fossil fuels and environmental pollution can thus be resolved.

Question 6.
Which compounds of hydrogen are known to you?
Answer:
Water, Acids, Alkalies, Carbohydrates, Oils, Sugar

Question 7.
Which properties of hydrogen are used in the following situations?
a. Balloons filled with hydrogen fly high in the air.
b. Used as fuel.
Answer:
a. Hydrogen has less density than air

b.

• High calorific value
• The possibility of pollution is very low.
• Ease in preparing.

Question 8.
Prepare a note on the possibility of using hydrogen as a fuel.
Answer:
Hydrogen as a fuel:
Hydrogen is a colourless, odourless, tasteless gas that is highly flammable. It is the most abundant element in the universe and is found in water, natural gas, and other sources. Hydrogen can be used as a fuel in a variety of ways, including in fuel cells, internal combustion engines, and gas turbines.

Advantages of Hydrogen as a Fuel

• Hydrogen is a clean fuel that produces no greenhouse gases when burned.
• Hydrogen has a high energy density, which means that it can be stored in a small space and used to generate a large amount of energy.
• Hydrogen can be used in a variety of engines and fuel cells, making it a versatile fuel option.
• Hydrogen can be produced from a variety of sources, including renewable energy sources such as wind, solar, and geothermal power.

Disadvantages of Hydrogen as a Fuel:

• Hydrogen is a flammable gas, which means that it must be stored and handled carefully.
• Hydrogen is currently more expensive to produce than other fuels, such as gasoline and diesel.
• Hydrogen infrastructure is not yet widely available, which can make it difficult to refuel hydrogen-powered vehicles.

Future of Hydrogen as a Fuel:
Despite the challenges, hydrogen has the potential to become a major source of energy in the future. As technology advances and the cost of producing hydrogen decreases, it is likely that hydrogen will become a more viable option for transportation and other applications.

• Almost all substances around us contain carbon.
• Carbon is found both in an elemental state and a combined state in nature.

Question 9.
Examine the periodic table and record the position, atomic number and electronic configuration of carbon in your science diary.
Answer:
Atomic number – 6
Electronic configuration – 2, 4
Period – 2
Group – 14

Question 10.
In diamond, each carbon atom is covalently bonded to how many surrounding carbon atoms?
Answer:
In diamond, each carbon atom is covalently bonded to four surrounding carbon atoms.

Question 11.
What is the valency of carbon?
Answer:
The valency of carbon is 4.

Question 12.
Are free electrons present in the diamond?
Answer:
No, free electrons are not present in diamonds.

Properties of diamond

• Hardness is very high.
• Do not conduct electricity.
• High thermal conductivity.
• High refractive index.

Question 13.
Find out the reason for using diamonds for making ornaments and cutting glasses.
Answer:
Diamonds are used for making ornaments and cutting glasses due to their unique properties:

• Hardness
• Durability
• Precision

Question 14.
Some uses of graphite are given below. Find out and write the characteristics and properties that facilitate it.
a) Used as an electrode in dry cell
b) Used to make pencil lead
c) Used as a lubricant
Answer:
a) Used as an electrode in dry cell
Property: Good electrical conductivity
Characteristic: Graphite’s layered structure allows for the easy movement of electrons, making it an excellent conductor of electricity. This property is essential for its use as an electrode in dry cells, where it facilitates the flow of electrical current.

b) Used to make pencil lead
Property: Soft and easily markable
Characteristic: Graphite’s layered structure allows the layers to slide over each other easily, making it soft and easily markable. This property is ideal for pencil lead, as it allows the graphite to leave a mark on paper with minimal pressure.

c) Used as a lubricant
Property: Low friction between layers.
Characteristic: The force of attraction between these layers is weak. Hence, one layer can slide over the other. So, graphite is used as a lubricant

Question 15.
List out and compare the properties of graphite and diamond
Answer:

Graphite	Diamond
Layered structure with hexagonal rings of carbon atoms	Three-dimensional network structure with tetrahedral arrangement of carbon atoms
Soft	Extremely hard
Good conductor	Insulator
High thermal conductivity	High thermal conductivity
Lubricant, pencils, electrodes	Cutting tools, jewellery, abrasives

Question 16.
What are the carbon compounds that you know?
Answer:
Carbon dioxide (CO₂)
Carbon monoxide (CO)
Carbonates (CO₃^2-)
Bicarbonates

Question 17.
Which of these compounds contain carbon and oxygen?
Answer:
Carbon dioxide (CO₂)
Carbon monoxide (CO₂)
Carbonates (CO₃^2-)
Bicarbonates

Question 18.
What is the major compound produced when carbon or carbon compounds burn in the air?
Answer:
Carbon dioxide (CO₂).
C + O₂ → CO₂
Preparation of carbon dioxide in the laboratory

Question 19.
What are the reactants used here?
Answer:
Calcium carbonate, Diluted HCl

Question 20.
Complete the chemical equation of the reaction.
Answer:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Question 21.
How can we identify that the gas formed here is CO₂?
Answer:
When the gas formed is passed through lime water (Ca(OH)₂), if it turns milky we can say that the gas is Carbon dioxide (CO₂).

Carbon dioxide gas is produced when calcium carbonate and dilute hydrochloric acid react with each other. When this gas is passed through lime water Ca(OH)2, it turns milky.

Uses of Carbon dioxide

• In fire extinguishers.
• To make soda water and soft drinks.
• In the manufacture of washing soda (Na₂CO₃.10H₂O), and baking soda (NaHCO₃).
• In the manufacture of chemical fertilisers like urea.
• In Carbogen (95% oxygen and 5% carbon dioxide), which is used for artificial respiration.
• Solid carbon dioxide is known as dry ice. It is used to create a smoky effect in stage shows etc.

Question 22.
Tick the properties of carbon dioxide from the following.
Answer:

• Coloured / colourless (√)
• Helps combustion / does not help combustion (√)
• Odourous / odourless (√)
• Denser than air (√) / lighter than air
• Aqueous solution – Acidic (√) / Basic

Question 23.
Match the columns A and B suitably.

A	B
Washing soda	NaHCO3
Carbon monoxide + Hydrogen	Na2CO3.10H2O
Baking soda	Producer gas
	Water gas

Answer:

A	B
Washing soda	Na2CO3.10H2O
Carbon monoxide + Hydrogen	Water gas
Baking soda	NaHCO3

 

• Nitrogen is the most abundant gas in the atmosphere.
• Nitrogen is present in all living beings.
• Nitrogen is an element that is essential for the growth of plants.

Question 24.
Under normal circumstances, atmospheric nitrogen is inert. Write down the reason.
Answer:
Inert behaviour refers to the tendency of a substance to resist chemical reactions. An inert substance is relatively unreactive and does not readily combine with other elements or compounds.

Reason:

• Triple Covalent Bond: Nitrogen atoms share three pairs of electrons, forming a very stable bond. This bond is much stronger than a single or double bond.
• High, Activation Energy: Breaking this triple bond requires a high activation energy, which is the minimum amount of energy needed to start a chemical reaction. This high energy barrier makes it difficult for nitrogen to react with other substances.

Question 25.
Which gas helps combustion?
Answer:
Oxygen.

Question 26.
Which gas helps to control the rate of combustion?
Answer:
Nitrogen.
Plants cannot absorb nitrogen directly from the atmosphere. It involves following steps

Step: 1
Through thunder and lightning: The triple bond in the nitrogen molecule is broken down during thunder and lightning, and nitrogen combines with atmospheric oxygen to form nitric oxide (NO). N₂ + O₂ → 2NO
The nitric oxide thus produced combines with more oxygen to form nitrogen dioxide (NO₂).
2NO + O₂ → 2NO₂
In the presence of oxygen, nitrogen dioxide dissolves in rainwater and reaches the soil as nitric acid (HNO₃).
4NO₂ + 2H₂O + O₂ → 4HNO₃
This nitric acid reacts with the minerals in the soil and forms nitrate salts which are absorbed by plants.

Step: 2
Through nitrogen fixation: Rhizobium bacteria in the roots of leguminous plants absorb atmospheric nitrogen and convert it into compounds that can be absorbed by the plants.

Question 27.
What are the methods by which we can supply the elements required for plant growth in large amounts?
Answer:
Use of organic fertilisers
Use of chemical fertilisers
Merits and demerits of chemical and organic fertilisers

Organic fertilisers	Chemical fertilisers
Eco friendly	Increase crop yield
Maintain the original nature of the soil	Easily available
Delay in making the nutrients available	Quick supply of the three nutrients (N, P, K)
Do not destroy microorganisms in the soil	Change the organic structure of the soil.
More quantity is required.	Increase the acidity of the soil.
Reduce the possibility of pollution.	Cause pollution.

Question 28.
Observe the picture of the preparation of oxygen in the laboratory.
a. Which is the reactant in this chemical reaction?
Answer:
Potassium permanganate

b. Heat the boiling tube containing potassium permanganate.
Insert a burning matchstick to the mouth of the boiling tube. What is the observation?
Answer:
The glowing matchstick flares up.

c. Complete the chemical equation of the reaction.
Answer:

d. Pick out from the following the physical properties suitable to oxygen. Write them down.
Answer:

Colour	Yes/ No √
Odour	Yes/ No √
Solubility in water	Soluble √ / Insoluble
Nature of combustion	Bums/ Helps to burn

Oxygen is manufactured by the electrolysis of water.

Question 29.
Haven’t you noticed that some metals lose lustre in the course of time due to their reaction with oxygen? Write the reason.
Answer:
Oxygen reacts with metals and non-metals to produce their oxides. Due to this, metals lose lustre in the course of time.

Question 30.
Prepare a note on the role of plants in maintaining the level of oxygen in the atmosphere. Present it in the class.
Answer:
Plants play a crucial role in maintaining the oxygen levels in our atmosphere. Through the process of photosynthesis, plants convert carbon dioxide into oxygen, a gas essential for the survival of most life on Earth.

Photosynthesis is a complex process that occurs in the chloroplasts of plant cells. It involves the use of sunlight, water, and carbon dioxide to produce glucose and oxygen. The chemical equation for photosynthesis is:
6CO₂ + 6H₂O + sunlight → C₆H₁₂O₆ + 6O₂

The oxygen produced by plants is released into the atmosphere, where it is used by animals, humans, and other organisms for respiration. During respiration, these organisms consume oxygen and release carbon dioxide back into the atmosphere. This creates a continuous cycle of oxygen and carbon dioxide exchange.

Question 31.
Conduct a seminar on ozone depletion and its solutions.
Answer:
Seminar: Ozone Depletion and Its Solutions

Introduction:
The ozone layer, a vital shield in Earth’s atmosphere, is facing a serious threat: depletion. This depletion has far-reaching consequences for human health, ecosystems, and the planet as a whole. In this seminar, we will explore the causes, effects, and potential solutions to ozone depletion.

Causes of Ozone Depletion:
Chlorofluorocarbons (CFCs): These man-made chemicals, once widely used in refrigerants, aerosols, and foam insulation, are the primary culprits in ozone depletion. When released into the atmosphere, CFCs rise to the stratosphere and react with ozone molecules, destroying them.

Effects of Ozone Depletion:
Increased UV Radiation: Ozone depletion allows more harmful ultraviolet (UV) radiation to reach Earth’s surface.
This can lead to:

• Skin cancer
• Cataracts
• Weakened immune systems
• Damage to plants and marine ecosystems

Climate Change: Ozone depletion can indirectly contribute to climate change by affecting atmospheric circulation and temperature patterns.

Solutions to Ozone Depletion:

• Raising awareness about the importance of the ozone layer and the consequences of its depletion.
• Encouraging people to choose products that are ozone-friendly.
• Strict regulations and enforcement are essential.
• Develop advanced monitoring techniques to track ozone levels.

Question 32.
Complete the table about chlorine after analysing the periodic table.
Answer:

Atomic number	17
Electronic configuration	2, 8, 7
Mass number/Atomic mass	35
No. of electrons	17
No. of neutrons	18
Valency	1
Name of the element family	Halogen

Question 33.
Analyse the given figure depicting chlorine preparation and write answers to the questions given below.

a. What are the chemicals required to prepare chlorine in the laboratory?
Answer:
Potassium permanganate (KMnO₄), Concentrated Hydrochloric acid (Con. HCl)

b. What are the products formed?
Answer:
Potassium chloride (KCl), Manganese chloride (MnCl₂), Water (H₂O), Chlorine gas (Cl₂).

Question 34.
The chlorine gas obtained is collected in the jar after passing it first through water and then through sulphuric acid. Why is it done so?
Answer:
A small amount of hydrogen chloride vapour is also released along with the chlorine gas. Hydrogen chloride vapours dissolve in water when passed through it. Sulphuric acid can absorb the water vapour in the chlorine gas. So it is passed through concentrated sulphuric acid.

PROPERTIES OF CHLORINE

• Greenish yellow in colour
• Pungent smelling.
• Denser than air.

Question 35.
Take two jars filled with chlorine gas. Take a coloured cloth and cut it into two pieces. Put a wet piece into one jar and the other dry piece of cloth into the second jar. What happens to the colour of the cloth in the two jars?
Answer:
The cloth in the first jar remains unchanged in colour. The colour of the wet cloth in the second jar disappears. This is because of the bleaching properties of chlorine. Chlorine bleaches coloured materials into colourless materials in the presence of moisture.
H₂O + Cl₂ → HCl + HOCl
HOCl → HCl + [O]
Water reacts with chlorine to form hydrochloric acid and hypochlorous acid. Hypochlorous acid is unstable. So, it decomposes to form hydrochloric acid and nascent oxygen. Nascent oxygen acts as a powerful oxidising agent to decolourise the coloured substances.

Question 36.
Some questions are given below. Find suitable answers from the box given. Write them down.
Nitrogen, Nitric oxide, Oxygen, Carbon, Carbon dioxide
a) Which gas is obtained when potassium permanganate is heated?
b) What is the product formed when nitrogen combines with oxygen at high temperatures?
c) Which reactant is used along with hydrogen for the manufacture of ammonia?
Answer:
a) Oxygen
b) Nitric oxide
c) Nitrogen

A thorough understanding of Kerala Syllabus 9th Standard Biology Textbook Solutions Chapter 3 Respiration and Excretion Notes Questions and Answers English Medium can improve academic performance.

SCERT Class 9 Biology Chapter 3 Notes Questions and Answers Respiration and Excretion

Std 9 Biology Chapter 3 Notes Pdf Kerala Syllabus English Medium Solutions Questions and Answers

Class 9 Biology Chapter 3 Let Us Assess Answers Respiration and Excretion

Question 1.
Which among the following given below is not the characteristic feature of an efficient gaseous exchange surface?
a) Thick cell wall
b) Proximity to blood capillaries
c) Moist membrane
d) Large surface area
Answer:
a. Thick cell wall

Question 2.
Redraw the diagram and label the parts.

Answer:
a) Trachea
b) Lungs
c) Diaphragm

Question 3.
Write the role of each of the following in the exchange of gases.
a) Plasma
b) RBC
c) Haemoglobin
d) Tissue fluid
Answer:
a) Plasma: The cells need oxygen for metabolism, which creates carbon dioxide as a waste product. The carbon dioxide is absorbed from the cells by the blood plasma (some of it binds to hemoglobin too) and is transported back to the lungs in the bloodstream.

b) RBC: Oxygen molecules attach to red blood cells, which travel back to the heart. At the same time, the carbon dioxide molecules in the alveoli are blown out of the body the next time a person exhales.

c) Haemoglobion: Haemoglobin is a molecule that is responsible for carrying almost all of the oxygen in the blood. It is composed of four subunits, each with a heme group plus a globin chain. A single haemoglobin molecule can carry four O₂ molecules along with the blood.

d) Tissue fluid: It is also called lymph or interstitial fluid helps to bring oxygen and nutrients to cells and to remove waste products from them.

Question 4.
Redraw the diagram, label the parts, and write their functions.

Answer:
a) Kidney: They remove waste products from the blood and produce urine. Kidneys control the levels of many substances in the blood. Kidneys help to control our blood pressure.
b) Ureters: The ureters are the part of the urinary system, The ureters’ role in the process is to carry urine from the kidneys to the bladder. Contractions in the ureter force urine away from the kidneys and into the bladder.
c) Urinary bladder: A hollow organ that stores urine from the kidneys before disposal by urination.
d) Urethra: The urethra acts as a passage to expel urine from the urinary bladder to the outside of the body.

Question 5.
Analyse the figure and answer the questions.

a) Write the names of the circulations X and Y.
b) Write the names of the blood vessels i, ii, iii, iv, y.
c) What is the role of these circulations in the exchange and transport of gases?
d) Explain the role of these circulations in the process of excretion.

a) X- Alveolar exchange of gases (Pulmonary circulation), Y – Systemic exchange of gases (Systemic circulation).

b) (i) Inferior vena cava, (ii) Superior vena cava, (iii) Pulmonary artery, (iv) Pulmonary vein, (v) Aorta

c) Pulmonary circulation moves blood between the heart and the lungs. It transports deoxygenated blood to the lungs to absorb oxygen and release carbon dioxide. The oxygenated blood then flows back to the heart. Systemic circulation moves blood between the heart and the rest of the body.

d) The pulmonary veins by pulmonary circulation return oxygenated blood to the heart, which releases CO₂ from the deoxygenated blood to the lungs for gaseous exchange. The release of CO₂ is a kind of excretion. In systemic circulation, blood with oxygen, nutrients, and hormones travels from the heart to the rest of the body. In the veins, the blood picks up waste products as the body uses up oxygen, nutrients, and hormones.

Extended Activities

Question 1.
Visit a primary health centre and conduct an interview with a doctor on diseases affecting the lungs and kidneys.
Answer:
(Hints: Some question answers regarding this topic are given below for your reference.)

1. Can you explain some common lung diseases and their symptoms?
Answer:
Pulmonary fibrosis and sarcoidosis are examples of lung tissue disease.

2. How can lifestyle choices, such as smoking or pollution exposure, impact lung health?
Answer:
Smoking is one of the biggest contributors to lung-related health problems. It damages the airways and structures of the lungs, leading to chronic conditions like chronic bronchitis and emphysema, as well as various types of lung cancer. Quitting smoking is the best way to improve lung health.

3. Can you discuss the relationship between lung health and kidney function?
Answer:
The function of the lungs is also closely linked to the kidneys, both in health and in disease. In fact, regulation of acid-base balance, control of blood pressure, and fluid homeostasis are closely linked to the interaction of the kidneys and lungs.

Question 2.
Construct models of the respiratory system, kidney, and related parts and display them in the class.
Answer:
Materials:
Respiratory System:

• Cardboard boxes (different sizes for various organs)
• Balloons (represent lungs)
• Straws (represent trachea and bronchi)
• Pipe cleaners (represent smaller airways)
• Tape
• Markers/ paints
  Labels

Kidneys:

• Play dough or modeling clay (in different colors)
• Beans or pebbles (represent nephrons)
• Pipe cleaners or string (represent ureters)
• Cardboard or plastic sheet (to create a base)
• Markers/ paints
• Labels

Hints for Construction:

Research: Before diving in, have students research the basic structure and function of the respiratory system and kidneys. This will guide their model building process.

Divide and Conquer: Break the class into groups, each focusing on either the respiratory system or the kidneys. This allows for focused work and better model representation.

Creative Representation: Emphasize creativity! Cardboard boxes can become lungs, balloons can represent their inflation, and straws can depict the airways. Playdough can be molded into kidney shapes, beans as nephrons, and pipe cleaners as ureters.

Labeling is Key: Labels are crucial for understanding. Encourage students to label the different parts of their models (lungs, trachea, bronchi, nephrons, ureters, etc.) with clear descriptions of their function.

Showcase and Explain: Once the models are built, each group can present their work to the class. This allows students to explain the structure and function of the respiratory system or kidneys using their model as a visual aid.

Connect the Systems: Discuss the connection between these two systems. Explain how the respiratory system provides oxygen needed by the kidneys for proper function, and how the kidneys filter waste products produced by the body during respiration.

Question 3.
Organize an awareness class on organ donation.
Answer:
(Hints: important points to be included in the awareness class are given below)

• The Critical Need for Organ Donation
• Types of Organs and Tissues Donated
• The Organ Donation and Transplantation Process
• Dispelling Myths and Misconceptions
• Legal Aspects of Organ Donation

By including these key points, you can create a comprehensive and informative organ donation awareness class that empowers individuals to make informed decisions and potentially save lives.

Question 4.
Organize and implement programs to make your home and school waste-free.
Answer:
Here are some programs you can implement to reduce waste at home and school:

At Home:
Reduce:

• Plan meals: Reduce impulse buys and food waste by planning meals and creating a grocery list.
• Buy in bulk (with caution): Purchase staples like grains or nuts in bulk to minimize packaging, but ensure you’ll use everything before it expires.
• Choose reusable products: Opt for reusable shopping bags, water bottles, and coffee mugs instead of disposable alternatives.
• Fix instead of toss: Try to repair broken items instead of throwing them away.
• Borrow or rent: Instead of buying something you’ll rarely use, consider borrowing it from a friend or renting it.

Reuse:

• Get creative with containers: Repurpose old jars, containers, and boxes for storage or crafting projects.
• Donate or sell unwanted items: Give pre-loved clothes, toys, or furniture a second life through donations or online marketplaces.
• Compost food scraps: Start a compost bin to turn food scraps into nutrient-rich fertilizer for your plants (check local regulations for composting restrictions).

Recycle:

• Learn your local recycling guidelines: Understand what materials your local program accepts and ensure proper sorting.
• Invest in recycling bins: Have designated bins for recyclables (paper, plastic, glass, etc.) to make sorting easier.
• Recycle electronics responsibly: Don’t throw away old electronics; many electronics stores or specialized facilities accept them for proper recycling.

At School:
Reduce:

• Paperless communication: Encourage online communication (school websites, email) for announcements and assignments whenever possible.
• Double-sided printing: Set printers to default to double-sided printing to reduce paper usage.
• Reusable lunchware: Promote reusable lunch containers and water bottles among students and staff.
• Reusable alternatives: Replace single-use items like paper towels with cloth towels in classrooms and kitchens.

Reuse:

• Refill stations: Install water bottle refill stations to discourage the use of disposable water bottles.
• Donate or sell old textbooks: Organize textbook donation drives or sales to give them a second life.
• Creative project materials: Encourage the use of recycled or repurposed materials for art projects.

Recycle:

• Clearly labeled recycling bins: Place clearly labeled recycling bins throughout the school for easy sorting of paper, plastic, and other recyclables.
• Composting programs: If feasible, explore starting a school composting program for food scraps (check local regulations).
• Educational campaigns: Organize workshops or presentations to educate students and staff about the importance of waste reduction and recycling.

Respiration and Excretion Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
What is the role of oxygen in respiration and burning?
Answer:
Oxygen is essential for the respiration of living organisms, and it supports combustion because it is essential for burning components. Burning of candles involves the addition of oxygen to carbon hydrogen which forms carbon dioxide, water vapour and energy in the form of heat and light. Respiration utilises oxygen from the atmosphere to break down glucose in food and release carbon dioxide.

Question 2.
What are the products of respiration and burning?
Answer:
Both processes require oxygen for the production of energy. They produce carbon dioxide as a byproduct.

Question 3.
What is the respiratory surface in human beings? How is it arranged in our body?
Answer:
The respiratory surface in humans is the alveoli. They are tiny ballon-shaped sacs found in the lungs which exchange carbon dioxide and oxygen. The human respiratory system is composed of a pair of external nostrils, nasal chamber, pharynx, larynx, and trachea (the windpipe), which is bifurcated when enters into two lungs, bronchi again divides into fine branches called bronchioles inside the lungs and at the ends of the bronchioles we can find some sac-like structures called the alveoli. Alveoli are covered by thin blood capillaries through this exchange of gases takes place.

Question 4.
Complete the by analyse the description and prepare a short note on the respiratory surface in human beings.
Answer:

• Nostrils
• Nasal Cavity
• Trachea
• Bronchi

The respiratory surface is the area or space where the gas exchanges take place, It should have larger surface-to-volume ratio, moist inner surface, and thin wall in contact with blood capillaries Gaseous exchange in the lungs takes place between alveoli and alveolar blood capillaries. The respiration surface in humans is the alveoli.

Question 5.
Don’t you involved in sports and physical exercises? Is there any change in the rate o ventilation (breathing) during such activities? Do the activity given below?

• Form a group of two children each.
• Take rest for 5 minutes. Both of them record the number of inspirations that happen within a minute during this time.
• Record the time using a stop watch.
• In an interval of one minute, record the number of inspirations two more times.
• Then record the number of inspirations of the two children after running for three minutes as mentioned above.
• Continue to record the number of inspirations in every minute till they reach the normal condition.
• Complete the table given below using the recorded results, draw a line graph and compare the rate of ventilation of both the children.

Answer:

Question 6.
Analyse the figure and illustration and complete the table.

Indicators	Inspiration	Expiration
Action of intercostal muscles
Movement of ribs
Change that occurs to the diaphragm
The volume of the thoracic cavity	Increases	Decreases
The pressure of air in the lungs
Movement of air

Answer:

Indicators	Inspiration	Expiration
Action of intercostal muscles	Contracts	Relaxes
Movement of ribs	Rises	Lowers
Change that occurs to the diaphragm	Contracts	Relaxes
The volume of the thoracic cavity	Increases	Decreases
The pressure of air in the lungs	Decreases	Increases
Movement of air	From outside to the lungs	From lungs to the outside

Question 7.
What are the characteristic features of the walls of the alveolus and blood capillaries?
Answer:
The walls of the alveoli are very thin, which makes it easier for oxygen and CO₂ to pass between the alveoli and very small blood vessels (capillaries). Both are made up of a single layer of cells.

Question 8.
What is the importance of moisture in the wall of the alveolus?
Answer:
Moisture is important for gas exchange because, in order for oxygen and carbon dioxide to effectively cross the alveoli membrane, they must dissolve in an aqueous-like solution. With the right amount of moisture, gas exchange occurs properly.

Question 9.
Mention the concentration level of O₂ and CO₂ in the alveolus and in the blood capillaries.
Answer:
During inspiration, the concentration of oxygen inside the alveoli is more than that in blood capillaries whereas the concentration of carbon dioxide in blood capillaries is more than that in alveoli.

Question 10.
How does the exchange of O₂ and CO₂ between the alveolus and the blood capillaries happen?
Answer:
The exchange of gases (oxygen and carbon dioxide) occurs between the alveoli and blood capillaries by simple diffusion.

Indicators	Glycolysis	Krebs cycle
Site of activity
Substances that take part in the chemical process
Products
Requirement of oxygen

Answer:

Indicators	Glycolysis	Krebs cycle
Site of activity	Cytoplasm	Mitochondria
Substances that take part in the chemical process	Glucose converted into pyruvic avid	Pyruvic acid is converted into carbon dioxide and water
Products	Pyruvic acid and 2 ATP molecules	Carbon dioxide, water and 28 ATPs
Requirement of oxygen	Oxygen is not required	Oxygen is required

Question 11.
Complete the chemical process illustrated below by including the missing reactants required for cellular respiration and the products formed.

Answer:

Question 12.
Photosynthesis and respiration are metabolic processes that take place in the living world. Compare these two processes and revise Table replacing glycolysis and Krebs cycle with photosynthesis and respiration respectively.
Answer:

Indicators	Photosynthesis	Respiration
Site of activity	Chloroplast	Mitochondria
Substances that take part in the chemical process	CO2 , Water, Sunlight, Chlorophyll	Glucose
Products	Cilucose, Water, Energy	CO2, Water, Energy
Requirement of oxygen	Oxygen not required	Oxygen is required

Question 13.
What are the conditions that lead to the decrease in the level of haemoglobin? What are the different types of anaemia? Find out.
Answer:
Conditions that lead to the decrease in the level of haemoglobin:

• Blood loss, which can occur because of: Bleeding in your digestive tract, such as from ulcers, cancers, etc.
• Frequent blood donation
• Heavy menstrual bleeding

Different types of anaemia:

• Haemolytic anaemia
• Vitamin deficiency anaemia
• Iron deficiency anaemia
• Sickle cell anaemia, etc.

Question 14.
What are the healthy habits that can be followed to prevent anaemia? Discuss.
Answer:

• Eat foods rich in iron and B vitamins.
• Eat fruits and vegetables high in vitamin C, which helps your body absorb iron.
• Ask your doctor about iron supplements if you don’t get enough iron in your diet.
• Get checked every year or 2 if you’re a woman of childbearing age who has heavy menstrual periods or a previous diagnosis of anaemia.

Question 15.
Analyse illustration and description. Now prepare a short note on respiration in plants.

Answer:
Plants do not have a respiratory system or separate organs for the transport of gases. But they have special features in leaf, stem, and root for the exchange of gases.

The small pores on the surface of leaves and young stems are called stomata. Stomata remain open in the daytime and close at night. During the day photosynthesis and respiration take place simultaneously in the mesophyll cells of leaves. Since the rate of photosynthesis is high during the day, oxygen formed is used for respiration and the excess oxygen is expelled. CO₂ formed as a result of respiration is utilized for photosynthesis and insufficient CO₂ is received from the atmosphere. At night when stomata close, respiratory gas exchange occurs through diffusion.

Lenticels are the lens-shaped openings found on the surface of mature stems and roots. They also help in gaseous exchange in plants. Plant roots have special cells that perform gaseous exchange with the air that is present in the soil particles.

Question 16.
Does anaerobic respiration take place in human beings? Find out.
Answer:
Yes. In humans, aerobic respiration takes place in all cells as part of energy production in mitochondria. But in certain conditions, our cells lack oxygen to produce enough energy. For example, when we are doing strenuous exercises or something like that, our muscle cells lack oxygen. In such situations, muscle cells respire without oxygen which results in the production of lactic acid in cells.

Question 17.
Mention the differences in the cellular respiratory processes in bacteria and yeast.
Answer:
Both bacteria and yeast respire anaerobically within the cytoplasm of the cell. In both cases, the respiratory substances are the same, ie, glucose. But the end products are different. In lactobacillus, glucose is anaerobically broken down with the production of 2 molecules of ATP.

This breaking produces lactic acid through an intermediate production of pyruvic acid. In the case of yeast cells, alcohol and CO₂ are the end products of anaerobic respiration. Here also glucose is first converted into pyruvic acid with the production of two ATP molecules and then to alcohol and CO₂.

Question 18.
Some situations in daily life are given below. Discuss how anaerobic respiration is beneficial in such situations and prepare short notes on the same.
a) Yeast is added to leaven the dough.
b) Curd is added to milk to prepare curd.
Answer:
a) Yeast is added to leaven the dough:
Yeasts ferment the sugar and produce carbon dioxide in the process. Fermentation is otherwise called anaerobic respiration. Carbon dioxide results in the leavening or rising of the dough to three times its original size.

b) Curd is added to milk to prepare curd:
A teaspoon of curd is added to milk contains the bacteria Lactobacillus. As the bacteria gets more lactose, it will produce lactic acid in the milk converting the entire milk into curd.

Question 19.
Which are the major excretory substances in our body?
Answer:
The main excretory product generated by the human body is urea. The urea is excreted out of the body via urine. Sweat is another excretory substance from skin.

Question 20.
If CO₂ is not eliminated on time, how does it adversely affect the body? Find out.
Answer:
CO₂ transported through our body in three different ways, such as bicarbonate ions, as Carbonic acid when combine with water, then as CO₂ as such. Whern excess CO₂ accumulated in the body it will combines with water molecules which are abaundant in our body results in the formation of excess amount of carbonic acid. Presence of carbonic acid increases the acid content in our body.

Question 21.
How does the formation of ammonia happens?
Answer:
Amino acids are formed by the breakdown of proteins in the cells. As a result of the metabolic activities of amino acids, several nitrogenous by-products are formed. The most harmful among these is ammonia.

Question 22.
Name the organ where urea is synthesised?
Answer:
Liver

Question 23.
How does Urea synthesis happens?
Answer:
The ammonia formed in tissues diffuses into blood through tissue fluids and blood transports it to the liver. In liver with the help of certain enzymes, ammonia combines with carbon dioxide and water to form urea.

Question 24.
Explain the elimination of urea.
Answer:
Kidneys play a major role in the excretion of urine which contains waste materials including urea. Urea is made when foods containing protein (such as meat, poultry, and certain vegetables) are broken down in the body. Urea is carried in the blood to the kidneys. This is where it is removed, along with water and other wastes in the form of urine

Question 25.
Which are the excretory organs in our body?
Answer:

• Kidney – Excretes water and salts through urine.
• Liver – Synthesises urea
• Skin – Secretes sweat
• Lungs – Eliminates Carbon dioxide

Question 26.
Analyze illustrations based on the hints and find out how the structure of kidneys and nephrons are suitable to remove waste materials and complete the given table.

Hints	Parts
Blood vessel which carries blood to the kidneys
Blood vessel which carries blood away from the kidneys
Ultra filters present in the kidneys
The double walled cup-shaped structure present at one end of the nephron
Network of minute capillaries present in the Bowman’s capsule
Blood vessel which carries bkxxl to the capillary network
Blood vessel which carries blood away from the capillary- network
The long tubule which connects the Bowman’s capsule and the collecting duct
The part where renal tubules enter, collects urine and carries it to the pelvis.
The initial part of ureter
Part that carry urine to urinary bladder.

Answer:

Hints	Parts
Blood vessel which carries blood to the kidneys	Renal artery
Blood vessel which carries blood away from the kidneys	Renal vein
Ultra filters present in the kidneys	Nephron
The double-walled cup-shaped structure present at one end of the nephron	Bowman’s capsule
A network of minute capillaries present in the Bowman’s capsule	Glomerulus
Blood vessel which carries blood to the capillary network	Afferent vessel
Blood vessel which carries blood away from the capillary network	Efferent vessel
The long tubule that connects the Bowman’s capsule and the collecting duct	Renal tubules
The part where renal tubules enter, collects urine and carries it to the pelvis.	Collecting duct
The initial part of the ureter	Pelvis
The part that carry urine to the urinary’ bladder	Ureters

Question 27.
Analyze the illustration and complete the given worksheet using the hints.


Answer:

Question 28.
Why all components in the glomerular filtrate are not present in urine? Find out the components that are reabsorbed and the components that are secreted.
Answer:
Because some of the components in the glomerular filtrate are reabsorbed into the blood and some are secreted into the renal tubules. so that all compoments in the glomerular fillrate are not present in urine.
Components that are reabsorbed: Glucose, amino acids, NaCl , and other essential salts
Components which are secreted: Hydrogen ions, potassium ions, etc

Question 29.
Is diagnosis of diseases possible through urine tests?
Answer:
Yes. Urine as a Sample to Diagnose Sexually Transmitted Diseases. Urine is extremely useful to diagnose infections of the reproductive tract.

Question 30.
Kidney stone, nephritis, uraemia etc. are some diseases affecting the kidneys. Prepare a presentation including the causes, symptoms, etc of these diseases and present it in the class.
Answer:

Diseases	Causes	Symptoms
Kidney stone	Not drinking enough water	Severe pain in your back or side, Blood in urine, A burning feeling when urinating.
Nephritis	Diabetes, genetic disorder link that affects the kidneys.	Decreased urine output, High blood pressure, urine appears dark, tea-coloured, or cloudy
Uraemia	Diabetes, High blood pressure	Nausea, Vomiting, Loss of appetite

Question 31.
Analyze the illustration and prepare a short note on how hemodialysis is done.

Answer:
Hemodialysis is a process of filtering the blood of a person whose kidneys are not working normally. Hemodialysis helps control blood pressure and balance important minerals, such as potassium, sodium, and calcium, in your blood. In this process, blood with a high quantity of waste materials is passed to the dialysis unit after adding heparin to prevent coagulation.

When blood flows through the dialysis unit the waste materials in the blood diffuse into the dialysis fluid. This fluid is removed in due course. Antiheparin is added to the purified blood and is returned to the body.

Question 32.
Why is the dialysis fluid removed from the dialysis unit in due course of time?
Answer:
If the dialysis fluid is not removed properly, it can cause swelling and increase your blood pressure, which makes your heart work harder. Too much fluid can build up in the lungs, making breathing difficult. Hemodialysis removes fluid as the blood is filtered through the dialysis machine.

Question 33.
When is kidney transplantation required?
Answer:
Kidney transplantation is done when a person whose kidneys no longer function properly.

Question 34.
Complete the given table by collecting information about the main excretory products and excretory organs of the organisms.

Organism	Main excretory product	Main excretory organ/mechanism
Amoeba	Ammonia	Contractile vacuole
Earthworm
Insects
Frog
Reptiles
Birds

Answer:

• The kidney is the main excretory organ in human beings, but it is different in various organisms.
• Plants do not have special excretory systems like animals.
• The Hydathodes are also known as the water stomata

The excretory material is different in each organism depending on its life processes. The kidney is the main excretory organ in human beings, but it is different in various organisms.

Organism	Main excretory product	Main excretory organ/Mechanism
Amoeba Earthworm Insects Fishes Frogs Reptiles Birds	Ammonia Urea Uric acid Ammonia Urea Uric acid Uric acid	Contractile vacuole Nephridia Malpighian tubules Kidney Kidney Kidney Kidney

Question 35.
Analyze illustration and prepare a short note on excretion in plants.

Answer:
Animals have extensive systems for excretion. But plants do not have special excretory systems like animals. The different plant structures that eliminate different forms of waste are leaves, stems, ageing tissues, xylem, roots, bark, fruits, and flowers.

The process of excretion in plants occurs in the following ways: Through the stomata of leaves and lenticels of stems, oxygen, carbon dioxide, and water vapour are expelled as gaseous wastes. Some waste materials are gathered in tree bark and leaves. The wastes are removed when the leaves and bark are shed.The Hydathodes are also known as the water stomata; they are found at the tips of the veins of the leaves and help in the removal of water and salts from plants.

Question 36.
Analyze illustration and prepare a short note on how the excretory organs help in maintaining homeostasis.

Answer:
The excretory organs includes of the kidneys, liver, skin, large intestine, and the lungs. The excretory system maintains homeostasis through the elimination of excess water and wastes from the body.
Refer to topic 9 for more information.

Maintenance of homeostasis is the sign of life. The excretory system maintains homeostasis through the elimination of excess water and wastes from the body. The excretory structures consists of the kidneys, liver, skin, large intestine, and the lungs.

Kidneys:

• Regulate Salt and water balance.
• Regulation of blood pressure.
• Regulation of pH
• Expels waste

Liver:

• Regulates metabolism.
• Neutralises toxins.

Skin:

• Regulation of temperature, salt, and water.

Lungs:

• Elimination of CO₂
• Regulation of O₂ level.
• Regulation of pH

Any change in the internal environment disrupts homeostasis. The major factors that may cause a change in the internal environment are:

• Improper lifestyles
• Wrong food habits
• Over-nutrition and under-nutrition,
• Lack of exercise
• Mental stress
• Alcoholism
• Smoking and the use of addictive drugs
• Pollution
• Lack of hygiene
• The abundance of pathogens
• Improper use of medicines, and contact with toxic substances.

The external environment is also as important as the internal environment. The External environment should be kept waste-free for the well-being of living organisms. Every individual should develop the right attitude towards this. Prepare a master plan for making the school and its surroundings waste-free on behalf of the health club and implement it as part of the school master plan.

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 5 Malayalam Medium അഭിന്നകഗുണനം can save valuable time.

Kerala SCERT Class 9 Maths Chapter 5 Solutions Malayalam Medium അഭിന്നകഗുണനം

Class 9 Maths Chapter 5 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 5 Malayalam Medium Textual Questions and Answers

Question 1.
ഒരേ വലുപ്പമുള്ള നാലു സമഭുജത്രികോണങ്ങളിൽ രണ്ടെണ്ണം നെടുകെ മുറിച്ചതും, രണ്ടെണ്ണം മുഴുവനായും ചേർത്തുവച്ച് ഒരു ചതുരമുണ്ടാക്കാം:

സമഭുജത്രികോണങ്ങളുടെയെല്ലാം വശങ്ങളുടെ നീളം 2 സെന്റിമീറ്റർ ആണെങ്കിൽ, ചതുരത്തിന്റെ ചുറ്റളവും പരപ്പളവും എത്രയാണ്?
Answer:
ത്രികോണങ്ങളുടെ വശങ്ങളുടെ അളവുകൾ പരിഗണിക്കുകയാണെങ്കിൽ;

എന്ന് കിട്ടും.
അപ്പോൾ ചതുരത്തിന്റെ വശങ്ങളുടെ നീളം;

വീതി = 2 സെ.മീ
നീളം = √3 + √3 = 2√3 സെ.മീ

ചതുരത്തിന്റെ ചുറ്റളവ് = 2(വീതി + നീളം)
= 2(2 + 2√3)
= 4 + 4√3
≈ 4+4 × 1.732
≈ 4 + 6.928
≈ 10.928 സെ.മീ

ചതുരത്തിന്റെ പരപ്പളവ് = വീതി × നീളം
= 2 × 2√3
= 4√3
≈ 4 × 1.732
≈ 6.928 സെ.മീ

Question 2.
ഒരു സമചതുരവും, അതിന്റെ വശങ്ങളുടെ രണ്ടു മടങ്ങ് നീളമുള്ള വശങ്ങളോടുകൂടിയ ഒരു സമഭുജത്രികോണവും ചുവടെ കാണുന്നതുപോലെ മുറിച്ചു മാറ്റിയടുക്കി ഒരു ലംബകമുണ്ടാക്കാം.

സമചതുരത്തിന്റെ വശങ്ങളുടെ നീളം 2 സെന്റിമീറ്റർ എന്നെടുത്താൽ, ലംബകത്തിന്റെ ചുറ്റളവും, പരപ്പളവും എത്രയായിരിക്കും?
Answer:
സമചതുരങ്ങൾ രണ്ടായി മുറിച്ചതിൽ ഒരു കഷ്ണത്തിന്റെ അളവുകൾ;

സമഭുജത്രികോണം രണ്ടായി മുറിച്ചതിൽ ഒരു കഷണത്തിന്റെ അളവുകൾ;

ലംബകത്തിന്റെ വശങ്ങളുടെ അളവുകൾ;

ലംബകത്തിന്റെ ചുറ്റളവ് = 2 + 2√3 + 2 + 2√2 + 2√3 + 2√2
= 4 + 4√3 + 4√2
= 4 (1 + √3 + √2)
≈ 4(1 +1.732 + 1.414)
≈ 4 × 4.146
≈ 16.584 m.

ലംബകത്തിന്റെ പരപ്പളവ് = സമചതുരത്തിന്റെ പരപ്പളവ് + സമഭുജത്രികോണത്തിന്റെ പരപ്പളവ്
= 22 + \(\frac{4 \times 2 \sqrt{3}}{2}\)
= 4 + 4√3
= 4 (1 + √3)
≈ 4 (1 + 1.732)
≈ 4 × 2.732
≈ 10.928 ചതു. സെ.മീ

Question 3.
രണ്ടു സമചതുരങ്ങൾ ചേർത്തുവച്ച രൂപമാണ് ചിത്രത്തിൽ. ഈ രൂപത്തിന്റെ താഴത്തെ വശത്തിന്റെ നീളം, സെന്റിമീറ്റർ വരെ കൃത്യമായി കണക്കാക്കുക.

Answer:
വലിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം = \(\sqrt{27}\)
= \(\sqrt{9 \times 3}\)
= √9 × √3
= 3 × √3
= 3 × 1,73
= 5.19 6m.

ചെറിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം = \(\sqrt{12}\)
= √4 × √3
= 2 × √3
= 2 × 1.73
= 3.46 സെ.മീ.

താഴത്തെ വശത്തിന്റെ നീളം
= വലിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം + ചെറിയ സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം
= 5.19 + 3.46
= 8.65 സെ.മീ.

Question 4.
രണ്ടു സമചതുരങ്ങൾ ഒരു മൂലയിൽ ചേർത്തുവച്ച രൂപമാണ് ചിത്രത്തിൽ. ചരിഞ്ഞ വരയുടെ നീളം കണ്ടുപിടിക്കുക.

Answer:
വശത്തിന്റെ നീളം 2 ആയിട്ടുള്ള ഒരു സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം 2 ആണെന്ന് നമുക്കറിയാം. അതുകൊണ്ട്,
വലിയ സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം = 3/2 സെ.മീ
ചെറിയ സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം = 22 സെ.മീ
ചരിഞ്ഞ വരയുടെ നീളം = 3√2 + 2√2 = 5√2 സെ.മീ

Question 5.
ചിത്രത്തിലെ മട്ടത്രികോണത്തിന്റെ മൂന്നാമത്തെ വശത്തിന്റെ നീളവും ചുറ്റളവും കണക്കാക്കുക.

Answer:
പൈഥാഗറസ് സിദ്ധാന്തം അനുസരിച്ച് മട്ടത്രികോണത്തിന്റെ മൂന്നാമത്തെ വശത്തിന്റെ നീളം
= \(\sqrt{50-18}\)
= \(\sqrt{32}\)
= \(\sqrt{16×2}\)
= √16 × √2
= 4 × √2
= 4√2 സെ.മീ .

വലിയ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം
= \(\sqrt{50}\)
= \(\sqrt{25 \times 2}\)
= \(\sqrt{25}\) × √2
= 5 √2

ചെറിയ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം = \(\sqrt{18}\)
= \(\sqrt{9×2}\)
= √9 × √2
= 3 × √2
= 3√2 സെ.മീ .
മട്ടത്രികോണത്തിന്റെ ചുറ്റളവ് = 4√2 + 5√2 + 3√2
= 12√2 സെ.മീ

Question 6.
ചുവടെയുള്ള സംഖ്യാ ജോടികളിൽ ചിലതിന്റെ ഗുണനഫലം എണ്ണൽ സംഖ്യയോ ഭിന്നസംഖ്യയോ ആണ് അവ കണ്ടുപിടിക്കുക.
(i) √3, √12
Answer:
√3 × √12 = √3 × \(\sqrt{3 \times 4}\)
= √3 × √3 × √4
= 3 × 2
= 6
ഗുണനഫലം എണ്ണൽസംഖ്യ.

(ii) √3, √1.2
Answer:
\(\sqrt{3} \times \sqrt{1.2}=\sqrt{3 \times 1.2}\)
= \(\sqrt{3.6}\)

(iii) √5, √8
Answer:
\(\sqrt{5} \times \sqrt{8}=\sqrt{5 \times 8}\)
= \(\sqrt{40}\)
= \(\sqrt{4 \times 10}\)
= √4 × √10
= 2 × √10
= 2√10

(vi) √0.5, √8
Answer:
√0.5 × √8 = \(\sqrt{0.5 \times 8}\)
= √4
= 2
ഗുണനഫലം എണ്ണൽസംഖ്യ.

(v) \(\sqrt{7 \frac{1}{2}}, \sqrt{3 \frac{1}{3}}\)
Answer:
\(\sqrt{7 \frac{1}{2}} \times \sqrt{3 \frac{1}{3}}=\sqrt{\frac{15}{2}} \times \sqrt{\frac{10}{3}}\)
= \(\sqrt{\frac{15}{2} \times \frac{10}{3}}\)
= \(\sqrt{5 \times 5}\)
= \(\sqrt{25}\)
= 5
ഗുണനഫലം എണ്ണൽസംഖ്യ.

(vi) \(\sqrt{\frac{2}{5}}, \sqrt{\frac{1}{10}}\)
Answer:
\(\sqrt{\frac{2}{5}} \times \sqrt{\frac{1}{10}}=\sqrt{\frac{2}{5} \times \frac{1}{10}}\)
= \(\sqrt{\frac{1}{25}}\)
= \(\frac{1}{5}\)
ഗുണനഫലം ഭിന്നസംഖ്യ.

Question 7.
(√2 + 1) (√-2 – 1) = 1 എന്ന് തെളിയിക്കുക. ഇത് ഉപയോഗിച്ച്:
i. \(\frac{1}{\sqrt{2}-1}\) രണ്ടു ദശാംശസ്ഥാനം വരെ കണക്കാക്കുക.
ii. \(\frac{1}{\sqrt{2}+1}\) രണ്ടു ദശാംശസ്ഥാനം വരെ കണക്കാക്കുക.
Answer:
(√2 + 1)(√2 – 1) = (√2)² – 1²
= 2 – 1
= 1

i. \(\frac{1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}-1}\)
= √2 + 1
= 1.41 + 1
= 2.41

ii. \(\frac{1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}+1}\)
= √2 – 1
= 1.41 – 1
= 0.41

Question 8.
ചിത്രത്തിൽ സമഭുജത്രികോണത്തിന്റെ വശങ്ങളുടെ നീളം മില്ലിമീറ്റർ വരെ കൃത്യമായി കണക്കാക്കുക.

Answer:
താഴെ തന്നിരിക്കുന്ന ത്രികോണം പരിഗണിക്കുക.

പൈഥാഗറസ് സിദ്ധാന്തം ഉപയോഗിച്ച്;
(2x)² – x² = 4²
4x² – x² = 16
3x² = 16
x² = \(\frac{16}{3}\)
x = \(\sqrt{\frac{16}{3}}\)
= \(\frac{\sqrt{16}}{\sqrt{3}}\)
= \(\frac{4}{1.732}\)
= 2.30 സെ.മീ

വശങ്ങളുടെ നീളം = 2 × 2.30
= 4.60 സെ.മീ

Question 9.
ചിത്രത്തിലെ ചുവന്ന ത്രികോണങ്ങളെല്ലാം സമഭുജമാണ്. പുറത്തെ സമചതുരത്തിന്റെയും, അകത്തെ സമചതുരത്തിന്റെയും വശങ്ങൾ തമ്മിലുള്ള അംശബന്ധം എന്താണ്?

Answer:
സമഭുജത്രികോണത്തിന്റെ ഒരു വശത്തിന്റെ നീളം ‘a’ എന്നെടുത്താൽ
പുറത്തെ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം = a + a√3 = a(1 + √3) = a (√3 + 1)
അകത്തെ സമചതുരത്തിന്റെ വശത്തിന്റെ നീളം = a + a√3 – (a + a)
= a + a√3 – 2a
= a√3 – a
= a (√3 − 1)
പുറത്തെ സമചതുരത്തിന്റെയും, അകത്തെ സമചതുരത്തിന്റെയും വശങ്ങൾ തമ്മിലുള്ള അംശബന്ധം

= \(\frac{\mathrm{a}(1+\sqrt{3})}{\mathrm{a}(\sqrt{3}-1)}\)
= \(\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
= (√3 + 1) (√3 – 1)

Question 10.
\(\sqrt{2 \frac{2}{3}}\) = 2\(\sqrt{\frac{2}{3}}\) എന്നും \(\sqrt{3 \frac{3}{8}}=3 \sqrt{\frac{3}{8}}\) എന്നും തെളിയിക്കുക. ഇതുപോലുള്ള മറ്റു സംഖ്യകൾ കണ്ടുപിടി ക്കാമോ?
Answer:

ഇതുപോലുള്ള മറ്റു സംഖ്യകളുടെ പൊതുരൂപം ഇങ്ങനെ വിശദീകരിക്കാം;

Question 11.
ചുവടെയുള്ള സംഖ്യാജോടികളിൽ ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ ചിലത് എണ്ണൽസംഖ്യയോ ഭിന്നസംഖ്യയോ കിട്ടും. അവ കണ്ടുപിടിക്കുക.
(i) √72, √2
(ii) √27, √3
(iii) √125, √50
(iv) √10, √2
(v) √20, √5
(vi) √18, √8
Answer:
(i) \(\frac{\sqrt{72}}{\sqrt{2}}=\frac{\sqrt{36 \times 2}}{\sqrt{2}}=\frac{\sqrt{36} \times \sqrt{2}}{\sqrt{2}}\)
= \(\sqrt{36}\)
= 6
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ എണ്ണൽസംഖ്യ കിട്ടും.

(ii) \(\frac{\sqrt{27}}{\sqrt{3}}=\frac{\sqrt{9 \times 3}}{\sqrt{3}}=\frac{\sqrt{9} \times \sqrt{3}}{\sqrt{3}}\)
= √9
= 3
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ എണ്ണൽസംഖ്യ കിട്ടും.

(iii) \(\frac{\sqrt{125}}{\sqrt{50}}=\frac{\sqrt{25 \times 5}}{\sqrt{25 \times 2}}\)
= \(\frac{\sqrt{25} \times \sqrt{5}}{\sqrt{25} \times \sqrt{2}}\)
= \(\frac{\sqrt{5}}{\sqrt{2}}\)

(iv) \(\frac{\sqrt{10}}{\sqrt{2}}=\frac{\sqrt{2 \times 5}}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{5}}{\sqrt{2}}\)
= √5

(v) \(\frac{\sqrt{20}}{\sqrt{5}}=\frac{\sqrt{4 \times 5}}{\sqrt{5}}\)
= \(\frac{\sqrt{4} \times \sqrt{5}}{\sqrt{5}}\)
= √4
= 2
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ എണ്ണൽസംഖ്യ കിട്ടും.

(vi) \(\frac{\sqrt{18}}{\sqrt{8}}=\frac{\sqrt{9 \times 2}}{\sqrt{4 \times 2}}=\frac{\sqrt{9} \times \sqrt{2}}{\sqrt{4} \times \sqrt{2}}\)
= \(\frac{\sqrt{9}}{\sqrt{4}}=\frac{3}{2}\)
ആദ്യസംഖ്യയെ രണ്ടാം സംഖ്യകൊണ്ട് ഹരിച്ചാൽ ഭിന്നസംഖ്യ കിട്ടും.

Question 12.
ചില സമഭുജത്രികോണങ്ങളുടെ ഒരു വശത്തിന്റെ നീളം ചുവടെ കൊടുത്തിരിക്കുന്നു. പരപ്പളവ് കണക്കാക്കുക.
i) 10 സെ.മീ
ii) 5 സെ.മീ
iii) √3 സെ.മീ
Answer:
സമഭുജത്രികോണങ്ങളുടെ പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) (വശO)²
i. പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × a
= \(\frac{\sqrt{3}}{4}\) × 10²
= \(\frac{\sqrt{3}}{4}\) × 100
= √3 × 25
~ 1.732 × 25
~ 43.3 സെ.മീ

ii. പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × 5²
= \(\frac{\sqrt{3}}{4}\) × 25
= √3 × 6.25
~ 6.25 × 1.732
~ 10.825 ചതു.സെ.മീ

iii. പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × (√3)²
= \(\frac{\sqrt{3}}{4}\) × 3
= √3 × 0.75
~ 1.732 × 0.75
~ 1.299 ചതു.സെ.മീ

Question 13.
വശങ്ങളുടെയെല്ലാം നീളം 6 സെന്റിമീറ്റർ ആയ സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.
Answer:
ഒരു സമഷഡ്ഭുജത്തിനെ അതിന്റെ വശങ്ങളുടെ അതേ നീളമുള്ള വശങ്ങളുള്ള ആറ് സമഭുജത്രികോ ണങ്ങളായി മുറിക്കാൻ പറ്റും.

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ്
= ആറ് സമഭുജത്രികോണങ്ങളുടെയും പരപ്പളവിന്റെ തുക
= 6 × ഒരു സമഭുജത്രികോണത്തിന്റെ പരപ്പളവ്
= 6 × \(\frac{\sqrt{3}}{4}\) × 6²
= 6 × √3 × 9
= 54 × √3
= 54 × 1.732
= 93.528 ചതു.സെ.മീ

Question 14.
ഉയരം 12 സെന്റിമീറ്റർ ആയ സമഭുജത്രികോണത്തിന്റെ ചുറ്റളവും പരപ്പളവും കണക്കാക്കുക.
Answer:
ഉയരം = 12 സെ.മീ
\(\frac{\sqrt{3}}{4}\) × വശം = 12
വശം = \(\frac{12 \times 2}{\sqrt{3}}\)
= \(\frac{3 \times 4 \times 2}{\sqrt{3}}\)
= \(\frac{4 \times \sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\)
= 8√3

പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × (വശം)
= \(\frac{\sqrt{3}}{4}\) × (8√3)²
= \(\frac{\sqrt{3}}{4}\) × 64 × 3
= √3 × 16 × 3
≈ 1.732 × 48
≈ 83.136 ചതു.സെ.മീ

ചുറ്റളവ് = 3 × വശം
= 3 × 8√3
= 24√3 സെ.മീ
≈ 24 × 1.732
≈ 41.568 സെ.മീ

Question 15.
സമാന്തരവശങ്ങൾ തമ്മിലുള്ള അകലം 6 സെന്റിമീറ്റർ ആയ സമഷഡ്ഭുജത്തിന്റെ ചുറ്റളവും പരപ്പളവും കണക്കാക്കുക.
Answer:
ഒരു സമഷഡ്ഭുജത്തിനെ അതിന്റെ വശങ്ങളുടെ അതേ നീളമുള്ള വശങ്ങളുള്ള ആറ് സമഭുജ ത്രികോണങ്ങളായി മുറിക്കാൻ പറ്റും.

അങ്ങനെനോക്കിയാൽ,
സമാന്തരവശങ്ങൾ തമ്മിലുള്ള അകലം = 2 × ഒരു സമഭുജത്രികോണത്തിന്റെ ഉയരം
6 = 2 × \(\frac{\sqrt{3}}{2}\) × സമഭുജത്രികോണത്തിന്റെ വശത്തിന്റെ നീളം
= 2 × \(\frac{\sqrt{3}}{2}\) × സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം
\(\frac{6}{2} \times \frac{2}{\sqrt{3}}\) = സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം

സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം = \(\frac{6}{\sqrt{3}}\)
= \(\frac{3 \times 2}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\)
= √3 × 2
= 2√3 സെ.മീ

സമഷഡ്ഭുജത്തിന്റെ ചുറ്റളവ് = 6 × സമഷഡ്ഭുജത്തിന്റെ വശത്തിന്റെ നീളം
= 6 × 2√3
= 12√3
≈ 12 × 1.732
≈ 20.784 സെ.മീ

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് = ആറ് സമഭുജത്രികോണങ്ങളുടെയും പരപ്പളവിന്റെ തുക
= 6 × \(\frac{\sqrt{3}}{4}\) × (2√3)²
= 6 × \(\frac{\sqrt{3}}{4}\) × 4 × 3
= 18 × √3
≈ 18 × 1.732
≈ 31.176 ചതു.സെ.മീ

Question 16.
വശങ്ങളുടെ നീളം 8 സെന്റിമീറ്റർ, 6 സെന്റിമീറ്റർ, 6 സെന്റിമീറ്റർ ആയ ത്രികോണത്തിന്റെ ഉയരവും പരപ്പളവും കണക്കാക്കുക.
Answer:

ഉയരം = \(\sqrt{6^2-4^2}\)
= \(\sqrt{36-16}\)
= \(\sqrt{20}\)
= \(\sqrt{4 \times 5}\)
= √4 × √5
= 2 × √5
= 2√5 സെ.മീ

പരപ്പളവ് = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 8 × 2√5
= 8√5 ചതു.സെ.മീ

Question 17.
ചില ത്രികോണങ്ങളുടെ വശങ്ങളുടെ നീളം ചുവടെ പറഞ്ഞിരിക്കുന്നു. ഓരോന്നിന്റെയും പരപ്പളവ് കണക്കാക്കുക.
i. 4 സെന്റിമീറ്റർ, 5 സെന്റിമീറ്റർ, 7 സെന്റിമീറ്റർ
ii. 4 സെന്റിമീറ്റർ, 13 സെന്റിമീറ്റർ, 15 സെന്റിമീറ്റർ
iii. 5 സെന്റിമീറ്റർ, 12 സെന്റിമീറ്റർ, 13 സെന്റിമീറ്റർ
Answer:
i. a = 4 സെ.മീ
b = 56 സെ.മീ
c = 7 സെ.മീ

ii. a = 4 സെ.മീ
b = 13 സെ.മീ
c = 15 സെ.മീ

iii. a = 5 സെ.മീ
b = 12 സെ.മീ
c = 13 സെ.മീ

Irrational Multiplication Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
തന്നിരിക്കുന്നവ ചെറുതിൽനിന്ന് വലുതിലേക്ക് എന്ന രീതിയിൽ എഴുതുക.
i. 3√5 and 4√3
ii. 2√5, 5√2 and 3√7
Answer:
(i) \(\sqrt{53}=\sqrt{3 \times 3 \times 5}=\sqrt{45}\)
4√3 = \(\sqrt{4 \times 4 \times 3}=\sqrt{48}\)
48 > 45 ⇒ \(\sqrt{48}>\sqrt{45}\)
⇒ 4√3 > 3√5
ചെറുതിൽനിന്ന് വലുതിലേക്ക് എന്ന രീതിയിൽ എഴുതിയാൽ; 3√5, 4√3

(ii) 2√5 = √2 × 2 × 5 = √20
5√2 = √5 × 5 × 2 = √50
3√7 = √3 × 3 × 7 = √63
20 < 50 < 63 ⇒ √20 <√50 < √63
⇒ 2√5 < 5√2 < 3√7
ചെറുതിൽനിന്ന് വലുതിലേക്ക് എന്ന രീതിയിൽ എഴുതിയാൽ; 2√5, 5√2, 3√7

Question 2.
x = √0.5, y = √32, z = √128
a) xy, yz, xz എന്നിവ കണ്ടെത്തുക.
Answer:

b) xy + yz + xz കണ്ടുപിടിക്കുക.
Answer:
xy + yz + xz = 4 + 64 + 8
= 76

c) y = 8x എന്ന് തെളിയിക്കുക
Answer:
8x = 8 × \(\sqrt{0.5}\)
= \(\sqrt{64 \times 0.5}\)
= \(\sqrt{32}\)
= y

Question 3.
√8 നെ \(\sqrt{4 \times 2}\) = 2√2 എന്ന് എഴുതാം
a) ഒരു പൂർണ്ണസംഖ്യയുടെയും √2 വിന്റെയും ഗുണനമായി √18 നെയും √32 നെയും എഴുതുക.
Answer:
\(\sqrt{18}=\sqrt{9 \times 2}\) = 3√2
\(\sqrt{32}=\sqrt{16 \times 2}\) = 4√2

b) √2 + √8 + √18 – √32 നെ ചെറുതാക്കി എഴുതുക.
Answer:
\(\sqrt{2}+\sqrt{8}+\sqrt{18}-\sqrt{32}\)
= √2 + 2√2 + 3√2 – 4√2
= 6√2 – 4√2
= 2√2

Question 4.
ചുറ്റളവ് 18 സെ.മീ ആയ സമഭുജത്രികോണത്തിന്റെ ഏരിയ എത്ര?
Answer:
ചുറ്റളവ് = 18
3 × വശം = 18

വശം = \(\frac{18}{3}\)
= 6 സെ.മീ

പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × (വശം)²
= \(\frac{\sqrt{3}}{4}\) × 6²
= √3 × 9
= 9√3 ചതു.സെ.മീ

Question 5.
38 ഉം 7/8 ഉം തമ്മിൽ ഗുണിക്കുക.
Answer:
3√8 × 7√8 = 3 × √8 × 7 × √8
= 3 × 7 × √8 × √8
= 21 × 8
= 168

A thorough understanding of Kerala Syllabus 9th Standard Biology Textbook Solutions Chapter 2 Digestion and Transport of Nutrients Notes Questions and Answers English Medium can improve academic performance.

SCERT Class 9 Biology Chapter 2 Notes Questions and Answers Digestion and Transport of Nutrients

Std 9 Biology Chapter 2 Notes Pdf Kerala Syllabus English Medium Solutions Questions and Answers

Class 9 Biology Chapter 2 Let Us Assess Answers Digestion and Transport of Nutrients

Question 1.
Which one of the following is related to the digestion of Fat?
a) Protease
b) Lipase
c) Amylase
d) Carbohydrase
Answer:
b) Lipase

Question 2.
An illustration related to the circulatory system of human beings is given below. Analyse it and answer the questions.

a) Which letter indicates Pulmonary artery?
Answer: A

b) Which blood vessel is indicated by the letter D?
Answer: Aorta

c) Does the blood that has entered the ventricles return to the atria? Why?
Answer:
No, the blood do not flow back to the atria. Because, tricuspid valve and bicuspid valve prevents the back flow of blood from ventricles to atria.

d) What is the importance of double circulation in human?
Answer:
The same amount of blood passes through the heart twice. This type of circulation is called double circulation. It includes pulmonary circulation and systemic circulation. Double circulation helps to maintain the level of oxygen.

Question 3.
A flowchart on the path of nutrients is given below. Observe it and answer the questions.

a) Name the blood vessels indicated by the letters A, B and C. A, B,C
Answer:
A – Portal vein
B – Hepatic vein
C – Venacava

b) Do all the nutrients absorbed from the small intestine have the same path? Explain.
Answer:
Nutrients like amino acids and glucose are absorbed into the blood capillaries of the villus and are transported to liver through portal vein. Then they are carried to venacava through hepatic vein. Substances like fatty acid and glycerol are absorbed by lacteal of villus. They are carried to venacava by lymph vessel. Venacava carries all these nutrients to the heart.

Question 4.
Which among the following processes takes place by utilising energy?
a) Entry of water into the root cells.
b) Entry of sucrose into the sieve tube.
c) Loss of water from leaves through transpiration.
d) Conduction of water molecules through xylem tubes.
Answer:
b) Entry of sucrose into the sieve tube

Question 5.
Redraw the figure given below, name and label the parts based on the indicators.

a) Part where odontoblast cells are seen
Answer: Pulp cavity

b) Tissue that holds the tooth in the gum.
Answer: Cementum

c) The living tissue by which tooth is made.
Answer: Dentine

Extended Activities

Question 1.
Conduct an awareness class on the topic ‘Bad habits and health of the heart’.
Answer:
Main Points

• Heart disease is the leading cause of death globally.
• While some risk factors are uncontrollable (age, genetics), many lifestyle habits significantly influence heart health:
• Excessive saturated and trans fats raise bad cholesterol, clogging arteries.
• High sodium intake increases blood pressure, putting strain on the heart.
• A sedentary lifestyle increases the risk of obesity, diabetes, and heart disease
• Smoking increases the risk of blood clots and heart attacks.
• Limit saturated and trans fats, processed foods, and added sugars
• Regular exercise strengthens your heart muscle and improves blood flow.
• Ensure adequate sleep for better overall health and stress management.

Question 2.
Organize a nutritious food fest under the auspices of the school health club using locally available food items.
Answer:

Digestion and Transport of Nutrients Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Which are the different stages of nutrition?
Answer:

• Ingestion
• Mechanical Digestion
• Chemical Digestion
• Absorption
• Transport
• Egestion of digestive waste

Question 2.
Compare the nutritional processes of amoeba, a unicellular organism and hydra, a multicellular organism and complete the table.

Hints	Amoeba	Hydra
Body structure	Unicellular
Means to help ingestion
Part where digestion takes place	Inside the cell
Egestion of digestive wastes

Answer:

Hints	Amoeba	Hydra
Body structure	Unicellular	Multicellular
Means to help ingestion	Pseudopodia	Mouth
Part where digestion takes place	Inside the cell	Outside the cell
Egestion of digestive wastes	Through the cell surface	Through the mouth

Question 3.
Which are the parts where mechanical digestion takes place?
Answer:

• Mouth
• Stomach
• Small intestine

Question 4.
How does mechanical digestion take place in these parts? prepare a note based on the indicators.
Indicators:

• Structure of tooth.
• Mechanical digestion in mouth, stomach and small intestine.

Answer:
Teeth facilitate the process of breaking down food into smaller pieces and grinding it during chewing. Enamel is the hardest substance in the human body, serving as the non-living outer covering of the tooth. Beneath the enamel is dentine, a living tissue that forms the bulk of the tooth. The innermost part of the tooth is the pulp cavity, which contains soft connective tissue called pulp, along with blood vessels, nerves, and odontoblast cells. The tooth is anchored in the gum socket by cementum, a calcium-containing connective tissue.

The strong peristalsis occurring in the stomach transforms food into a paste form. The circular muscles of the stomach maintain food within the stomach for a sufficient duration. The small intestine involves mechanical processes, namely peristalsis and segmentation, to help in the movement of food and the blending of food with digestive fluids.

Question 5.
What are the things to be taken care for proper dental care? Conduct an interview with a doctor, prepare a poster and exhibit it.
Answer:
Interview
Interviewer: What are the most important things people can do to maintain a good oral health?
Doctor:

• Brush your teeth twice a day
• Flossing once a day removes plaque and food particles from between the teeth, where a brush can’t reach
• Schedule dental checkups and cleanings at least once a year.

Interviewer: Are there any additional tips you recommend for optimal dental health?
Doctor:

• Limit sugary drinks and foods
• Don’t use tobacco products
• Drink plenty of water
• Consider using mouthwash

Poster

Question 6.
Why doesn’t food enter trachea while swallowing it?
Answer:
Tongue compresses the food into balls with the help of palate. Uvula closes the nasal cavity that opens to the pharynx. Posterior part of the tongue allows food to move over the epiglottis into the oesophagus. Trachea rises up and is closed by the epiglottis.

Question 7.
What is the reason for saying that one should not talk while eating food? Find out.
Answer:
Talking while eating can cause food to enter the trachea instead of oesophagus. This will trigger coughing or choking.

Question 8.
How far is the structure of villus suitable for the absorption process? Prepare note according to the indicators.
Indicators:

• Villus and surface area of absorption.
• Lacteal and absorption.
• Blood capillaries and absorption.

Answer:
Villi increase the surface area of absorption of nutrients to a great extent within the small intestine. Absorption of simple nutrients and 90 percent of water takes place through villi. Villi are lined with single-layered epithelial cells, which are the primary surface for nutrient absorption.

Blood capillaries within the villus, formed from an arterial branch, absorb glucose, fructose, galactose, and amino acids, and then unite to form veins that exit the villus. Additionally, the lacteal, a branch of the lymph vessel, absorbs fatty acids and glycerol into the lymph.

Question 9.
Complete the illustration related to the blood components.

Answer:

Question 10.
Find out the functions of different proteins in plasma.
Answer:
Albumin – Regulates blood pressure
Globulin – Helps in defence
Fibrinogen – Plays a major role in the coagulation of blood

Question 11.
How is the quantity of tissue fluid regulated in the space between cells? Find out.
Answer:
Along with the formation of tissue fluid, it absorbs into the lymph capillaries. So there is no increase of pressure in the intercellular space.

Question 12.
Complete the flowchart using the hints.
Hints:

• Portal vein
• Lacteal of villus
• Amino acids
• Glycerol
• Hepatic vein
• Lymph vessel

Answer:

Question 13.
Which are the phases included in a cardiac cycle?
Answer:

• Atrial systole
• Ventricular systole
• Joint diastole

Question 14.
How the function of heart is maintained in people with a functionless pacemaker. Find out.
Answer:
In people with functionless pacemaker, artificial pacemaker is implanted for the functioning.

Question 15.
Apart from the wrist, which are the other parts of the body where we can feel pulse? Find out.
Answer:
Neck, Groin, Foot

Question 16.
Complete the chart

SI . No	Name of Children	Rate of heart beat		Pulse rate
		At rest	After doing exercise	At rest	After doing exercise
1
2

Answer:

SI No	Name of Children	Rate of heart heat		Pulse rate
		At rest	After doing exercise	At rest	After doing exercise
1.	Megha	72/m in	120/min	72/min	120/min
2.	Nikhil	74/min	125/min	74/min	125/min

Question 17.
What is the reason for variations of blood pressure? How does it affect the body? Find out.
Answer:
The normal rate of blood pressure is 120/80 mmHg. The disease condition in which the blood pressure increases above the normal rate is called hypertension. This happens due to many reasons. Unhealthy habits such as excess use of salt and fat, smoking, lack of exercise etc. The condition in which the blood pressure rate goes below the prescribed rate is called hypotension. These hypotension and hypertension may lead to stroke or heart attack.

Question 18.
How does the body utilise each nutrient? Find out.
Answer:
Carbohydrates are broken down into simple sugars in the digestive system. Glucose enters the bloodstream and is transported to cells, where it’s used for immediate energy production or stored for later use as glycogen in the liver and muscles.

Proteins are broken down into amino acids during digestion. Amino acids are then absorbed into the bloodstream and used to build, repair, and maintain various body structures and functions.

Fats are broken down into fatty acids and glycerol during digestion. Fatty acids are then transported to cells for energy storage or used to build cell membranes and signaling molecules.

Vitamins are absorbed in small amounts from food and transported throughout the body. They work in conjunction with enzymes to regulate numerous metabolic processes.

Students rely on Kerala Syllabus 9th Standard Chemistry Notes Pdf Download Chapter 7 Non Metals Extra Questions and Answers to help self-study at home.

Kerala Syllabus Std 9 Chemistry Chapter 7 Non Metals Extra Questions and Answers

Question 1.
a) What are the chemicals required for the laboratory preparation of Hydrogen?
b) Hydrogen-filled balloons rise up in the air. Why?
Answer:
a) Zinc and Hydrochloric acid
b) The density of hydrogen is less than air.

Question 2.
Hydrogen is used as a fuel because of its high calorific value.
a) What is calorific value?
b) Write any other advantage of hydrogen fuel.
c) Hydrogen is not commonly used as domestic fuel. Give reason
Answer:
a) The calorific value of a fuel is the heat energy released from one unit mass of that fuel on complete combustion.
b) Availability is plenty / High calorific value/pollution free.
c) Bum with an explosion, it is very difficult to store and transport.

Question 3.
a) What are the chemicals required to prepare hydrogen in the laboratory?
b) Write the balanced chemical equation of this reaction?
c) Which type of reaction is this?
Answer:
a) Dilute hydrochloric acid and Zinc
b) Zn + 2HCl → ZnCl₂ + H₂
c) Displacement reaction

Question 4.
List the uses of hydrogen
Answer:

• For the industrial production of ammonia and methanol.
• To saturate unsaturated oils.
• As a fuel.

Question 5.
a) Which fuel has the highest calorific value?
b) What will be the product formed when hydrogen burns in the air?
Answer:
a) Hydrogen
b) Water (H₂O)

Question 6.
Write the reaction between Hydrogen and chlorine.
Answer:
Hydrogen combines with chlorine in the presence of sunlight to form hydrogen chloride.
H₂2 + Cl₂ 2HCl

Question 7.
Write any three properties of Hydrogen.
Answer:
Insoluble in water, Hydrogen gas bums with a pop sound, Hydrogen is a colourless and odourless gas

Question 8.
Carbon is an element that shows allotropy.
a) What is allotropy?
b) Non-crystalline allotropes of carbon are commonly called __________
Answer:
a) Different forms of the same element having different physical properties but with the same chemical properties are known as Allotropes, and this phenomenon is called Allotropy.

b) Amorphous carbon.
The structure of two crystalline allotropes of carbon is given below,

Question 9.
The structure of two crystalline allotropes of carbon is given below.
a) Which among these is the structure of a diamond?


b) Which of these crystalline structures has no free electron?
c) Graphite is used as a lubricant. Why?
Answer:
a) Figure B
b) Figure B (Diamond)
c) Layers are held together by weak van der Waals forces. Hence, one layer can slide over another.

Question 10.
Carbon monoxide and carbon dioxide are two compounds formed by the combination of carbon and oxygen.
a) Write the balanced chemical equation showing the formation of carbon monoxide
b) Which are the gases mixed with carbon monoxide to prepare water gas and producer gas respectively?
c) Inhaling of excess carbon monoxide leads even to death. Why?
Answer:
a) 2C + O₂→ 2CO
b) Water gas → CO + H₂, Producer gas → CO + N₂
c) It reacts with the haemoglobin in the blood and forms carboxyhaemoglobin. As a result, the oxygen-carrying capacity of blood decreases, leading even to death.

Question 11.
When eggshell is added to Hydrochloric acid, a gas is produced.
a) Identify this gas.
b) Write the chemical equation of this reaction.
Answer:
a) Carbon dioxide
b) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Question 12.
Carbon dioxide (CO₂) is an important oxide of carbon.
a) Name the process by which CO₂ is utilised by plants.
b) Write an environmental problem due to the tremendous increase in the amount of CO₂ in the atmosphere.
c) What is dry ice? Write any one use.
Answer:
a) Photosynthesis
b) Climate change/ Greenhouse effect
c) The solid form of carbon dioxide.
Used as refrigerant / To create foggy effects in stage shows.

Question 13.
Writ any three uses of carbon dioxide.
Answer:
As gaseous fuel, To produce industrial fuels like water gas and producer gas, As a reducing agent in metallurgy.

Question 14.
a) Why does graphite conduct electricity?
b) Write any two properties of graphene
c) What are Buckyballs and Buckytubes?
Answer:
a) Graphite conducts electricity due to delocalised electrons and layered structure.

b) High electrical conductivity
High thermal conductivity

c) Fullerene has a hollow spherical structure consisting of pentagons and hexagons. They are
known as Bucky balls.
Fullerene with a cylindrical structure is used as carbon nanotubes. These are called Buckytubes.

Question 15.
What are the different ways in which nitrogen is obtained by plants?
Answer:

• By decomposition
• Through fertilisers
• By nitrogen fixing by bacteria

Question 16.
“It is said that lightning is a boon to plants”. Justify the statement.
Answer:
During lightning, the triple bond in nitrogen breaks, and nitrogen combines with the atmospheric oxygen to form nitric oxide (NO)
N₂ + O₂ → NO
Nitric oxide thus combines with oxygen to form nitrogen dioxide (NO₂)
2NO + O₂ → 2NO₂
Nitrogen dioxide dissolves in rainwater in the presence of oxygen and reaches the soil as nitric acid (HNO₃)
4NO₂ + 2H₂O + O₂ → 4HNO₃
Nitric acid reacts with the minerals in the soil to form nitrate salts, which the plants absorb. So, it is said that lightning is a boon to plants.

Question 17.
Can you list out which other means are there for getting a greater amount of elements for plants?
Answer:

• Use of organic fertilisers
• Use chemical fertilisers
• Bio decomposition
• Through nitrogen fixation

Question 18.
Write any two advantages of using organic fertilizers.
Answer:
Eco-friendly, Preserve the nature of the soil.

Question 19.
Nitrogen is inert in nature. Because it contains ………………. in the molecule.
(a) single bond (b) double bond (c) triple bond (d) ionic bond
Answer:
Triple bond

Question 20.
Classify the following gases into molecules having single bond, double bond and triple bond (a) Nitrogen (b) Chlorine (c) oxygen
Answer:
Single bond – Chlorine, Double bond – Oxygen, Triple Bond – Nitrogen

Question 21.
List the merits and limitations of the application of organic fertilisers.
Answer:
Merits:

• Eco friendly
• Preserve the nature of the soil

Limitations:

• Biodegradation needs time, so it cannot easily be absorbed by plants in time.
• The presence of microorganisms is necessary.
• Storage and transportation are not easy

Question 22.
Heat potassium permanganate in a dry boiling tube. When a glowing match stick is shown into
this boiling point tube, it flares up.
a) The presence of which gas is indicated by the flaring up of the glowing match stick?
b) Write down the chemical equation of the reaction taking place in the boiling tube.
c) To which category does this reaction belong?
Answer:
a) Oxygen
b) 2KMnO₄ + Heat → K₂MnO₄ + MnO₂ + O₂
c) Decomposition

Question 23.
Suggest a method to prepare oxygen in the laboratory
Answer:
Oxygen can be produced in the laboratory by the strong heating of potassium permanganate. 2KMnO₄ + heat → K₂MnO₄ + MnO₂ + O₂
Oxygen can also be prepared in the laboratory, by adding a small pinch of manganese dioxide in hydrogen peroxide (H₂O₂)
2H₂O₂ → 2H₂O + O₂

Question 24.
a) What is ozone?
b) Where is ozone found in the atmosphere?
c) What are chlorofluorocarbons (CFCs)?
Answer:
a) Ozone is an allotrope of oxygen. Three oxygen atoms combine to form an ozone molecule, O3.
b) In the stratosphere.
c) Chlorofluorocarbons are a class of compounds containing chlorine fluorine and carbon.

Question 25.
How CFC causes ozone layer depletion.
Answer:
Chlorofluorocarbons released into the atmosphere reach the stratosphere and break down by the action of ultraviolet radiations and release chlorine. This chlorine decomposes ozone molecules into oxygen. This disturbs the equilibrium in the ozone-oxygen cycle process.

Question 26.
What is the importance of ozone in the atmosphere?
Answer:
Ozone acts like a shield, protecting us from the sun’s harmful rays. It forms and breaks down in a continuous cycle to maintain a balance.

• The sun emits harmful ultraviolet (UV) radiation.
• UV radiation breaks apart oxygen molecules (O₂) into oxygen atoms (O).
• These oxygen atoms combine with other oxygen molecules to form ozone (O₃)
• Ozone absorbs more harmful UV radiation, protecting Earth’s surface.
• Some ozone molecules break down back into oxygen, but the formation of new ozone balances this process.
• This constant cycle helps maintain a stable level of ozone in the atmosphere, protecting us from harmful UV rays.

Question 27.
What are the uses of oxygen?
Answer:

• It’s the breath of life for all living organisms.
• For combustion, oxygen is necessary.
• As an oxidant in rocket fuels.
• For artificial respiration.
• For decomposition of wastes.

Question 28.
How can we reduce the rate of depletion of the ozone layer?
Answer:
Today, the use of CFC is being controlled in most countries. Harmful CFCs are replaced nowadays with safer substances. This has helped in reducing the rate of depletion of the ozone layer.

Question 29.
The picture given below shows the laboratory preparation of chlorine. Analyse the picture and answer the following questions.

a) What are the chemicals required for the preparation of chlorine gas?
b) Why is chlorine gas passed through water?
c) Which substance is used to remove moisture from chlorine?
d) How is bleaching powder prepared?
Answer:
a) KMnO₄ & HCl
b) To remove the traces of hydrogen chloride vapours that come out along with chlorine.
c) Sulphuric acid (H₂SO₄)
d) By passing dry chlorine gas over dry slaked lime.

Question 30.
List out the chlorine compounds you are familiar with.
Answer:
Hydrogen chloride (HCl), Sodium chloride (NaCl), Potassium chloride (KCl)

Question 31.
In the laboratory preparation of chlorine, why is chlorine gas passed through water?
Answer:
Hydrogen chloride vapours coming along with chlorine is removed by passing it through water.

Question 32.
Write the balanced chemical equation for the preparation of chlorine. Write the name of the reactant involved in the preparation of chlorine.
Answer:
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂
Potassium permanganate and Con. HCl are the reactants.

Question 33.
Write the uses of chlorine.
Answer:
For bleaching.
For the preparation of insecticides.
For removing stains in the fabrics.
For purification of water.
For the preparation of bleaching powder.

Question 34.
How is bleaching powder produced, and what is the chemical process involved in its bleaching action?
Answer:
Bleaching powder is prepared by passing dry chlorine gas over dry slaked lime. Chlorine gas reacts
with the water to give hydrochloric acid and hypochlorous acid (HClO).
Cl₂ + H₂O → HCl + HOCl
Hypochlorous acid decomposes and liberates atomic oxygen
HOCl → HCl + [O] This atomic oxygen oxidises coloured substances

Question 35.
Which of the following is used to purify water?
(a) Hydrogen
(b) Oxygen
(c) Nitrogen
(d) Chlorine
Answer:
(d) Chlorine

Question 36.
Chlorine gas is passed through concentrated sulphuric acid. Why?
Answer:
Sulphuric acid can absorb the water vapour present in chlorine gas when it passes through the concentrated sulphuric acid (as a drying agent).

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 7 Malayalam Medium ന്യൂനസംഖ്യകൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 7 Solutions Malayalam Medium ന്യൂനസംഖ്യകൾ

Class 9 Maths Chapter 7 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 7 Malayalam Medium Textual Questions and Answers

Question 1.
ചുവടെക്കൊടുത്തിരിക്കുന്ന പട്ടിക പൂർത്തിയാക്കുക.

Answer:
6 + (-10) = 6 – 10
= -(10 – 6)
= -4

-6 + 10 = 10 – 6
= 4

-6 + (-10) = -6 – 10
= -(6 + 10)
= -16

-6 + 6 = 6 – 6
= 0

6 + (-6) = 6 – 6
= 0

x	y	x + y
6	-10	-4
-6	10	4
-6	-10	-16
-6	6	0
6	-6	0

Question 2.
ചുവടെപ്പറയുന്ന ഓരോ കണക്കിലും പറഞ്ഞിരിക്കുന്ന തരത്തിലുള്ള രണ്ടു ജോടി x, y കണ്ടുപിടി ക്കുക.
(i) x അധിസംഖ്യ, y ന്യൂനസംഖ്യ, x + y = 1
(ii) x ന്യൂനസംഖ്യ, y അധിസംഖ്യ, x + y = 1
(iii) x അധിസംഖ്യ, y ന്യൂനസംഖ്യ, x + y = 1
(iv) x ന്യൂനസംഖ്യ, y അധിസംഖ്യ, x + y = 1
Answer:
(i) x = 2, y = -1 ഉം x = 5, y = -4 ഉം
(ii) x = -3, y = 4 ഉം x = -6, y = 7 20
(iii) x = 1, y = -2 ഉം x = 3, y = -4 ഉം
(iv) x = -3, y = 2 ഉം x = -5, y = 4 ഉം

Question 3.
ചുവടെയുള്ള പട്ടിക പൂർത്തിയാക്കുക.

Answer:
ആദ്യത്തെ വരി
(x + y) + z = (2 + 4) + (-5)
= 6 – 5
= 1

x + (y + z) = 2 + (4 + (-5))
= 2 + (4 – 5)
= 2 + (-1)
= 2 – 1
= 1

രണ്ടാമത്തെ വരി
(x + y) + z = ( (2 + (-4)) + 5
= (2 – 4)+ 5 = -2 + 5
= 5 – 2 = 3

x + (y + z)
= 2 + (-4 + 5)
= 2 + (5 – 4)
= 2 + 1
= 3

മൂന്നാമത്തെ വരി
(x + y) + z = (-2 + ·4) + (-5)
= (4 – 2) – 5 = 2 – 5
= -(5 – 2)
= -3

x + (y + z) = -2+(4 + (-5))
2 + (4 – 5) )
= -2 + – (5 – 4)
= -2 + (-1)
= -2 – 1
= -(2 + 1)
= -3

നാലാമത്തെ വരി
(x + y) + z = (2 + (-4)) + (-5)
= (2 – 4) + (-5)
= -2 – 5
= -(2 + 5)
= -7

x + (y + z) = 2 + (-4 + (-5))
= 2 + (-4 – 5)
= 2 + -(4 + 5)
= 2 + (-9)
= 2 – 9
= -(9 – 2)
= -7.

അഞ്ചാമത്തെ വരി
(x + y) + z = (-2 + 4) + 5
= (4 – 2) + 5
= 2 + 5
= 7

x + (y + z)
= -2 + (4 + 5)
= -2 + 9
= 9 – 2
= 7

ആറാമത്തെ വരി
(x + y) + z = (-2 + (-4)) + 5
= -(2 + 4) + 5
= -6 + 5
= 5 – 6
= -1

x + (y + z) = -2 + (-4 + 5)
= -2 + (5 – 4)
= -2 + 1
= 1 – 2
= -1

ഏഴാമത്തെ വരി
(x + y) + z = (-2 + (-4)) + (-5)
= (-2 – 4) – 5
= -(2 + 4) – 5
= -6 – 5
= -(6 + 5)
= -11

x + (y + z) = -2 + (-4 + (-5))
= -2 + (-4 – 5)
= -2 – 9
= -(2 + 9)
= -11

x	y	z	(x + y) + z	x +(y + z)
2	4	-5	1	1
2	-4	5	3	3
-2	4	-5	-3	-3
2	-4	-5	-7	-7
-2	4	5	7	7
-2	-4	5	-1	-1
-2	-4	-5	-11	-11

Question 4.
x, y, z ഇവ പല അധിസംഖ്യകളായും ന്യൂനസംഖ്യകളായും എടുത്ത്, x – (y – z) ഉം (x – y) + z ഉം കണക്കാക്കുക. രണ്ടും ഒരേ സംഖ്യയാണോ?
Answer:
x = 1, y = -1, z = 2 എന്നെടുക്കുക.
x – (y – z) = 1 -(-1 – 2)
= 1 – (-3)
= 1 + 3 = 4

(x – y) + z = (1 – (-1)) + 2
= (1 + 1) + 2
= 2 + 2
= 4
രണ്ടും ഒരേ സംഖ്യയാണ്.

Question 5.
ചുവടെപ്പറയുന്ന ഓരോ കണക്കിലും, പറഞ്ഞിരിക്കുന്ന തരത്തിലുള്ള രണ്ടു ജോടി x, y കണ്ടുപിടി ക്കുക.
(i) x അധിസംഖ്യ, y ന്യൂനസംഖ്യ, x – y = 1
(ii) x ന്യൂനസംഖ്യ, y അധിസംഖ്യ, x – y = 1
Answer:

Question 6.
(i) a, b, c, d ഇവ അടുത്തടുത്തുള്ള നാലു എണ്ണൽ സംഖ്യകളോ, അല്ലെങ്കിൽ അവയുടെ ന്യൂനങ്ങ ളായോ എടുത്ത് b- – c + d കണക്കാക്കുക.
(ii) സംഖ്യകൾ ഏതായാലും ഇതു പൂജ്യമാകുന്നത് എന്തുകൊണ്ടാണെന്ന് ബീജഗണിതമുപയോഗിച്ച് വിശദീകരിക്കുക.
(iii) a – b = c + d എന്നതിനുപകരം a + b – c – d ആയാലോ?
(iv) a – b + c – d ആയാലോ?
Answer:
i) a = 1, b = 2, c = 3, d = 4 എന്നെടുക്കുക.
a – b – c + d = 1 – 2 – 3 + 4
= -1 – 3 + 4
= -4 + 4
= 0

ii) Let a = x, b = x + 1, c = x + 2, d = x + 3 എന്നെടുക്കുക.
a – b – c + d = x – (x + 1) – (x + 2) + (x + 3)
= x – x – 1 – x – 2 + x + 3

iii) a + b – c – d = 1 + 2 – 3 – 4
= 3 – 3 – 4
= 0 – 4
= -4

iv) a – b + c – d = 1 – 2 + 3 – 4
= -1 + 3 – 4
= 2 – 4
= -2

Question 7.
x, y, 2 ആയി പല അധിസംഖ്യകളും ന്യൂനസംഖ്യകളും എടുത്ത് (x + y) ഉം xz + yz ഉം കണക്കാക്കുക. എല്ലാറ്റിലും (x + yz = xz + yz എന്ന സമവാക്യം ശരിയാകുന്നുണ്ടോ എന്ന് പരിശോധിക്കുക.
Answer:
x = 1, y = 2, z = -1 എന്നെടുത്താൽ,
(x + y)z = (1 + 2)(-1) = 3 × (-1) = -3.
xz + yz = 1 × (-1) + 2 × (-1) = -1 + (-2) = -1 − 2 = -3.
ഇവിടെ, (x + y)z = xz + yz.

x = -1, y = 3, z = -4 എന്നെടുത്താൽ,
(x + y)z = (-1+3)(-4) = 2 × (-4) = -8.
xz + yz = -1 × (-4) + 3 × (-4) = 4 + (-12) = 4 – 12 = -8.
ഇവിടെ, (x + y)z = xz + yz.

Question 8.
(x + y)(u + v) = xu + xv + yu +v എന്ന സമവാക്യം x, y, u, v ഇവയ്ക്കു പകരം -x, -y, -u, -v എടുത്താലും ശരിയാകുമെന്ന് തെളിയിക്കുക.
Answer:
(-x + (-y))(-u + (-v)) = (-x- y)(-u – v)
= -(x + y) × -(u + v)
= (x + y)(u + v)
= xu + xv + yu + yv

Question 9.
ചുവടെയുള്ള സമവാക്യങ്ങളിലെല്ലാം x ആയി പറഞ്ഞിട്ടുള്ള സംഖ്യകൾ എടുക്കുമ്പോൾ, y ആയി കിട്ടുന്ന സംഖ്യ കണ്ടുപിടിക്കുക.
Answer:
(i) y = x², x = -1
(ii) y = x² + 3x + 2, x =’-1
(iii) y = x² + 3x + 2, x = -2
(iv) y = (x + 1)(x + 2), x = -1
(v) y = (x + 1)(x + 2), x = -2
Answer:
i) y = (-1)²
= -1 × -1
= 1

ii) y = (-1)² + 3 × (-1) + 2
= 1 + (-3) + 2
= 1 – 3 + 2
= -2 + 2
= 0

iii) y = (-2)² + 3 × (-2) + 2
= 4 + (-6) + 2
= 4 – 6 + 2
= -2 + 2
= 0

iv) y = (-1 + 1)(-1 + 2)
= 0 × 1
= 0

v) y = (-2 + 1)(-2 + 2)
= -1 × 0
= 0

Question 10.
y = x² + 4x + 4 എന്ന സമവാക്യത്തിൽ x ആയി പല അധിസംഖ്യകളും, ന്യൂനസംഖ്യകളും എടുത്ത് y കണക്കാക്കുക. എന്തുകൊണ്ടാണ് x എന്തായാലും y അധിസംഖ്യയോ പൂജ്യമോ തന്നെ ആകുന്നത്?
Answer:
x = -1, എന്നെടുത്താൽ, y = (-1)² + 4 × (-1) + 4 = 1 – 4 + 4 = -3 + 4 = 1.
x = 1, എന്നെടുത്താൽ, y = 1² + 4 × 1 + 4 = 1 + 4 + 4 = 9.
x = 2, എന്നെടുത്താൽ, y = 2² + 4 × 2 + 4 = 4 + 8 + 4 = 16.
x = -2, എന്നെടുത്താൽ, y = (-2)² + 4 × (-2) + 4 = 4 – 8 + 4 = -4 + 4 = 0.
ഇവിടെ, y = x² + 4x + 4 = (x + 2)²
പൂജ്യമല്ലാത്ത ഏതൊരു സംഖ്യയുടെയും വർഗ്ഗം അധിസംഖ്യ ആയിരിക്കും. പൂജ്യത്തിന്റെ വർഗ്ഗം പൂജ്യം തന്നെയാണ്. ചുരുക്കിപറഞ്ഞാൽ ഏതൊരു സംഖ്യയുടെയും വർഗ്ഗം പൂജ്യമോ അധിസംഖ്യയോ ആയിരിക്കും. അതുകൊണ്ടാണ് x എന്തായാലും y അധിസംഖ്യയോ പൂജ്യമോ തന്നെ ആകുന്നത്.

Question 11.
എണ്ണൽ സംഖ്യകൾ, അവയുടെ ന്യൂനങ്ങൾ, പൂജ്യം ഇവയെ എല്ലാം പൊതുവായി പൂർണ്ണസംഖ്യകൾ എന്ന് പറയാം.
(i) x + y = 25 എന്ന സമവാക്യം ശരിയാകുന്ന എത്ര ജോടി പൂർണ്ണസംഖ്യകൾ കണ്ടുപിടിക്കാം?
(ii) x – y = 25 എന്ന സമവാക്യം ശരിയാകുന്ന എത്ര ജോടി പൂർണ്ണസംഖ്യകൾ കണ്ടുപിടിക്കാം?
Answer:
i) 0² + 5² = 25 → (0, 5)
5² + 0² = 25 → (5,0)
0² + (-5)² = 25 → (0, -5) .
(-5)² + 0² = 25 → (-5, 0)
3² + 4² = 25 → (3, 4)
42 + 3² = 25 → (4, 3)
(-3)² + (-4)² = 25 → (-3,-4)
(-4)² + (-3)² = 25 → (-4,-3) ‘
(-3)² + 4² = 25 → (-3, 4)
4² + (-3)² = 25 → (4, -3)
(-4)² + 3² = 25 → (-4, 3)
3² + (4)² = 25 → (3,-4)
പന്ത്രണ്ടു ജോടി പൂർണ്ണസംഖ്യകൾ.

ii) 5² – 0² = 25
(-5)² – o² = 25
13² – 12² = 25
(-13)² – 12² = 25
13² – (-12)² = 25
(-13)² – (-12)² = 25
ആറു ജോടി പൂർണ്ണസംഖ്യകൾ : (5, 0), (-5, 0), (13, 12), (-13, 12), (13, -12), (-13, -12)

Question 12.
1 × 2 × 3 × 4 × 5 = 120. ആണല്ലോ. (-1) × (-2) × (-3) × (-4) × (-5) എത്രയാണ്?
Answer:
(-1) × (-2) × (-3) × (-4) × (-5) = 2 × (-3) × (-4) × (-5)
= -6 × (-4) × (-5)
= 24 × (-5)
= -120

Question 13.
(1 × 2 × 3 × 4 × 5) + [(-1) × (-2) × (-3) × (-4) × (-5)] എത്രയാണ്?
Answer:
1 × 2 × 3 × 4 × 5 = 120
(-1) × (-2) × (-3) × (-4) × (-5) = -120
(1 × 2 × 3 × 4 × 5) + [(-1) × (-2) × (-3) × (-4) × (-5)]
= 120 + (-120)
= 120 – 120
= 0

Question 14.
y = \(\frac{1}{x}\) എന്ന സമവാക്യത്തിൽ, x ആയി 2, -2, \(\frac{1}{2}\), –\(\frac{1}{2}\) എന്നീ സംഖ്യകളെടുത്താൽ y ആയി കിട്ടുന്ന സംഖ്യകൾ കണക്കാക്കുക.
Answer:
x = 2, എന്നെടുത്താൽ, y = \(\frac{1}{2}\)
x = -2, എന്നെടുത്താൽ, y = \(\frac{1}{-2}\) = -(\(\frac{1}{2}\))
x = \(\frac{1}{2}\), എന്നെടുത്താൽ, y = \(\frac{1}{\frac{1}{2}} = 2\) = 2
x = –\(\frac{1}{2}\), എന്നെടുത്താൽ, y = \(\frac{1}{\frac{-1}{2}} = 2\) = 2

Question 15.
y = \(\frac{1}{x-1}+\frac{1}{x+1}\) എന്ന സമവാക്യത്തിൽ x = -2 എന്നെടുക്കുമ്പോഴും x = –\(\frac{1}{2}\) എന്നെടുക്കുമ്പോഴും y ആയി കിട്ടുന്ന സംഖ്യകൾ കണക്കാക്കുക.
Answer:
x = -2 എന്നെടുത്താൽ,
y = \(\frac{1}{-2-1}+\frac{1}{-2+1}\)
= \(\frac{1}{-3}+\frac{1}{-1}\)
= \(-\left(\frac{1}{3}+\frac{3}{3}\right)\)
= \(-\left(\frac{4}{3}\right)\)

x = –\(\frac{1}{2}\) എന്നെടുത്താൽ,
y = \(\frac{1}{-\left(\frac{1}{2}\right)-1}+\frac{1}{-\left(\frac{1}{2}\right)+1}\)
= \(\frac{1}{-\frac{3}{2}}+\frac{1}{\frac{1}{2}}\)
= \(-\frac{2}{3}+\frac{2}{1}=\frac{2}{1}-\frac{2}{3}\)
= \(\frac{4}{3}\)

Question 16.
z = \(\frac{x}{y}-\frac{y}{x}\) എന്ന സമവാക്യത്തിൽ, x, y ആയി ചുവടെപ്പറയുന്ന സംഖ്യകളെടുക്കുമ്പോൾ 2 ആയി കിട്ടുന്ന സംഖ്യകൾ കണക്കാക്കുക.
(i) x = 10, y = -5
Answer:
Z = \(\frac{10}{-5}-\frac{-5}{10}\)
= -2 – \(\frac{-1}{2}\)
= -(2 –\(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(ii) x = -10, y = 5
Answer:
Z = \(\frac{-10}{5}-\frac{5}{-10}\)
= – 2 – \(\frac{1}{-2}\)
= -(2 – \(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(iii) x = -10, y = -5
Answer:
Z = \(\frac{-10}{-5}-\frac{-5}{-10}\)
= 2 – \(\frac{1}{2}\)
= \(\frac{3}{2}\)

Class 9 Maths Chapter 7 Malayalam Medium Intext Questions and Answers

Question 1.
ചുവടെ പറഞ്ഞിരിക്കുന്ന കണക്കുകൾ ചെയ്യുക.
(i) 6 – 8
Answer:
6 – 8 = -(8 – 6)
= -2

(ii) -6 + 8
Answer:
-6 + 8 = 8 – 6
= 2

(iii) -6 – 8
Answer:
-6 – 8 = -(6 + 8)
= -14

(iv) 2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(\frac{5}{2}-\frac{7}{2}=-\left(\frac{7}{2}-\frac{5}{2}\right)\)
= \(-\left(\frac{7-5}{2}\right)=-\left(\frac{2}{2}\right)\)
= -1

(v) -2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
= \(-\frac{5}{2}+\frac{7}{2}\)
= \(\frac{7}{2}-\frac{5}{2}=\frac{2}{2}\)
= 1

(vi) -2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(-\frac{5}{2}-\frac{7}{2}\)
= \(-\left(\frac{5}{2}+\frac{7}{2}\right)=-\left(\frac{12}{2}\right)\)
= -6

Question 2.
സ്ഥാനങ്ങൾ അടയാളപ്പെടുത്തിയ വരയിൽ, ആദ്യസ്ഥാനവും സ്ഥാനമാറ്റവും വേറെ സംഖ്യകളാക്കിയ ഒരു പട്ടികയാണ് ചുവടെ.

ചിത്രം വരച്ച് അവസാനസ്ഥാനങ്ങൾ കണക്കാക്കി പട്ടികയിൽ എഴുതുക.
Answer:
ആദ്യത്തെ വരി;

രണ്ടാമത്തെ വരി

മൂന്നാമത്തെ വരി

നാലാമത്തെ വരി

അഞ്ചാമത്തെ വരി

ഏഴാമത്തെ വരി

എട്ടാമത്തെ വരി;

ന്യൂനസംഖ്യകൾ

സ്ഥാനമാറ്റം	ആദ്യസ്ഥാനം	അവസാനസ്ഥാനം
7	3 വലത്തേയ്ക്ക്	10
3	7 വലത്തേയ്ക്ക്	10
-7	3 വലത്തേയ്ക്ക്	-4
-3	7 വലത്തേയ്ക്ക്	4
7	3 ഇടത്തേയ്ക്ക്	4
3	7 ഇടത്തേയ്ക്ക്	-4
-7	3 ഇടത്തേയ്ക്ക്	-10
-3	7 ഇടത്തേയ്ക്ക്	-10

Negative Numbers Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
തന്നിരിക്കുന്ന നമ്പറുകൾ ഉപയോഗിച്ച് z = x – y എന്ന സമവാക്യത്തിലെ 2 ന്റെ മൂല്യം കണക്കാക്കുക.
) x = 7, y = 2
b) x = -3, y = -6
c) x = -8, y = 3
d) x = -4, y = 9
Answer:
a) x = 7, y = 2
z = x – y
= 7 – 2
= 5

b) x = -3 y = -6
z = x- y = -3- (-6)
= -3 + 6
= 3

c) x = -8 y = 3
z = x – y
= -8 – 3
= -11

d) x = -4 y = 9
z = x – y
= -4 – 9
= -13

Question 2.
x = 4, y = -3, z = 8 ആയാൽ തന്നിരിക്കുന്നവ കണക്കാക്കുക.
a) (x + y) + z
b) x + (y + z)
c) xyz
d) (x + y)z
e) xy + xz
Answer:
a) (x + y) + z
= (4 – 3) + 8
= (4 – 3) + 8
= 1 + 8
= 9

b) x + (y + z)
= 4 + (-3 + 8)
=4 + 5
= 9

c) yz = 4 × (-3) × 8
= -12 × 8
= -96

d) (x + y)z = (4 – 3) × 8
= 1 × 8
= 8

e) xy + xz = (4 × (-3)) + (4 × 8)
= -12 + 32
= 20

Question 3.
x ന് തന്നിരിക്കുന്ന മൂല്യങ്ങൾ എടുത്ത് y = x² + 9x – 5 കണക്കാക്കുക.
a) x = 1
b) x = -3
c) x = 0
Answer:
a) x = 1
y = x² + 9x – 5
= 1² + (9 × 1) – 5
= 1 + 9 – 5
= 10 – 5
= 5

b) x = -3
y = x² + 9x – 5
= (-3)² + (9 × (-3)) – 5
= 9 – 27 – 5
= -23

c) x = 0
y = x² + 9x – 5
= 0² + (9 × 0) – 5
= -5

Question 4.
x = -1 ആയാൽ y = x⁴ + x³ + x² + x + 2 കണക്കാക്കുക.
Answer:
y = (-1)⁴ + (-1)³ + (-1)² + (-1) + 2
= 1 + (-1) + 1 + (-1) + 2
= 1 – 1 + 1 – 1 + 2
= 2.

Question 5.
തന്നിരിക്കുന്ന സമവാക്യങ്ങൾ സർവ്വസമവാക്യങ്ങൾ ആണോ എന്ന് പരിശോധിക്കുക. x = 1, 2, 3 എന്നും x = -1, -2, -3 എന്നും എടുത്ത് ഓരോന്നിലും നിന്നും കിട്ടുന്ന പാറ്റേൺ എഴുതുക.
a) -x + (x + 3)
b) (x + 2) – (x + 3)
c) -x – (x + 1) + 2x + 1
Answer:
x = 1 ആയാൽ,
a) -x + (x + 3) = -1 + (1 + 3)
= -1 + 4
= 3

b) (x + 2) – (x + 3) = (1 + 2) – (1 + 3)
= 3 – 4
= -1

c) -x – (x + 1) + 2x + 1 = -1 – (1 + 1) + 2 + 1
= -1 – 2 + 2 + 1
= 0

ഇതുപോലെ, x = 2, 3 എടുക്കുമ്പോഴും ഇവ സർവ്വസമവാക്യങ്ങൾ ആണെന്ന് കാണാം. ഇനി, if x = -1 ആയാൽ,
a) -x + (x + 3) = -(-1) + (-1 + 3)
= 1 + 2
= 3

b) (x + 2) – (x + 3) = (-1 + 2) – (-1 + 3)
= 1 – 2
= -1

c) -x-(x + 1) + 2x + 1 = -(-1) – (-1 + 1) – 2 + 1
= 1 – 0 – 1
= 0
ഇതുപോലെ, x = -2, -3 എടുക്കുമ്പോഴും ഇവ സർവ്വസമവാക്യങ്ങൾ ആണെന്ന് കാണാം.

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 8 Malayalam Medium വൃത്തങ്ങളുടെ അളവുകൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 8 Solutions Malayalam Medium വൃത്തങ്ങളുടെ അളവുകൾ

Class 9 Maths Chapter 8 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 8 Malayalam Medium Textual Questions and Answers

Question 1.
വ്യാസം 2 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവ് ഏകദേശം 6.28 മീറ്ററാണെന്നു അളന്നു കണ്ടു പിടിച്ചു.
i) വ്യാസം 4 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവെത്രയാണെന്ന് അളക്കാതെ എങ്ങനെ കണക്കാക്കും?
ii) വ്യാസം 1 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവോ?
iii) വ്യാസം 3 മീറ്ററായാലോ?
Answer:
i) വ്യാസം 2 മീറ്ററായുള്ള വൃത്തത്തിന്റെ ചുറ്റളവ് 6.28 ആയതിനാൽ വ്യാസം 4 മീറ്ററായുള്ള വൃത്ത ത്തിന്റെ ചുറ്റളവ് 6.28 ന്റെ ഇരട്ടി ആയിരിക്കും. അതായതു 2 × 6.28 = 12.56

ii) വ്യാസം 1 മീറ്ററായാൽ വൃത്തത്തിന്റെ ചുറ്റളവ് = \(\frac{6.28}{2}\) = 3.14 മീ

iii) വ്യാസം 1 മീറ്റർ ആയാൽ വൃത്തത്തിന്റെ വ്യാസം 3.14
∴ വ്യാസം 3 മീറ്റർ ആയാൽ വൃത്തത്തിന്റെ വ്യാസം = 3 × 3.14 = 9.42 മീ

Question 2.
ഒരു കമ്പി വളച്ച് 4 സെന്റിമീറ്റർ വ്യാസമുള്ള വൃത്തമുണ്ടാക്കി. ഇതിന്റെ പകുതി നീളമുള്ള കമ്പി വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസമെന്തായിരിക്കും ?
Answer:
വൃത്തങ്ങളുടെ ചുറ്റളവുകൾ മാറുന്നത് വ്യാസങ്ങൾ മാറുന്ന അതെ തോതിലാണ് .ആയതിനാൽ കമ്പിയുടെ നീളം പകുതിയാക്കി അത് വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസവും പകുതിയായി കുറയുന്നു .
അതായതു വൃത്തത്തിന്റെ വ്യാസം 2 സെന്റിമീറ്റർ ആയിരിക്കും.

Question 3.
ചുവടെ വരച്ചിരിക്കുന്ന വൃത്തങ്ങളുടെയെല്ലാം ചുറ്റളവ് കണക്കാക്കുക.

Answer:
ചിത്രം: 1

ചിത്രത്തിൽ കാണിച്ചിരിക്കുന്നതുപോലെ സമഷഡ്ഭുജത്തെ ആറു തുല്ല്യ സമഭുജത്രികോണങ്ങളാക്കി ഭാഗിക്കുന്നു
ഇവിടെ ത്രികോണം OAB പരിഗണിച്ചാൽ,
OA എന്ന വശം വൃത്തത്തിന്റെ ആരത്തിനു തുല്ല്യമാണ്
OA = r = 2 സെമീ
ആയതിനാൽ,
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π
= 2 × π × 2
= 47 സെമീ

ചിത്രം: 2

ചിത്രത്തിലെ സമചതുരം ABCD എന്ന് പരിഗണിച്ചാൽ, വികർണ്ണം AC വൃത്തത്തിന്റെ വ്യാസത്തിനു തുല്യമാണ്
ആയതിനാൽ,
AB = BC = 2 സെമീ, കോൽ B = 90 °
AC = \(\sqrt{2^2+2^2}=\sqrt{8}=2 \sqrt{2}\) സെമീ
വൃത്തത്തിന്റെ ആരം = \(\frac{1}{2}\) × 2√2 = 2 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × √2 = 2√2 π സെമീ

ചിത്രം: 3

ചിത്രത്തിൽനിന്നും,
PR = \(\sqrt{2^2+(1.5)^2}=\sqrt{6.25}\) = 2.5 സെമീ
വൃത്തത്തിന്റെ ആരം = \(\frac{1}{2}\) × 2.5 = 1.25 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × 1.25 = 2.5π സെമീ

Question 4.
ഒരു വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും 4 സെന്റിമീറ്റർ അകലെയുള്ള ഞാണിന്റെ നീളം 6 സെന്റി മീറ്ററാണ്. വൃത്തത്തിന്റെ ചുറ്റളവ് കണക്കാക്കുക.
Answer:
വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും ഞാണിലേക്കുള്ള ദൂരം = 4 സെമീ
ഞാണിന്റെ നീളം = 6 സെമീ

മുകളിൽ പറഞ്ഞ ചിത്രത്തിൽ നിന്നും AC യുടെ നീളം AB യുടെ പകുതിയാണ്
അതായത്,
AC = \(\) = 3 സെമീ
ത്രികോണം AOC പരിഗണിച്ചാൽ,
(OA)² = (AC)² + (OC)²
r² = 4²+ 3²
r² = 16 + 9 = 25
r = √25 = 5 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × r = 2π × 5 = 10 സെമീ

Question 5.
ചുവടെയുള്ള ചിത്രത്തിൽ, വൃത്തത്തിലെ മൂന്നു ബിന്ദുക്കൾ മൂലകളായി പാദവും ഉയരവും 4 സെന്റിമീറ്ററായ സമപാർശ്വത്രികോണം വരച്ചിരിക്കുന്നു.

വൃത്തത്തിന്റെ ചുറ്റളവ് കണക്കാക്കുക.
Answer:

മുകളിൽ പറഞ്ഞിരിക്കുന്ന ചിത്രത്തിൽ O വൃത്തകേന്ദ്രമാണ്
OA = OB = OC = r
CD = 4 സെമീ ആയതിനാൽ,
OD = CD – OC = 4 – r സെമീ
AD = 2 സെമീ

ത്രികോണം AOD യിൽ
(AO)² = (AD)² + (OD)²
r² = 2² + (4 – r)²
r² = 4 + 16
8r = 20
r = \(\frac{20}{8}=\frac{5}{2}\) = 2.5 സെമീ

വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π × r
= 2π × 2.5
= 5π സെമീ

Question 6.
ചുവടെയുള്ള ചിത്രങ്ങളിലെല്ലാം, വൃത്തങ്ങളുടെ കേന്ദ്രങ്ങൾ ഒരേ വരയിലാണ്. ആദ്യത്തെ രണ്ടു ചിത്രങ്ങളിൽ, ചെറിയ വൃത്തങ്ങൾക്ക് ഒരേ വ്യാസമാണ്.

എല്ലാ ചിത്രങ്ങളിലും, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് എന്നു തെളിയിക്കുക.
Answer:
ചിത്രം: 1

ചെറിയ വൃത്തങ്ങളുടെ വ്യാസം തുല്ല്യമാണ്
വ്യാസം d എന്ന് എടുത്താൽ
ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π × വ്യാസം = πd
ചെറിയ രണ്ടു വൃത്തങ്ങളുടെ ചുറ്റളവ് = 2πd
വലിയ വൃത്തത്തിന്റെ വ്യാസം = d + d = 2d
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π × 2d = 2πd
അതായത്, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്

ചിത്രം: 2

ചെറിയ വൃത്തങ്ങളുടെ വ്യാസം d എന്ന് കരുതിയാൽ
ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πd
ചെറിയ മൂന്നു വൃത്തങ്ങളുടെ ചുറ്റളവ് = 3πd
വലിയ വൃത്തത്തിന്റെ വ്യാസം = d + d + d = 3d
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π × 3d = 3rd
അതായത്, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്

ചിത്രം: 3

ചിത്രത്തിൽ നിന്നും ചെറിയ വൃത്തങ്ങളുടെ വ്യാസം p, q, r എന്നു കരുതാം
അങ്ങനെയെങ്കിൽ,
ആദ്യത്തെ ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πp
രണ്ടാമത്തെ ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πq
മൂന്നാമത്തെ ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = πr
മൂന്നു ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവ് = πp + πq + πr = π(p + q + r)
വലിയ വൃത്തത്തിന്റെ വ്യാസം = (p + q + r)
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = π (р + q + r)
അതായത്, ചെറിയ വൃത്തങ്ങളുടെ ചുറ്റളവുകളുടെ തുകയാണ് വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്

Question 7.
ചുവടെയുള്ള ചിത്രത്തിൽ, ഒരേ കേന്ദ്രമായ രണ്ടു വൃത്തങ്ങൾ വരച്ചിരിക്കുന്നു.

വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്, ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവിനേക്കാൾ എത്ര കൂടുതലാണ്?
Answer:
ചെറിയ വൃത്തത്തിന്റെ ആരം എന്ന് കരുതാം
വലിയ വൃത്തത്തിന്റെ ആരം = r + 1 സെമീ
ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = 2πr സെമീ
വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π (r + 1) = 2πr + 2π സെമീ
അതായതു, വലിയ വൃത്തത്തിന്റെ ചുറ്റളവ്, ചെറിയ വൃത്തത്തിന്റെ ചുറ്റളവിനേക്കാൾ 2π കൂടുതൽ ആണ്

Question 8.
ഒരു വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും 3 സെന്റിമീറ്റർ അകലെയുള്ള ഞാണിന്റെ നീളം 4 സെന്റിമീറ്ററാണ്. വൃത്തത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.
Answer:

cerardo = \(\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}\)
പരപ്പളവ് = πr² = π(√13)² = 13π m

Question 9.
ചുവടെയുള്ള ചിത്രങ്ങളിൽ, വൃത്തത്തിന്റെയും സമബഹുഭുജത്തിന്റെയും പരപ്പളവുകൾ തമ്മിലുള്ള വ്യത്യാസം രണ്ടു ദശാംശസ്ഥാനങ്ങൾ വരെ കണക്കാക്കുക?

Answer:
ചിത്രം 1
സമചതുരത്തിന്റെ വികർണ്ണത്തിന്റെ നീളം = 2.5 സെമീ

സമചതുരത്തിന്റെ പരപ്പളവ് = വികർണ്ണത്തിന്റ വർഗ്ഗത്തിന്റെ പകുതിയാണ്
= \(\frac{2.5^2}{2}=\frac{6.25}{2}\)
= 3.125 ച.സെമീ

വൃത്തത്തിന്റെ വ്യാസം = 2.5 സെമീ
വൃത്തത്തിന്റെ ആരം = 1.25 സെമീ

വൃത്തത്തിന്റെ പരപ്പളവ് = πr²
= π(1.25)²
= 4.91 ച.സെമീ

പരപ്പളവുകൾ തമ്മിലുളള വ്യത്യാസം = 4.91 – 3.125
= 1.79 ച.സെമീ

ചിത്രം 2
സമഷഡ്ഭുജം 6 സമഭുജത്രികോണങ്ങൾ ചേർന്നതാണ്.
സമഭുജത്രികോണത്തിന്റെ ഒരുവശം = വൃത്തത്തിന്റെ ആരം = 2 സെമീ
ഒരു സമഭുജത്രികോണത്തിന്റെ പരപ്പളവ് = \(\frac{\sqrt{3}}{4}\) × (വശം)²
= \(\frac{\sqrt{3}}{4}\) × (2)²
= √3
= 1.73 ച. സെമീ

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് = 6 × 1.73
= 10.38 ച.സെമീ

വൃത്തത്തിന്റെ പരപ്പളവ് = πr² = 1(2)
= 12.56 ച.സെമീ

പരപ്പളവുകൾ തമ്മിലുളള വ്യത്യാസം = 12.56 – 10.38
= 2.18 ച.സെമീ

Question 10.
ഒരു സമചതുരത്തിന്റെ നാലു മൂലകളിൽക്കൂടിയും, ഒരു ചതുരത്തിന്റെ നാലു മൂലകളിൽക്കൂടിയും വൃത്തങ്ങൾ വരച്ചത് ചുവടെ കാണിച്ചിരിക്കുന്നു.

രണ്ടു വൃത്തങ്ങളുടെയും പരപ്പളവ് കണക്കാക്കുക.
Answer:
ചിത്രം 1
വ്യാസം = സമചതുരത്തിന്റെ വികർണം
വ്യാസം = \(\sqrt{(3)^2+(3)^2}\)
= \(\sqrt{2 \times 9}\)
= 3√2 സെമീ

വൃത്തത്തിന്റെ പരപ്പളവ് = πr²
= π × \(\left(\frac{3 \sqrt{2}}{2}\right)^2\)
= 4.5 π ച.സെമീ

ചിത്രം 2
= \(\sqrt{(4)^2+(2)^2}\)
= \(\sqrt{20}\) സെമീ

ആരം = \(\frac{\sqrt{20}}{2}\) സെമീ

പരപ്പളവ്= π × (\(\frac{\sqrt{20}}{2}\))²
= 5 ച.സെ.മീ

Question 11.
ഒരു സമചതുരം വരച്ച്, അതിന്റെ നാലു മൂലകൾ കേന്ദ്രമായും, വശത്തിന്റെ പകുതി ആരമായും വൃത്തങ്ങൾ വരയ്ക്കുക. ആദ്യത്തെ സമചതുരത്തിന്റെ അതേ വലുപ്പമുള്ള നാലു സമച തുരങ്ങൾ ചേർന്ന സമചതുരം വരച്ച്, അതിനുള്ളിൽ കൃത്യമായി ചേർന്നിരിക്കുന്ന വൃത്തം വരയ്ക്കുക.

വലിയ വൃത്തത്തിന്റെ പരപ്പളവ്, നാലു ചെറുവൃത്തങ്ങളുടെയും പരപ്പളവുകളുടെ തുക യാണെന്ന് തെളിയിക്കുക.
Answer:
ചിത്രം 1
ചെറിയ വൃത്തത്തിന്റെ ആരം = r സെമീ
ഒരു ചെറിയ വൃത്തത്തിന്റെ പരപ്പളവ് = πr² ച.സെമീ
4 ചെറിയ വൃത്തത്തിന്റെ പരപ്പളവ് = 4 × πr² ച.സെമീ
= 4πr² ച.സെമീ

ചിത്രം 2
വൃത്തത്തിന്റെ ആരം = രണ്ടു ചെറിയ വ്യത്തങ്ങളുടെ ആരങ്ങളുടെ തുകയാണ് = 2r
വൃത്തത്തിന്റെ പരപ്പളവ് = (2r)²
= 4πr² ച.സെമീ
ചെറിയ നാല് വ്യത്തങ്ങളുടെ പരപ്പളവുകളുടെ തുകയാണ് വലിയ വ്യത്തത്തിന്റെ പരപ്പളവ്

Question 12.
ചുവടെയുള്ള രണ്ട് ചിത്രങ്ങളിലെയും സമചതുരങ്ങൾക്ക് ഒരേ വലുപ്പമാണ്. ചിത്രങ്ങളിലെ പച്ച ഭാഗങ്ങൾക്ക് (ഷെയ്ഡ് ചെയ്ത ഭാഗം) ഒരേ പരപ്പളവാണെന്നു തെളിയിക്കുക.

Answer:
ചിത്രം 1
സമചതുരത്തിന്റെ വശം = 2a സെമീ
സമചതുരത്തിന്റെ പരപ്പളവ് = (2a)² = 4 a² ച.സെമീ
ഒരു വൃത്തത്തെ നാലു തുല്യഭാഗങ്ങളാക്കി സമചതുരത്തിന്റെ നാലു മൂലകളിൽ വെച്ചിരിക്കുന്നു അതിനാൽ, വൃത്തഭാഗത്തിന്റെ ആരം = a സെമീ
വൃത്തഭാഗത്തിന്റെ പരപ്പളവ് = \(\frac{\pi a^2}{4}\) ച.സെമീ
സമചതുരത്തിന്റെ നാലു മൂലകളിൽ വെച്ചിരിക്കുന്നു വൃത്തഭാഗത്തിന്റെ പരപ്പളവ്
= 4 × \(\frac{\pi a^2}{4}\) = πa² ച.സെ.മീ
പച്ചഭാഗത്തിന്റെ (ഷെയ്ഡ് ചെയ്ത ഭാഗം) പരപ്പളവ് = (4 a² – πa²) ച.സെമീ

ചിത്രം 2
സമചതുരത്തിന്റെ വശം = 2a സെമീ
സമചതുരത്തിന്റെ പരപ്പളവ് = (2a)² = 4 a² ച.സെമീ
വൃത്തത്തിന്റെ ആരം = a സെമീ
വൃത്തത്തിന്റെ പരപ്പളവ് = πa² ച.മീ
പച്ചഭാഗത്തിന്റെ (ഷെയ്ഡ് ചെയ്ത ഭാഗം) പരപ്പളവ് = (4a² – πa²) ച.മീ

Question 13.
ഒരു സമഷഡ്ഭുജത്തിന്റെ മൂലകൾ കേന്ദ്രമായി വൃത്തഭാഗങ്ങൾ വരച്ച്, ചുവടെക്കാണുന്ന രൂപം വെട്ടിയെടുക്കുന്നു. മുറിച്ചെടുത്ത രൂപത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.

Answer:
ചിത്രത്തിൽ സമഷഡ്ഭുജത്തിന്റെ ആറു മൂലകളിലും ഓരോ വ്യത്തഭാഗങ്ങൾ കാണാൻ കഴിയും. സമഷഡ്ഭുജത്തിന്റെ ഓരോ മൂലയുടെയും കോൺ 120° ആണ്
ഇതിൽ നിന്നും മൂന്നു വൃത്തഭാഗങ്ങൾ ചേർന്നാൽ ഒരു പൂർണ്ണ വൃത്തം ലഭിക്കും
ചിത്രത്തിൽ ഇത്തരത്തിൽ ആറു വൃത്തഭാഗങ്ങളാണ് ഉള്ളത്
അതിനാൽ ഇതിൽനിന്നും രണ്ടു പൂർണ്ണ വൃത്തങ്ങൾ ലഭിക്കും
അതായത്, ആറു വൃത്തഭാഗങ്ങൾ രണ്ടു പൂർണ്ണ വൃത്തങ്ങൾക്ക് തുല്യമായി കാണാൻ കഴിയും വൃത്തഭാഗത്തിന്റെ ആരം = 1 സെമീ
വൃത്തത്തിന്റെ പരപ്പളവ് = πr²
= π ച.സെമീ

രണ്ടു വൃത്തത്തിന്റെ പരപ്പളവ് = 2 × πr²
= 2π ച.സെമീ

സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് = \(\frac{1}{2}\) × 2 × √3 × 6
= 6√3 ച.സെമീ

മുറിച്ചെടുത്ത രൂപത്തിന്റെ പരപ്പളവ് = സമഷഡ്ഭുജത്തിന്റെ പരപ്പളവ് – രണ്ടു പൂർണ്ണ
വൃത്തത്തിന്റെ പരപ്പളവ്
= (6√3 – 2π) ച.മീ

Question 14.
ഒരു സമചതുരത്തിനുള്ളിൽ, ചുവടെക്കാണുന്നതുപോലെ വൃത്തഭാഗങ്ങൾ വരയ്ക്കുന്നു.

ചിത്രത്തിൽ നീലനിറം കൊടുത്തിരിക്കുന്ന (ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ്, സമചതുരത്തിന്റെ പരപ്പളവിന്റെ പകുതിയാണെന്നു തെളിയിക്കുക.
Answer:
സമചതുരത്തിന്റെ വശം = a സമീ
സമചതുരത്തിന്റെ പരപ്പളവ് = a² ച.സെമീ
ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ് കാണാൻ സമചതുരത്തിന്റെ പകുതി ഭാഗത്തിൽ
വരുന്ന അർദ്ധവ്യത്തത്തിന്റെ പരപ്പളവും കൂടാതെ ബാക്കി വരുന്ന ഭാഗത്തിലുള്ള രണ്ടു കാൽ വ്യത്തങ്ങൾക്കിടയിലുള്ള ഭാഗത്തിന്റെ പരപ്പളവു കണ്ടു പിടിക്കുകയാണ് വേണ്ടത്,
അർദ്ധവ്യത്തത്തിന്റെ പരപ്പളവ് = \(\frac{\pi a^2}{8}\) ച.സെമീ
രണ്ടു കാൽ വൃത്തങ്ങൾക്കിടയിലുള്ള ഭാഗത്തിന്റെ പരപ്പളവ് = [\(\frac{a^2}{2}\) – 2 × \(\frac{1}{4}\) × π × \(\left(\frac{a}{2}\right)^2\)] ച.സെമീ
ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ് = \(\frac{\pi a^2}{8}+\frac{a^2}{2}-\frac{\pi a^2}{8}=\frac{a^2}{2}\) ച.മീ

Circle Measures Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ഒരു കമ്പി വളച്ച് 6 സെന്റിമീറ്റർ വ്യാസമുള്ള വൃത്തമുണ്ടാക്കി. ഇതിൻറെ ഇരട്ടി നീളമുള്ള കമ്പി വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസമെന്തായിരിക്കും ?
Answer:
വൃത്തങ്ങളുടെ ചുറ്റളവുകൾ മാറുന്നത് വ്യാസങ്ങൾ മാറുന്ന അതെ തോതിലാണ്. ആയതിനാൽ കമ്പിയുടെ നീളം ഇരട്ടിയാക്കി അത് വളച്ചുണ്ടാക്കുന്ന വൃത്തത്തിന്റെ വ്യാസവും ഇരട്ടിയാക്കുന്നു. അതായതു വൃത്തത്തിന്റെ വ്യാസം 12 സെന്റിമീറ്റർ ആയിരിക്കും.

Question 2.
വ്യാസം 3 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവ് ഏകദേശം 9.42 മീറ്ററാണെന്നു അളന്നു കണ്ടു പിടിച്ചു.
i) വ്യാസം 6 മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവെത്രയാണെന്ന് അളക്കാതെ എങ്ങനെ കണക്കാക്കും?
ii) വ്യാസം 1\(\frac{1}{2}\) മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവോ?
Answer:
i) വ്യാസം 2 മീറ്ററായുള്ള വൃത്തത്തിന്റെ ചുറ്റളവ് 9.42 ആയതിനാൽ വ്യാസം 6 മീറ്ററായുള്ള വൃത്തത്തിന്റെ ചുറ്റളവ് 9.42 ന്റെ ഇരട്ടി ആയിരിക്കും. അതായതു 2 × 9.42 = 18.84 മീ
ii) വ്യാസം 1\(\frac{1}{2}\) മീറ്ററായ വൃത്തത്തിന്റെ ചുറ്റളവ് = \(\frac{9.42}{2}\) = 4.7 മീ

Question 3.
20 സെമീ നീളമുള്ള ഒരു കമ്പി വളച്ചൊരു വൃത്തം ഉണ്ടാക്കിയിരിക്കുന്നു. അതെ കമ്പിയുടെ പകുതി വെട്ടി മറ്റൊരു വൃത്തമായി വളച്ചാൽ പുതിയ വൃത്തത്തിന്റെ വ്യാസം എത്രയായിരിക്കും.
Answer:
കമ്പിയുടെ ആകെ നീളം = 20 സെമീ
കമ്പിയുടെ പകുതി നീളം = \(\frac{20}{2}\) = 10 സെമീ

കമ്പിയുടെ നീളം വൃത്തത്തിന്റ ചുറ്റളവിനു തുല്യമായിരിക്കും
അതായത്,
10 = π × വ്യാസം
വ്യാസം = \(\frac{10}{\pi} /latex] സെമീ

Question 4.
ചുറ്റളവ് 81 സെമീ ആയ വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്നും 2 സെമീ അകലെയുള്ള ഞാണിന്റെ നീളം കണ്ടെത്തുക.

Answer:
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2πг
8π = 2πr
r = [latex]\frac{8}{2}\) = 4 സെമീ

ത്രികോണം POR പരിഗണിച്ചാൽ,
(OP)² = (OR)² + (PR)²
4² = 2² + (PR)²
PR² = 16 – 4 = 12
PR = √12 = 2√3 സെമീ

Question 5.
ചിത്രത്തിലെ സമഷഡ്ഭുജത്തിന്റെ ചുറ്റളവ് 24 സെമീ ആണ്

i) വൃത്തത്തിന്റെ ആരമെത്ര?
ii) വൃത്തത്തിന്റെ ചുറ്റളവെത്ര?
Answer:
സമഷഡ്ഭുജത്തിന്റ ചുറ്റളവ് = 6 × ഒരു വശത്തിന്റെ നീളം
24 = 6 × ഒരു വശത്തിന്റെ നീളം
ഒരു വശത്തിന്റെ നീളം = \(\frac{24}{6}\) = 4 സെമീ

ചിത്രത്തിൽ കാണിച്ചിരിക്കുന്നതുപോലെ സമഷഡ്ഭുജത്തെ ആറുതുല്ല്യ സമഭുജത്രികോണങ്ങളാക്കി
മാറ്റുന്നു
വശത്തിന്റെ നീളം = വൃത്തത്തിന്റെ ആരം
വൃത്തത്തിന്റെ ആരം = 4 സെമീ
വൃത്തത്തിന്റെ ചുറ്റളവ് = 2π
= 2π × 4
= 8 സെമീ

Question 6.
ആരം 8 സെന്റിമീറ്റർ ആയ വൃത്തത്തിൻറെ ചുറ്റളവും പരപ്പളവും കണക്കാക്കുക.
Answer:
വൃത്തത്തിന്റെ ആരം = 8 സെമീ
ചുറ്റളവ് = 2 πr = 2 × π × 8 = 16 π സെമീ
പരപ്പളവ് = πr² = π × 8² = 64 ച.സമീ

Question 7.
ഒരു വൃത്തത്തിന്റെ വ്യാസം 10 സെന്റീമീറ്ററാണ്. വൃത്തത്തിന്റെ പരപ്പളവ് കണ്ടുപിടിക്കുക.
Answer:
വൃത്തത്തിന്റെ വ്യാസം = 10 സെമീ
ആരം = 5 സെമീ
പരപ്പളവ് = πr² = π × 5² = 25 ച.മീ

Question 8.
i) ആരം 15 സെന്റിമീറ്ററായ ഒരു ചക്രത്തിന്റെ ചുറ്റളവ് എന്ത്? ഈ ചക്രം 5 തവണ കറങ്ങിയാൽ എത്ര ദൂരം സഞ്ചരിക്കും?
ii) ഈ ചക്രത്തിന്റെ രണ്ട് മടങ്ങ് ആരമുള്ള ചക്രം 5 തവണ കറങ്ങുകയാണെങ്കിൽ എത്ര ദൂരം സഞ്ചരിക്കും
Answer:
i) ചക്രത്തിന്റെ ആരം = 15 സെമീ
ചുറ്റളവ് = 2πr = 2 × π × 15 = 30 π സെമീ
ചക്രം 5 തവണ കറങ്ങിയാൽ സഞ്ചരിക്കുന്ന ദൂരം = 5 × 30 π = 150 π സെമീ

ii) ചക്രത്തിന്റെ ആരം = 2 × 15 = 30 സെമീ
ചുറ്റളവ് = 2πr = 2 × π × 30 = 60 7 സെമീ
ചക്രം 5 തവണ കറങ്ങിയാൽ സഞ്ചരിക്കുന്ന ദൂരം = 5 × 60 π = 300 π സെമീ

Question 9.
ചിത്രത്തിന്റെ AB വലിയ അർധവ്യത്തത്തിന്റെ വ്യാസമാണ്. AC, CD, DB ഇവ വ്യാസങ്ങളായ അർധവൃത്തങ്ങൾ വരച്ചിരിക്കുന്നു AC = 4 സെന്റിമീറ്റർ, CD = 4, DB = ഭാഗത്തിന്റെ പരപ്പളവ് എത്രയാണ്?

Answer:
AB വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² = π(3.5)² = 12.25π ച.സെ.മീ
AC വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² = π(2)² = 4π ച.സെ.മീ
CD വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² = π(1) = π ച.സെ.മീ
DB വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് = πr² – π(0.5)² = 0.25π ച.സെ.മീ
ഷെയ്ഡ് ചെയ്ത ഭാഗത്തിന്റെ പരപ്പളവ് = AB വ്യാസമായാ അർധവ്യത്തത്തിന്റെ പരപ്പളവ് – (AC വ്യാസമായ അർധവൃത്തത്തിന്റെ പരപ്പളവ് + CD വ്യാസമായ അർധവ്യത്തത്തിന്റ പരപ്പളവ്+ DB വ്യാസമായ അർധവ്യത്തത്തിന്റെ പരപ്പളവ്)
= 12.25π – (4π + π + 0.25π) = 7π ച.സെ.മീ

Students rely on Kerala Syllabus 9th Standard Chemistry Notes Pdf Download Chapter 5 Chemical Kinetics Extra Questions and Answers to help self-study at home.

Kerala Syllabus Std 9 Chemistry Chapter 5 Chemical Kinetics Extra Questions and Answers

Question 1.
Reactions can be classified as Direct combination, decomposition, simple displacement, double decomposition, and Redox reactions. State which of the following types takes place in the reactions given below.
(i) Cl₂ + 2KI → 2KCl + I₂
(ii) 2Mg + O₂ → 2MgO
(iii) AgNO₃ + HCl → AgCl + HNO₃
(iv) 4HNO₃ → 4NO₂ + 2H₂O + O₂
Answer:
(i) Cl₂+ 2KI → 2KCl + I₂: Simple displacement
(ii) 2Mg + O₂ → 2MgO: Direct combination
(iii) AgNO₃ + HCl → AgCl + HNO₃: Double decomposition
(iv) 4HNO₃ → 4NO₂ + 2H₂O + O₂: Decomposition reaction

Question 2.
Give balanced equations for reactions involving the evolution of a gas in addition of dilute acid to:
(a) sodium sulphite
(b) calcium carbonate
Answer:
(a)Na₂SO₃ + H₂SO₄ [dil.] → Na₂SO₄ + H₂O + SO₂ [g]
Sulphur dioxide gas has evolved.

(b) CaCO₃ + 2HCl [dil.] → CaCl₂ + H₂O + CO₂ [g]
Carbon dioxide gas has evolved.

Question 3.
Define the following types of chemical changes or reactions with a suitable example of each.
(a) Direct combination reaction or synthesis
(b) Decomposition reaction
(c) Displacement reaction or substitution reaction
(d) Double decomposition reaction
Answer:
(a) A chemical reaction in which two or more elements or compounds react to form one new compound is called a Direct combination reaction or synthesis.
For example,
Hydrogen bums in the air to give water:
2H₂ [g] + O₂ [g] → 2H₂O [1]

(b) A chemical reaction in which a chemical compound decomposes into two or more simpler
substances (elements and/or compounds) is called a Decomposition reaction.
For example,
Calcium carbonate, on heating, decomposes to Calcium Oxide and Carbon dioxide:
CaCO₃ \(\xrightarrow{\Delta}\) CaO + CO₂

(c) A chemical reaction in which an element or radical has replaced another element in a compound is known as a Displacement reaction or Substitution reaction.
For example,
Magnesium displaces Copper from Copper [II] sulphate solution:
Mg + CuSO₄ → MgSO₄ + Cu

(d) A chemical reaction in which both reactants [compounds] are decomposed to give two new compounds by exchanging their radicals is known as a Double decomposition reaction.
For example,
Silver nitrate + Potassium chloride → Silver chloride + Potassium nitrate
AgNO₃ + KCl → AgCl + KNO₃

Question 4.
Define a displacement reaction with a suitable example. State how it is represented. Give a reason why zinc displaces hydrogen from dilute sulphuric acid, but copper does not.
Answer:
A chemical reaction that takes place when an element [or radical] has replaced another element in a compound is known as a Displacement reaction It is represented as :
X + YZ → Y + XZ
Eg: Magnesium + Copper [II] sulphate → Magnesium sulphate + Copper
Mg + CuS0₄ → MgS0₄ + Cu
As a more electropositive metal displaces a less electropositive metal from its aq. soln. therefore, zinc, being more electropositive, is placed above [H] in the activity series and displaces hydrogen from dilute sulphuric acid, whereas copper, being less electropositive, is placed below [H] in the electrochemical series and cannot displace hydrogen from sulphuric acid.

Question 5.
Differentiate between a decomposition reaction and a double decomposition reaction.
Answer:

Decomposition reaction	Double decomposition reaction
A chemical reaction in which a compound decomposes to give two new elements / a new compound & an element / two new compounds is called a Decomposition reaction.	A chemical reaction in which both reactants [compounds] are decomposed to give two new compounds by exchanging their radicals is called a Double decomposition reaction. It is represented as XY + AB → XB + AY
For example: 2HgO → 2Hg + O2	For example: CaCl2 + Na2CO3 → 2NaCl + CaCO3↓

Question 6.
Match the chemical reactions in List I with the appropriate answer in List II.

List I	List II
AB → A + B	Endothermic reaction
X+Y– + A+B– → X+B– + A+Y–	Decomposition reaction
X + YZ → XZ + Y	Double displacement reaction
X + Y → XY – Δ (on heating)	Displacement reaction

Answer:

List I	List II
AB → A + B	Decomposition reaction
X+Y– + A+B– → X+B– + A+Y–	Double displacement reaction
X + YZ → XZ + Y	Displacement reaction
X + Y → XY – Δ (on heating)	(on heating)

Question 7.
Analyse the chemical equations and answer the following questions.
NaOH + HCl → NaCl + H₂O
(a) What is the salt formed in this reaction?
(b) What is the name of this type of reaction?
Answer:
(a) Sodium chloride
(b) Double displacement reaction

Question 8.
(a) What is the chemical, required to prepare hydrogen in the laboratory
(b) Write the balanced chemical equation of this reaction
(c) Which type of reaction is this
(Decomposition, Combination, Displacement, Double decomposition)
Answer:
(a) Dilute hydrochloric acid and Zinc
(b) Zn + 2HCl → ZnCl₂ + H₂
(c) Displacement reaction

Question 9.
A small quantity of sodium chloride solution is taken in a test tube. One drop of silver nitrate solution is added to this.
(a) What do you observe?
(b) NaCl + AgNO₃ → X + NaNO₃
Answer:
(a) A white curdy precipitate of silver chloride (AgCl) is formed
(b) Silver chloride (AgCl)

Question 10.
How does a physical change differ from a chemical change?
Answer:
Both physical and chemical changes involve a change in matter, but they differ in the way the matter is altered.

Physical change	Chemical change
The composition remains the same	Composition Changes
No new substances are formed	New substances are formed with different properties
Reversible	Irreversible
It may involve energy transfer	Involves energy transfer
Example: Melting of ice into water	Example: Burning wood

Question 11.
What are the methods usually adopted to make firewood burn faster?
Answer:

• Provide more air
• Split up into small pieces
• Make firewood dry

Question 12.
Describe an experiment to prove that the nature of the reactants affects the rate of a chemical reaction.
Answer:
Materials required for the experiment. Zn, Mg, dil. HCl and test tubes.
Procedure
Take an equal volume of dil. HCl in two test tubes. Add Zn to one and Mg of the same mass to the other. Hydrogen gas is produced in both test tubes. The rate of reaction is faster in the test tube containing Mg. This is because Mg is more reactive than Zn.

Question 13.
Why does the rate of reaction increase when concentration increases?
Answer:
As the concentration of reactants increases, the number of molecules per unit volume and the number of effective collisions increase. Consequently, the rate of reaction increases.

Question 14.
What is the relation between the rate of reaction and surface area? Write an experiment to prove it.
Answer:
Take an equal volume of dil. HC1 in two beakers. Add a small piece of marble into one and marble powder of equal mass into the other. The reaction rate is greater when marble powder is used. The rate of reaction increases when surface area increases.

Question 15.
Why does the rate of reaction increase when surface area increases?
Answer:
When solids are made into small pieces or powder, their surface area increases. As a result, the number of molecules undergoing effective collision also increases. Hence, the rate of reaction increases.

Question 16.
Write an experiment to prove the relation between temperature and rate of reaction.
Answer:
Materials required: Sodium thiosulphate, hydrochloric acid, water, boiling tube, spirit lamp – Procedure: Prepare a dilute solution of sodium thiosulphate in a beaker. Take equal volumes of this solution in two boiling tubes. Heat one boiling tube for some time. Add dilute hydrochloric acid in equal amounts in both boiling tubes.
Observation: Reaction is faster in a heated test tube.

Question 17.
What are catalysts?
Answer:
Catalysts are substances that alter the rate of chemical reactions without undergoing anypermanent chemical change.

Question 18.
The chemical reaction between marble and dilute HC1 is given.
CaCO₃ + 2HCl → CaCl₂ + H₂O + X
(a) Which gas is formed here? How can you identify this gas?
(b) Suggest any two ways you would choose to increase the rate of this chemical reaction.
Answer:
(a) Carbon dioxide (CO₂). Show a burning splinter in the gas. It will get extinguished.
OR
Show a glass rod dipped in clear lime water into the gas. The lime water will turn milky

(b) 1) Use powdered marble
2) Use concentrated HCl
3) Heat the mixture.

Question 19.
Sulphur pieces do not react with cold, concentrated nitric acid. But sulphur powder reacts.
(a) Explain the reason why the rate of chemical reaction is increased here.
(b) Suppose you want to increase the rate of reaction again. Which way would you choose? Give reason
Answer:
(a) When surface area increases, the rate of effective collision increases. So, the rate of reaction increases.

(b) Heat the mixture.
When temperature increases, more molecules will get threshold energy. So rate of effective collision and rate of reaction increases.

Question 20.
Small amounts of phosphoric acid are usually added to hydrogen peroxide to prevent its decomposition.
a. What is the function of phosphoric acid here?
b. By which name are these types of substances known?
c. Which substance would you add to increase the rate of decomposition of hydrogen peroxide?
Answer:
(a) Slow down the rate of decomposition of hydrogen peroxide
(b) Negative catalyst
(c) Manganese dioxide

Question 21.
The experiments conducted by two students are given below.
Experiment 1:
2 mL of sodium thiosulphate solution is taken in a test tube and heated, and 2 mL of diluted HC1 solution is added.
Experiment 2:
2 mL of sodium thiosulphate solution is taken in a test tube, and to it, 2 mL of diluted HC1 solution is added.
(a) In which experiment is the precipitate formed quickly? Justify your answer.
(b) Write the balanced equation for the reaction.
Answer:
(a) In a heated test tube. (Experiment 1) When temperature increases, more molecules get threshold
energy. So, the rate of effective collision increases, and the rate of reaction increases.

(b) Na₂S₂O₃ + 2HCl → 2NaCl + SO₃ + S

Question 22.
Some materials available in the laboratory are given below. Magnesium ribbon, marble powder, marble pieces, dilute HCl, concentrated HCl.
a. Which materials will you choose for the preparation of more CO₂ in less time?
b. Write the balanced chemical equation of the reaction.
Answer:
(a) Marble powder and concentrated HCl
(b) CaCO₂ + 2HCl → CaCl₂ + H₂O + CO₂

Question 23.
What are the factors that influence the speed of chemical reactions?
Answer:

• Nature of the reactants
• Concentration of the reactants
• Total surface area
• Temperature
• Presence of catalyst

Question 24.
Find a reason for the following.
a. Splitted-up logs burn fast.
b. It is said that drinking water along with food is not good.
c. Certain medicines are kept in refrigerators.
Answer:
(a) When logs are split up, the total surface area increases. So, the number of collisions per unit area increases, and the speed of chemical reaction (burning) increases.

(b) Intake of water will dilute the digestive enzymes and hydrochloric acid produced in the stomach which help the digestive process. So it is better to avoid drinking water half an hour before and half an hour after food.

(c) A rise in temperature may cause the decomposition of certain chemicals. Such substances are kept in refrigerators.

Question 25.
Match the following suitably.

Activity	Catalyst
To reduce the decomposition of hydrogen peroxide	Manganese dioxide
Industrial production of sulphuric acid	Phosphoric acid
Industrial production of ammonia	Vanadium pentoxide
To speed up the decomposition of potassium permanganate	Iron

Answer:

Activity	Catalyst
To reduce the decomposition of hydrogen peroxide	Phosphoric acid
Industrial production of sulphuric acid	Vanadium pentoxide
Industrial production of ammonia	Iron
To speed up the decomposition of potassium permanganate	Manganese dioxide

Question 26.
When, eggshell is added to Hydrochloric acid, a gas is produced
(a) Identify this gas
(b) Write the chemical equation of this reaction.
(c) If we add powdered eggshell, what happens to the rate of the above chemical reaction? Why?
Answer:
(a) Carbon dioxide (CO₂)

(b) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

(c) The reaction becomes faster. When the solid reactants are made into small pieces or-powders, their surface area increases. More reactant molecules come in contact and take part in collision, resulting in an increase in the number of effective collisions. Thus, the speed of the reaction increases.

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 1 Refraction of Light Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 1 Notes Solutions Refraction of Light

SCERT Class 9 Physics Chapter 1 Notes Solutions Kerala Syllabus Refraction of Light Questions and Answers

Class 9 Physics Chapter 1 Let Us Assess Answers Refraction of Light

Question 1.
Ray diagram showing the path of light through mediums A and B is given.

a) In which medium will the speed of light be less A or B ?
b) Which will be the optically denser medium? Justify your answer.
Answer:
a) Speed of light will be less in medium B.
b) Bis the optically denser medium. Here, when light enters from A to B, the light ray deviates towards the normal. It happens when light enters from an optically rarer medium to a denser medium. So B will be the optically denser medium.

Question 2.
Complete the given diagram. Mark the angle of incidence and angle of refraction.

Answer:

Question 3.
Light passes from medium X to Y. Here, the angle of refraction is greater than the angle of incidence.
a) In which medium is the speed of light more?
b) Which is the medium of larger refractive index?
c) Draw the path of light.
Answer:
a) Y
b) X
c)

Question 4.
Refractive index of different mediums are given in the table.

Medium	Refractive index
Crown glass	1.52
Glycerine	1.47
Sunflower oil	1.47
Water	1.33
Flint glass	1.62

a) In which medium does light travel with maximum speed?
b) Will a ray of light entering obliquely from glycerine to sunflower oil deviate? Explain the reason.
c) Light is transmitted from glass to each medium listed in table above If light is incident at an angle of 30°, which medium will have the largest angle of refraction? Why?
Answer:
a) In water
b) No, As speed of light is same in both these mediums, (as refractive index of both mediums are same) light does not undergo refraction.
c) In water. Among these, water has least refractive index.

Question 5.
Observe the figure. Ray of light incident on two mediums is depicted.

a) Which is the medium of higher optical density? Why?
b) Which is the medium of greater refractive index?
Answer:
a) Medium 1 is the medium of higher optical density.
The angle of refraction in medium 1 is less than that in medium 2. That is when light enters from air to medium 1, the refracted ray bends more towards the normal.
b) Medium 1 has greater refractive index. Observe the figure.

Question 6.
Observe the figure.

a) Which figure indicates total internal reflection?
b) Which figures indicate refraction?
Answer:
a) Fig (i)
b) Fig(ii) and Fig (iii)

Question 7.
The critical angle of glass is 42o. Choose the angle of incidence for which total internal reflection takes place.
a) 40°
b) 49°
c) 38°
d) 42°
Answer:
49°

Question 8.
We can see many small prisms in the reflectors used in motorcycles. Describe the benefits of using them.
Answer:
Many small prisms are used in reflectors used in motor cycles. Light falling on the prisms undergoes total internal reflection and reflects back without loosing the intensity of light. The image can be seen with greater visual clarity.

Question 9.
Observe the table.

Medium	Refractive index
Air	1.0003
Kerosene	1.33
Turpentine oil	1.44
Crown glass	1.47
Diamond	1.52

a) Choose the medium in which light has the least speed.
b) The speed of light in air is 3 × 10⁸ m/s. What is the speed of light in kerosene?
c) When a ray of light enters obliquely from air to diamond, will the refracted ray deviate towards the normal, or away from the normal? Justify your answer.
Answer:
a) Diamond

c) The refracted ray deviates towards the normal. Diamond has greatest optical density. As light enters obliquely from air to diamond, it deviates towards the normal.

Question 10.
The path of light through mediums A, B, C and D are given. Choose the correct figures. (The optical density of the mediums are in the order A<B<C<D.)

Answer:
Fig(b), (d), (e)

Question 11.
Speed of light in ethanol is lesser than that in methanol. Which medium has lower refractive index? Why?
Answer:
Methanol has lower refractive index. Speed of light in methanol is higher than that in ethanol.

Class 9 Physics Chapter 1 Extended Activities Answers Refraction of Light

Question 1.
Make periscopes using prisms instead of mirrors. Exhibit them.
Answer:
Hint:
Periscopes are devices used to see over or around objects by using a series of mirrors or prisms. Usually, periscopes use mirrors angled at 45 degrees, but prisms can also be used.

Materials required

• 2 right-angled prisms
• 2 PVC pipes or cardboard tubes (30-40 cm each)
• Strong glue
• Cutter
• Sandpaper

Steps:
1. Prepare Tubes

• Cut two PVC pipes or cardboard tubes to desired length.
• Smooth edges with sandpaper.

2. Fix Prisms

• Attach a prism inside each tube end with the hypotenuse facing the tube opening.
• Secure with hot glue or adhesive.

3. Join Tubes

• Align tubes at a right angle.
• Securely join them.
• Optionally, paint inside black to reduce reflections.
• Ensure all parts are securely fixed.
• For demonstration,position the periscope to look over obstacles and explain how prisms helps in providing a clear image in periscopes.

Question 2.
Find out the critical angle in different mediums like glycerine, water, coconut oil, glass etc., and compare them. Prepare a report including different stages like objective, materials, method of study, results etc.
Answer:
Short Report on Critical Angle in Various
Media
Objective
To determine the critical angle of various substances, such as glass, coconut oil, glycerine and water.

Material	Refractive index
Glass	1.52
Coconut oil	1.45
Glycerine	1.47
Water	1.33

Method of study
Formula to find critical angle
Snell’s law helps us to find the critical angle (θ_c). The critical angle is the angle of incidence above which total internal reflection occurs for light travelling from a denser to a rarer medium.
Snell’s law states that n₁sin (θ₁) = n₂sin(θ₂)
When θ₂ = 90° (at critical angle), sin(θ₂) = 1
So we have sin (θ_c) = \(\frac{n_2}{n_1}\)
θ_c = sin¹\(\left(\frac{n_2}{n_1}\right)\)

where n₁ is the refractive index of the denser medium and n₂ is the refractive index of the rarer medium (air). Refractive index of air is taken as n = 1
Calculating critical angle
Critical angle for glass

Substances	Critical angle
Glass	41.14°
Coconut oil	43.60°
Glycerine	42.86°
Water	48.75°

It can be observed that water has the largest critical angle, indicating that it has the lowest refractive index among the mediums studied and glass has the smallest critical angle, indicating it has the highest refractive index. Because of their same refractive indices, glycerine and coconut oil have relatively close critical angles.

Conclusion
The critical angle is one important property in optics that varies depending on the refractive index of the medium. The critical angles for glycerine, water, coconut oil, and glass were computed and compared in this study. Applications requiring light transmission and total internal reflection, such as fibre optics and different optical devices, require an understanding of these angles.

Question 3.
Can the angle of refraction be 90° when light enters from an optically rarer medium to a denser medium? Do an activity and write it down in the science diary.
Answer:
Objective
To find out if angle of refraction can be 90° when it moves from air (less dense) to water or glass (denser).
Assumption
Angle of refraction cannot be 90° when it goes from a less dense to a denser medium.

Materials required

• Laser pointer or flashlight
• Protractor
• Clear container (like a glass tank)
• Water
• Piece of glass or plastic
• Pen or pencil

Procedure

• Fill the clear container with water and place it on a table.
• Put the piece of glass or plastic next to the container.
• Shine the laser or flashlight at the water surface at different angles (0°, 30°, 45°, 60°, etc.)
• Measure the angle of incidence and the angle of refraction.
• Record Observations

Results

• When light goes from air to water, the angle of refraction is always less than the angle of incidence.
• The same happens with the glass or plastic.

Snell’s law states that n₁sin (θ₁) = n₂sin(θ₂) Here n, and n, are the refractive indices of the rarer and denser media, respectively, and θ₁, and are the θ₂, angles of incidence and refraction.
When light passes from a rarer to a denser medium, the refractive index of the denser medium (n₂) is greater than that of the rarer medium (n₁). Thus, the sine of the angle of refraction sin(θ₂) (cannot be greater than 1, limiting θ₂ to less than 90°.

Conclusion
Light cannot bend at a 90° angle when it goes from a less dense medium (like air) to a denser medium (like water or glass). This is because the refraction law shows it is impossible.

Refraction of Light Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Have you ever observed the straw in a glass of lemonade? Does it appear bent? What could be the reason?

Answer:
The straw in a glass of lemonade appears bent because of the phenomenon of light called refraction.

Question 2.
Don’t you think the light rays reflected from the straw dipped in lemonade travel through different mediums before they reach our eyes?
Answer:
Yes

Activity
Fill three fourth of a trough with water. Add two or three drops of milk into it. Cover the trough with a transparent sheet of paper. Fill the remaining space in the trough with a smoke. Then flash the light from a laser torch such that light should fall obliquely on the surface of water as shown in figure below.

Observe the path of light and record the observations in the science diary. Ray diagram showing the path of light travelling from air to water

Question 3.
Which are the mediums through which light travels within the trough?
Answer:
Air, water
Light is travelling from air to water

Question 4.
How will the path of light ray be, when it passes through only one medium?
Answer:
The path of light will be in a straight line.

Question 5.
Is there any deviation in the direction of the ray of light when it enters obliquely from one medium to another?
Answer:
Yes, there is deviation.

Question 6.
Where does the ray of light undergo a change in the direction?
Answer:
At the surface of water.

Activity
Allow light from a laser torch to fall perpendicular to the surface of water taken in a trough.

Question 7.
Is there a deviation in the path of the ray of light? Record your observation.
Answer:
No, there is no deviation in the path of the ray of light because light is incident normally.

When light travels through a single medium the path of light will be a straight line. A ray of light undergoes a deviation at the surface of separation when it enters obliquely from one medium to another. There will be no deviation in the path of the ray of light incident normally.

Activity
Observe the speed and a direction of a toy car moving from a smooth surface to a rough surface.

Question 8.
Is there a change in the direction of motion of the car?
Answer:
Yes, the direction of motion of the car changes.

Question 9.
Where does this change occur?
Answer:
This change occurs at the surface of separation of the two media.

Question 10.
Did the car move with the same speed on both surfaces?
Answer:
No, the car moves with different speed on both surfaces.
Note:The change in direction is caused by the change in the speed of the car as it moves from one surface to the other.

Reason for the change in the direction of light when it passes from one medium to another
The speed of light is different in different mediums. The change in the speed of light causes the change in the direction of the ray of light, when it passes from one medium to another. Examine the table showing the speed of light through various mediums.

Medium	Speed of Light (approximate)
Air	3 × 108 m/s
Water	2. 25 × 108 m/s
Glass	2 × 108 m/s
Diamond	1.25 × 108 m/s

By examining the table we can understand that the speed of light differs in various mediums due to the difference in their optical densities.

The ability of a medium to influence the speed of light through the medium is its optical density.

The speed of light will be lower in a medium of higher optical density (optically denser medium). The speed of light will be higher in a medium of lower optical density (optically rarer medium). Note that optical density has no relation with material density.

Question 11.
Arrange the mediums in the table shown below in the increasing order of their optical densities.

Medium	Speed of Light (approximate)
Air	3 × 108 m/s
Water	2. 25 × 108 m/s
Glass	2 × 108 m/s
Diamond	1.25 × 108 m/s

Answer:
Air <Water<Glass<Diamond

Activity to understand about refraction
A straw is kept obliquely in a glass as shown in figure below. Then fill the glass with water.

Question 12.
What difference do you observe?
Answer:
The straw appears bent when water is poured.

Question 13.
Why does the straw appear bent?
Answer:
It is due to refraction of light that the straw appears bent.

Activity
A straw is kept in a trough. Observe the path of the rays of light from the straw falling on the eyes before and after filling the trough with water.

Question 14.
Is there a deviation for the ray of light falling on the eye from point B of the straw before pouring water?
Answer:
After the water is poured, the ray of light from the immersed part of the straw entering air from water is deviated at the surface of separation.

Question 15.
Doesn’t the ray of light appear to come from C, though it is actually coming from B?
Answer:
The straw appears bent, as the actual position B seems to have risen to the position C.

Refraction

When a ray of light enters obliquely from one medium to another of different optical densities, it undergoes a deviation at the surface of separation of the mediums. This phenomenon is refraction. The ability of light to undergo refraction in a medium depends on the optical density of the medium. The role of refractive index is very important in designing various optical devices.

Refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in the medium.

Refractive index (n) = \(\frac{\text { The speed of light in vacuum (c) }}{\text { The speed of light in the medium (v)}}\) n = \(\frac{c}{v}\) The speed of light in vacuum (c) = 3 × 108 m/s

Question 16.
The speed of light in different mediums is given in the table below. Find the refractive index of each medium and complete the table.

Medium	Speed of light (v)	Refractive index (n)
Air	3 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1
Glass	2 × 108 m/s
Water	2.25 × 108 m/s

Answer:

Medium	Speed of light (v)	Refractive index (n)
Air	3 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1
Glass	2 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1.5
Water	2.25 × 108 m/s	n = \(\frac{c}{v}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{2.25 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 1.33

Question 17.
If the refractive index of diamond is 2.4, what is the speed of light that passes through it?
Answer:
Refractive index of diamond, n = \(\frac{c}{v}\) = 2.4
V = \(\frac{c}{v}\) = \(\frac{3 \times 10^8}{2.4}\)
v = 1.25 × 10⁸ m/s

Question 18.
How is the speed of light related to refractive index?
Answer:
As refractive index increases, speed of light decreases. The speed of light is less in a medium of higher refractive index. The speed of light is higher in a medium of less refractive index.

Question 19.
Write down how optical density is related to refractive index?
Answer:
Optical density will be higher in a medium having higher refractive index.
Note: A ray of light entering obliquely from one medium to another with different optical densities undergoes deviation at the surface of separation of the two mediums.

Question 20.
Is the deviation of light the same when it enters from air to water and from water to air when the angle of incidence is the same?
Answer:
No, the deviation of light is not the same when it enters from air to water and from water to air when the angle of incidence is the same.
Ray diagrams showing refraction
Observe the figures.

Incident ray:The ray falling on the surface of separation of the two mediums is the incident ray. Refracted ray:The ray which undergoes refraction is the refracted ray.
Angle of incidence: The angle between incident ray and the normal(NN’) .
Angle of refraction: The angle between refracted ray and the normal is the angle of refraction.

Question 21.
Find the incident ray, refracted ray, angle of incidence and angle of refraction from each figure and complete the table.

Answer:

	Fig.(1)	Fig (2)
Incident ray	AB	PQ
Refracted ray	BC	QR
Angle of incidence (i)	∠ABN	∠PQN
Angle of refraction (r)	∠CBN	∠NQR
The angle of refraction is greater than/ less than the angle of incidence	The angle of refraction is less than the angle of incidence	The angle of refraction is greater than the angle of incidence

Question 22.
How does the direction of light change when a ray of light enters obliquely from air to water? (deviates towards the normal/deviates away from the normal)
Answer:
deviates towards the normal

Question 23.
How does the direction of light change when a ray of light enters obliquely from water to air? (deviates towards the normal / deviates away from the normal)
Answer:
deviates away from the normal

Question 24.
Observe the figures that show light passing through different pairs of mediums.

Find answers to the following questions based on the optical densities of mediums.
a) Which are the figures in which the ray of light enters obliquely from an optically denser medium to a rarer medium?
Answer:
Fig.(4), Fig.(6)

b) In this case, to which direction does the refracted ray deviate? (towards the normal / away from the normal)
Answer:
away from the normal

c) Choose the figures in which the refracted ray deviates towards the normal.
Answer:
Fig.(3), Fig.(8)

d) In which situation will a ray of light deviate towards the normal as it passes from one medium to another?(from an optically denser medium to a rarer medium/from an optically rarer medium to a denser medium)
Answer:
from an optically rarer medium to a denser medium.

e) There is no refraction of light in Fig.(5) and Fig.(7). What may be the reason?
Answer:
When a ray incident normally at the surface of separation of mediums, it does not undergo refraction.

Note: When light enters from one medium to another, the incident ray, the refracted ray and the normal at the point of incidence, will be on the same plane. When light enters from an optically rarer medium to a denser medium, it deviates towards the normal. When light enters from an optically denser medium to a rarer medium, it deviates away from the normal. A ray incident normally at the surface of separation of mediums does not undergo refraction.

Allow light from a laser torch to fall on a glass slab.

Observe the path of light.

Question 25.
At which points do refraction occur for the ray of light?
Answer:
Refraction occurs at the points Q and R.

Question 26.
Draw the ray diagram showing the refraction of a ray of light in a glass slab
Answer:

Note: When light passes from air to glass it moves towards the normal and when it moves from glass to air it bends away from the normal. In a glass slab, the incident ray (PQ) and emergent ray (RS) are parallel to each other. The emergent ray is just shifted to the side, but it remains parallel to the original path.

Question 27.
Allow light to incident normally on the glass slab. Does refraction occur?
Answer:
No, the ray incident normally on the glass slab does not undergo refraction.

Put a coin in a vessel. Walk backwards looking at the coin. When the coin disappears, ask one of your classmates to pour water into the vessel.

Question 28.
What do you observe? Analyse the figure given above and find out the reason.
Answer:
On pouring water, the ray of light coming from the coin undergoes refraction and reaches the eye. Hence, the coin becomes visible again.

Question 29.
Place a glass slab on the letters in a textbook. The letters appear raised. What may be the reason? Find out.
Answer:
The ray of light coming from the letters undergoes refraction through the glass slab and reaches the eye. So the letters appear raised.

Question 30.
Observe the figure given below.

You can see a coin lying under the water in a trough. Try to take the coin out by looking through any one side of the trough.
Can you do it with ease? What may be the reason?
Answer:
We cannot do it with ease. When light coming from the coin enters from an optically denser medium to a rarer medium, it deviates away from the normal. So by looking through any one side of the trough we cannot locate the actual position
of the coin.

Question 31.
The bottom of a pond appears elevated when viewed from a distance than from a nearer point. Why?
Answer:
When light coming from the bottom of the pond enters from an optically denser medium to a rarer medium, it deviates away from the normal. As the distance from the pond and the observer increases, the angle of refraction made by the incident ray increases. So the bottom of a pond appears elevated.

Question 32.
People who engage in bow fishing aim at a point slightly below the perceived position of the fish. Why?
Answer:
When the reflected light coming from the fish enters from an optically denser medium to a rarer medium, it is deviated away from the normal at the surface of separation. This refracted light appears to come from a position just above the actual position. So people who engage in bow fishing aim at a point slightly below the perceived position of the fish.

Observe the figures.

Question 33.
Does the ray of light from the star reach our eyes by travelling in a straight line?
Answer:
No, the ray of light from the star does not reach our eyes by travelling in a straight line.

Question 34.
An illustration of the path of a ray of light from a distant star through the Earth’s atmosphere is given in Fig.(9) .Doesn’t the ray of light undergo irregular deviations? What may be the reason?
Answer:
Yes. The ray undergoes irregular deviations.
Stars appear as point sources of light because they are at a greater distance from the Earth compared to the planets. The light coming from the stars reaches our eyes by traversing through the atmosphere. The optical density of the medium through which the light travels goes on changing as the physical conditions (pressure, temperature etc.) of the layers of atmosphere change continuously. Hence, the light undergoes an irregular refraction. Therefore, when the light rays from the stars reach the eyes after refracted several times, the star cannot be seen continuously at the same position. This is the reason for the twinkling of stars.

Question 35.
Even after the Sun has passed the western horizon, the Sun is visible for some more time. Similarly, the Sun can be seen a few seconds before it reaches the eastern horizon in the morning. What is the reason? Analyse the figure given below and record the explanation in the science diary.

Answer:
This is due to atmospheric refraction. The temperature difference causes density difference in the layers of atmosphere. Due to this, optical density of the medium through which the light travels goes on changing. Light enters to the atmosphere undergoes continuous refraction. We can see the apparent position of the Sun just before and just after the actual Sunrise and actual Sunset.

Question 36.
Will refraction occur whenever light enters obliquely from one medium to another?
Answer:
No

Question 37.
When light passes through the fibres of the decorative lamps as shown in figure, it emerges only through the fibre tips. What may be the reason?
Answer:
It is due to total internal reflection of light.
Activities to understand total internal reflection

Activity 1
Allow light from a laser torch to fall at different angles on the surface of water taken in a trough.

Question 38.
Which are the mediums through which light passes inside the trough?
Answer:
water, air

Question 39.
Does the light enter from an optically denser medium to a rarer medium or from an optically rarer medium to a denser medium?
Answer:
Here, light enters from an optically denser medium to a rarer medium (That is from water to air).

Question 40.
Increase the angle of incidence gradually. What do you observe?
Answer:
Here the light enters from an optically denser medium to rarer medium. The angle of refraction goes on increasing with the increase in the angle of incidence. The incident ray reflects back to the water completely when the angle of incidence is above a specific value.

Question 41.
When does the light reflect completely to the same medium, without refraction?
Answer:
The incident ray reflects completely to the same medium when the angle of incidence is above a specific value.

Activity 2

Draw a circle on a chart paper. Mark angles on the paper as if two protractors are placed together. Keeping this on the surface of a table, place a thick semicircular glass slab on the circle drawn on the chart paper as shown in figure given above. Allow the light from a laser torch to fall on this slab at different angles.

Question 42.
Observe the angle of incidence and angle of refraction in each case. Draw the ray diagram.
Answer:

Question 43.
From which medium to which does the ray of light enter?
(from an optically denser medium to a rarer medium/from an optically rarer medium to a denser medium)
Answer:
From an optically denser medium to a rarer medium.

Question 44.
What is the change in the angle of refraction when the angle of incidence increases?
Answer:
The angle of refraction increases, when the angle of incidence increases.

Question 45.
What is the angle of incidence when the angle of refraction is 90o ?
Answer:
42°.

Question 46.
What peculiarity do you notice when the light is incident at an angle greater than this angle of incidence?
Answer:
The incident ray is reflected back completely to the same medium without undergoing refraction.

Critical Angle When a ray of light enters from an optically denser medium to a rarer medium, the angle of incidence at which the angle of refraction becomes 90° is the critical angle. In the glass-air pair, the critical angle is 42°.Total Internal Reflection When a ray of light enters from an optically denser medium to a rarer medium, at an angle of incidence greater than the critical angle, the ray is reflected back completely to the same medium without undergoing refraction. This phenomenon is total internal reflection.

Question 47.
The path of light through different mediums is given. Analyse the figures and answer the following questions.

Air
(a) Which of the above figures represent total internal reflection?
Answer:
Fig. (a), (e)

b) What is the critical angle of glass in this case?
Answer:
42°

c) Will total internal reflection occur for a ray of light entering from water to air at an angle of 500? Why? What is the critical angle for the water-air pair?
Answer:
Total internal reflection will occur for a ray of light entering from water to air at an angle of 50°. Because the angle of incidence is greater than the critical angle of water. The critical angle for the water-air pair is 48.6°

d) What are the two conditions required for total internal reflection to occur?
Answer:
The conditions required for total internal reflection to occur are

• The ray should enter from an optically denser medium to a rarermedium.
• The angle of incidence should be greater than the critical angle.

Note: If only one medium is mentioned while stating the critical angle, the second medium will be either air or vacuum.

Question 48.
Observe the figure.

The bottom of the aquarium is seen above the surface of water. What may be the reason?
Answer: The light coming from the bottom of the aquarium undergoes internal reflection at the surface of water. Hence, the bottom is seen above the surface of water. The distance from the surface of water to the bottom of the aquarium is the same as the distance from the surface to the image caused by total internal reflection.

Fibres in decorative lamp

One end of each optical fibre in the decorative lamp is connected to a suitable source of light. Light rays from this source travel through the fibre. While travelling through the fibre, it makes an angle of incidence greater than the critical angle with the walls of the fibre. Hence The ray should enter from an optically denser the light undergoes successive total internal reflection and emerges through the other end.

Optical illusions caused by total internal reflection in our surroundings.

Question 49.
During summer season there seems to be water logging on roads when viewed from a distance. What may be the reason?

Answer: During summer season the layers of air closer to the road have a low optical density as it is warmer than the upper layers. The optical density of the air increases gradually as we go higher. When light rays coming from the surrounding objects pass through different layers of air with different optical densities, they undergo refraction and then total internal reflection. Such deviated light rays fall on our eyes. Hence their image appears to have formed on roads. This phenomenon is known as mirage. Such familiar images are usually seen on the surface of water. That is why water seems to be logged on roads when viewed from a distance

Reflector

A large number of small prisms are fixed inside the reflectors in the tail lamps of vehicles. The rays of light incident on a prism get reflected back due to total internal reflection. It can be seen even in dim light.

Question 50.
How does a ray of light incident on a prism get reflected? Explain on the basis of the critical angle of glass.

Answer:
The light ray is incident normally on the side PQ. Hence, there is no refraction. The critical angle of glass is 42o. The angle of incidence at A and B is 45°. Hence the light falling on A undergoes total internal reflection and reaches B. There it undergoes total internal reflection again and comes out of the reflector. The same process happen in other prisms in the reflector as well.

Two equilateral prisms are used in periscope. Due to total internal reflection, the image can be seen with greater visual clarity.

The outer coating of optical fibre with higher refractive index is known as cladding and inner coating is known as the core. Core has a lower refractive index. Light rays passing along the core at an angle greater than the critical angle undergoes total internal reflection. Thousands of optical signals can be sent simultaneously through single cable without the loss of intensity. Such signals can be sent to distant places with the speed of light. This is the reason for using optical fibre cables in communication.

Uses of optical fibre cables
Optical fibres are used in the medical field and communication field.

Question 51.
The use of optical fibres in medical field is increasing day by day. Collect information increasing day by day. Collect information regarding this and present in your class.
Answer:
Hint:
These are some of the applications in medical field which make use of optical fibres
Endoscopy uses optical fibres to obtain the image of the inside of the body.

An optical fibre bundle can be inserted into the body by a physician. Some transport light into the body, while others return light that has been reflected off interior body surfaces. This enables the physician to look clearly into the body to diagnose conditions like cancer

Picture showing endoscope used to view inside of stomach.
Optical fibres helps to observe what is happening during keyhole surgery.
Laser Surgery
Optical fibres help deliver laser light for precise surgical procedures. The fibres guide laser light to the target area for cutting or repairing tissues.
Uses: Common in eye surgeries (like LASIK), skin treatments, and removing tumors.
Comparison of total internal reflection and reflection from a plane mirror

Question 52.
Statements regarding total internal reflection and reflection from plane mirror are given. Tabulate them suitably.

• Occurs only when a ray of light enters from an optically denser medium to a rarer medium at an angle greater than the critical angle.
• The ray of light is not completely reflected.
• Reflection occurs on a surface at any angle of incidence.
• The ray of light is completely reflected.

Answer:

Reflection from a plane mirror	Total internal reflection
The ray of light is not completely reflected	The ray of light is completely reflected.
Reflection occurs on a surface at any angle of incidence.	Occurs only when a ray of light enters from an optically denser medium to rarer medium at an angle greater than the critical angle.