## KSEEB Solutions for Class 9 Science Chapter 8 Motion

Students can Download KSEEB Solutions for Class 9 Science Chapter 8 Motion, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka Board Class 9 Science Chapter 8 Motion

### KSEEB Solutions for Class 9 Science Chapter 8 Intext Questions

Our Displacement Calculator helps determine your engine’s size by its bore and stroke. This is a very useful tool for building high-performance racing engines.

Question 1.
An object has moved through a distance can it have zero displacement? If yes, support your answer with an example.

In the figure distance of an object moved from 0 to A is 60 km and distance travelled is 60 km and displacement is 60 km. From 0 to A and back to B, distance travelled is 60 km + 25 km = 85 km. But displacement (35 km) is not equal to distance travelled (85 km). If we observe like this, displacement of a body is zero, but distance travelled is not zero, starting from point ‘0’ and returning back to ‘O’ final position is mixing with initial point. Hence displacement is zero.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
A farmer moves along the boundary of a square field of side 10 m in 40s
2 minutes 20 seconds means 140 seconds.
Distance travelled by farmer
= $$\frac{40}{40}$$ × 140 m
= 140 m.
∴ Farmer moves for 2 minutes 20 seconds. It means $$\frac{140}{40}$$ = 3.5 rounds
From the original point, he moves for 2 minutes 20 seconds.
Case i)
Original point means any point in the corner of the field. In this case he moves for 2 min 20 seconds from the diagonal of the field.
Displacement is equal to diagonal of the field displacement
= $$\sqrt{10^{2}+10^{2}}$$
= 14.1 m.
Case ii)
Original point means any point in the middle of the side of the field.
∴ Displacement is equal to any side of the field = 10 m.
It means displacement is between any original point ie in between 14.1 m and 10 m.

Question 3.
Which of the following is true for displacement?
a) It cannot be zero.
b) Its magnitude is greater than the distance travelled by the object.
b) Its magnitude is greater than the distance travelled by the object.

Question 4.
Distinguish between speed and velocity

 Speed Velocity 1. Scalar quantity Vector quantity 2. Distance travelled in a given time. Distance travelled along with a path with the given time. 3. It is always is + ve. It is + ve or – ve

Average velocity calculator is really nice tool to use and it will help you solve all your problem related to average velocity.

Question 5.
Under what condition (s) is the magnitude of the average velocity of an object equal to its average speed?
If the speed of the object changing uniformly, the magnitude of the average velocity of an object equal to its average speed.

Question 6.
What does the odometer of an automobile measure?
Odometer measures the speed of the vehicle.

Question 7.
What does the path of an object look like when it is in uniform motion?
When the object is in uniform motion, it covers equal distance at equal intervals of time.

Question 8.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, which is 3 × 10 ms-1.
Time taken by the signal to reach the ground station from the spaceship = 5 min = 5 × 60 = 300 seconds
Speed of the signal = 3 × 108 m/s.
Speed = $$\frac{\text { Total distance travelled }}{\text { Total time taken }}$$
∴ Distance travelled = speed × time taken
= 3 × 108 × 300
= 9 × 1010m
∴ Distance of change of velocity = 9 × 1010 m

A versatile acceleration calculator with which you can calculate the acceleration given initial and final speed and acceleration time.

Question 9.
When will you say a body is in

1. Uniform acceleration?
2. non-uniform acceleration?

1. If the object covers equal distances in equal intervals of time, it is added to be in uniform motion.
2. If the object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion.

Question 10.
A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5s. Find the acceleration of the bus.
Initial velocity of the bus u = 80 kmh-1

Final velocity, V = 60 km/h
time taken to decrease the velocity of bus, t = 5 seconds
acceleration, a = 5 seconds

= -1.112 ms-2.

Question 11.
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-1 in 10 minutes. Find its acceleration.
Initial speed of the train u = 0

∴ acceleration of a train = 0.0185 ms-2.

Question 12.
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
(i) The nature of the distance-time graphs for the uniform motion of an object is a straight line.

(ii) The nature of tire distance time graph for non-uniform motion is curved line.

Distance-time graph for a car moving with non-uniform speed

Question 13.
What can you say about the motion of an object whose distance-time graph is straight line parallel to the time axis?
This graph indicates the object is at rest.

Question 14.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
This graph indicates the object is in uniform motion.

Question 15.
What is the quantity which is measured by the area occupied below the velocity-time graph?
The quantity which is measured by the area occupied below the velocity-time graph is length = l

Question 16.
A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
(a) Initial velocity of the bus u = 0 (Bus is at rest).
acceleration, a = 0.1 ms-2
time t = 2 minutes = 120 seconds
Let the speed of bus be ‘V’

∴ v = 12m/s
(b) As per third law of motion:
v2 – u2 = 2as
(12)2 – (0)2 = 2(0. 1)s
∴ s = 720 m.
the speed of acquired by bus = 12 m/s.
the distance travelled = 720 m.

Question 17.
A train is travelling at a speed of 90 kmh-1 Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will go before it is brought to rest.
Let the initial speed of the train be u = 90 Km/h
= 25 m/s.
Final speed of the train, v = 0 (train comes to rest)
acceleration a = 0.5 ms-2
As per 3rd law of motion
v2 = u2 + 2as
(0)2 = (25)2 + 2(0.5)s
s = train travelled distance

Train will travel 625 km before it is brought to rest.

Question 18.
A trolley, while going down an inclined plane has an acceleration of 2cms-2 what will be its velocity 3s after the start?
Initial velocity of a trolley, u = 0 (at rest)
acceleration, a = 2 cms-2 = 0.02 m/s2 time t = 3S
As per 1 st law of motion
v = u + at
Here V means velocity of a trolley after 3 s
V = 0 + 0.02 × 3
= 0.06 m/s.
∴ = 0.06 m/s.

Question 19.
A racing car has a uniform acceleration of 4ms-2, what distance will it cover in 10s after start?
Initial velocity of a racing car, u = 0 (at rest)
acceleration, a = 4 m/s2
time t = 10 s
As per second law of motion

Here S means distance travelled by car after 10S

∴ Distance covered by car after 10 S = 200 m.

Question 20.
A stone is thrown in a vertically upward direction with a velocity of 5ms-1 If the acceleration of the stone during its motion is 10ms-2 in the downward direction, what will he the height attain by the stone and how much time will it take to reach there?
The initial velocity of a stone, u = 5 ms-1.
Final velocity of a stone v = 0 (at rest)
If the acceleration of the stone during its motion is = -10ms-2.
Let the maximum height be ‘s’
v = u + at
0 = 5 + (-10)t

time taken by stone = 0.5s
As per 3rd Law of motion
v2 = u2 + 2as
(0)2 = (5)2 + 2(-10)s

The height attained by the store = 1.25 m.

### KSEEB Solutions for Class 9 Science Chapter 8 Textbook Exercises

Question 1.
An athlete completes one round of a circular track of diameter 200 m. in the 40s. What will be the distance covered and the displace¬ment at the end of 2 minutes 20s?
Diameter of circular track, d = 200 m.
r = $$\frac{\mathrm{d}}{2}$$ = 100m
Circumference = 2πr = 2p(100) = 200πcm.
in 40s, athlete covers distance 200πm
in 1s, distance covered by athlete = $$\frac{200 \pi}{40}$$ m
Athlete runs for 2 minutes 20s means 140s
∴ Displacement of all athlete is equal to the diameter of circle
= $$\frac{200 \times 22}{40 \times 7}$$ × 140
= 2200 m
∴ Displacement = 200 m.

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 M road in 2 minutes 30 seconds and then turns around and jogs 100 M back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
a) from A to B distance covered by Joseph = 300 M.
Time taken to cover this distance
= 2 min 30 sec. = 150 seconds

Displacement = Nearest distance between A and B = 300 m.
Time taken = 150s
Average velocity = ygy = 2 m/s.

b) From A to C :

Total distance covered = Distance between A to B + distance between B to C
= 300 + 100 = 400 m.
Total time taken to cover from A to B + Time taken to cover from B to C
= 150 + 60= 210s.

Displacement from A to C
= AC + AB – BC
= 300 – 100
= 200m
Total time taken = Time taken to travel from A to B + Time taken to travel from B to C
= 210 s
∴ Average velocity = $$\frac{200}{210}$$ =0.952 ms-1 .

Question 3.
Abdul, while during to school, computes the average speed for his trip to be 20Kmh-1 on his return trip along the some route, there is less traffic and the average speed is 30 Kmh-1. What is the average speed for Abdul’s trip?
i) Average speed of Abdul’s trip while driving to school = 20 Kmh-1
Average speed = $$\frac{\text { Total distance }}{\text { Total time taken }}$$
Total distance = Distance travelled to reach school = t1
∴ 20 = $$\frac{\mathrm{d}}{\mathrm{t}_{1}}$$
t1 = $$\frac{d}{20}$$ …………… (i)
While returning from school
Total distance = Distance travelled while returning from school = d Now total time taken = t2.

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms-2 for 8.0 s. How far does the boat travel during this time?
Initial velocity of motor boat, u =0
(at rest) ,
Accieratlon of a motor boat a = 3m/s2
Time taken, t = 85
According to second equation of motion, s
S = 0 +$$\frac{1}{2}$$ × (8)2 = 96m.
∴ Motorboat travels = 96m.

Question 5.
A driver of a car travelling at 52kmh-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3kmh-1 in another car applies his brakes slowly and stop In 10s. on the same graph paper. Plot the speed versus time graphs for the two cars, which of the two cars travelled farther after the brakes were applied?
initial velocity of A car u1 = 52 kmh-1
= 14.4 m/s.
Time taken to stop car t1 = 5 s.
After 5s car comes to rest Initial velocity of B car, u2 = 3Kmh-1
= 0.833 m/s. = 0.83 m/s.
Time taken to stop car t2 = 10s
After 10s car comes to rest.
Graph:

Area of ∆OPR = Area of ∆OSQ
When compared car A to car B, A car travelled father. After brake it travels 52 Kmh-1

Question 6.
Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions.

a) Which of the three is travelled the fastest?
b) Are all three ever at the same point on the road?
c) How far has C travelled when B passes A?
d) How far has B travelled by the time passes C?

Slope of B object is greater than slopes A and C. This is travelling fastest.

b) All three objects never meet at a point, hence these are not at the same point.

c) There are 7 boxes in X-axis = 4 Km
1 box = $$\frac{4}{7}$$ Km
In the beginning, C object is 4 boxes away from 0
= $$\frac{16}{7}$$ km
Distance to C from origin = Distance of C when it moves from B to A =8 Km

∴ Distance travelled by C when B passes A = 5.714 km.

d) Distance B travelled by the time it passes C = 9 boxes

∴ B has travelled 5.143 k.m. by the time when it passes C.

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms-1 with what velocity will it strike the ground? After what time will it strike the ground?
Distance ball covers, S = 20 m
acceleration, a = 10Kmh-1 = 10 m/s
initial velocity, u = 0 (at rest)
When ball strikes, its final velocity,
According to 3rd law of motion
v2 = u2 + 2as
v2 = 0 + 2(10)(20)
v = 20 m/s.
According to 1 st Law of motion v = u + at
Time taken by ball to strike the ground, t,
20 = 0 + 10 (t)
∴ t = 2s
∴ 2s is required to ball for striking the ground and velocity is 20 m/s

Question 8.
The speed-time graph for a car is shown in figure

a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by car during the period.
b) Which part of the graph represents uniform motion of the car?
a) Shade lightly in the graph about distance travelled by car in 4s.
b) Shade with red colour about 6 cm to 10 cm. It represents the uniform motion of the car.

Question 9.
State which of the following situations are possible and give an example for each of these;
a) an object with a constant acceleration but with zero velocity.
b) an object moving in a certain direction with an acceleration in the perpendicular direction.
a) This is possible
Eg: when a ball is thrown to a height its velocity is zero.
(For a ball G = 9.8 m/s2)

b) This is possible
Eg: when a car moves in a curved line, its acceleration is perpendicular to the given direction)

Question 10.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Satellite is moving in circular orbit radium of satellite r = 42250 km
time t = 24 Hrs = 3600 × 24 S

∴ Speed of the satellite when it moves in circular orbit = 3.07 kms-1.

### KSEEB Solutions for Class 9 Science Chapter 8 Additional Questions

Fill in the blanks with suitable words.

Question 1.
The two physical quantities required to know final position of an object are _________ and _________
Distance travelled, displacement.

Question 2.
S.I. unit of speed is ________
m/s

Question 3.
Distance travelled by the object in unit time is ________
Speed.

Question 4.
If acceleration is in the direction of velocity it is _________ acceleration.
positive.

Question 5.
When an object moves in a circular path with uniform speed, its motion is called ________
uniform circular motion.

Question 1.
Give an express10n for centripetal force?
$$f=\frac{m v^{2}}{r}$$

Question 2.
What is uniform circular mot10n?
Body moving in a circular path arbitrary any instant along a tangential to the posit10n of the body on the circular path at that instant or time.

Question 3.
A body falls freely. What is constant?
When the body falls freely. It has a constant acceleration.

Question 4.
Define uniform speed.
The speed of an object is said to be uniform speed if it travels equal distances in equal intervals of time.

## Karnataka Board Class 9 Science Chapter 11 Work and Energy

### KSEEB Solutions Class 9 Science Chapter 11 Intext Questions

Question 1.
A force of 7N acts on an object. The displacement is, say 8m, in the direction of the force. Let us take it that they force acts on the object through the displacement. What is the work done in this case?
Work done on an object = 7N × 8m = 56 Nm or 56 J.

Question 2.
When do we say that work is done?
Two conditions need to be satisfied for work to be done:

1. a force should act on an object and
2. the object must be displaced.

Question 3.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Let a constant force F act on an object. Let the object be displaced through a distance S in the direction of the force. Let W be work done. We define work to be equal to the product of the force and displacement.
Work done = force × displacement W = F × S.

Question 4.
Define 1 J of work.
1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

Question 5.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?
Force exerted, F = 140N.
Displacement, S = 15m
Work done in ploughing the field
W = F × S
= 140 × 15
= 2100J
= 2.1 × 103 J.

Free Kinetic Energy Calculator – calculate kinetic energy step by step.

Question 6.
What is the kinetic energy of an object?
Objects in motion possess energy, we call this energy kinetic energy.

Question 7.
Write an expression for the kinetic energy of an object.
KE = $$\frac{1}{2}$$ mv2.

Question 8.
The kinetic energy of an object of mass, m moving with a velocity of 5 ms-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Kinetic energy, k = $$\frac{1}{2}$$ mv2.
Where m = mass of the object
V = velocity of the object.
Here mass (m) is the same in both cases (∵ object is same)
$$\frac{\mathbf{k}_{1}}{\mathbf{k}_{2}}=\left(\frac{\mathbf{v}_{1}}{\mathbf{v}_{2}}\right)^{2}$$
Initial kinetic energy k1 = 25J
Initial velocity V2 = 5 ms-1
New kinetic Energy K2 = ?
New velocity v2 = 3v1 = 3 × 5 = 10 ms-1

∴ When velocity is increased three times its K.E. is 225J.

Question 9.
What is power?
Power is defined as the rate of doing work or the rate of transfer of energy w
P = $$\frac{w}{t}$$

Question 10.
Define 1 watt of power.
Power is 1 W when the rate of consumption of energy is 1 JS-1.

Question 11.
A lamp consumes 1000 J of electrical energy in 10s. What is its power?
Power = 1000 J Work w = 1000J
Time = 10 s
Power of lamp, P = w/t
= 1000/10 = 100W
∴ Power of a lamp = 100 KW.

Question 12.
Define average power.
We obtain average power by dividing the total energy consumed by the total time taken.

### KSEEB Solutions for Class 9 Science Chapter 11 Textbook Exercises

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’

1. Suma is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Food grains are getting dried in the sun.
7. A sailboat is moving due to wind energy.

1. Work is being done by Seema because she displaces the water by applying the force.
2. No work is being done by the donkey because the direction of force i.e the load is vertically downward and displacement is along the horizontal. If displacement and force are perpendiculars then no work is done.
3. Work is done because the windmill is lifting the water i.e., it is changing the position of water.
4. No work is done because there are no force and displacement.
5. Work is done because the engine is changing the position of the train.
6. No work’ is done because there is no force and no displacement.
7. Work is done because the force acting on the boat is moving it.

Question 2.
An object was thrown at a certain angle to the ground moves in a curved path and falls back to the ground the initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Work done by the force of gravity, W = mgh.
Where h = difference in height of initial and final positions of the object.
According to the question, the initial and final positions of the object lie in the same horizontal line. So h = 0.
∴ Work done W = mg × 0 = 0

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
In the case given in the question, the battery has chemical energy which is converted into energy. Electric energy provided to the bulb further converted into light energy.

Question 4.
A certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done by the force.
Mass, m = 20 kg.
Initial velocity, u = 5 ms-1
Final velocity v = 2ms-1
Work done by the force = change in kinetic energy
= Final kinetic energy – Initial kinetic energy

Question 5.
A mass of 10kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal what is the work done on the object by the gravitational force? Explain your answer.
Work done by gravitational force W = mgh
Where h = Difference in the heights of initial and final positions of the object.
Here both the initial and final positions are on the same horizontal line.
So there is no difference in height i.e., h = 0.
∴ work done W = mg × 0 = 0.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
The total mechanical energy remains constant / as P.E. of the freely falling object decreases. Its kinetic energy and the kinetic energy increases on account of an increases its velocity) the law of conservation of energy is not violated.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?

1. Muscular energy into kinetic energy
2. the kinetic energy of the rotation of the which into kinetic energy of the bicycle.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
When we push a huge rock and fall to move it. The energy spent in doing so is absorbed by the.rock. This energy is converted into potential energy of the configuration of the rock which results in its deformation. However t his deformation is not visible on account of the huge size of the rock.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is this in Joules?
Energy consumed w = 250 units
= 250 kwh
= 250 × 1000 W × 3600 S
= 250 × 1000 J/S × 3600 S
= 9 × 108 J.

Question 10.
An object of mass 40 kg is raised to a height of 5m above the ground what is its potential energy? If the object is allowed to fall, find its kinetic energy when it is halfway down.
Mass m = 40 kg.
height h = 5 m.
Potential energy PE = mgh = 40 × 9.8 × 5 = 1960J.
KE at half way down = PE at halfway down = $$\mathrm{mg} \frac{\mathrm{h}}{2}$$
= 40 × 9.8 × $$\frac{5}{2}$$ = 980 J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

The satellite round the earth moves in a circular orbit. Here the force of gravity acts towards the centre of the earth and displacement of the satellite is along the tangent of the circular path that means therefore and displacement are perpendicular to each other.
So, work done, W = F.S cosθ
= F × S cos 90°
= F × S × 0
= 0
That is, no work is done by the force of gravity.

Question 12.
Can there be displacement of an object in the absence of any force acting on it? think. Discuss this question with your friends and teacher.
If an object moves with a constant velocity (i.e, there is no acceleration) then no force acts on it. As the object is moving ie it is displaced from one position to another position.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
The person has no movement ie his displacement is zero. So the person had done no work (∵ work is done only when the object is displaced).

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Time = 10 h = 10 × 60 min
= 10 × 60 × 60 S

Energy = power × time
= 1500 × 10 × 60 × 60
= 5.4 × 107 J.

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate why does the bob eventually comes to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
When the pendulum oscillates in the air, the air friction opposes its motion. So some part of the kinetic energy of the pendulum is used to overcome this friction. With the passage of time, the kinetic energy of the pendulum goes on decreasing and finally becomes zero. The kinetic energy of the pendulum is transferred to the atmosphere. So energy is being transferred ie it is converted into one form to another. So here is no violation of the law of conservation of energy.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
The work done on the object to bring the object to rest
= change in kinetic energy
= Final kinetic energy – Initial Kinetic energy
Here final kinetic energy is zero because the object is brought to rest.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at the velocity of 60km/h?
Mass, m = 1500 kg.
Initial velocity, u = 60 kmh-1
= 60 × $$\frac{5}{18}$$ = 16.67 ms-1
Final velocity, v = 0
(∵ the car comes to rest)
Work done to stop the car = change in kinetic energy
= Final kinetic energy – Initial Kinetic energy

Question 18.
In each of the following a force, F is acting on an object of mass M. The direction of displacement is from west to east shown by the longer arrow. Observe the diagram carefully and state whether the work done by the force is negative, positive or zero.

Cas i) The force and displacement are perpendicular to each other. So S = 90°
Work done = FScos θ°
= FScos 90°
= FS × 0 = 0
(∵ cos90° = 0)
Cas ii) The force and displacement are in the same direction so θ = 0°
Work done = FScos θ°
= FScos 90°
= FS × 1 = FS
(∵ cos θ° = 1)
That is the work done is positive.
Cas iii) The force and displacement in opposite direction θ = 180°
Workdone = FScos θ°
= FScos 180°
= FS × -1 = FS
(∵ cos 180° = -1)
That is workdone is negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Yes, I agree with Soni, the acceleration of an object can be zero even when several forces are acting on it if the resultant of all the forces acting is zero.

Question 20.
Find the energy in kWh consumed in 10 hours by four devices of power 500 w each.
Total power p = 500w × 4 = 2000w.
Time, t = 10h
Energy = p × t.
= 2000w × 10h
= 2kwh × 10h = 20kwh.
∴ Energy = 20kwh.

Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
P.E. of the configuration of the body and the ground (the body may be deformed and the ground may at the place of collision).
This process energy in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy charges is called dissipation of energy.

### KSEEB Solutions for Class 9 Science Chapter 11 Additional Questions and Answer

Fill in the blanks:
Question 1.
Work done = ___________
force, displacement

Question 2.
An object having the capability to do work is said to possess __________
energy.

Question 3.
__________ is the energy possessed by an object due to its motion.
Kinetic energy

Question 4.
P.E
mgh.

Question 5.
Larger unit of energy is called
kilojoule.

Question 1.
Can there be displacement of an object in the absence of any force acting on it?
In the absence of any force on the object i.e., F = 0, ma, (as F = ma) since m ≠ 0 and a = 0. In such a case the object is either at rest or in a state of uniform motion in a straight line in the latter case there is a displacement . of the object without any force acting on it.

Question 2.
How does a bullet pierce a target?
A bullet moves with large velocity and as such possesses a lot of kinetic energy the work in piercing the target is derived from the kinetic energy of the bullet.

Question 3.
Why do some engines require fuels like petrol and diesel?
Internal combustion heat engines use the chemical energy of fossil fuel (petrol and diesel) for their operation. These engines first convert the chemical energy of the fuels into heat energy. Which is later on converted into mechanical energy.

Question 4.
Calculate the work done by a body. By the force of 5 N makes to move through a distance of 12 m.
W = F x s = 5 x 12 = 60 joules

Question 5.
When force 6 N applied on a wall, the wall remains in the same position, calculate the work done.
W = F x s
W – 6 x s = 6 x 0 = 0 Joules No work is done on the body

Question 6.
Electricity is the most convenient form of energy. Why?
It can be converted into other forms of energy easily.
It can be produced by different means. It is ecofriendly

Question 7.
Calculate the power experienced by a source that can do work of 50 joules in 5 seconds.
Power = $$\frac{\text { Workdone }}{\text { Time taken }}$$
$$=\frac{50}{5} \frac{\text { Joules }}{\text { Second }}$$ = 10 J/s = 10 watts

## KSEEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Students can Download Kannada Chapter 3 Atoms and Molecules Questions and Answers, Summary, Notes Pdf, Siri KSEEB Solutions for Class 9 Science, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka Board Class 9 Science Chapter 3 Atoms and Molecules

### KSEEB Solutions for Class 9 Science Chapter 3 Intext Questions

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water
Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water
5.3 + 6 ➝ 8.2 + 2.2 + 0.9
= 11.3 g = 11.3 g
Weight of reactants is equal to weight of products. This observation is in agreement with the law of conservation of mass.

Question 2.
Hydrogen and Oxygen combine in the ratio or 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas. Because in water the ratio of the mass of hydrogen to the mass of oxygen is always 1 : 8.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
The relative number and kinds of atoms are constant in a given compound. This postulate is the result of the law of conservation of mass.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Atoms combine in the ratio of small whole numbers to form compounds. This postulate explains the law of definite proportions.

Question 1.
Define the atomic mass unit.
One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12.

Question 2.
Why is it not possible to see an atom with naked eyes?
Atoms are very small, they are smaller than anything that we can imagine or compare with. Therefore it is not possible to see an atom with naked eyes.

Question 1.
Write down the formulae of:

1. sodium oxide
2. alluminium chloride
3. sodium sulphide
4. magnesium hydroxide

1. sodium oxide: Na2O
2. aluminium chloride: Al2Cl3
3. sodium sulphide : NaS
4. magnesium hydroxide: Mg(OH)2

Question 2.
Write down the names of compounds represented by the following formulae:

1. Al2(SO4)3
2. CaCl2
3. K2SO4
4. KNO3
5. CaCO3

1. Al2(SO4)3: Aluminium sulphate
2. CaCl2: Calcium chloride
3. K2SO4: potassium sulphate
4. KNO3: potassium nitrate
5. CaCO3: calcium carbonate

Question 3.
what is meant by the term chemical formula?
The chemical formula of a compound is a symbolic representation of its composition.

Question 4.
How many atoms are present in a
i) H2S molecule and
ii) PO43- ion?
i) H2S molecule:

ii) PO43- ion

Question 1.
Calculate the molecular masses of
H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
i) H2
= 2 × H
= 2 × 1
= 2u

ii) O2
= 2 × O
= 2 × 16
= 32u

iii) Cl2
= 2 × Cl
= 2 × 35.5
= 7lu

iv) CO2
= 1 × 12 + 2 × O
= 1 × 12 + 2 × 16
= 12 + 32
= 44u

v) CH4
= 1 × 12 + 4 × 1
= 12 + 4
= 16u

vi) C2H6
= 2 × 12 + 6 × 1
= 24 + 6
= 30u

vii) C2H4
= 2 × 12 + 4 × 1
= 24 + 4
= 28u

viii) NH3
= 14 × 1 + 3 × 1
= 14 + 3
= 17u

ix) CH3OH
= 12 × 1 + 3 × 1 + 16 × 1 + 1 × 1
= 12 + 3 + 16 + 1
= 32u

Question 2.
Calculate the formula unit masses of ZnO, Na2O, K2CO3 given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and O = 16u.
i) ZnO
(Automic mass of Zn) + (Automic mass of O)
= 65 + 16 = 81u.

ii) Na2O
(2 × Automic mass of Na) + (Automic mass of O)
= 2 × 23 + 1 × 16
= 46 + 16
= 62u.

iii) K2CO3
(2 × Atomic mass of K)+ (Atomic mass of C) + (3 × Atomic mass of oxygen)
= 2 × 39 + 12 × 1 + 16 × 3
= 78 + 12 + 48
= 138u.

Question 1.
If one mole of carbon atoms weighs 12 gms, what is the mass (in gms) of 1 atom of carbon?
Number of moles = n
Given mass = m
molar mass = M
Given the number of particles = N
Avogadro number of particles = N0
i) Atomic mass of carbon = 12u.
Atomic mass of one mole of carbon=12g
But Atomic mass of carbon = 12 g mass of 6.022 × 1023 carbon atoms = 12g.

∴ Mass of carbon atoms
= 1.9926 x 10-23 gm.

Question 2.
Which has more number of atoms, 100 g, 100 gms of sodium or 100 gms of sodium or 100 gms of iron (given, the atomic mass of Na=23u, Fe = 56u).
Atomic mass of sodium = 23u (data)
It means the gram atomic mass of sodium = 23 gm
Now atoms present in 23 gm sodium = 6.022 × 1023

It means Number of atoms in 100 gm sodium = 1.6753 × 1024
∴ 100 gm sodium has more number of atoms rather than 100 gms of iron.

### KSEEB Solutions for Class 9 Science Chapter 3 Textbook Exercises

Question 1.
A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Mass of Boron = 0.096 g (Data)
Mass of Oxygen = 0.144g (Data)
Given mass = 0.24 g (Data)
∴ The percentage composition of Boron

∴ The percentage of O2 = $$\frac{0.144}{0.24}$$ × 100
= 60%.

Question 2.
When 3.00 g of carbon is burnt in 8.009 of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50,000 g of oxygen? Which law of Chemical combination will govern your answer?
3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
But when 3 g of carbon is burnt in 50 g of oxygen, only 3 g of carbon reacts with 8 g of oxygen.
Remaining 42 gm of oxygen will not react.
11 gm of carbon dioxide is produced.
∴ Our answer obeys law of constant proportion.

Question 3.
What are polyatomic ions? Give examples.
A group of atoms carrying a charge is known as a polyatomic ion.

Question 4.
Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
(a) Magnesium chloride : MgCl2
(b) Calcium oxide : CaO
(c) Copper nitrate : Cu(NO3)2
(d) Aluminium chloride : AlCl3
(e) Calcium carbonate : CaCO3.

Question 5.
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
(a) Quick lime: Calcium, carbon, oxygen
(b) Hydrogen bromide: Hydrogen, Bromine
(c) Baking powder: Sodium, Bicarbonate
(d) Potassium sulphate: Potassium, Sulphur, Oxygen.

Question 6.
Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus Molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid HNO3.
(a) Ethyne C2H2
Molar Mass = 2 × 12 + 2 × 1
= 24 + 2
= 26 g.

(b) Molar mass of Sulphur molecule
= 8 × 32
= 256 g.

(c) Molar Mass of Phosphorus molecule = 4 × 31
(Atomic mass of Phosphorus) = 124 g.

(d) Molar mass of Hydrochloric acid = HCl
= 1 + 35.5
= 36.5 g.

(e) Molar mass of HNO3
= 1 + 14 + 3 × 16
= 15 + 48
= 63 gm.

Question 7.
What is the mass of –
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of Aluminium = 27)
(c) 10 moles of Sodium Sulphite (Na2SO3)
(a) Mass of 1 mole of nitrogen= 14g.

(b) Mass of 4 moles of Aluminium
= 4 × 27
= 108 g.

(c) Mass of 10 moles of Sodium sulphite
= 10 × [2 × 23 + 32 + 3 × 16]
= 10 × 126
= 1260 gm.

Question 8.
Convert into mole:
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
(a) 32gm of oxygen means 1 mole
12gm oxygen means $$\frac{12}{32}$$ mole
∴ 12g of oxygen gas = 0.375 mole
(b) 18 gm of water means = 1 mole
20 gm of water means $$\frac{20}{18}$$ mole = 1.11 mole
(c) 22 g of carbon dioxide means
$$\frac{22}{44}$$ = 0.5 mole.

Question 9.
What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
(a) Mass of 1 mole of oxygen = 16 g
Mass of 0.2 mole of oxygen = 0.2 × 16
= 3.2 g.
(b) Mass of 1 molecule of water= 18 gm.
Mass of 0.5 mole of water = 0.5 × 18
= 9 gm.

Question 10.
Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
One mole of Sulphur (S) = 8 × 32
= 256 gm.
256 g of solid sulphur = 6.022 × 1023 molecules.

= 3.76 × 1022
(Approximate)

Question 11.
Calculate the number of aluminium ions present in 0.0519 of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u).
One mole of Aluminium oxide
= 2 × 27+ 3 × 16
= 54 + 48
= 102 g.
102 g of Al2O3 = 6.022 × 1023 moles (aluminium oxide)

It means Aluminium present in 0.051 gm
= 3.011 × 1020 Aluminium oxide molecules
Number of Al ions in one mole of Al2O3 = 2
∴ Number of Al Ions In 3.011 × 1026 molecules
0.051 Al2O3
= 2 × 3.011 × Number of Al ions In 3.011 × 1020
= 6.022 × 1020.

### KSEEB Solutions for Class 9 Science Chapter 3 Additional Questions

Question 1.
Write the symbols of the following:
a) Iron
c) Zinc
d) Oxygen
e) Chlorine
a) Iron = Fe
c) Zinc = Zn
d) Oxygen = O
e) Chlorine = Cl

Question 2.
What is a molecule?
The smallest particle of an element or a compound that is capable of independent existence and shows all the properties of that substance.

Question 3.
What is an ion?
An ion is a charged species present in metals and non-metals.

Question 4.
What is molecular mass?
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance.

Question 5.
What is a mole?
One mole of any species (atoms, molecules, ions, or particles) is that quantity in number having a mass equal to its atomic or molecular mass in grams.

### KSEEB Class 9 Science Atoms and Molecules Additional Questions and Answers

Question 1.
An element Z forms an oxide with formula Z2O3. What is its valency?
Valency is 3+

Question 2.
Mention the elements present in (1) quick line (2) sodium hydrogen carbonate.,
(1) Quick line (calcium oxide) (CaO) element present are calcium and oxygen.
(2) Sodium hydrogen carbonate (NaHCO3) elements are sodium, hydrogen, hydrogen, carbon and oxygen.

Question 3.
Calculate the total number of ions in 0.585 g of sodium chloride.
Gram formula mass of NaCl = 23 + 35-5 = 58.5 g
58.5 g of NaCl have ions = 2 × NA
58.5 g of NaCl have ions
$$=2 \times \mathrm{N}_{\mathrm{A}} \times \frac{0.585}{58 \cdot 5}=0 \cdot 02 \times \mathrm{N}_{\mathrm{A}}$$
= 0.02 × 6.022 x 1023
= 1.20 × 1022 ions

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.4.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches :
2, 3, 4, 5, 0, 1. 3, 3, 4, 3
Find the mean, median and mode of these scores.
Solution:
Number of goals scored by a team:
(i) 2, 3, 4, 5, 0, 1, 3, 3, 4, 3

= 2.8

(ii) If number of goals are arranged in ascending order,
0, 1, 2, 3, 3. 3, 3, 4, 4. 5
No. of observations =10 (even number)

= 5th observation.
= 3

= 5 + 1
= 6th observation ➝ 3

= 3

(iii) Mode is the value of the observation which occurs most frequently.
Here 3 is repeated 4 times.
∴ Mode = 3.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded : 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Solution:
In a mathematics test given to 15 students, the following marks (out of 100) are recorded :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60

Mode= 54.8

(ii) When marks are arranged in ascending order,
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Number of observations = 15 (odd number)

∴ Median = 52.

(iii) Mode: It is the repeated value.
Here 52 is repeated three times.
∴ Mode = 52.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63. find the value of x,
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Median = 63.
Number of observations = 10 (even number)

∴ x + 1 = 63 ∴ x = 63 – 1
∴ x = 62.

Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Solution:
14, 25, 14, 28, 18, 17, 18, 14, 23, 22; 14, 18
Mode = ?
Arranging the scores in ascending order,
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Mode is the repeated value.
∴ Here 14 is repeated 4 times.
∴ Mode = 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table :

Solution:

∴ The men salary of workers is Rs. 5083.33.

Question 6.
Give one example of a situation in which
(i) The mean is an appropriate measure of central tendency.
(ii) The mean is not an appropriate measure of central tendency but the median is appropriate of central tendency.
Solution:
(i) Each score is in lesser difference, it is easy to calculate the average than the median.
Ex: monthly salary of 5 persons, 10000, 10100, 10200, 10300, 10400
Mean of the scores is 10200.

(ii) If scores have more differences, it is easy to calculate the median than the mean.
Ex.: Marks obtained by 7 students in Mathematics :
2, 10, 20, 15, 4, 23, 3
Ascending Order : 2, 3, 4, 10, 15, 20, 23
Median =10 but the mean is 11.

We hope the KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.4, drop a comment below and we will get back to you at the earliest.

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.3.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %).

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) Scale: x axis ➝ 1 horizontal axis = 1 cm.
y axis ➝ 5% = 1 cm.

(ii) Reproductive health conditions.
(iii) Nerve and skin.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys In different sections of Indian society is given below.

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution:
(i) Scale: x axis ➝ 1 horizontal = 1 cm.
y axis ➝ 100 girls = 1 cm.

(ii) After verification we came to know that Strength of Girls is more in ST section and it is less in Urban section.

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections :

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats ?
(ii) A Political Party won the maximum number of seats.
Solution:
(i) Scale: x axis ➝ 1 horizontal(Party) = 1 cm.
y axis ➝ 10 Parties = 1 cm.

(ii) ‘A’ political Party won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data ?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long ? Why ?
Solution:
(i) Scale: x axis ➝ 1 Class interval = 1 cm.
y axis ➝ 2 leaves = 1 cm.

(ii) Frequency Polygon.
(iii) No. Because from 145 to 153, length of leaves is 153.

Question 5.
The following table gives the life times of 400 neon lamps:

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours ?
Solution:
(i) Scale: x axis ➝ 1 Class interval = 1 cm.
y axis ➝ 10 Bulbs = 1 cm.

(ii) 184 lamps have a life time of more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them :

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
(i) Scale: x axis ➝ 5 marks = 1 cm.
y axis ➝ 2 frequency = 1 cm.

Section B students scored less marks in great numbers.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below :

Represent the data of both the teams on the same graph by frequency polygons.
(Hint: First make the class intervals continuous.)
Solution:
Class interval is continuous,
1 – 6, 7 – 12 Here the difference is 1.
∴ $$\frac{1}{2}$$ = 0.5 is lower limit is subtracted,
0.5 is taken upper limit, it becomes
0.5 – 6.5
6.5 – 12.5.

Scale: x-axis ➝ 3.5 mark = 1 cm.
y – axis ➝ 1 frequency = 1 cm

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows :

Draw a histogram to represent the data above.
Solution:
Scale: x-axis ➝ 1 year = 1 cm.
y – axis ➝ 1 child = 1 cm.

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows :

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie
Solution:
(i) Scale: x-axis ➝ 1 letter = 1 cm.
y – axis ➝ 4 surnames = 1 cm.

(ii) Class interval which has maximum number of surnames is : 6 – 8.

We hope the KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.3, drop a comment below and we will get back to you at the earliest.

## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6

(Assume π = $$\frac{22}{7}$$ unless stated otherwise)

Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm3 = 1 l)
Solution:
The circumference of the base of cylindrical vessel, C = 132 cm.
height, h = 25 cm
C = 2πr = 132

r = 21 cm
∴ Volume of Cylinder, V = πr2h

= 34650 cm3
1000 cubic cm. = 1 litre
∴ 34650 cm3 = … ? …

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Solution:
Let the inner diameter of a cylindrical vessel be d1 and outer diameter be d2.

height of pipe, h = 35 cm.
∴ volume of pipe, V = $$\pi r_{2}^{2} h-\pi r_{1}^{2} h$$

= 110 × 52
∴ V = 5720 cm3
Mass of pipe of 1 ccm is 0.6 gm.
∴ Mass of 5720 cm3. … ? …
= 5720 × 0.6
= 3432 Kg.

Question 3.
A soft drink is available in two packs –
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much ?
Solution:
Length of rectangular tin, l = 5 cm,
height, h = 15 cm.
∴ Volume of rectangular tin, V = l × b × h
V = 5 × 4 × 15
= 300 cm3
Diameter of plastic cylinder, d = 7 cm,
∴ r = $$\frac{7}{2}$$
Volume of cylinder, v = $$\pi \mathrm{r}^{2} \mathrm{h}$$

= 11 × 7 × 5
= 385 cm3.
∴ Plastic cylinder’s volume is great.
= Volume of cylindrical tin – Volume of rectangular tin.
= 385 – 300
= 85 cm3.
∴ Volume of plastic tin is greater than rectangular tin by 85 cm3.

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(ii) its volume. (Use π = 3.14)
Solution:
Curved surface area of cylinder = 94.2 cm2
height, h = 5 cm.
r = ?
V = ?
(i) Curved Surface area of cylinder, L.S.A.
A= 2πrh
94.2 = 2 × 3.14 × 5 × r

= 3.01 cm.
= 3 cm.
(ii) Volume of cylinder, V = πr2h
= 3.14 × (3)2 × 5
= 3.14 × 9 × 5
= 141.3 cm3.

Question 5.
It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m2, find
(i) inner curved surface area of the vessel.
(iii) capacity of the vessel.
Solution:
(i) Cost of painting for cylindrical vessel = Rs. 2200
Cost of painting is at the rate of Rs. 20 per m2.
∴ Curved Surface area of vessel = $$\frac{2200}{20}$$
= 110 m2

(ii) Depth of vessel, h = 10 m
r = ?
Curved surface area of vessel, V = 2πrh

(iii) Capacity of the Vessel = Volume of Vessel
V = πr2h

V = 96.25 m3.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?
Solution:
Volume of a cylindrical vessel = 15.4 litres
1000 c.cm. = 1 litre.

height, h = 1 m,
Curved surface area =?
Volume of cylinder, V = πr2h
0. 0154 = $$\frac{22}{7}$$ × r2 × 1

r = 0.7 m
Now, curved surface area of cylinder vessel,

= 0.44 × 1.07
A = 0.4708 m2

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Solution:
Diameter of pencil, d = 7 mm = 0.7 m.

length of pencil, h = 14 cm
diameter of pencil, d = 1 mm = 0.1 cm

Volume of wood in pencil, V

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
Diameter of pencil, d = 7 cm

height of soup, h = 4 cm
Volume of soup given for 1 patient is 154 cm3
Volume of soup given for 250 patients .. ? ….
= 154 × 250 = 38500 cm3
For 1000 cm3, 1 litre
For 38500 cm3 ….. ? …
= $$\frac{38500}{1000}$$
= 38.5 litres soup.

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.2.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The blood groups of 30 students of Class VIII are recorded as follows :
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B. O
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
The blood groups of 30 students of class VIII are as follows

i) Out of 30 students only three students have blood group AB. This is rarest blood group.
ii) Most common blood group students are 12. Hence ‘O’ is the maximum blood group.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows :

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 – 5 (5 not included). What main features do you observe from this tabular representation ?
Solution:
Size of the class interval = 5
First class interval is 0 – 5 (5 is not included).

More number of people, i.e. 5 to 15 Km come.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:

(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about ?
(iii) What is the range of this data ?
Solution:
Size of class interval = 2
(i) Frequency distribution table :

(ii) Relative humidity is more, these data are taken during rainy season.
(iii) Maximum humidity = 99.2
Minimum humidity = 84.0
∴ Range = Maximum – Minimum
= 99.2 – 84.9
= 14.3

Question 4.
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows :

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table ?
Solution:
Maximum height : 150 cm.
Minimum height : 173 cm.
Size of class interval = 5
(i) Frequence distribution table for grouped data:

(ii) Referring to above table, 50% students have height less than 165 cm.
Height of majority students is 160 – 165 cm.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per millon ?
Solution:
(i) Grouped frequency distribution table:

(ii) In 8 days the concentration of sulphur dioxide is more than 0.11 parts per million.

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows :

Prepare a frequency distribution table for the data given above.
Solution:
Frequency distribution table :

Question 7.
The value of n upto 50 decimal places is given below :
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits ?
Solution:
Frequency distribution table :

(ii) Digits which have more frequency are 3 and 9.
‘0’ is the digit which has less frequency.

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows :

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.
(ii) How many children watched television for 15 or more hours a week ?
Solution:
(i) Size of Class Interval = 5

(ii) Number of children watched TV for 15 days or more hours a week is 2.

Question 9.
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows :

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Solution:
Size of the class interval = 0.5
Class interval : 2 – 2.5, 2.5 – 3 …….
Grouped Frequency distribution table:

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Statistics Ex 14.2 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Statistics Exercise 14.2, drop a comment below and we will get back to you at the earliest.

## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Solution:
Match box is in the form of cuboid.
Its length, l = 4 cm.
height, h = 1.5 cm.
∴ Volume of Cuboid, V = l × b × h
= 4 × 2.5 × 1.5
V = 15 cm3.
Volume of 1 match box is 15 cm3.
Volume of 12 match boxes …?… .
= 15 × 12
= 180 cm3.

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold ? (1 m3 = 1000 l)
Solution:
Length of cuboidal water tank, l = 6 m
height, h =4.5 m
∴ Volume of cuboidal water tank, V= l × b × h
= 6 × 5 × 4.5
V= 135 m3.
1 m3 = 1000 l.
135 m3 = ? = 135 × 1000
= 135000 lit.
∴ Number of litres of water tank = 1350000 litres.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid ?
Solution:
Volume of cuboidal vessel, V = 380 m3.
length, l = 10 m
breadth, b = 8 m height, h = ?
Volume of vessel, V= l × b × h
380 = 10 × 8 × h
380 = 80 h

∴ h = 4.75 m.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3.
Solution:
Length of cuboidal pit, l = 8 m
breadth, b = 6 m height, h = 3 m
∴ Volume of pit, V = l × b × h
= 8 × 6 × 3
= 144 m3.
∴ Cost of digging a cuboidal 1 m3 is Rs. 30
Cost of digging a cuboidal pit for 144 m3 … ? …
= 144 × 30
= Rs. 4320.

Question 5.
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution:
Volume of cuboidal tank, V = 50000 lit.
1 c.c = 1000 litres
∴ Volume of tank = $$\frac{50000}{1000}$$ = 50 c.c
length of tank, l = 2.5 m
height, h = 10m
Volume of cuboid, V = l × b × h

∴ b = 2m.

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last ?
Solution:
Length of cuboid water tank, l = 20 m
height, h = 6 m
Volume, V = ?
Volume of water tank, V = l × b × h
= 20 × 15 × 6
= 1800 m3.
1m3 = 1000 litres
∴ 1800m3 = 1800 x 1000
= 1800000 litres.
Water required daily per man= 150 litres
∴ Quantity of water required for 4000 men?
= 4000 × 150
= 600000 litres.
Quantity of water for 1 day = 600000
Number of days for consuming 1800000 litres … ? …

= 3 Days.

Question 7.
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown
Solution:
Length of rectangular godown, l1 = 40 m
height, h1 = 10 m
∴ Volume of godown, V= l1 × b1 × h1
= 40 × 25 × 10
= 10000 m3.
Length of wooden crate, l2 = 1.5 m
height, h2 = 0.5 m
∴ Volume of wooden crate. V = l2 × b2 x h2
= 1.5 × 1.25 × 0.5
= 0.9375 m3.
For 0.9375 m3, 1 crate

= 10666.66

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Each side of a cube, a = 12 cm.
∴ Volume of cube, V = a3
= (12)3
= 1728 cm3.
Solid cube is cut into 8 cubes of equal volume.
Volume of each small cube,

∴ l3 = 216
∴ l = 6 cm.
∴ Each side of small cube is 6 cm.
Ratio of outer area :

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution:
Depth of river, h = 3m
1 Km. = 1000 m
∴ 2 Km. = 2000 m.
Speed of the river per hour is 2 Km.
For 60 minutes, 2000m
For 1 minute …?… = $$\frac{2000}{60}$$

∴ Depth of river in 1 minute, l = $$\frac{100}{3}$$ m.
Volume of water flows in 1 minute,
V = l × b × h
= $$\frac{100}{3}$$ × 40 × 3
= 4000 m3
∴ In One minute, 4000 m3 water reaches the sea.

## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4

(Assume π = $$\frac{22}{7}$$, unless stated otherwise)

Question 1.
Find the surface area of a sphere of radius :
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm.
Solution:

Surface area of the sphere = $$4 \pi r^{2}$$

= 22 × 3 × 21
= 1386 cm2.

(ii) r = 5.6 cm. = $$\frac{56}{10}$$cm.
Surface area of the sphere = $$4 \pi r^{2}$$

= 394.24 cm2.

(iii) r = 14 cm.
Surface area of the sphere = $$4 \pi r^{2}$$

= 88 × 28
= 2464 cm2.

Question 2.
Find the surface area of a sphere of diameter :
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m.
Solution:
(i) Diameter, d = 14 cm.

Surface area of a sphere = $$4 \pi r^{2}$$

= 616 cm2.
(ii) Diameter, d = 21 cm.

Surface area of a sphere = $$4 \pi r^{2}$$

= 1386 cm2.
(iii) Diameter, d = 3.5 m. = $$\frac{7}{2}$$

Surface area of a sphere = $$4 \pi r^{2}$$

= 38.5 m2.

Question 3.
Find the total surface area of a hemisphere of radius 10 cm.
(Use π = 3.14 )
Solution:
r = 10 cm.
Total surface area of a hemisphere = $$3 \pi r^{2}$$

= 942 cm2.

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Radius of spherical balloons be r1 and r2, r1 = 7 cm. and r2 = 14 cm.
Ratio of surface area

∴ Ratio of surface areas of the balloons = 1 : 4.

Question 5.
A hemispherical bowl made of brass has an inner diameter of 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2.
Solution:
Inner diameter of hemispherical bowl = 10.5 cm.

Curved surface area of the bowl = $$2 \pi r^{2}$$

= 173.25 cm2.
Cost of tin-plating for 100 cm2 is Rs. 16.
Cost of tin-plating for 173.25 sq.cm. …?…

= Rs. 27.72.

Question 6.
Find the radius of a sphere whose surface area is 154 cm2.
Solution:
Total surface area of a sphere= 154 cm2

∴ r = 3.5 cm

Question 7.
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Let the diameter of earth be d’ unit
∴ Diameter of Moon = $$\frac{d}{4}$$
radius of the earth = $$\frac{\mathrm{d}}{2}$$
radius of the moon = $$\frac{1}{2} \times \frac{\mathrm{d}}{4}=\frac{\mathrm{d}}{8}$$
Ratio of surface area,

= 1 : 16

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Thickness of the hemispherical bowl = 0.25 cm.
inner radius, r = 5 cm.
∴ Outer Surfce area= 5 + 0.25 = 5.25 cm.
Outer Curved Surface area = 2πr2

= 173.25 cm2.

Question 9.
A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder
(iii) ratio of the areas obtained in (i) & (ii).

Solution:
(i) If radius of a sphere is ‘r’ cm, its Surface area = 4πr2
(ii) height of cylinder, h = diameter of sphere
h = r + r
∴ h = 2r
∴ Curved surface area of cylinder = 2πrh
= 2πr × h
= 2πr × 2r
= 4πr2

= 1 : 1

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.1.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.1

Question 1.
Give five examples of data that you can collect from your day-to-day life.
Some of the examples of data that we can collect from our day-to-day life are as follows :

1. Number of TV viewers in the city.
2. Number of Colleges in the city.
3. Number of sugar factories in the city.
4. Measuring the height of students in the classroom.
5. Number of children below 15 years in India.

Question 2.
classify the data in Q.1 above as primary or secondary data.