## Kerala State Syllabus 9th Standard Maths Solutions Chapter 12 Proportion

### Proportion Textual Questions and Answers

Textbook Page No. 182

Question 1.
A person invests 10000 rupees and 15000 rupees in two different schemes. After one year, he got 900 rupees as interest for the first amount and 1200 rupees as interest for the second amount.
i. Are the interests proportional to the investments?
ii. What is the ratio of the interest to the amount invested in the first scheme? What about the second?.
iii. What is the annual rate of interest in the first scheme? And in the second?
i. The ratio between the amounts invested = 10000:15000 = 2 : 3
The ratio between the interests = 900 : 1200 = 3 : 4
Since the ratios are different. So, interests will be not proportional to the investments.

ii. The ratio between the amount invested and interest in scheme 1
= 10000 : 900= 100 : 9
The ratio between the amount invested and interest in scheme 2
= 15000 : 1200 = 25 : 2

iii. Rate of interest in the first scheme
= 900/10000 × 100 = 9%
Rate of interest in the second scheme =
1200/15000 × 100 = 8%

Question 2.
The area of A0 paper is one square metre. Calculate the lengths of the sides of A4 paper correct to a millimetre, using a calculator.
The area of A1 paper is half of the area of A0 paper. The area of A2 paper is half of the area of A1 paper. Let con-sider the area of the A0 paper be 1 square metre, the area of A1 paper is
1/2 square metre , the area of A2 paper is 1/4 square metre , the area of A3 paper is 1/8 square metre , and the area of A4 paper is 1/16 square metre ,
If the length of A4 paper is √2 x and breadth x
length of A4 paper : breadth = √2 : 1
Area of A4 paper = length x breadth
√2x × x = √2x2
This is 1/16 m2, therefore √2 x2 = 1/16

Question 3.
In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10 : 3: 12 . When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is it calcium carbonate?
The ratio between calcium, carbon and oxygen in calcium carbonate = 10 : 3: 12 The ratio of calcium, carbon and oxygen in the given compound = 60 : 20: 70 = 6 : 2: 7
Since this ratio is not equal to the ratio of calcium, carbon and oxygen in calcium carbonate, it is not calcium carbonate.

Textbook Page No. 185

Question 1.
For each pair of quantities given below, check whether the first is proportional to the second. For proportional quantities, calculate the constant of proportionality.
i. Perimeter and radius of circles.
ii. Area and radius of circles.
iii. The distance travelled and the number of rotations of a circular ring moving along a line.
iv. The interest got in a year and the amount deposited in a scheme in which interest is compounded annually.
v. The volume of water poured into a hollow prism and the height of the water level.
i. The perimeter of a circle is n times its diameter. That is 2n times the radius.
∴Perimeter = 2πr
∴ The perimeter and radius are proportional.
The constant of proportionality is 2π

ii. The area of a circle is π times the square of the radius.
∴ Area = πr2
∴ Area and radius are not proportional.

iii. When the ring rotates once, the distance travelled is equal to its perimeter. When it rotates twice the distance travelled is twice its perimeter. When it rotates ‘n’ times, the distance travelled is ‘n ‘ times the perimeter of the ring.

iv. If the amount deposited is P and the rate of interest is R.
Annual interest = I = PNR,
I = P × R (N = 1)
The amount and interest are propor-tional.
Constant of proportionality is rate of interest, R.

v. Volume of a hollow prism
= base × height
Volume, of water and height of water level, are proportional. Constant of proportionality is the base area.

Question 2.
During rainfall, the volume of water falling in each square metre may be considered equal.
i. Prove that the volume of water falling in a region is proportional to the area of the region.
ii. Explain why the heights of rainwater collected in different sized hollow prisms kept near one another are equal.
i. The volume of water falling in each square metre are equal.
Let the volume of the rain falling on the 1 square metre be k The volume of the rain falling on the 2 square metre = 2k
The volume of the rain falling on the 3 square metre = 3k
The volume of the rain falling on the x square metre y = kx Here x and y are proportional.

ii. Let x be the base area of the vessel and h be the height of the water collected, then the volume of water y = xh.
y/x =h
The volume of water collected is different in hollow prisms having different base area.

But $$\frac { volume }{ Area }$$ is always equal to the height of water level. So the height at which rainwater collected is same.

Question 3.
When a weight is suspended by a spring, the extension is proportional to the weight. Explain how this can be used to mark weights on a spring balance.

Mark the pointer when no weight is hanging on it. Then mark the point when a constant weight is hanging on it. For example when we hang 1 kg on it a 2 cm extension is made. So we mark 1, 2, 3, 4 from the points marked first, i.e.,
2, 4, 6 and 8.
These distances are divided into 10 equal parts, then we can mark the points 1.1 kg and 2.4 kg etc.

Use this online balance the equation calculator to convert the unbalanced equation to a balanced chemical equation that is not beneficial for productivity.

Question 4.
In the angle shown below, for different points on the slanted line, as the distance from the vertex of the angle changes, the height from the horizontal line also changes.

i. In the picture, perpendicular to the horizontal lines are drawn from points on the slanted line.
Δ ABP, Δ ACQ, Δ ADRand Δ AES are similar triangles. (Because these are a right-angled triangle with is common to all)
∴ Ratio of their sides are equal. That is sides are proportional.

Let k be the constant of proportionality
PB = k × AB, QC = k × AC
RD = k × AD, SE = k × AE ,
i.e., the change in height is proportional to the distance.

The common ratio calculator of a geometric sequence, divide any two consecutive terms of the sequence.

Textbook Page No. 189

Question 1.
i. Prove that for equilateral triangles, area is proportional to the square of the length of aside. What is the constant of proportionality?
ii For squares, is area proportional to square of the length of a side? If so, what is the constant of proportionality?
i. Δ ABC is an equilateral triangle
Consider their sides are ‘a’,
Draw a perpendicular line CD to AB
from C. In right-angled triangle ADC, according to the Pythagoras theorem AD2 + DC2 = AC2

ii. If x be the one side of a square then its area y = x2
Area of the square is proportional to the square of their sides.
Constant of proportionality = 1

Question 2.
In rectangles of area one square metre, as the length of one side changes, so does the length of the other side. Write the relation between the lengths as an algebraic equation. How do we say this in the language of proportions?
Area of the rectangle is the sum product of length and breadth.
Let × be the length and y be the breadth, then
x × y = 1 m2

x = 1/y = m
Example when y = 3 m
x = 1/3 meter
when y = 4 m
x = 1/4 meter
The algebraic equation is x = 1/y
i.e., length of rectangle is proportional to the reciprocal of the breadth, i.e., length of rectangle is proportional to its breadth.

Question 3.
In triangles of the same area, how do we say the relation between the length of the longest side and the length of the perpendicular from the opposite vertex? What if we take the length of the shortest side instead?
Let ‘a’ be the large side, h be the length of perpendicular from opposite vertices, ‘A’ be the area, then

Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.
i.e., The length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

Question 4.
In regular polygons, what is the relation between the number of sides and the degree measure of an outer angle? Can it be stated in terms of proportion?
The sum of the exterior angles of all polygon is 360°. If ‘n’ is the number of sides.
Measure of an exterior angle = $$\frac{\text { Sum of exterior angles }}{\text { No. of sides }}$$
One outer angle = 360/n
(n = number of sides)
If the measure of an exterior angle is ‘x’.
x= $$\frac{1}{n} \times 360^{0}$$
One outer angle and number of sides are inversely proportional. The constant of proportionality is 1/n.

Question 5.
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.
i. The rate of water flow and the height of the water level.
ii The rate of water flow and the time taken to fill the tank.
i. Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then
x = Ay
Height of the water level in the tank is proportional to the rate of the water flowing.

ii. If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in ‘t’ second is given by
C = V × t
$$V=\frac{C}{t}=C \times \frac{1}{t}$$
That is the rate of water flow and the time taken for filling the tank are inversely proportional. C is the constant of proportionality.

### Proportion Exam oriented Questions and Answers

Question 1.
the weight of 6 spheres of same size made of the same metal is 14 kg. When 9 more spheres are added the weight is 35 kg. Check whether the number of spheres and their weights are proportional.
Weight of 6 spheres = 14kg
Ratio of number and weight = 6 : 14 = 3 : 7
Total number of sphers are added = 15
Total weight = 35kg
Ration of number and weight = 15 : 35 = 3 : 7
Since the ratios are equal, the number of sphers and their weights are propor-tional.

Question 2.
Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Ratio of investments = 60000 : 100000 = 6 : 10 = 3 : 5
Ratio of profit divided
1800 : 3000 = 18 : 30 = 3 : 5
Ratio of investments and Ratio of profit divided are equal. Hence they are proportional.

Question 3.
The two sides of a triangle having perimeter 10 m are 2$$\frac { 1 }{ 2 }$$ m and 3 $$\frac { 1 }{ 2 }$$ m.
What is the ratio of the length of the three sides of triangles?
Perimeter of triangle = 10 m
First side = 3$$\frac { 1 }{ 2 }$$ m = $$\frac { 5 }{ 2 }$$ m
Second side = 3$$\frac { 1 }{ 2 }$$ m = $$\frac { 7 }{ 2 }$$ m
Third side = 4 m Ratio of three sides of a triangle
$$=\frac{5}{2}: \frac{7}{2}: 4 = 5: 7: 8$$

Question 4.
150 litres of water is flowing through a pipe in 6 minutes. If 200 litres of water flows through it in 8 minutes check whether the quantity of water and time of flow are proportional ?
Quantity of water flowing in 6 minutes = 150 litres
Ratio between the quantity of water and timeofflow= 150: 6 = 25: 1
Quantity of water flowing in 8 minutes = 200 litres
Ratio between quantity of water and times of flow = 200 : 8 = 25: 1
Since the radios are equal, the amount of water flowing and the time of flow are proportional.

Question 5.
Unniyappam was made using 1 kg rice, 250 g plantain and 750g jaggery. Find the ratio between the ingredients.
Rice = 1
kg= 1000g
Plantain = 250 g, Jaggery = 750 g.
Ratio between the ingredients
= 1000 : 250 : 750 = 4 : 1 : 3

Question 6.
Sathyan got Rs. 500 after working for 6 hours. Gopi got Rs. 400 after working for 4$$\frac { 1 }{ 2 }$$ hours. Are the wages obtained proportional to the working time ?
Ratio of working hours = 6 : 4 $$\frac { 1 }{ 2 }$$
= 12 : 9 = 4 : 3
Ratio of wages = 800 : 600 = 8 : 6 = 4 : 3
Since the ratios are equal, the working hours are proportional to the wages.

Question 7.
Are the length and breadth of a square having same perimeter inversely proportional?
for a square the perimeter is 20 cm. so,
Let x be the length and y be the breadth then possible values of x and y are
x — y
9 — 1
8 — 2
7 — 3
Here xy is not a constant term i.e. the changes in the x and y is not in the firm of xy = kept
hence length and breadth are not in inversely proportional.

Question 8.
The weight of an object having mass 5 kg is 49 Newton. The weight of another object having mass 15 kg is 147 Newton. Check whether the mass and weight are proportional ? What is the constant of proportionality? What is the weight of an object having mass 8 kg ?
Weight of 5 kg object = 49 N
Ratio between mass and weight = 5 : 49
Weight of 15 kg massed object = 147N
Ratio between mass and weight = 15:147 = 5:49
Mass and weight are proportional.
Weight of 8 kg massed object = 8 × 9.8 = 78.4N

Constant of proportionality = 9.8

Question 9.
The perimeter of a triangle is 60 cm. Sides are in the ratio 3: 4: 5. Then j find the length of the sides.
Ratio of sides = 3 : 4 : 5
Therefore, the sides are the $$\frac { 3 }{ 12 }$$, $$\frac { 4 }{ 12 }$$ and $$\frac { 5 }{ 12 }$$ part of the perimeter .
The perimeter is 60 cm, then the length of sides are

$$60 \times \frac{5}{12}=25 \mathrm{cm}$$

Question 10.
Will the dye which contains 10 L blue colour and 15 L white colour and another dye which contain 12L blue colour and 17 L white colour have j the same colour? Why?
In the first dye , blue colour: white colour = 10 : 15 = 2 : 3
In the second dye, blue colour: white colour = 12: 17
The ratios are not same. Hence the both will not have same colour.

Question 11.
The face perimeter of some vessels in the shape of square prisms are equal. 12 litres of water can be filled in the vessel having height 15 cm. 16 litres of water can be filled in the vessel having height 20 cm. Check whether the volumes of the vessel and height are proportional. What is the constant of proportionality?
Since the face perimeters are equal,
Volume = face perimeter × height
Volume of the vessel having height 15 cm = 12 litres.
Ratio of height to volume =15 : 12 = 5 : 4 Volume of the vessel having height 20 cm = 16 litres
Ratio of height to volume =20 : 16 = 5 : 4 Since the ratios are equal, height and volume are proportional.
$$\frac { volume }{ height }$$ = $$\frac { 4 }{ 5 }$$ constant of proportionality
Volume of the vessel having height 35 cm
= 35 × $$\frac { 4 }{ 5 }$$ = 28 liters.

Question 12.
The ratio of carbon, sulphur and potassium nitrate to make gun powder is 3 : 2: 1. How much quantity of each is required to make 1.2 kg of gun powder ?

Question 13.
Raju got Rs. 400 after working for 8 hours. Damu worked for 6 hours and got Rs. 300. Are the wages obtained proportional to the work time?
Ramu working hours = 8 hour
Wages Raju obtained = Rs. 400
Damu working hours = 6 hour
Wages obtained by Damu = Rs. 300

∴ Wages obtained are proportional to the work time.

Question 14.
150 L of water flows through a pipe for 6 minutes. 200 L of water flows for 8 minutes through the same pipe. Are the time and amount of water flowing proportional?
Ratio of quantity of water
= 150 : 200 = 15 : 20 = 3 : 4
Ratio oftime =6 : 8 = 3 : 4
The ratios are same , hence they are proportional.

Question 15.
A car with 5L of petrol travels a distance of 75 km. What is the proportionality constant between the distance travelled and the quantity of petrol? How much petrol is needed for travelling 180 km?
Taking the distance travelled as x and quantity of petrol as y, then the constant of proportionality

The quantity of petrol needed to travel 180 km= 12 litre

Question 16.
The angles of a triangle are in the ratio 1 : 3: 5. How much is each angle of the triangle?
Sum of the angles of a triangle = 180°
The angles of the traingle are $$\frac { 1 }{ 9 }$$, $$\frac { 3 }{ 9 }$$ and $$\frac { 5 }{ 9 }$$
parts of 180° , hence the angles are

## Kerala Syllabus 9th Standard Physics Solutions Chapter 3 Motion and Laws of Motion

You can Download Motion and Laws of Motion Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Physics Solutions Part 1 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 9th Standard Physics Solutions Chapter 3 Motion and Laws of Motion

### Motion and Laws of Motion Textual Questions and Answers

Question 1.
What will be the result when a man tries to move a vehicle by pushing it, standing inside the vehicle? The vehicle moves/the vehicle doesn’t move.
The vehicle doesn’t move.

Question 2.
What if the same vehicle is pushed from outside.
Yes, the vehicle moves
Internal forces can not move an object, but only an external unbalanced force can cause motion.

Newton’S First Law Of Motion

Every object continues in its state of rest or of uniform motion along a straight line unless an unbalanced external force acts on it. This is Newton’s First law of motion.

Question 3.
What is inertia of rest?
Inertia of rest is the tendency of a body to remain in its state of rest or its inability to change its state of rest by itself.

Question 4.
Write down an activity to show inertia of rest.

Place a bottle filled with water on a thick rough pa¬per as shown. Pull the paper suddenly to one side. The bottle falls in the opposite direction fo the motion of paper. This is because of inertia of rest.

Question 5.
What is inertia of motion?
Inertia of motion is defined as the inability of a body to change it state of motion by itself.

Question 6.
Write down an activity to show inertia of motion.

Place a bottle on a thick paper with rough surface. Bring the bottle into motion by pulling the paper slowly. Gradually increase the speed of pulling stop pulling when the bottle gains a certain speed. The bottle falls in the direction of motion of the pa-per, due to inertia of motion of the bottle.

Mass Transfer MCQ PDF arranged chapterwise! Start practicing now for exams, online tests.

Question 7.
Find out reasons for the Situations
a) Place some carom board coins in a pile. Using the striker, strike out the coin at the bottom. What do you observe? What is the reason?
b) When a running bus is suddenly stopped, passengers, standing in the bus show a tendency to fall forward.
c) Place a small brick on a plank. When the plank
is pulled suddenly the brick remains in the same position as before.
d) When a bus moves forward suddenly from rest, the standing passengers tend to fall backward.
e) Accidents that happen to passengers who do not wear seat belts are more fatal.
a) Only the coin at the bottom is thrown away. Others will remain in the previous stage.
b) The passengers start to move forward due to the tendency to continue in its state of motion.
c) It is due to the tendency to continue in the state of rest.
d) The passengers tend to fall backward due to the tendency to continue in their state of rest.
e) The passengers inside the vehicle have the tendency to move forward due to inertia of motion.
Inertia: The inability of a body to change its state of rest or of uniform motion along a straight line by its itself.
Inertia of rest: Inertia of rest is defined as the inability of a body to change its state of rest by itself.
Inertia of motion : Inertia of motion is defined as the inability of a body to change its state of motion by itself. Eg. When a running bus is suddenly stopped, the standing passengers fall forward due to inertia of motion. When a bus at rest starts suddenly, the standing passengers fall backward due to inertia of rest.

Question 8.
Expand the table by finding more situations from daily life.

 Inertia of rest Inertia of motion 1. When the branch of a mango tree is shaken mangoes fall. 1. A running athlete cannot stop himself abruptly at the finishing point. 2. When the carpet is trapped, dust particle scatters. 2. A man stepping down from a slowly moving bus stops after few steps of running. 3. When the coin on a Cardboard placed over a glass is struck, it falls into the glass 3. In hammer throw, before the hammer is let go off, it is whirled along a circular path.

Question 9.
Find out the reasons
a) An athlete doing a long jump-start his run from a distance.
b) A running elephant cannot change its direction suddenly.
a) This activity helps to cover long distances by utilizing inertia of motion.
b) Mass of elephant is greater. So inertia of motion also be greater. Also, it cannot be able to change its direction suddenly.

Mass And Inertia

Question 10.
It is dangerous for loaded vehicles to negotiate a curve in the road without reducing speed. What is the reason?
Loaded vehicles possess more inertia of motion. As ” mass increases, inertia also increases.

Question 11.
It is more difficult to roll a filled tar drum than an empty drum.
a) Which of two has a greater mass?
b) Which has greater inertia?
a) The mass of drum filled with tar will be greater,
b) Inertia will be greater to tar filled drum
Conclusion: Inertia depends on mass. When the mass increases inertia also increases. When the mass decreases inertia also decreases.

Question 12.
If a tennis ball (mass 58.5 g) and a cricket ball (mass 163 g) are the reach a certain distance when hit with a cricket bat, which is to be hit with greater force? The tennis ball / the cricket ball?
The cricket ball

Question 13.
Will the change of velocity be the same in both the cases?
Velocities are different in both cases due to the difference in masses.

The inertia of an object depends upon its mass. When the mass increases, inertia also increases.

Momentum

Momentum is a characteristic property of moving objects. It is measured as the product of the mass of the body and its velocity.

Momentum = mass × velocity
Unit of momentum is kg m/s

Question 14.
A car of 1000 kg moves with a velocity of 10 m/s. On applying brakes it comes to rest in 5s. Then what are its initial momentum and final momentum?
m = 1000 kg
u = 10 m/s
v = 0
t = 5s
Initial momentum = mu
= 1000 × 10 =10000 kg m/s
Final momentum = mV
= 1000 × 0 = 0 Kg m/s

Question 15.
A hockey ball of mass 200 g hits on a hockey stick with a velocity 10 m/s. Calculate the change in momentum if the ball bounces back on the same path with the same speed.
m = 200g = 200/1000 = 0.2 kg
Initial momentum = mu
= 0.2 × 10 = 2 kg m/s
Final momentum = mv
= 0.2 × 70 = 2 kgm/s
change in momentum = mv – mv = – 2 – 2
= -4 kg m/s

Question 16.
A loaded lorry of mass 12000 kg moves with a velocity of 12 m/s. Its velocity becomes 10 m/s after 5 s.
a) What is the initial momentum and what is the final momentum?
b) What is the change in momentum?
c) What is the rate of change of momentum?
a) Initial momentum = mu
=12000 × 12 = 144000 kg m/s Final momentum = mv
=12000 × 10 = 120000 kg m/s
b) Change in momentum = mv – mu
= 120000 – 144000 = –24000 kg m/s
c) Rate of change in momentum = $$\frac { mv – mu }{ t }$$
= $$\frac { -24000 }{ 5 }$$
= – 4800 kgm/s2

Newton’S Second Law Of Motion

The rate of change of momentum of a body is directly proportional to the unbalanced external force acting on it.

Equation for Calculating Force:
According to second law of motion F ∝ ma
F = kma (k – a constant)
k = 1 ∴ F = 1 × ma
F = ma,

ie. F- Force, m – mass, a – acceleration.
Unit of force is Newton (N)
Another unit is dyne.

Question 17.
A constant force is applied for 2 s on a body of mass 5 kg. As a result, if the velocity of the body is changed from 3 m/s to 7 m/s, find out the value of the applied force.

Question 18.
A car moving with a speed of 108 km/h comes to rest after 4s on applying brake, if the mass of the car including the passengers is 1000 kg, what will be the force applied when brake is applied?
Initial velocity of the car u = 108 km/h

= 30 m/s
Final velocity v = 0
Mass m = 1000 kg
Time t = 4s
According to newton’s second law
F = ma

= – 7500 N
The negative sign indicates that the applied force is opposite to the direction of motion.

Question 19.
Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2s. Find out the force applied. If the duration for which force acts is extended to 5s, what will be the velocity of the object then?
u = 3 m/s
v = 7 m/s
t = 2s
m = 5 kg
According to Second Law of Motion F = ma

If we substitute the values in the equation v = u + at, velocity can be calcualted when the time for force is extended to 5s.
v = 3 + (2 × 5) = 13 m/s

Question 20.
Velocity-time graph of an object of mass 20 g, moving along the surface of a long table is given below.

What is the frictional force experienced by the object?
From the graph
Initial velocity u = 20 m/s
Final velocity v = 0m/s
t = 10 s
m = 20
g = 20/100 kg
F = ma

= -0.04 N
The negative sign shows that the frictional force is acting opposite to the direction of motion of the object.

Question 21.
m1 and m2 are the masses of two bodies. When a force of 5 N is applied on each body, m1 gets an acceleration of 10 m/s2 and m2, 20 m/s2. If the two bodies are tied together and the same force is applied, find the acceleration of the combined system.

Impulse

Impulsive force is a very large force acting for a very short time.
Impulse of force is the product of the force and the time.
Impulse = Force × time
Impulse – Momentum Principle
According to Newton’s second law of motion,
$$F=\frac{m(v-u)}{t}$$
F x t = m(v – u)
F × t= mv – mu
ie. impulse = change Is momentum This is known as impulse-momentum principle. It states that a change in momentum of an object is equal to the impulse experienced by it.

Question 22.
Explain the following situation by relating force and time.
When the change in momentum is a constant, the force acting on a body will be inversely proportional to the time taken. As time increase, the force acting decreases and as time decreases, the force acting increases.

Question 23.
During a pole vault jump, the impact is reduced by falling on a foam bed.
When the change in momentum is a constant, the force acting on a body will be inversely proportional to the time taken. As time increase, the force acting decreases and as time decreases, the force acting increases.

Question 24.
Hay or sponges are used while packing glassware. This helps to avoid breaking of glasswares due to collision.
When the change in momentum is a constant, the force acting on a body will be inversely proportional to the time taken. As time increase, the force acting decreases and as time decreases, the force acting increases.

Question 25.
Karate experts move their hands with great speed to chop strong bricks.
When the change in momentum is a constant, the force acting on a body will be inversely proportional to the time taken. As time increase, the force acting decreases and as time decreases, the force acting increases.

Newton’S Third Law Of Motion

Pass a long string through a straw and tie the string between two windows of the classroom Paste an inflated balloon on the straw.

Question 26.
Inflate a balloon and release it suddenly. What happens?
Release of air causes the balloon to move in the opposite direction. Releasing air from the balloon is action and the movement of the balloon is reaction.

Question 27.
What do you observe?
The cork pushed out due to the pressure of steam. The tube moves backward.

Question 28.
If the action is the cork being pushed out due to the force exerted by the steam on it, what is the reaction?
The reaction is the backward movement of boiling tube.

Question 29.
Write down the action and reaction while we are walking on a floor?
When we are walking through a floor, we applies a force on the floor. This is action. The floor applies a force in the opposite direction. This is reaction.
Rocket Propulsion:

• Escaping of hot gases from the jet of rocket is action.
• The force exerted by these gases on the rocket is reaction.

Question 30.
Are the action and reaction equal and opposite?
Action and reaction are equal and opposite.
Newton’s Third law of Motion For every action there is an equal and opposite reaction.

Question 31.
Examine the following situations and complete the table.

 Situation Action (FJ Reaction (FJ 1. A man jumps from a boat to ’ the shore. The man exerts a force on the boat. The boat moves backward. The boat exerts an equal force on the man. The man lands on the shore. 2. A bullet is fired from a gun the bullet exerts a force on the gun So it moves backward The gun exerts an equal force to the bullet, so it moves forward 3. A boat is rowed. The man applied force on the water The boat moves forward.

Conclusions:

• As a result of the applied force by a second body to a body, reaction will occur at the second body.
• Action and reaction are equal and opposite.
• Since action and reaction takes place in two different bodies, they do not cancel each other.

Law Of Conservation Of Momentum

Observe the figure and answer the following questions.

Question 32.
Move the first marble slightly back and roll forward. What happens?
One marble will be thrown off from the other end and reaches the previous position.

Question 33.
Bring the two marbles into contact and let them roll. What happens?
Two marbles will move from the other end.

Observe the figure and answer the following questions.

Question 34.
Total momentum before collision = ………….
m1 u1 + m2 u2.

Question 35.
Total momentum after collision =…….
m1 v1 + m2 v2

Question 36.
Initial momentum of A=
m1 u1

Question 37.
Final momentum of A =
m1 v1

Question 38.
Change in momentum of A=
m1 v1 – m1 u1

Question 39.
Rate of change of momentum of A =
$$\frac{m_{1} v_{1}-m_{1} u_{1}}{t}$$

Question 40.
Initial momentum of B =……
m2 u2

Question 41.
Final momentum of = …………….
m2 v2

Question 42.
Change in momentum of B = ………
m2 v2 – m2 u2.

Question 43.
Rate of change of momentum of B = ……………
$$\frac{\mathrm{m}_{2} \mathrm{v}_{2}-\mathrm{m}_{2} \mathrm{u}_{2}}{\mathrm{t}}$$

According to IInd Law of motion, rate of change of momentum is directly proportional to the external force.
Force exerted by B on A

Law Of Conservation Of Momentum

In the absence of an external force, the total momentum of a system is a constant.

Question 44.
A bullet is fired with a velocity v from a gun of mass M. What will be the recoil velocity of the gun if the mass of the bullet is m?
According to low of Conservation of Momentum, total momentum ofthe gun and the bullet before firing and their total momentum after firing will be equal,
ie. Total momentum before firing = 0 + 0 =0
Total momentum after firing = MV + mv
Accoriding to Law of Conservation of Momentum 0 =MV + mv
MV = -mv
V = $$\frac { -mv }{ M }$$
The recoil velocity ofthe gun V = $$\frac { -mv }{ M }$$. The negative
sign indicates that the gun moves in the opposite direction of motion ofthe bullet.

Question 45.
Suppose a child of mass 40 kg running on a horizontal surface with a velocity of 5m/s jumps on a stationary skating board of mass3 kg while running as shown in the figure. If there is no other force acting horizontally (assuming the frictional force on the wheels to be zero), calculate the velocity of the combined system of child and the skating board.
Suppose the velocity of the board while moving is u. Total momentum of the child and the skating board before jumping will be
= 40 kg × 5 m/s + 3 kg × 0 m/s
= 200 kg m/s Total momentum when the system starts moving (body and skating board)
= (40 + 3) kg × u m/s
= 43 × u kgm/s.
According to the Law of Conservation of Momentum, Total momentum remains constant.
43 u = 200
u = 200/43 = + 4.65 m/s

Circular Motion

Motion of a body through a circular path is known as circular motion.

Question 46.
Does the velocity of an object moving with a uniform speed along a circular path change?
Yes

Question 47.
How does this change in velocity happen?
Due to change in speed/change in direction/due to change in both speed and direction.
Due to change in direction.

• Whirling of a stone tied to a string is a type of circular motion.
• The force we apply form the center of the circle reaches the object through the string. The acceleration which a body in circular motion experiences towards the center of the circle through the radius is centripetal acceleration. The force that creates centripetal acceleration is a centripetal force.
Centripetal force (Fc) = mv2/t
m – mass, v-velocity, r- radius of the circular path
• In the absence of centripetal force, circular moving body thrown off through the tangent.
• If a body moving along a circular path covers equal distances in equal intervals of time, it is said to be in uniform circular motion.

Question 48.
In hammer throw, before the hammer is let go off, why is it whirled around along a circular path?
It is to get initial momentum. Also helpful to cover long distances through the tangent.

Question 49.
How does the speed of a giant wheel in an amusement park?
The motion of giant wheel is controlled by mechanically. So its speed is uniform except when starting and stopping.
Examples for uniform circular motion:

• Motion of needles in a watch.
• Whirling of a stone tied to a string.
• Movement of the leaf of an electric fan except when starting and stopping.
• Circular motion of earth’s artificial /satellites.

The acceleration experienced by an object in a circular motion, along the radius, towards the center of the circle, is known as centripetal acceleration. The force that creates a centripetal acceleration is called centripetal force. Centripetal acceleration and centripetal force are directed towards the center.
If an object moving along a circular path covers equal distances in equal intervals of time, it is said to be in uniform circular motion.

Let Us Assess

Question 1.
Observe the figures given below. Answer the following questions.

a) When the card is suddenly struck off, what happens to the coin? Explain.
b) What is the law to which this property is related?
c) How is this property related to the mass of the object?
a) The coin falls into the glass due to inertia of rest.
b) Newton’s 1st law of motion.
c) Mass increases inertia increases.

Question 2.
What are the balanced forces acting on a book at rest on a table?
a) Downward force exerts by the book on the table (i.e., weight of the book).
b) Upward force applies by the table on the book (reaction).

Question 3.
To remove the dust from a carpet, it is suspended and hit with a stick. What is the scientific principle behind it?
Inertia of rest. The dust in the carpet shows the tendency to continue in its state of rest.

Question 4.
Acar and a bus are traveling with the same velocity. Which has greater momentum? Why?
Bus. Because when mass increases momentum increases.

Question 5.
On the basis of Newton’s third law of motion, explain the source of force that helps to propel a rocket upward.

• It is due to the reaction force by the escaping gas from rocket.
• Escaping gases form the rocket is action.
• The force exerted by escaping gases on the rocket is reaction.

Question 6.
A car-travels with a velocity of 15 m/s. The total mass of the car and the passengers in it is 1000 kg. Find the momentum of the car.
Momentum p = mv
= 1000 × 15
= 15000 kg m/s

Question 7.
Give reasons:
a) When a bullet is fired from a gun, the gun recoils.
b) When a bus at rest suddenly moves forward, the passengers, standing in the bus, fall backward.
c) We slip on a mossy surface.
a) The gun recoils due to the reaction force applied by the shot to the gun. Forward movement of the shot is action and the Backward movement of the gun is reaction.
b) Inertia of rest is the reason. The passengers tends continue in state of rest.
c) Absence of reaction force is the cause for this.

Extended Activities

Question 1.
Prepare and present an experiment to illustrate inertia of rest.
Make a pile of coins on a table. Strikes off the lowest coin by a knife quickly. Only that particular coin thrown off and the others remains in the previous manner. This is due to the tendency of coins to remain in its state of rest.

Question 2.
Find out situations from our daily life to explain the law of conservation of momentum and note them down.

• Recoil of a gun during firing.
• Rocket propulsion.
• Bomb explosion.

### Motion and Laws of Motion More Questions and Answers

Question 1.
Fill in the blanks.
a) As the time interval decreases, rate of change of momentum
b) The force required to produce an acceleration of 1 m/s2 on a body of mass 1 kg is
a) Increases
b) 1 Newton (1N)

Question 2.
Correct the statement if any wrong.
An object moving with uniform speed along a circular path undergoes velocity change due to the change in speed.
An object moving with a uniform speed along a circular path undergoes velocity change due to change in direction.

Question 4.
a) What is meant by momentum?
b) Write down the equation for calculating momentum.
c) A body of mass 50 kg starts from rest. If its velocity changes to 15 m/s after 10 seconds, calculate the change in momentum? What will be the rate of change of momentum?
a) The measurement of the quantity of motion of a body is called momentum
b) Momentum p = mv

Question 5.
a) Define circular motion and uniform circular motion.
b) What is the relation between centripetal force and centripetal acceleration?
a) The motion of an object through a circular path is said to be circular motion.
If an object moving along a circular path covers equal distance in equal intervals of time if is said to be uniform circular motion.
b) The force requires to produce centripetal acceleration is centripetal force.

Question 6.
a) State Newton’s II law of motion?
b) Derive the equation for force on the basis of this law.
a) The rate of change of momentum of a body is directly proportional to the unbalanced external force acting on it.
b) According to II law.

F = kma (k – a constant)
The force required to produce an acceleration of 1 m/s2 on a body of mass 1 kg is 1N
1 = k × 1 × 1
k = 1
∴ F = 1 × ma
F = ma

Question 7.
A car of mass 1000 kg runs with a velocity of 10 m/s. What is the momentum of this car?
Momentum p = mv
m = 1000 kg
v = 10 m/s
∴ P = 1000 × 10
= 10000 kg m/s

Question 8.
A loaded lorry of mass 1500 kg moves with a velocity of 12 m/s. Within a small interval of the time, the velocity becomes 10 m/s.
a) What is the initial momentum of the lorry?
b) What is its final momentum?
c) What is the change in momentum?
a) Initial momentum P = mv, m = 1500 kg
= 1500 × 12, v1 = 12m/s
= 18000 kg m/s, v2 = 10 m/s
b) Final momentum p = mv2
= 1500 × 10 = 15000 kg m/s
c) Change in momentum = mv2 – mv1
=15000 – 18000
= – 3000 kg m/s
Rate of change of momentum
= $$\frac { change in momentum }{ Time }$$
= $$\frac { m(v – u) }{ t }$$

## Kerala State Syllabus 9th Standard Maths Solutions Chapter 3 Pairs of Equations

### Kerala Syllabus 9th Standard Maths Pairs of Equations Text Book Questions and Answers

Textbook Page No. 36

Do each problem below either in your head, or using an equation with one letter, or two equations with two letters:

Question 1.
In a rectangle of perimeter one metre, one side is five centimetres longer than the other. What are the lengths of the sides?
Shortest side = x
Longest side = x + 5
Perimeter = 1 m = 100 cm
2(x + x + 5) = 100
2x + 5 = 50; 2x = 45;
x = 22.5
∴ Shortest side = 22.5
Longest side = 22.5 + 5 = 27.5

Question 2.
A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of the boys. How many girls and boys are there in the class?
Number of boys = x
Number of girls = x + 4
2(x – 8) = x + 4
2x – 16 = x + 4
2x – x = 4 +16; x = 20
∴ Number of boys = 20
Number of girls = 24

Question 3.
A man invested 10000 rupees, split into two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?
If one part is x then
the remaining part is 10000 – x
$$x\times \frac { 8 }{ 100 } +\left( 10000-x \right) \times \frac { 9 }{ 100 } =100$$
8x + 90000 – 9x = 87500
90000 – 87500 = x
2500 = x
one part = 2500 and
remaining part = 7500

System of Linear Equations Calculator … This calculator will solve the system of linear equations.

Question 4.
A three and a half metre long rod is to be cut into two pieces, one piece is to be bent into a square and the other into an equilateral triangle. The length of their sides must be the same. How should it be cut?
Total length = 3½ m
Since the sides of a square and equilateral triangle are equal, all these 7 sides are equal.
∴ Length of one side
$$3\frac { 1 }{ 2 } \div 7=\frac { 7 }{ 2 } \div 7=\frac { 1 }{ 2 }$$m
Length of the rod for the square
$$= 4\times \frac { 1 }{ 2 }$$ = 2m
Length of the rod for the equilateral triangle = $$3\times \frac { 1 }{ 2 }$$ = $$1\frac { 1 }{ 2 }$$m

Question 5.
The distance travelled in t seconds by an object starting with a speed of u metres/second and moving along a straight line with speed increasing at the rate of a metres/second every second is given by ut + $$\frac { 1 }{ 2 }$$ at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does its speed change?
If t = 2
ut + $$\frac { 1 }{ 2 }$$at²= 10
2u + 2a= 10
u + a = 5 — (1)
If t = 4
4u + 8a = 28
u + 2a = 7 — (2)
from (1) and (2)
a = 2
∴ u = 3

Textbook Page No. 40

Question 1.
Raju bought seven notebooks of two hundred pages and five of hundred pages, for 107 rupees. Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees. What is the price of each kind of notebook?
Cost of 200 page note book = x
Cost of 100 page note book = y
7x + 5y= 107 …………(1)
5x + 7y = 97 …………(2)
(1) × 5 $$\Rightarrow$$ 35x + 25y = 535 …………(3)
(2) × 7 $$\Rightarrow$$ 35x + 49y = 679 …………(4)
(4) – (3) $$\Rightarrow$$ 24y = 144
y = $$\frac{144}{24}$$ = 6
Substitute y = 6 in equation (1)
7x + 30 = 107; 7x = 77
x = $$\frac{77}{7}$$ = 11
Price of the 200 pages notebook = Rs. 11
Price of the 100 pages notebook = Rs. 6

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Question 2.
Four times a number and three times number added together make 43. Two times the second number, subtracted from three times the first give 11. What are the numbers?
Let the first number = x and
the second number = y
4x + 3y = 43 …………(1)
3x – 2y = 11 …………(2)
(1) × 3 $$\Rightarrow$$ 12x + 9y= 129 …………(3)
(2) × 4 $$\Rightarrow$$ 12x – 8y = 44 …………(4)
(3) -(4) $$\Rightarrow$$ 17y = 85; y = $$\frac{85}{17}$$ = 5
Substitute y = 5 in equation (1)
4x + 3y = 43
4x + 15 = 43
4x = 43 – 15 = 28
∴ x = $$\frac{28}{4}$$ = 7, y = 5
First number = 7
Second number = 5

Question 3.
The sum of the digits of two – digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
If the numbers are x and y
x + y = 11 …………(1)
10x + y + 27 = 10y + x
9x – 9y = -27
X – y = -3 …………(2)
(1) + (2) 2x = 8; x = 4
x + y = 11
4 + y = 11
y = 7
∴ Required number is 47.

Question 4.
Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be double. What are their ages now?
Ramu’s present age = x
Rahim’s present age = y
4 years back,
Ramu’s age = x – 4
Rahim’s age = y – 4
3(x – 4) = y – 4
3x – 12 = y – 4
3x – y = 8 ……….(1)
After 2 years,
Ramu’s age = x + 2
Rahim’s age = y + 2
2(x + 2) = y + 2
2x + 4 = y + 2
2x – y = -2 ……….(2)
(1) – (2) $$\Rightarrow$$ x = 10
3x – y = 8; 30 – y = 8; y = 22
x = 10, y = 22
Ramu’s present age = 10
Rahim’s present age = 22

Question 5.
If the length of a rectangle is in-creased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?
length = x; breadth = y
(x + 5)(y – 3) = xy – 5
xy – 3x + 5y – 15 = xy – 5
– 3x + 5y = + 10
3x – 5y = -10 ………..(1)
(x + 3)(y + 2) = xy + 50
xy + 2x + 3y + 6 = xy + 50
2x + 3y = 44 ………..(2)
(2) × 1 $$\Rightarrow$$ 6x-10y = -20 ……….(3)
(3) × 2 $$\Rightarrow$$ 6x + 9y = 132 …………(4)
(3)- (4) $$\Rightarrow$$ -19y = -152
y = $$\frac{-152}{-19}$$ = 8, 2x + 3y = 44
2x + 24 = 44; 2x = 20; x = 10
∴ x = 10, y = 8
Length of the rectangle = 10 m
Breadth of the rectangle = 8m

Textbook Page No. 42

Question 1.
A 10 metre long rope is to be cut into two pieces and a square is to be made using each. The difference in the areas enclosed must be 1$$\frac{1}{4}$$ square metres. How should it be cut?
Length of one piece = x m
Length of other piece = (10 – x) m

∴ Rope is divided into 6 m and 4 m.

Question 2.
The length of a rectangle is 1 metre more than its breadth. Its area is 3$$\frac{3}{4}$$ square metres. What are its length and breadth?
Length = x
x = y + 1; x – y = 1
$$x y = 3 \frac{3}{4} =\frac{15}{4}$$
(x + y)² = (x – y)² + 4xy
1² + 4 × $$\frac{15}{4}$$ = 1 + 15 = 16
x – y = 1; x + y = 4
2x = 5; x = 5/2 = 2.5
y=1.5
∴ Length = 2.5 m

Question 3.
The hypotenuse of a right triangle is 6$$\frac{1}{2}$$ centimetres and its area is 7$$\frac{1}{2}$$ square centimetres. Calculate the lengths of its perpendicular sides.
The perpendicular sides are x and y. Given that,

From (3) and (4)
2x = 24/2 = 12
∴ x = 6
6 – y = 7/2
∴ y = 5/2 = 2.5
∴ Perpendicular sides = 6 and 2.5

### Kerala Syllabus 9th Standard Maths Pairs of Equations Exam Oriented Text Book Questions and Answers

Question 1.
There are some oranges in a bag. When 10 oranges more added in the bag; the numbers become 3 times of the oranges initially had. Then how many oranges were there in the bag initially.
Let the number of oranges initially taken = x
X + …….. = 3X
3X – X = ……..
2X = ……..
X = ……… /2 = ……..
x + 10 = 3x; 3x – x = 10 ; 2x = 10;
x = $$\frac{10}{2}$$ = 5

Question 2.
A box contains some white balls and some black balls. The number of black balls is 8 more than the number of white balls. The total number of balls is 4 times the number of white balls. Find out the number of white balls and the number of black balls.
Number of white balls = x
Number of black balls = ……… + 8
Total number of balls = ……. ‘ x;
i. e. (x) + (x + 8) = ………. x;
2x + 8 = ……. x;
8 = …… x – 2x
= …… x; x = 8/ …….
white balls = ……….
black balls = ……… + 8 = ……..
Number of white balls = x
Number of black balls = x + 8
Total number of balls = 4 ‘ x;
(x) + (x + 8) = 4x ; 2x + 8 = 4x;
8 = 4x – 2x = 2x ; x = $$\frac{8}{2}$$ = 4
white balls = 4
black balls = 4 + 8 = 12

Question 3.
The sum of two numbers is 36 and the difference is 8. Find the numbers.
Let x, y be the numbers
x + y = 36
x – y = 8
(x + y) + (x – y) = ……. + ……
2x = …..
x = …… /2 = ……..
x – y = 8
……. – y = 8
…….. – 8 = y
x + y = 36; x – y = 8
(x + y) + (x – y) = 36 + 8
2x = 44; x = $$\frac{44}{2}$$ = 22
x – y = 8; 22 – y = 8;
22 – 8 = y; y = 14
numbers 22, 14

Question 4.
The cost of 2 pencils and 5 pens is Rs 17, two pencils and 3 pens of the same rate is Rs 11. Find out the prices of a pencil and a pen.
Let the price of pencil = x;
Price of pen = y;
∴ 2x + 5y = …….. 2x +….. y = …..
2x + …… y = 11
(2x + 5y) (-2x + ….. y) = -11
…… y = …….
y = $$\frac{…..}{……}$$
2x + 5x ….. = 17;
2x = 17 – ……..; x = ……. /2; = ……..
2x + 5y = 17; 2x + 3y = 11;
(2x + 5y) – (2x + 3y) = 17 – 11; 2y = 6
y = $$\frac{6}{2}$$ = 3; 2x + 5 × 3 = 17;
2x = 17 – 15;
x = $$\frac{2}{2}$$ = 1
Price of pencil = Rs 1
Price of pen = Rs 3

Question 5.
Twice of a number added with thrice of another number gives 23. Four times the first number and 5 times the second number when added gives 41. Find out the numbers.
First number = x
Second number = y
∴ 2x + 3y = ……
4x + 5y = ……
2(2x + 3y) = 2 …….
4x + 6y = …….
(4x + 6y) – (4x + 5y) = ( …… ) – ( ….. )
y = …….; 2x + 3x ……. = …….
2x = ( …… ) – ( ……. ); x = ………./2
2x + 3 y = 23
4x + 5y = 41
2(2x + 3y) = 2 × 23; 4x + 6y = 46
(4x + 6y) – (4x + 5y) = 46 – 41;
y = 5
2x + 3 × 5 = 23
2x = 23 – 15;
x = $$\frac{8}{2}$$ = 4
Price of pencil = Rs 4
Price of pen = Rs 5

Question 6.
Rama spends Rs 97 to buy 4 two hundred page note books and 5 hundred page note books. Geetha spends Rs 101 to buy 5 two hundred page note book and 4 one hundred page note books. What is the prices of two types of note books?
Let the cost of two hundred page note books = x
The cost of one hundred page note books = y
(1) 4x + 5y = 97
(1) × 5 $$\Rightarrow$$ 20x + ……..y = ……..
(2) 5x + 4y = 101
(2) × 4 $$\Rightarrow$$ 20x + …….y = …….
……y = ( ….. ) – ( ……. );
y = …….. /……..
4x + 5x( ……. ) = 97
4x = 97 – ( ……. )
x = ……../4
(1) 4x + 5y = 97
(2) 5x + 4y = 101
(1) × 5 → 20x + 25y = 485
(2) × 4 → 20x + 16y = 404
9y = 485 – 404
y = $$\frac{81}{9}$$ = 9
4x + 5 × 9 = 97
4x = 97 – 45 = 52
Cost of two hundred page note book = Rs 13
Cost of one hundred page note book = Rs 9

Question 7.
6 years back the age of Muneer was 3 times the age of Mujeeb. After 4 years the age of Muneer becomes twice the age of Mujeeb. Find the age of two of them now.
Age of Mujeeb 6 years back = x
Age of Muneer 6 years back = 3x
After 4 years
3x + 4 = 2(x + 4)
3x + 4 = 2x + 8
3x – 2x = 8 – 4;
x = 4
Age of Mujeeb 6 years back = 4 + 6 = 10
Age of Muneer 6 years back = 3(4 + 6) = 18 years.

Question 8.
The cost of 4 chairs and 5 tables is Rs 6600 and the cost of 5 chairs and 3 tables is Rs 5000 at the same prices. What are the prices of a table and a chair?
Cost of a chair = Rs a
Cost of a table = Rs b
4a + 5b = 6600 ¾ …………(1)
5a + 3b = 5000 ¾ …………(2)
(1) × 5 → 20a + 25b = 33000
(2) × 4 → 20a + 12b = 20000
(20a + 25b) – (20a + 12b) = 33000 – 20000
(1) $$\Rightarrow$$ 13b = 13000
b = $$\frac{13000}{13}$$ = Rs 1000
4a + 5b = 6600;
4a + 5 × 1000 = 6600
4a = 6600 – 5000 = 1600
a = $$\frac{1600}{4}$$ = Rs 400
Cost of a table = Rs 1000
Cost of a chair = Rs 400

## Kerala State Syllabus 9th Standard Maths Solutions Chapter 8 Polynomials

### Polynomials Textual Questions and Answers

Textbook Page No. 123

Question 1.
In rectangles with one side 1 centimetre shorter than the other, take the length of the shorter side as x centimetres.
i. Taking their perimeters as p(x) centimetres, write the relation between p(x) and x as an equation,
ii. Taking their areas as a(x) square centimetres, write the relation between a(x) and x as an equation.
iii Calculate p(l), p(2), p(3), p(4), p(5). Do you see any pattern?
iv. Calculate a(l), a(2), a(3), a(4), a(5). Do you see any pattern?
Let x be the shorter side, then the other side will be (x+ 1).
i. Perimeter = 2[x + (x + 1)] = 2(2x + 1) = 4x + 2
That is, p(x) = 4x + 2

ii. Area = x(x + 1)
a(x) = x2 + x
Area, a(x) = x2 + x

iii. p(x) = 4x + 2
p(1) = 4 × 1 + 2 = 6
p(2) = 4 × 2 + 2 = 10
p(3) = 4 × 3 + 2= 14
p(4) = 4 × 4 + 2= 18
p(5) = 4 × 5 + 2 = 22
Perimeter is a sequence increasing by 4.

iv. a(x) = x2 + x
a(1) = 12 + 1 = 2 = 1 × 2 = 2
a(2) = 22 + 2 = 6 = 2 × 3 = 6
a(3) = 32 + 3 = 12 = 3 × 4 = 12
a(4) = 42 + 4 = 20 = 4 × 5 = 20
a(5) = 52 + 5 = 30 = 5 × 6 = 30
Area is the product of x and the number one more than x.

Question 2.
From the four corners of a rect-angle, small squares are cut off and the sides are folded up to make a box, as shown below:

i. Taking a side of the square as x centimetres, write the dimensions of the box in terms of x.
ii. Taking the volume of the box as vfojcubic centimetres, write the relation between v(x) and x as an equation.
iii. Calculate $$\mathrm{V}\left(\frac{1}{2}\right) \quad, \mathrm{V}(1), \quad \mathrm{V}\left(1 \frac{1}{2}\right)$$
If 1cm is the length of the small squares they are cut off, then the length of the maked box by folding it up = 7 – 1 – 1 = 5 cm Width = 5 – 1 – 1 = 3 cm, Height = 1 cm
If 2 cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – 2 – 2 = 3 cm Width = 5 – 2 – 2 = 1 cm, Height = 2cm
i. If x cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – x – x = (7 – 2x) cm
Width = 5 – x – x = (5 – 2x) cm
Height = x cm

ii. If volume of the box be v(x), then
v(x) = Length × Width × Height v(x) = (7 – 2x) (5 – 2x)x cm3

v (1) = (7 – 2 × 1) (5 – 2 × 1) 1 = (7 – 2) (5 – 2) × 1 = 5 × 3 × 1 = 15

Question 3.
Consider all rectangles that can be made with a 1-metre long rope. Take one of its sides as x centimetres and the area enclosed as a(x) square centimetres.
i. Write the relation between a(x) and x as an equation.
ii. Why are the numbers a(10) and a(40) equal?
iii. To get the same number as a(x), for two different numbers as x, what must be the relation between the numbers?
i. 1 m = 100 cm
If one side is x cm, then the other side is 50 – x.
Area of rectangle = a(x) = x(50 – x)
= 50x – x2 cm2
a(x) = 50x – x2

ii. a(10) = 50 × 10 – 102 = 500 – 100 = 400
a(40) = 50 × 40 – 402 = 2000 – 1600 = 400
10 and 40 are the sides of rectangle, so a(10) and a(40) are same .

iii. Numbers must be sides of rectangle. If we add the two numbers together we get the sum as half the length of the wire (50 cm) used to make it.

Textbook Page No. 126

Question 1.
Write each of the relations below in algebra and see if it gives a polynomial. Also, give reasons for your conclusion.
i. A 1-metre wide path goes around a square ground. The relation between the length of a side of the ground and the area of the path.

ii A liquid contains 7 litres of water and 3 litres of acid. More acid is added to it. The relation between the amount of acid added and the change in the percentage of acid in the liquid.

iii. Two poles of heights 3 metres and 4metres are erected upright on the ground, 5 metres apart. A rope is to be stretched from the top of one pole to some point on the ground and from there to the top of the other pole:

The relation between the distance of the point on the ground from the foot of a pole and the total length of the rope.
i. Let the side of the ground be x then the side of the square including the path is x + 2 metres.
Area of path = Area of large square – Area of small square.
a(x) = (x + 2)2 – x2 = x2 + 4x +4 – x2
= 4x + 4
Here a(x) is a polynomial. Here x is multiplied by 4 and 4 is added to it.

ii. Ratio of acid in the first fluid = 3/10 litre that is 30%
If x litres of acid is added to it, then change in the amount of acid is
$$=\frac{3+x}{10+x}$$
Change in the percentage of acid is = $$\frac{3+x}{10+x} \times 100 \%$$
$$\mathrm{b}(\mathrm{x})=\frac{300+100 x}{10+x} \%$$

iii.

In figure, AB = x m, BD = 5 – x
Total length of the wire = BC + BE

It is not a polynomial since it involves square root of variable x.

Factor by Sum and Difference of cubes calculator – find factors of polynomials using Sum and Difference of cubes, step-by-step online.

Question 2.
Write each of the operations below as an algebraic expression, find out which are polynomials and explain why.
i. Sum of number and it’s reciprocal
ii. Sum of a number and its square root.
iii. Product of the sum and difference of a number and its square root.
i. Let the number be x, then the reciprocal is 1/x
sum = $$x+\frac{1}{x}$$
This is not a polynomial, because here the operation of reciprocal is involved.

ii. Let the number be x, then the square root is √x
sum = x + √x
This not a polynomial because here the square root is taken.

iii. Let x be the number
(x + √x) (x – √x) = x2– x
This is a polynomial.

Question 3.
Find p(1) and p(10) in the following polynomials,
i. p(x) = 2x + 5
ii. p(x) = 3x2 + 6x + 1
iii. p(x) = 4x3 + 2x2 + 3x + 7
i. p(x) = 2x + 5
p(1) = 2 × 1 + 5 = 2 + 5 = 7
p(10) = 2 × 10 + 5 = 20 + 5 = 25

ii. p(x) = 3x2 + 6x + 1
p(1) = 3 × 12 + 6 x 1 + 1
= 3 + 6 + 1 = 10
p(10) = 3 × 102 + 6 × 10 + 1 = 300 + 60 + 1 = 361

iii. p(x) = 4x3 + 2x2 + 3x + 7
p(1) = 4 × 13 + 2 × 12 + 3 × 1 +7
=4 + 2 + 3 + 7 = 16
p(10) = 4 × 103 + 2 × 102 + 3 × 10 + 7
= 4000 + 200 + 30 + 7 = 4237

Is this a Polynomial Calculator is a free online tool that displays the addition, subtraction, multiplication, and division of two polynomials.

Question 4.
Find p(0), p(1) and p(-1) in the fol-lowing polynomials,
i. p(x) = 3x + 5
ii. p(x) = 3x2 + 6x + 1
iii. p(x) = 2x2 – 3x + 4
iv. p(x) = 4x3 + 2×2 + 3x + 7
v. p(x) = 5x3 – x2 + 2x – 3
i. p(x) = 3x + 5
p(0) = 3 × 0 + 5 = 5
p(1) = 3 × 1 + 5 = 8
p(-1) = 3 x-1 + 5 = 2

ii. p(x) = 3x2 + 6x + 1
p(0) = 3 × 02 + 6 × 0 + 1 = 1
p(1) = 3 × 12 + 6 × 1 + 1 = 10
p(-1) = 3 × (-1)2 + 6 × (-1) + 1 = -2

iii. p(x) = 2x2 – 3x + 4
p(0) = 2 × 02 – 3 × 0 + 4 = 4
p(1) = 2 × 12 – 3 × 1 + 4 = 3
p(-1) =2 × (-1)2 – 3 × (-1) + 4 = 9

iv. p(x) = 4x3 + 2x2 + 3x + 7
p(0) = 4 × 03 + 2 × 02 + 3 × 0 +7 = 7
p(1) = 4 × 13 + 2 × 12 + 3 × 1 + 7 = 16
p(-1) = 4 × (-1)3 + 2 × (-1)2 + 3 × (-1) +7 =2

v. p(x) = 5x3 – x2 + 2x – 3
p(0) = 5 × 03 – 02 + 2 × 0 – 3 =-3
p(1) = 5 × 13 – 12 + 2 × 1 – 3 = 3
p(-1) = 5 × (-1)3 – (-1)2 + 2 × (-1) – 3 = -11

Question 5.
Find polynomials p(x) satisfying
each set of conditions below.
i. First degree polynomials with p(1) = 1 and p(2) = 3
ii. First degree polynomials with p(1) = -1 and p(-2) = 3
iii. Second degree polynomials with p(0) = 0, p(1) = 2 and p(2) = 6.
iv. Three different second degree polynomials with p(0) = 0 and p(1) = 2.
i. General form of a first degree poly-nomial is
p(x) = ax + b
Let p(1) = 1, then a × 1 + b = l
a + b = 1 ………. (1)
Let p(2) = 3, then a × 2 + b = 3
2a + b = 3 …….. (2)
(1) × 2, 2a + 2b = 2 …… (3)
(3) – (2), b = -1
From (1), a + -1 = 1,
a = 1 + 1 = 2 Polynomial p(x) = 2x – 1

ii. General form of a first degree poly-nomial is p(x) = ax + b
Let p(1) = -1, then a × 1 + b = -l
a + b = -1 ……….. (1)
Let p(-2) =3 , then a × (-2) + b = 3
-2a + b = 3 ………. (2)
(1) × 2, 2a + 2b = -2 ………. (3)
(2) + (3), 3b = 1, b = 1/3
From (1), a = 1/3 = -1

iii. General form of a second degree polynomial is
p(x) = ax2 + bx + c
Letp(0) = 0, then a × 02 + b × 0 + c = 0
0a + 0b + c = 0.
c = 0 (1)
Letp(1) = 2 ,then a × 12 + b × 1 + c = 2
a + b + 0 = 2
a + b = 2 ………. (2)
Letp(2)= 6, then a × 22 + b × 2 + c = 6
4a + 2b = 6
2a + b = 3 (3)
(3) – (2), a = 1
From (2), 1 + b = 2, b = 2 – 1 = 1
Polynomial p(x) = x2 + x

iv. General form of a second degree polynomial is p(x) = ax2 + bx + c
Letp(0) = 0, then a x 0 + b x 0 + c = 0
0 + 0 + c = 0
c = 0
Letp(1) = 2, then a × 12 + b × 1 + c = 2
a + b + c = 2
a + b = 2
Selecting a and b such that a + b = 2 will give different polynomials.
a = 1, b = 1
a= 3, b =-l
a=4, b =-2
Three different second degree poly¬nomials are
p(x) = x2 + x
p(x) = 3x2 – x
p(x) = 4x2 – 2x

### Polynomials Exam Oriented Questions and Answers

Question 1.
In a polynomial p(x) = 2x3 + ax2 – 7x + b.
p(1) = 3, p(2) = 19. Then find the value of ‘a’ and ‘b’?
p(x) = 2x3 + ax2 – 7x + b
We have p(1) = 3.
p(1) = 2 x 13 + a x 12 – 7 x 1 + b = 3
2 + a – 7 + b = 3
a + b – 5 = 3
a + b =8 ………. (1)
We have p(2) = 19.

p(2 ) = 2 x 23 + a x 22– 7 x 2 + b = 19
16 + 4a – 14 + b = 19
4a + b + 2 = 19
4a + b = 17…….. (2)
From equation (1), (2)
(2) – (1) 4a + b = 17
$$\frac{a+b=8}{3 a=9}$$
a = 9/3 = 3
From equation (1)
a + b = 8
3+b =8
b =5

Question 2.
In a polynomial p(x) = 2x3 + 9x2 + kx + 3, p(-2) = p(-3). Find the value of k.
p(x) = 2x3 + 9x2 + kx + 3
p(-2) = 2(-2)3 + 9(-2)2 + k(-2) + 3
= -16 + 36 – 2k + 3 = 23 – 2k
p(-3) = 2(-3)3 + 9(-3)2 + k(-3) + 3
= -54 + 81 – 3k + 3 = 30 – 3k
We have p(-2) = p(-3),
23 -2k = 30 -3k
-2k + 3k = 30 – 23
k = 7

Question 3.
From the polynomial p(x) = 2x – 3x +1, find p(0), p(1) and p(-1).
p(x) = 2x2 – 3x +1
p(0) = 2(0)2 – 3(0) + 1 = 1
p(1) = 2(1)2 – 3(1) +1 = 2 – 3 + 1 = 0
p(-1) = 2(-1)2 – 3(-1) + 1 = 2 + 3 + 1 = 6

Question 4.
In a polynomial p(x) = 2x3 – 7x2 + kx + 20,
p(2) = p(3).
a. Find the value of k.
b. Using the value of k, write the polynomial.
c. Find p(1).
a. p(2) = 2 × 22 – 7 × 22 + k × 2 + 20
= 16 – 28 + 2k + 20 = 8+ 2k
p(3) = 2 × 33 – 7 × 32+ k × 3 + 20
= 54 – 63 + 3k + 20 = 11 +3k
P(2) = P(3)
8 + 2k = 11 + 3k
k = -3

b. p(x) = 2x3 – 7x2 -3x + 20

c. p(1) = 2 – 7 – 3 + 20 = 12

Question 5.
Simplify the followi ng
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
ii. (3x + 4)2 – (2x – 1) (3x + 4)
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
= (3x + 4) [(2x + 1 + 4x + 3)]
= (3x + 4)[6x + 4]
= (3x x (6x) + (3x) x (4)+ 4x(6x) + 4 x 4
= 18x2 + 12x + 24x + 16
= 18x2 + 36x + 16

ii. (3x + 4)2 – (2x – 1) (3x + 4)
= (3x + 4) [3x + 4 – (2x – 1)]
= (3x + 4)[3x + 4 – 2x + 1] = (3x + 4) (x + 5)
= 3x2 + 15x + 4x + 20
= 3x2 + 19x + 20

Question 6.
7x3 – 4x2 – x + 4 is a polynomial.
a. Write the terms of the polynomial.
b. Write the coefficient of x2.
c. Write the constant terms of the polynomial.
d. What is the degree of the polynomial ?
a. Terms = 7x2, -4x2, -x, 4
b. Coefficient of x2 = -4
c. Constant term = 4
d. Degree of the polynomial = 3

Question 7.
In the polynomial p(x)=3x2 – ax + 1,
Find ‘a’ satisfying p(1) = 2.
p(x)=3x2 – ax + 1
p(1) = 3(1)2 – (a × 1) + 1 = 3 – a + 1 =4 – a
Given p(1) = 2
That is, 4 – a = 2 , a = 4 – 2 = 2

Question 8.
If p(x)= x3 + 2x2 – 3x + 1 and q(x) = x3 – 2x2 + 3x + 5
a) Find p(x) + q(x). What is its de-gree?
b) Find p(x) – q(x). What is its de-gree?
a) p(x) = x3 + 2x2 – 3x + 1,
q(x) = x3 – 2x2 + 3x + 5
p(x) + q(x)
= x3 + 2x2 – 3x + 1 + x3 – 2x2 + 3x + 5
= 2x3 + 6
Degree = 3

b) p(x) – q(x)
= x3 + 2x2 – 3x + 1 – (x3 – 2x2 + 3x + 5)
= x3 + 2x2 – 3x + 1 -x3 + 2x2 – 3x – 5
= 4x2 – 6x – 4
Degree = 2

Question 9.
A right-angled triangle of perpendicular sides 3 cm and 4 cm are ex-tended equally then get another large right-angled triangle. Write the algebraic form of the hypotenuse of the large right-angled triangle.
Hypotenuse 2 = base 2 + height 2
(According to pythagorus theorem)
Let the perpendicular side be x cm Length of other sides = 3 + x cm, 4 + x cm

Question 10.
In the polynomial p(x)= 3x2 – 4x + 7, check whether p(1) + p(2) = p(3) and P(2) × p(3) = p(6).
p(x) = 3x2 – 4x + 7
p(1) = 3(1)2 – 4(1) + 7 = 3 – 4 + 7 = 6
p(2) = 3(2)2 -4(2) + 7 = 12 – 8 + 7 = 11
p(3) = 3(3)2 – 4(3) +7 = 27 – 12 + 7 = 22
p(l) + p(2) ≠ p(3)
p(6) = 3(6)2 -4(6) + 7 = 108 – 24 + 7 = 91
P(2) × p(3) ≠ p(6)

Question 11.
In the polynomial p(x)= 2x2 + ax2 – 7x + b,
p(1) = 3 and p(2) = 19. Find a and b.
p(x)=2x2 + ax2 – 7x + b
p(l)=2 x 13+ a x 12 – 7 x 1 +b
=2 + a – 7 + b =a + b + 2 – 7 = a + b – 5
p(1) = 3, a + b – 5 = 3
a + b = 8……… (1)
p(2) = 2 x 23 + a x 22 – 7 x 2 + b
= 16 + 4a – 14 + b = 4a + b + 2
p(x) = 19
4a + b + 2 = 19
4a + b = 17 ……….. (2)
(2) – (1)
3a = 9, a = 3 a + b – 5 = 3
From equation (1),
a + b = 8
3 + b = 8
b = 5

Question 12.
If p(x) = x2 + 3x + 1 and q(x) = 2x – 4, then
i. Find the degree of the polyno-mial p(x) q(x).
ii. If the degree of p(x) × r(x) is 5, then find the degree of the polynomial r(x) .
iii. If p(x) is a third degree poly-nomial and q(x) is a fourth de-gree polynomial then find the degree of p(x) × q(x).
i. p(x) = x2 + 3x + 1 , q(x) = 2x – 4
p(x) × q(x) = (x2 + 3x + 1) (2x – 4)
The degree of p(x) x q(x) is 3.

ii. p(x) = x2 + 3x + 1 is a second degree polynomial.
If the degree of the polynomial p(x) × r(x) is 5, then p(x) × r(x) must have a term of x5.
x2 × x3 = x5
So, r(x) is a third-degree polynomial.

iii. If p(x) is a third-degree polynomial, then p(x) must have a term of x3.
q(x) is a fourth-degree polynomial, then q(x) must have a term of x4
Then in p(x) × q(x), must have a term of x3 × x4= x7
So p(x) × q(x) is a seventh-degree polynomial. In general,
If p(x) is an mth degree polynomial and q(x) is an n,h degree polynomial then p(x) × q(x) is an (m + n),h degree polynomial.

Question 13.
If p(x) = 4x2 + 3x + 5 and q(x) = 3x2 – x – 7, then find p(x) + q(x) and p(x) – q(x).
p(x) + q(x) = (4x2 + 3x + 5) + (3x2 – x – 7)
= 4x2 + 3x2 + 3x – x + 5 – 7
= 7x2 + 2x – 2
p(x) – q(x) = (4x2 + 3x + 5) – (3x2 – x – 7)
= 4x2 + 3x + 5 – 3x2 + x + 7
= x2 + 4x + 12

Question 14.
In polynomial p(x) = ax3+ bx2 + cx + d, p(l) = p(-l). Prove that a + c = 0.
p(x) = ax3 + bx2 + cx + d
p(l) = a(1)3 + b(1)2 + c(1) + d
= a+ b + c + d
p(-1) = a(-1)3 + b(-1)2 +c(-1) + d
= -a + b – c + d
p(1) = p(-1) is given ie, a + b + c + d = -a + b – c + d
2a + 2c = 0; 2(a + c ) = 0;
ie, a + c = 0

Question 15.
If we divide a polynomial by (x-2) we get quotient as x2 + 1 and remainder as 5. Find the polynomial.
Polynomial = (x2 + 1) (x – 2) + 5
= x3 + x – 2x2 – 2 + 5 ,
= x3 – 2x2 + x + 3

## Kerala State Syllabus 9th Standard Maths Solutions Chapter 13 Statistics

### Statistics Textual Questions and Answers

Textbook Page No. 193

Plus Two Teacher Salary Kerala Question 1.
The weight of 6 players in a volleyball team are all different and the average weight is 60 kilograms.
i. Prove that the team has at least one player weighing more than 60 kilograms.
ii. Prove that the team has at least one player weighing less than 60 kilograms.
i. Total weight of 6 players = 60 × 6 = 360 kg
The team contains players having weights 60, less than 60 or greater than 60. If the weights of all the players are less than 60, the average will also be less than 60. This is not possible. Therefore there will be at least one player having weight greater than 60.

ii. If the weight of all the players are greater than 60, the average will also be greater than 60. Therefore there will be at least one player having weight less than 60.

Statistics 9th Class Question 2.
Find two sets of 6 numbers with average 60, satisfying each of the conditions below:
i. 4 of the numbers are less than 60 and 2 of them greater than 60.
ii. 4 of the numbers are greater than 60 and 2 of them less than 60.
Total sum = 60 × 6 = 360
i. 20, 30, 40, 50, 100, 120

ii. 5, 15, 70, 80, 90, 100
Other ways are also possible.

Chapter 13 Maths Class 9 Question 3.
The table shows the children in a class, sorted according to the marks they got for a math test.

 Marks Children 2 1 3 2 4 5 5 4 6 6 7 11 8 10 9 4 10 2

Calculate the average marks of the class.
Total number of children is 45. Repeated addition can be written as multiplication.

Average mark = $$\frac { Total }{ Number }$$ = $$\frac { 297 }{ 45 }$$ = 6.6

10th Maths Solution Question 4.
The table below shows the days in a month sorted according to the amount of rainfall in a locality

 Rainfall(mm) Days 54 3 56 5 58 6 55 3 50 2 47 4 44 5 41 2

What is the average rainfall per day during this month?

 Rainfall(mm) Days Total 54 3 54 × 3 = 162 56 5 56 ×5 = 280 58 6 58 × 6 = 348 55 3 55 × 3 = 165 50 2 50 × 2=100 47 4 47 × 4 =188 44 5 44 × 5 = 220 41 2 41 × 2 = 82 Total 30 1545

The average of the rain fall per day during that month = $$\frac { Total rain fall }{ Number of days }$$
= $$\frac { 1545 }{ 30 }$$ = 51.5mm

Solutions Maths Question 5.
The details of rubber sheets a farmer got during a month are given below.

 Rubber (Kg) Days 9 3 10 4 11 3 12 3 13 5 14 6 16 6

i. How many kilograms of rubber did he get a day on average in this month?
ii. The price of rubber is 120 rupees per kilogram. What is his average income per day this month from selling rubber?

i. Average Quantity of rubber per day = $$\frac { 381 }{ 30 }$$ = 12.77kg
ii. . If the price is Rs. 120 per kg, then average incomeperday = 12.7 × 120 = Rs.1524

Textbook Page No. 197

Statistics Class 9 Question 1.
Find different sets of 6 different numbers between 10 and 30 with each number given below as mean:
i. 20
ii. 15
iii. 25
i. The mean of 6 numbers is 20.
ie; sum = 6 x 20= 120 (Write 3 pairs with sum 40)
i.e., 15, 25, 18, 22, 19, 21

ii. The mean of 6 numbers is 15
i.e; sum =6 x 15=90
(Write 30 pairs with sum 3)
12, 18, 13, 17, 14, 16

iii. Mean is 25
sum = 25 x 6 = 150
(Write 50 pairs with sum 3)
22, 28, 23, 27, 24, 26

HSSLive Statistics Question 2.
The table below shows the children in a class, sorted according to their heights.

 Height(cm) Number of children 148 – 152 8 152 – 156 10 156 – 160 15 160 – 164 10 164 – 168 7

What is the mean height of a child in this class?

 Height (cm) No. of children Mid Value Total Height 148 -152 8 150 150 × 8 = 1200 152 -156 10 154 154 × 10 = 1540 156 -160 15 158 158 × 15 = 2370 160 -164 10 162 162 × 10 = 1620 164 -168 7 166 160 × 7 = 1162 Total 50 7892

Mean height = $$\frac { Total height }{ No of children }$$
= $$\frac { 7892 }{ 50 }$$ = 157.84 cm

Natural Rubber Sheets Question 3.
The teachers in a university are sorted according to their ages, as shown below.

 Age Number of Persons 25 – 30 6 30 – 35 14 35 – 40 16 40 – 45 22 45 – 50 5 50 – 55 4 55 – 60 3

What is the mean age of a teacher in this university?

 Age No.of persons Midvalue Total 25 – 30 6 27.5 165 30 – 35 14 32.5 455 35 – 40 16 37.5 600 40 – 45 22 ‘42.5 935 45 – 50 5 47.5 237.5 50 – 55 4 52.5 210 55 – 60 3 57.5 172.5 Total 70 2775

Mean age = $$\frac { Total age }{ No of persons }$$
= $$\frac { 2775 }{ 70 }$$ = 39.64

HSSlive Maths Question 4.
The table below shows children in a class sorted according to their weights.

 Weight (kg) Number of children 21 – 23 4 23 – 25 — 25 – 27 7 27 – 29 6 29 – 31 3 31 – 33 1

The mean weight is calculated as 26 kilograms. How many children have weights between 23 and 25 kilograms?
Let’s prepare the table for finding the mean by considering the number of children in the group 23 to 25, as ‘x’.

 Weight (kg) No.of children Mid value Total weight 21 – 23 4 22 22 × 4 = 88 23 – 25 X 24 24 × x = 24x 25 – 27 7 26 26 × 7 = 182 27 – 29 6 28 28 × 6 = 168 29 – 31 3 30 30 × 3 = 90 31 – 33 1 32 32 × 1 = 32 Total 21 + x 560 + 24x

Mean weight = 26 kg

i.e; the number of children having weight between 23 to 25 is 7.

### Statistics Exam oriented Questions and Answers

Question 1.
The details of rubber sheets got for a month by a farmer are given in the table.

 Rubber (kg) No. of days 7 3 8 4 9 5 10 6 11 ? 12 4 13 3

During this month he got an average of 10 sheets per day. If so in how many days did he get 11kg per day ?

 Rubber (kg) Days Total weights (kg) 7 3 7 × 3 = 21 8 4 8 × 4= 32 9 5 9 × 5 = 45 10 6 10 × 6 = 60 11 X 11 × x= 11x 12 4 12 × 4 = 48 13 3 13 × 3 = 39 Total 25 + x 245 + 11x

Let ‘x’ be the number of days in which he got 11 kg rubber sheet.

∴ He got 11 kgs of sheets for 5 days.

Question 2.
In a factory, there are workers belonging to four categories. The average income and the number of workers in each category are given. What is the mean income when all the workers in the four categories are combined?

 Class No.of workers Average income (Rs) I 12 6000 II 16 8000 III 8 9500 IV 4 11000

Mean income = $$\frac { 320000 }{ 40 }$$
= Rs. 8000

Question 3.
The daily wages of 10 workers in a factory are given below.
400, 350, 450, 500, 400, 500, 350, 500, 350, 450
If one more person is joined, the mean becomes Rs. 450. What is the daily wage of the new person?
Total wages of 10 workers = 4250 Total wages of 11 workers= 11 × 450 = Rs. 4950
Wage of the 11th person = 4950 – 4250

Question 4.
Find 10 different numbers between 10 and 30 whose mean is 20.
Given mean is 20
Sum 20 × 10 = 200.
We have to find 10 different numbers whose sum is 200 (for this find 5 pairs of sum 40)
(15, 25) (16, 24) (17, 23) (18, 22) (19, 21)
The numbers are 15, 16, 17, 18, 19, 21, 22, 23, 24, 25

Question 5.
A table categorizing the workers in an office on the basis of their salary is given below.

 Salary (Rs) Number of workers 15000 -18000 1 18000 – 21000 3 21000 – 24000 5 24000 – 27000 4 27000 – 30000 1 30000 – 33000 1

Find the mean of salary.

Mean income = $$\frac { 349500 }{ 15 }$$
= R.s 23300

Question 6.
i. Find the mean of natural numbers from 1 to 100.
ii. What is the mean of even numbers from 1 to 100? What is the mean of odd numbers?
iii. What is the difference between the means of the first 100 even numbers and odd numbers?
iv. What is the difference between the means of the first 200 even numbers and 200 odd numbers?
Sum of the natrural numbers from 1 to n = $$\frac n{ n + 1 }{ 2 }$$

Sum of the first 100 odd numbers
= 1002 = 100 × 100

In general, the difference between the means of n even numbers and n odd numbers is always 1.

Question 7.
A table tabulating the players in a cricket team on the basis of their age is given below.

 Age Number of players 21 1 22 2 25 3 26 3 29 2 30 1

Calculate the mean age of the players?

Mean age of players = 306/12 = 25.5

## Kerala Syllabus 9th Standard Chemistry Solutions Chapter 3 Redox Reactions and Rate of Chemical Reactions

You can Download Redox Reactions and Rate of Chemical Reactions Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 3 Redox Reactions and Rate of Chemical Reactions

### Redox Reactions and Rate of Chemical Reactions Textual Questions and Answers

Activity -1

Given below is a table showing the mass of the reactants and the products when hydrogen combines with chlorine to form hydrogen chloride in two situations.

HSSlive chemistry redox reactions Question 1.
Write down the total mass of the reactants and also the total mass of product in the above experiment.
a) Situation 1: …………………
b) Situation 2: ……………….
a) Total mass of reactants 73g products 73g
b) Reactants 146g products 146 g

Activity-2

Given below is a table showing the mass of the reactants and the products when hydrogen combines with oxygen to form water, in two situations.

what is redox reaction class 10 Question 2.
Write down the total mass of the reactants and the total mass of product in the above experiment.
a) Situation 1:
b) Situation 2:
a) Reactants 18 g, Product 18 g
b) Reactants 36 g, Products 36 g

Question 3.
What is the relation between the total mass of the reactants and the total mass of products?
Both are equal.

State The Law Of Conservation Of Mass

In a chemical reaction mass is neither created nor destroyed. This is the law of conservation of mass.
This law was proposed by the scientist Antoine Lavoisier. That is, the total mass of the reactants will be equal to the total mass of the products.

Activity-3

redox reaction and rate of chemical reaction Question 4.
A piece of magnesium is burned in air. What do you observe?
It burns with sparkling light

Name the reducing agent in given below chemical reaction Question 5.
What is the white powder formed?
Magnesium oxide (MgO)

Question 6.
Which are the reactants here?
Mg and O2

chemical reaction equalizer Question 7.
Which is the product?
MgO

redox reactions questions and answers Question 8.
Note down the number of atoms in the reactant side and the number of atoms in the product side in the table below.

Question 9.
Is the number of atoms of each element equal on both sides?
No

Question 10.
The number of which atom is not equal on both sides?
O

Question 11.
How many product molecules are needed to equalize the number of oxygen atoms on both sides?
2

Question 12.
How will you represent two molecules of magnesium oxide?
2 MgO

Question 13.
Now, is the number of magnesium atoms equal on both sides?
No

Question 14.
How many magnesium atoms are needed in the reactant side to equalize the number of magnesium atoms on both sides?
2

Question 15.
Then how can you rewrite the above equation?
2 Mg + O2 2MgO

Question 16.
Is the total number of atoms of each element in the molecules present in the reactant side and that in the product side equal in this equation?
Yes

Question 17.
What is balancing of equations?
Equalizing the number of atoms of each element in molecules in the reactant side and that in the product side is called balancing of equation.
Zn + HCl → ZnCl2 + H2
Tabulate the total number of atoms of each elements in the reactant side and that in the product side of the above reaction.

Question 18.
How many Zn atoms are there in the reactant side and in the product side?
1 each

Question 19.
Which are the atoms showing a difference in their number?
H and Cl

Question 20.
In order to make them equal on both sides how many molecules of HCI should be taken as reactant.
2
Now rewrite the equation
Zn + HCl → ZnCl2 + H2

Question 21.
Balance the chemical equation H2 + O2 → H2O
Step I: H2+ O2 → H2O
Step II: 2H2 + O2 → 2H2O

Question 22.
Balance the chemical equation Al + O2 → Al2O2
Step I: 2Al + O2 → Al2O3
Step II: 2Al+ 3O2 → 2Al2O3

Question 23.
Balance the chemical equation H2 + O2 → H2O
Step I: H2 + O2 → 2H2O
Step II: 2H2 + O2 → 2H2O

Question 24.
Some chemical equations are given below. Note down the number of reactant atoms and that of product atoms in the table given below.
1. C + O2 → CO2
2. N2 + H2 → NH3
3. 2H2 O2 → 2H2O + O2
4. SO2 + O2 → SO3
5. BaCl2 + H2SO4 → BaSO+2HCl

 No. Reactant atoms Product atoms 1 C-1, O-2 C-1, O-2 2 N-2, H-2 N-1,  H-3 3 H-4, O-4 H-4, O-4 4. S-1, O-4 S-1, O-3 5. H-2, S-1, O-4, H-2, S-1,O-4, Ba-1 Cl-2 Ba-1, Cl-2

Balance the equations which are unbalanced
1. N2 + 3H2 → 2NH3
2. 2SO2 + O2 → 2SO3

Oxidation And Reduction

Question 25.
The electronic configuration of magnesium and chlorine are 2, 8, 2 and 2, 8, 7 respectively. How many electrons does a magnesium atom donate? What charge will it get?
2, two positive (2+)

Question 26.
Let us complete the equation for this process,
Mg → Mg2+ + ……………
Mg → Mg2+ + 2e

Question 27.
How many electrons are accepted by each chlorine atom? What will be the charge acquired by each atom?
One 1 Cl1-

Question 28.
Complete the equation of this process.
Cl + 1e
Cl + 1e → Cl1-

Question 29.
What are oxidation and reduction?
Oxidation is the process of loss of electrons. Reduction is the process of gain of electrons. The atom which loss electron is called the reducing agent and the atom which gains electron is called the oxidizing agent.

Question 30.
In the above chemical reaction, which atom is oxidized?
Sodium

Question 31.
Which atom is reduced?
Chlorine

Question 32.
Which is the oxidizing agent in this chemical reaction?
Chlorine (Cl)

Question 33.
Which is the reducing agent in this chemical reaction?
Sodium (Na)

Question 34.
Analyze the following equations and list the oxidized atom, reduced atom, oxidizing agent and reducing agent.
a) Mg + F2 → MgF2
b) 2Na + Cl → NaCl
a) Oxidized atom — Mg
Equation of oxidation — Mg → Mg2+ + 2e
Reduced atom — F
Equation of reduction — F + e → F
Oxidizing agent — F
Reducing agent — Mg

b) Oxidized atom — Na
Equation of oxidation — Na → Na + 1e
Reduced atom — Cl
Equation of reduction — Cl+ 1e → Cl
Oxidizing agent — Na
Reducing agent — Cl

Question 35.
Analyse the following equations and complete the Table given below.
1. Mg → Mg2+ + 2e
F+ le → F
2. Na → Na++1e
Cl + 1e → Cl
3. Fe → Fe2++ 2e
O + 2e → O2

 Equation of oxidation Reducing agent Equation of reduction Oxidizing agent Mg → Mg2++ 2e– Mg F + 1e–→ F– F Na → Na++1e– Na Cl + 1e → Cl–Cl Cl Fe → Fe2++ 2e– Fe O + 2e– → O2- O

Oxidation Number

Oxidation number of an atom in a molecule is the formal charge assigned on the atom if all the bonds in the substance arc considered to be ionic.

Question 36.
Consider the equation H2 +Cl2 → 2HCl What is the oxidation number of hydrogen in H2?
O

Question 37.
What is the oxidation number of chlorine in Cl2?
O

Question 38.
In this reaction, does the oxidation number of hydrogen increase or decrease?
Increases

Question 39.
What change takes place in the oxidation number of chlorine?
Decreases

Question 40.
What are oxidation and reduction on the basis of change in oxidation number?
The process in which the oxidation number increases is called oxidation.
The process in which the oxidation number decreases is called reduction.

Question 41.
During the formation of hydrogen chloride, which atom was oxidized?
H

Question 42.
Which is the reducing agent?
Cl

Question 43.
Which atom was reduced, during this reaction?
Cl

Question 44.
Which is the oxidizing agent?
H

Question 45.
Analyze oxidation numbers in the given equation and list the oxidizing agent and reducing agent in the table given below.

Oxidizing agent – O
Reducing agent-C

Question 46.
Analyze the oxidation number and note down the oxidizing agent and reducing agent in the following reaction.

a) Oxidation number of which atom is increased?
b) The oxidized atom is
c) Oxidation number of which atom is decreased?
d) The reduced atom is
e) Oxidizing agent
f) Reducing agent
a) Na
b) Na
c) Cl
d) Cl
e) Na
f) Cl

Question 47.
Analyze the following equation fill-up the blanks.

Question 48.
The oxidation number of zinc increases/decreases from … to…
0 to +2

Question 49.
The oxidized atom is…
Zn

Question 50.
The oxidation number of hydrogen increases/decreases from …. to
+1 to 0

Question 51.
The reduced atom is …
H
Here HCI is the oxidizing agent and Zn is the reducing agent. „

Question 52.
How do you determine the oxidation number of sulfur in H2 SO4?
Oxidation state of hydrogen = +1
Oxidation state of oxygen = -2
Let the oxidation state of sulphur be ‘x’ We know the sum of oxidation states of all atoms in a compound is zero. Therefore,
[2×(+1)] + x + (4x – 2) = 0
(+2) + x + (-8) = 0
x – 6 = 0
x = +6

Question 53.
Find the oxidation number of Mn in KMnO4 (oxidation number of K is +1, oxidation number of O is -2)
Oxidation state of potassium = +1
Oxidation state of oxygen = -2
Let the oxidation state of Mn be ‘x’.
1 x (+1) + x + 4(-2) = 0
(+1) + x + (-8) = 0
x – 7 =0
x = +7

Question 54.
Find the oxidation number of Mn in MnO2, Mn2O3 and Mn2O7
MnO2
Mn + 2x -2 = 0
Mn + 4 = 0
Mn = +4

Mn2O3
2Mn + 2x -2 = 0
2Mn + -4 = 0
2 Mn = +4
Mn = +2

Mn2O7
2Mn + -2×7 = 0
2Mn + -14 = 0
2Mn = +14
Mn = +7

Question 55.
What is Redox reaction?
The process of oxidation and reduction take place simultaneously. Hence these two reactions together are known as redox reaction.
Examples:
H2 + Cl2 → 2HCl
Mg + F2 → MgF2
2Na + Cl2 → 2NaCl

Question 56.
What are the methods usually adopted to make firewood burn faster? ‘
1. Provide more air
2. Split up into small pieces
3. Make firewood dry

Question 57.
Describe an experiment to prove that nature of the reactants affects the rate of chemical reaction.
Materials required for the experiment.
Zn, Mg, dil. HCl arid test tubes.
Procedure:
Take equal volume of dil. HCl in two test tubes. Add Zn to one and Mg of same mass to the other. Hydrogen gas is produced in both the test tubes. Rate of reaction is faster in the test tube containing Mg.

Question 58.
Write an experiment to prove that concentration of reactants affect the rate of reactions.
Materials required: Mg, dil. HCl, Con. HCl and test tubes.
Procedure: Take magnesium ribbons of equal mass in two test tubes. Add concentrated HCl to one test tube and dilute HCl to the other in equal volume

Question 59.

Test tube 1: ……………
Test tube 2: …………..
Test tube 1: Reaction is faster in con. HCl
Test tube 2: Reaction rate is slow

Question 60.
Why rate of reaction increases when concentration increases?
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 61.
What is the relation between rate of reaction and surface area? Write an experiment to prove it.
Take equal volume of dil. HCI in two beakers. Add a small piece of marble into one and marble powder of equal mass into the other. Reaction rate is greater when marble powder is used. Rate of reaction increases when surface area increases

Question 62.

Is there any difference in the rate of reaction in the two beakers?
Reaction rate is greater when powdered marble is used.

Question 63.
What about the concentration of acid in both the reactions?
Same

Question 64.
Is there any difference in the mass of the marble?
No

Question 65.
What about the surface area of marble?
Different

Question 66.
In which case is the chance for greater number of acid molecules to get in contact with marble, in a given time?
If marble is powdered

Question 67.
What is the change in the rate of collision when surface area increases?
When surface area increases rate of reaction also increases.

Question 68.
What will happen to the rate of this reaction if marble is further crushed or powdered?
Will increase

Question 69.
Why rate of reaction increases when surface area increases?
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 70.
Write an experiment to prove the relation between temperature and rate of reaction.
Materials required: Sodium thiosulphate, hydrochloric acid, water, boiling tube, spirit lamp. Procedure: Prepare 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 the boiling tubes.
Observation: Reaction is faster in heated test tube.

Question 71.
In which of the boiling tubes is the precipitate formed faster?
In heated test tube

Question 72.
What is the color of the precipitate formed?
Pale yellow

Question 73.
What is Threshold Energy?
To take part in a chemical reaction, molecules should attain certain minimum kinetic energy. This energy is called threshold energy.

Question 74.
Write an experiment to prove the influence of catalysts in a chemical reaction.
Take some hydrogen peroxide solution in a test tube. Show a glowing incense stick into the test tube.
What is the observation? Is there any difference occurring in the way in which the incense stick bums? No . ‘
Now add some manganese dioxide (MnO2) into the test tube. Again show the glowing incense stick. Observation:
Glowing incense stick flares up now.

It indicates that when manganese dioxide is added, the rate of reaction increases and oxygen is formed faster. Filter the solution using a filter paper when the reaction is completed.

The substance remaining in the filter paper is manganese dioxide itself When examined carefully it becomes clear that there is no change in its amount or property. The presence of manganese dioxide has increased the rate of the reaction. Manganese dioxide acts as a catalyst in this reaction.

Question 75.
What are catalysts?
Catalysts are substances which alter the rate of chemical reactions without themselves undergoing any permanent chemical change.

### Redox Reactions and Rate of Chemical Reactions Additional Questions and Answers

Magnesium combines with chlorine to form magnesium chloride. The equation is given below
Mg + Cl2 → MgCl2

Question 1.
Write down the electronic configuration of magnesium and chlorine
Mg – 2, 8, 2
Cl – 2, 8,7

Question 2.
How many electrons are donated by magnesium? Write the chemical equation for this process
2, Mg → Mg2+ +2e

Question 3.
How many electrons are accepted by chlorine?
1, charge – ve

Question 4.
Write the chemical equation.
Cl + 1e → cl

Question 5.
Explain oxidation, reduction, oxidizing agent and reducing agent?

• Oxidation is the process by which the removal of electrons.
• Reduction is the process by which the addition of electrons.
• In the above process, magnesium reduces chlorine by donating electrons to it hence magnesium is the reducing agent. In this process since chlorine oxidizes magnesium by accepting electrons it is considered as the oxidizing agent.

Question 6.
From the given chemical equation identify the oxidizing agent and the reducing agent by writing the chemical equation of oxidation and reduction
(i) Mg + F2 → MgF2
(ii) 2Na + Cl2 → 2NaCl
(i) Mg → Mg2+ + 2e (Oxidation)
F + 1e → F (Reduction)
Here magnesium is the reducing agent and fluorine is the oxidising agent.
(ii) Na → Na+ + 1e (Oxidation)
Cl + 1e → Cl (Reduction)
Here sodium is the reducing agent and chlorine is the oxidizing agent.

Question 7.

In the formation of hydrogen chloride does the oxidation number of hydrogen increase or decrease?
The oxidation number of hydrogen increases from 0 to +1

Question 8.
Whether the oxidation number of chlorine decrease or increase?
The oxidation number of chlorine decreases from 0 to — 1.

Question 9.
Based on oxidation number which is the oxidizing agent and reducing agent?
H2 is the reducing agent and Cl2 is the oxidizing agent.

Question 10.
What is meant by Redox reaction?
A process in which the increase in oxidation number is the oxidation and decrease in oxidation number is the reduction. A process in which both oxidation and reduction takes place is the redox reaction. The molecule in which the atom whose oxidation number decreases during the process is the oxidizing agent and that in which oxidation number of an atom increase is the reducing agent.

Question 11.
In the following chemical reactions which is the oxidizing agent which is the reducing agent? Analyze the oxidation number and find out?

Reducing agent – C
Oxidising agent – Cl2

Question 12.
Given below is the equation of the reaction between zinc and hydrochloric acid also indicating the oxidation number.

Analyze and find the following:
In this process, oxidation number of zinc increases from 0 to +2.
Oxidation number of hydrogen decrease from +1 to 0 The reducing agent is zinc and oxidising agent is HCl.

Question 13.
Calculate the oxidation number of‘S’ in H2 SO4?
Oxidation number of hydrogen = +1
Oxidation number of oxygen = -2
Let the oxidation number of ‘S’ be x
The sum of the oxidation numbers of all the Elements in a compound is zero,
Hence
[2x (+1] + x + 4 x (-2] = 0
2 + x – 8 = 0
x = +8 – 2
x = +6
ie., oxidation number of ‘S’ in H2 SO4 is +6.

Question 14.
Find the oxidation number of the ions in ionic compound?
The charge of ions in ionic compound is the oxidation number. Eg. In FeCl2 the oxidation number of Fe is +2 and FeCl3 is +3.

Let Us Assess

Some chemical equations are given below.
C + O2 → CO2
CH4 + 2O2 → CO2 + 2H2O
N2 + O2 → NO
CaCO2 → CaO + CO2
H2 + l2 → Hl
Fe + HCl → FeCl2 + H2
CO2 + C → CO
a) Which of these are balanced equations?
b) Balance the unbalanced equations.
c) Which among these are redox reactions?
a) C + O2 → CO2
CH4 + 2O2 → CO2 +2 H2O
CaCO3 → CaO +CO2
b) N2 + O2 → 2NO
H2 + l2 → 2Hl
Fe + 2HCl → FeCl2 + H2
CO2 + C2CO

Question 2.
The chemical reaction between marble and dilute HCl is given.
CaCO3 + 2HCl → CaCl2 + H2O +CO2
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. Explain the reason.
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 3.
Sulfur pieces do not react with cold concentrated nitric acid. But sulfur 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 you would choose? Give reason.
a) When surface area increases 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 4.
Small amounts of phosphoric acid is usually added to hydrogen peroxide to prevent its decomposition?
a) What is the function of phosphoric acid here?
b) By which name are these type of substances known?
c) Which substance would you add to increase the rate of decomposition of hydrogen peroxide?
a) Slow down the rate of decomposition of hydrogen peroxide
b) Negative catalyst
c) Manganese dioxide

Extended Activities

Question 1.
Find the oxidation number of the elements which are underlined in the compounds given below. Among these find out the elements which show variable oxidation numbers.
MnO2, Mn2O7, K2Cr2O7, KCrO3, MnCl2, MgO, MgCl2, Al2O3,AlCl3
(Hint. Oxidation number 0 = -2, Cl = -1, K = +1)
MnO2 Let x be the oxidation number of Mn

Question 2.
Some apparatus and chemicals are given.
Zn, Mg, dilute HCl, CaCO3, test tube, water.
a) Design an experiment to prove that the nature of reactants can influence the rate of reaction.
b) Write the equations for the chemical reactions.
c) Write the expression for the rate of the reaction.
a) Take equal volume of dil HCl in two test tubes. Add a piece of Mg m to one of them. Add a piece of Zn of equal mass to the other.
Observation:- Mg reacts faster with dil. HCl than Zn. This is due to the difference in the chemical nature of Mg and Zn.
b) Mg + 2HCl → MgCl2 +H2
Zn + 2HCl → ZnCl2 + H2
c) Rate of a chemical reaction $$=\frac{\text { amount of reactants used up }}{\text { time }}$$
= $$\frac{\text { amount of product formed }}{\text { time }}$$

Question 3.
The experiments conducted by two students are given below.
Experiment 1: 2 mL of sodium thiosulphate solution is taken in a test tube, heated and to it 2 mL of HG solution is added.
Experiment 2:2 mL of sodium thiosulphate solution is taken in a test tube and to it 2 mL of HC1 solution is added.
a) In which experiment is the precipitate formed quickly? Justify your answer.
b) Write the balanced equation for the reaction.
a) ln heated test tube.
When temperature increases more molecules get threshold energy. So rate of effective collision increases and rate of reaction increases.
b) Na2 S2O3 + 2HCl → 2NaCl + SO2 + S

Question 4.
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.
a) Marble powder and concentrated HCl CaCO3 + 2 HCl → CaCl2 + H2O + CO2

## Kerala Syllabus 9th Standard Social Science Solutions Part 1 Chapter 6 India, the Land of Synthesis

You can Download India, the Land of Synthesis Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Social Science Solutions Part 1 Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 9th Standard Social Science Solutions Part 1 Chapter 6 India, the Land of Synthesis

### India, the Land of Synthesis Textual Questions and Answers

India the Land of Synthesis Questions 1.
The sufis helped to maintain relations among different religious during ………. and ………. periods.
Sultanate and Mughal

Bhakti Movement in Kerala Question 2.
Complete the table.

 Sufi saints Place 1. Baba Farid Ayodhan 2. 3. 4. 5.

 Sufi saints Place 1. Baba farid Ayodhan 2. Sayyid Muhammed Jesudaras Gulbarga 3. Shah alam Bukhari Gujarat 4. Bahaudheen Sakaria Multan 5. Shaikh Shihabudeen Suhrawardi Sathette

Class 9 Social Science India, the Land of Synthesis Kerala SCERT Solutions Question 3.
Mention the features of early Bhakti traditions.

• Its propagators were the saints who were Bhakti poets.
• Provided representation to the lower castes and women.
• Rendered contributions in the field of culture and music.

Question 4.
Where did Bhakti movement emerged at first?

Kerala Chronicle Question 5.
Discuss the role played by Alwars and Nayanars in propagating the Bhakti movement in South India.
The Bhakti movement emerged at first in Tamil Nadu. This movement was led by the Alwars and the Nayanars. The always were Vaishnavites and the Nayanars, Saivites. They composed and propagated devotional songs. Both the Alwars and the Nayanars opposed caste inequalities. Among them there were women saints also. The woman saint Andal was one of the famous Vaishnavite saints. Another woman saint Karakkal Ammayarwasa Saivite.

Question 6.
Prepare a flow chart showing ideas of Virasaiva movement.

Synthesis Pride Question 7.

• Basavanna
• Allamaprabhu

Popular Literature Syllabus Question 8.
Examine the role of the idea of Kabir in promoting religious harmony among different sections of people.
A part of the Doha (meaningful couplets) of Kabir, who was an important figure in the Bhakti Movement is given above. He reminds that the Hindu and the Muslim are two wares made out of the same soil and he tried to promote brotherhood between Hinduism and Islam. The aim of Kabir was to propagate a religion of love uniting people of all castes and religions.

He strongly opposed all discriminations on the basis of caste, religion, class, family and wealth. He focused on the unity of humanity and vehemently opposed caste system and untouchability. He rejected idol worship, pilgrimages and bathes in holy rivers. He was not in favor of forsaking family life for accepting sainthood. The ideas of Kabir developed through discussions and deliberations on the basis of the Bhakti – Sufi traditions.

Keraliya Kalakal in Malayalam Language Question 9.
Name the women who were active in Bhakti movement.

• Mirabai
• Lalded
• Bhahinabai

9th Standard Social Science Book in Kannada Question 10.
Prepare a note on the ideas of Guru Nanak.
Guru Nanak was another preceptor who focused on the idea of monotheism. He tried to combine the ideas of Hinduism and Islam. He was a spokesman of the ideals of religious tolerance and universal brotherhood. He opposed the rituals of both Hinduism and Islam. He said that one has to maintain purity in character and behavior to reach God. He vehemently opposed idol worship and pilgrimages.

He promoted a middle path which permitted spiritual life along with the responsibilities of a householder. The prayer songs known as ‘Shabad’ were the contribution of Nanak. He strongly opined that all human beings are equal and hence there is no need of caste distinction among them. He instructed his followers to have food from the common kitchen (lunger). He also instructed that the people from all castes could take food from the lunger. The ideas of Nanak later led to the emergence of Sikhism.

Question 11.
What are the changes brought about by the Bhakti movement in the social system of India?
Even though the Bhakti Movement evolved with the aim of dedication to God, it created certain qualitative changes in the Indian social system. The propagators of the Bhakti cult, through their words and songs, maintained that everyone was equal in front of God.

This strengthened the concept of social equality against caste discriminations. The idea of gender equality also began to take shape. The idea of social equality got wide popularity since the Movement was led by low caste people like Thiruppana Alwar, Kabir, and Lalded. Regional languages developed. Evil practices were questioned.

Question 12.
Complete the table.

 Languages Works/Branches of literature Authors Bengali Marathi Oriya Telugu Tamil Kannada Malayalam

Question 13.
Conduct a seminar on the topic the cultural life of medieval India/Areas to be considered:
1. Regional languages
2. Music
3. Painting and architecture.
Growth of Regional Languages:
The propagators of the Bhakti – Sufi movements spread their ideas through their vernacular languages. This was to enable the common people to understand them. This resulted in the growth of the regional languages. The two forms of Hindi – Braj (Vrije language) and Awadhi were used as spoken languages. Many languages such as Punjabi, Kannada, Telugu, Oriya, Assamese, Marathi, Bengali, Sindhi and Malayalam evolved during this period. In almost all parts of the country, Persian was the court language of the medieval period. Hence many Persian words can be found in many regional languages.

Out of the relationship between Persian and Hindi, a new language, Urdu, came into being. Urdu is the most important example of the cultural synthesis of India during the medieval period. Urdu, which was spoken in the Deccan, was. strongly influenced by the languages such as Telugu and Marathi. Gradually Urdu became the most popular spoken language of the towns. In the Western Coast, merchants used Arabic. Soon it began to influence the regional languages of that area. Because of the influence of the Arabic language, a new mixed dialect, Arabi – Malayalam, evolved in Kerala.

People also used Sanskrit during the Middle Ages. Sanskrit was used in the court of Vijayanagara on festive occasions. Sanskrit prevailed as the language of higher education. The popular literature of Sanskrit such as the Puranas, Ramayana and the Mahabharata were available in the regional languages. The Persian literature influenced some poets and writers of the period. The most notable Persian writer of the period was Amir Khusrau. His love towards India and the pride of being an Indian is clearly revealed through the content of his works.

During this period a large number of Indian works were translated into Persian language,. Ramayana, Mahabharata, Atharvaveda, Upanishads, etc. belong to this category. The historic chronicle of Kalhana, Rajatarangini, was translated into Persian during the reign of the Kashmir king, Zain-ul-abin.

Hindi received tremendous impetus during the medieval period. Ramananda and Kabir delivered their moral advice in Hindi. The dohas of Kabirwere the finest example of Hindi literature. The Sursagar of Surdas, Ramacharitamanes of Tulsidas, the Bhajans of Mirabai and Padmavati of Malik Muhammed Jayasi enriched Hindi language.

Music:

It was during the medieval period that music was enriched with different forms. Considerable progress was attained by the Carnatic music, the musical style of South India. The famous Carnatic musician of the period was Naik Gopal. A large number of musicians migrated from Deccan to North India during the medieval period. During the same period, a new music style evolved in North India, due to the influence of Persian music. It came to be known as the Hindustani music. Amir Khusrau, who lived in the Sultanate period was one of the famous Hindustani musicians of the Sultanate period.

Amir Khusrau received training in different aspects of Indian music from the Carnatic musician, Naik Gopal. The medieval rulers were patrons of this category of music. New forms of music emerged during this period. The techniques of Persia were experimented in Indian music. The singing styles such as Khayal and Thumri got popularity. Many musical instruments like sitar, sarangi, tabala, etc.also became popular. The special interests of some Sufis in singing also were responsible for this. It was through the Sufi saints that Khayal and Gazal became widespread in India.

The musical works of Sanskrit were translated into Persian during this period. The work on Music Ragadarpan was translated into Persian during the regime of Firoz Shah Thuglaq. Emperor Akbar was a great lover of music. Abul Fazl states that 36 musicians were patronized by Akbar. The most famous among them was Tanzen. The Durbar raga, which enriched the Hindustani music was his contribution. The work Sangeeta Ratnakaram authored by Sarngadevan of the 13th century was an important contribution of medieval India to music. The camatic music of South India was influenced by the Persian and Arabic music during the Vijayanagara rule.

Architecture and Painting:

Architecture and painting attained progress in India during the medieval period. Itwasduringthe medieval period that the new styles of architecture such as Dravida, Nagara, Indo- Persian, etc. developed. The blending of different sculptural styles was a specialty of the age.

The painting style that came into vogue during the Mughal period is known as miniature painting. This is the style of drawing pictures like the pictures in books. It was a combination of both Indian and Persian styles.

The Mughal Emperor Humayun contacted Persian artists, and Mir Sayyid Ali and Abdu Samad wer brought to his court. The Mughal style of painting was developed by them.

Painting competitions were organized in the royal institutes (Kharkhana) during the region of Akbar. people from different sections participated in them. The stories of Mahabharata were completely compiled into a series of paintings called Raznama by Daswant Who lived during the period of Akbar. Indian subjects and natural scenes gradually became the themes of painting. The painters of the period showed their proficiency in drawing the pictures of birds and animals. The quality of paintings increased considerably during the reign of Jahangir. Bishandas and Abul Hassan were the greatest painters of the period. Kalyandas was a painter during the period or Shajahan.

There were groups of painters during the medieval period. They were from many parts of India such as Gujarat, Kashmir, Deccan, etc. Their paintings were influenced by their respective regional styles. Their paintings were based on the themes of the puranas and stories of both India and Persia. The wall paintings of that time were also remarkable. The wall paintings on the Tanjore temple during the period of the Cholas were an, important feature of medieval painting. The Rajasthani style was another style developed during this period. It was a combination of the traditional and the Mughal styles of painting.

The cultural synthesis shaped in medieval India brought many changes in the history of India. The greatest models of this synthesis wer Din-i-llahi and Sikhism emerged from among the people. The new styles introduced in the fields of architecture, painting, literature, and music enriched the mixed culture of India. People from different regions of India, followers of different religions and institutions did make their contributions to this cultural synthesis.

Question 14.
Explain the features of Sufi movement in the medieval period.
The Sufis were those who showed reluctance to luxurious life and gave predominance to spiritual life. They consider devotion as a means to reach close to the God.
They preached to respect all human beings. The Sufi saint was known as Pir or Shaikh and his follower Murid. The residences of the Sufis are known as Khanqahs. The devotional songs reciting in the Sufi centers are the Qawwalis.

Question 15.
Match the following table

 A B Ezhuthachan Kannada Appar Marathi Amoghavarshan Tamil Varnana Pandit Malayalam

 A B Ezhuthachan Malayalam Appar Tamil Amoghavarshan Kannada Varnana Pandit Marathi

Question 16.
How did Urdu language emerge?
Out of the relationship between Persian and Hindi, Urdu came into being.

## Kerala Syllabus 9th Standard English Solutions Unit 3 Chapter 3 Climate Change is not Hysteria – It’s a Fact

You can download Climate Change is not Hysteria – It’s a Fact Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard English Solutions Unit 3 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 9th Standard English Solutions Unit 3 Chapter 3 Climate Change is not Hysteria – It’s a Fact

### Climate Change is not Hysteria – It’s a Fact Textual Questions and Answers

Let’s revisit and reflect

Climate Change is Not Hysteria It’s a Fact Question and Answer Question 1.
Why did the people march in the streets of New York?
The people marched in the streets of New York to show their concern for climate change and demanding solutions for the climatic crisis.

Climate Change is Not Hysteria Its a Fact Notes Question 2.
What is the attitude of human beings towards climate change?
Human beings don’t take climatic change as a serious issue. They think it is just an imaginary thing and it would go away somehow.

Question 3.
What are the ‘undeniable climate events’ that are happening now?
The undeniable climate events are: extreme weather events, increased temperatures and melting ice sheets from West Antarctic and Greenland.

Climate Change is Not a Hysteria Lesson Plan Question 4.
What is the difficult task that we face?
The difficult task we face is finding solutions for the climate change.

Question 5.
What happens when an ecosystem collapses?
When an ecosystem collapses, the economy itself will die.

Question 6.
What is the scope of renewable energy in future?
The scope of renewable energy is good. It is achievable and it is a good economic policy.

Climate Change is Not Hysteria It’s a Fact in Malayalam Question 7.
Solving the crisis is a question of our survival. Explain.
Clean air and water and a liveable climate are essential for our survival. But we have the crisis of climate change threatening us. We have to solve this. crisis

Climate Change is Not Hysteria It’s a Fact Question 8.
Why is the present the ‘most urgent of times’?
The present is the most urgent of times because there is a wanton destruction of our collective home. Unless we take action now, it will be too late.

Activity -1

Your school is celebrating National Science Day on the 28th of February. You are the convenor of the Science Club. Prepare a notice.

ST. GEORGE HIGHER SECONDARY SCHOOL
EDAPPALLY
SCIENCE CLUB

15 January 2019

NOTICE

The Science Club is celebrating the National Science Day on 28th February 2019. The Inaugural meeting will be at 10.00 a.m. The famous Environmentalist Fr. Joy Peenickaparambil has kindly agreed to be our Chief Guest. Our Municipal Counsellor Jose Pathadan and K.J. Joseph will offer felicitations. Our winners of the National Level Science Competitions will be honored in the meeting.

There will be an exhibition showing some of the highlights from the lives of C.V. Raman and Dr. APJ Abdul Kalam. There will also be extensive coverage of the Lunar Missions undertaken by India.
Sd /
Rahul Binoy Convenor

Activity – 2

a) Prepare a few posters to make people aware of the necessity for preserving nature and protecting the environment.
Captions for the Posters:

• Avoid Using Plastic Bags
• Don’T Use Dangerous Pesticides Like Endosulfan
• Don’T Litter The Place
• Don’T Cut Down Trees
• Plant Trees

b) It has been decided to invite a famous ecologist to inaugurate the Science Exhibition in your school. As the school leader, you are asked to invite him. Prepare an e-mail to invite the scientist.
mehboobsaithu@hotmail.com
Sub: Invitation to inaugurate science exhibition
Dear Sir,
I am Riya of Class IX and I am the school leader of S.N. Higher Secondary School, Irinjalakuda. We are planning to hold a science exhibition to promote the study of science among students. The exhibition will have exhibits that show the growth of science in India. There will be pictures of the Indian Scientists and their short biographies. We will also trace the origin and growth of India’s space programmes.

As a leading personality in the scientific field, we request you to inaugurate the Exhibition. The time for the inaugural meeting is 10.00 a.m. on Wednesday, the 30th July 2019. The Venue is the school auditorium.

We hope you will kindly accept our invitation and grace the occasion with your presence.
Tanking you,
Yours sincerely,
Trishelle

Hysteria Meaning in Malayalam Question 1.
The scientist has accepted your invitation. At the inaugural function of the science exhibition, you have to make the welcome speech. Prepare the script.
Distinguished Chair and eminent guests,
As the school leader, I have great pleasure in making this welcome address. First and foremost I welcome our Chief Guest Mr. Mehboob Saithu who is a well-known figure in our midst. Although his concentration is on Agriculture, he has special interests in the scientific field. We have read many of his articles about Science and Scientific Growth and how science has revolutionized agriculture. I can say he continues with the Green Revolution making our country selfsufficient in food. He also takes a keen interest in the space missions of India.

I also welcome our eminent speakers on this occasion. Mrs. Sonia Varghese is well known to you as the Chairperson of the Municipality. Mr. Ahmed Sultan is also well-known to you as the prominent businessman of our town. I heartily welcome both of them.

Now I welcome the headmaster and the teachers to this inaugural meeting. In fact welcoming them would be almost redundant as they are the real hosts of this event, But for the sake of formality, I welcome them. I also welcome all of you for coming and making this event successful.

Thank you all.

Language Activities

a) Look at the sentences given below.
We must discuss this calmly.
The commotion dies slowly.
I strongly oppose this plan.
Identify the structure of the sentences and write them below

 Noun Phrase Verb phrase VP Constituents We must discuss. this calmly Aux+Verb+ NP+Adverb The commotion dies slowly VP + Adverb I strongly oppose this plan. Adv+V+NP(object)

b) Complete the table given below using adverbs from the play, ‘Listen to the Mountain.’ How do they enrich the play?

 curiously proudly contemptuously doubtfully haughtily certainly

c) Study the following sentences from the play, ‘Listen to the Mountain.’
1. Rudrappa and Kannan are talking excitedly.
2. Dixit and Sagar look around in panic.
3. They are starting work tomorrow
The adverb excitedly, around and tomorrow shows how, where and when things happened.

d) Write some sentences with adverbs :

 How When Where? She drives fast. They came yesterday Place it there He walks slowly The meeting will start soon Keep it here. Kareena dances well. My father is coming tomorrow. He went everywhere with his dog.

Activity 2

1. We are building a hotel here. A five star hotel.
2. We are bringing the labourers, of course. A few hundred.
3. They are playing football.
4. He is reading a novel.
Identify the verb phrases and fill in the blanks appropriately.

 VP Function 1. are building 1. 2. 2. planned future action 3. 3. continuous action 4. 4.

 VP Function 1. are building 1. planned future action, 2. are bringing 2. planned future action 3. are playing 3. continuous action 4. is reading 4. action going on now

a) Identify the planned future actions from the sentences
1. The Prime Minister is arriving tomorrow to visit the flood-affected areas.
2. Geetha is going to the market.
3. We are going on a tour next week.
4. Rahul and Kabeer are playing chess
1. is arriving tomorrow.
2. are going on a tour next week.

b) Rewrite the sentences using am/is/ are+ verb +ing.
1. Trucks and lorries will arrive with cement, bricks, and marble tomorrow.
2. The villagers will stage a peaceful protest the next day
3. We will submit a petition to the Collector tomorrow
1. Trucks and lorries are arriving with cement, bricks, and marble tomorrow.
2. The villagers are staging a peaceful protest the next day.
3. We are submitting a petition to the Collector tomorrow.

Activity – 3

Which of the following statements are true? Tick the correct ones. Justify your answer.
1. The time of the action is mentioned in both the sentences.
2. The tense form used in each sentence is different.
3. The actions in both the sentences are complete.
1. False.
2. True.
3. True

a) Look at the following words and expressions. Some of them go only with sentences in the simple past tense. And others with the present perfect. Arrange them under the given titles.

 With sentences in present perfect With sentences in simple past ever yesterday many times one year ago before previous year never last week already at that moment yet that day so far one day

Note : Some of the words like before, never, once, so far, etc. can be used with present perfect as well as the simple past.

Activity – 4

The grandmother in the play ‘Listen to the Mountain’ gives certain instructions to the headmaster.
You must talk to the children.
Find out what they feel about it.
Don’t put ideas into their heads.
Just tell them everything.
These instructions can be converted into requests of different types.
e.g. You must talk to the children.
Could you please talk to the children?
Would you mind talking to the children?

a) Convert the other instructions into requests in different ways.
Would you mind finding out what they feel about it?

You should not put ideas into their heads.
You should desist from putting ideas into their heads.
Be careful not to put ideas into their heads.

Could you please tell them everything?
Would you mind telling them everything?

How would the sentences be then?
You had better talk to the children.

c) There are many people who are not concerned about the environment. How would you advise them?

You had better plant more trees.
Water scarcity is a serious problem, …………………… not wastewater.
We get vegetables with toxic residue from the market ………
Water scarcity is a serious problem. → You had better not wastewater.
We get vegetables with toxic residue from the market. → You had better grow your own vegetables.
You are getting low marks. → You had better spend more time with your books.

Activity – 5

a) Read the paragraph given below and guess the meaning of the words underlined.

Sagar called on Narayan and spoke to him of his decision to put up a new hotel in Dharmagiri. But Narayan could foresee its dangers and he put it across to his mother. His ninety-year-old mother couldn’t put up with this news. So she told Narayan to do something to put an end to Sagar’s programme. Hearing this, Narayan explained the dangers to Sagar and he turned down Sagar’s proposal. Yet, Sagar was not ready to cancel his project. Therefore, Narayan called a meeting of the villagers and they together put their mind to start a protest.

 A. Word/phrase B. Meaning called on visited put up build put it cross communicated put up with tolerate put an end stop turned down ejected

b) Look at this expression.
go away – neglect
Find out similar expressions and their meanings from the lessons ‘Listen to the Mountain’ and ‘Climate Change is not Hysteria’.
From Listen to the Mountain:
knocked down — demolished
pour into — come in large numbers
get things moving — make things happen
put up — build
bring in — get
come up — start, grow
get on — continue, proceed

Climate Change is not hysteria – it’s a fact:
looked at — considered
go away — vanish, disappear
depend on — rely on

c) Read the headlines given below. Identify the phrasal verbs and guess their meanings. You may refer to a dictionary.

 Headlines Phrasal Verbs Meaning Jazz legend passes away in sleep passes away dies Youngsters urged to give up smoking give up stop Bomb goes off in town goes off blasts Government ready to take on new projects take on start Prime Minister calls on President to discuss security issues calls on visit

d) The following is a questionnaire enquiring into the study habit of one among your friends. Discuss the meaning of the phrasal verbs used in the questions and write down the answers. Add a few questions of your own, using phrasal verbs.
1. At what time do you usually get up?
2. How much time do you take to review your previous day’s lessons?
3. Which subjects are hard to keep up with?
4. Do you jot down notes in the class?
5. Do you look up unfamiliar words in a dictionary?
6. Do you finish off your work in time?
7. Do you cheer up your friends when they are down?
8. How do you while away your leisure time?
9. Do you stay up late, the night before the exam?
1. get up — wake up
2. take to — use.
3. keep up with — understand, study, follow
4. jot down — write quickly
5. lookup — search, find out
6. finish off — complete
7. cheer up — encourage
8. while away — pass, spend
9. stay up — keep awake

e) Collect a few phrasal verbs and use them in sentences of your own. Topics: family, everyday life, sports, study, etc.

 Phrasal verbs Sentences Put up with Narayan cannot put up with the behavior of Sagar put up We are putting up another building close to our home. get up I get up at 6 in the morning. get into I got into some trouble the other day with my boss. take away Problems take away the pleasure of life. put on I  try to put on a different dress each day. close down Our shop was closed down last week. prefer to I prefer coffee to tea bring up The matter was brought up for discussion. insist on My father insists on hard work.

Activity 6

Let’s edit

Here is an excerpt from a speech by Yugratna Srivastava, a 13-year-old Indian girl, which was delivered at the U N Summit on Climate Change on 22 September 2009. Some errors have been made by a pupil while copying it. These are underlined. Correct the errors.

The Himalayas are melting, the polar bears are dying. Two of every five people (a) doesn’t have access to clean drinking water. The earth’s temperature (b) are increasing. We (c) have losing the untapped information and potential of plant species. The Pacific’s water level (d) risen. Is this what we (e) are go-to hand over to our future generations? We (f) receive a clean and healthy planet from our ancestors and we are (g) gift a damaged one to our successors. Is there any justice in this? Honorable Excellencies, we need to call for action now. We (h) have protect the earth not just for us but for our future generations.
a. don’t have
b. is increasing
c. have lost
d. has risen
e. are going
h. have to protect

## Kerala Syllabus 9th Standard English Solutions Unit 2 Chapter 3 Tolstoy Farm

You can download Tolstoy Farm Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard English Solutions Unit 2 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 9th Standard English Solutions Unit 2 Chapter 3 Tolstoy Farm

### Tolstoy Farm Textual Questions and Answers

Tolstoy Farm 9th Class Question 1.
Who were the inmates of the Tolstoy farm?
The inmates of the Tolstoy Farm were people of different religions. They were Hindus, Muslims, Parsis, and Christians.

Tolstoy Farm Question 2.
Why couldn’t Gandhiji appoint special teachers for Indian students?
He could not appoint special teachers because he could not afford to pay them decent salaries. Moreover, he did not like the existing system of education. He wanted to experiment something new.

Active Maths 2 Solutions Chapter 3 Question 3.
What did Gandhiji regard as the proper foundation for the education of the children?
He regarded character building as the proper foundation for the education of the children.

Labour India Class 2 Kerala Syllabus Question 4.
Who assisted Gandhiji in literary training?
Mr. Kallenbach and Sjt. Pragji Desai.

Question 5.
What training was given to children of all ages in the farm?
The children of all ages in the farm were given moral training.

HSSlive Training 9th Class Question 6.
What were the different types of works in the Tolstoy Farm?
The different kinds of work in the Tolstoy Farm were cooking, digging pits, felling timber, lifting loads and gardening.

Teaching in the Tolstoy Farm Summary Question 7.
Why was illness scarce on the farm?
Illness was scarce in the Farm because the inmates got good exercise by doing the various jobs there. The jobs included cooking digging pits, felling timber, lifting loads and gardening.

Question 8.
How did Gandhiji introduce vocational training in Tolstoy Farm?
He introduced vocational training in the Tolstoy Farm by teaching the inmates how to make shoes and also carpentry

Question 10.
What made learning a cheerful experience for children in the farm?
Active participation in the work by the teachers made learning a cheerful experience for the children in the Farm.

Let’s revisit and reflect

Question 1.
In ‘Tolstoy Farm’, there is reference to various skills. Identify them and fill in the bubbles.

cooking, shoemaking, carpentry, gardening

Question 2.
You must be familiar with Gandhiji’s concept of education:
‘By Education, I mean an all-round drawing out of the best in the child and man, body, mind and spirit.’
How far is it true with the learning experiences in the Tolstoy Farm? Write your answer in a short paragraph
Gandhiji said, “By education, I mean an all-round drawing out of the best in the child and man, body, mind, and spirit.” This is quite true with the learning experiences in the Tolstoy Farm. The priority in the Farm was culture of the heart and the building of character. Moral training was common there. Kallenbach and Sjt. Pragji Desai gave the students literary training. Training of the body was also essential.

There were no servants in the Farm and all the work was to be done by the inmates. Kallenbach was fond of gardening and he had some experience in it. Those who were not busy in the kitchen had to help in the garden. Children happily did the job of digging pits, cutting trees and lifting loads. Some were not happy and tried to avoid work. They were also given training in shoe-making and carpentry. Thus the body, mind and spirit of the inmates were taken care of.

Activity -1

‘…there were no servants on the farm and all the work, from cooking down to scavenging, was done by the inmates’, says Gandhiji.

I think this practice is very good because it will bring dignity of labour. It will show that any work can be done by anybody and there is nothing called ‘high’ or ‘low’ work. It also will make the people healthy as they get plenty of physical exercises.

Haritha Keralam, Organic Farming, Biodiversity Park and Swatch Bharath Drive are some of the programmes in schools. They require the active participation of all students. They also include work that some people may hesitate to do.

Question 2.
Do these programmes promote dignity of labor among children?
Yes, they do.

Question 3.
What are your views? Conduct a group discussion and prepare a write-up based on the main points.
Haritha Keralam is an Umbrella Mission which includes Waste Management, Organic Farming, and Water Resources Management. It has an ambitious outlook to address the issues of piling waste, impending drought and health hazards due to the consumption of pesticide-treated vegetables and in general, the agricultural dependency of the State. The Haritha Keralam Missiorraimsto integrate the 3 most important and inter-related sectors:

Household-level segregation and safe disposal of organic waste through feasible options like composting, biogas, arrangements for institutional waste disposal, re-use, recycling and safe disposal of non-degradable and electronic waste are given priorities. Rejuvenation of tanks, ponds, streams, and rivers are the focus in the water resource sector.

The thrust in promoting organic agriculture will be to produce safe to eat vegetables and fruits to make the state self-sufficient within the next 5 years.

Biodiversity is a contraction of biological diversity. Biodiversity reflects the number, variety, and variability of living organisms. It includes diversity within species (genetic diversity), between species (species diversity), and between ecosystems (ecosystem diversity).

Swatch Bharat Drive is a nation-wide campaign in India. It aims to clean up the streets, roads, and infrastructure of India’s cities, towns, and rural areas. The objectives of Swatch Bharat include eliminating open,n defecation through the construction of household-owned and community-owned toilets and establishing an accountable mechanism of monitoring toilet use.

Run by the Government of India, the mission aims to achieve an “open-defecation free” India by 2 October 2019, the 150th anniversary of the birth of Mahatma Gandhi, by constructing 90 million toilets in rural India. The mission has two thrusts: Swatch Bharat Abhiyan (‘rural’), which operates under the Ministry of Drinking Water and Sanitation; and Swatch Bharat Abhiyan (‘urban’), which operates under the Ministry of Housing and Urban Affairs.

Activity 2

Gandhiji says, ‘It was my intention to teach every one of the youngsters some useful manual vocation.’
List out the vocational skills mentioned in the text and complete the following:

 Vocational skills acquired from Tolstoy Farm The jobs related to the vocational skills Carpentry Carpenter, Furniture Designer, Interior Decorator Shoemaking Cutting leather, stitching, making shoes and repairing them Gardening Preparing the land, Planting, nurturing seedlings, watering, giving manure, protecting the plants from harmful insects and others. Cooking Cooking different kinds of food, cleaning, knowledge of diet

Everyone can benefit out of this type of vocational education as it provides an opportunity to learn a skill or trade. Discuss. Everyone can benefit out of this type of vocational education as it provides an opportunity to learn a skill or trade. It helps one to earn a living. It helps one to have self-employment. One does not have to go from one place to another looking for a job. In self-employment, you become your own master. You can also work when it is convenient to you as the working hours are not fixed. Learning trades bring in self-sufficiency.

A child learning such vocational skills will be able to do various electrical, carpentry and plumbing repairs himself. It not only bring you monetary profit but also satisfaction. Vocational education is essential for a country like India where there are lakhs of people who have no employment. By using their skills wisely people can have better lives and they can help in the overall prosperity of the country. Learning such skills promote the dignity of labor. In India we certainly divide workers as white-collar and blue-collar. People who do white-collar jobs are supposed to be better than those who do blue-collar jobs. This is a wrong idea. Any honest job has its own dignity.

Activity 3

Collect pictures, photographs, newspaper cuttings, cartoons, caricatures and other materials related to the main events in the life of Gandhiji. Using these, prepare an album about Gandhiji. You may give suitable captions and short descriptions, wherever required.

Language activity

a) Read the following sentences from ‘Maternity’.
1. The child was condemned to die of hunger.
2. The women gave vent to various cries of horror.
3. She pressed the baby to her heart.
4. Mikali’s heart trembled with joy.

 1. The child was condemned to die of hunger 2. The women gave vent to various cries of horror. 3. She pressed the baby to her heart. 4. Mikali’s heart trembled with joy.

b) Now, read the passage below and identify the subjects and predicates.

The little child was crying for milk. Mikali had no money in his pocket to buy milk. The poor boy approached the women around. All the women in the camp who saw the child gave vent to cries of horror. A kind Chinese woman finally gave it milk.

 Sentence Subject (Noun phrase) Predicate (Verb phrase) 1. The little child was crying for milk The little child was crying for milk 2. Mikali had no money in his Mikali had no money in his 3. The poor boy approached the women around, The poor boy approached the women around, 4. All the women in the camp who saw the child gave vent to cries of horror. All the women in the camp who saw the child gave vent to cries of horror. 5. A kind Chinese woman finally gave it milk. A kind Chinese woman finally gave it milk.

In a typical sentence, a Noun Phrase (NP) is immediately followed by a Verb Phrase (VP). A noun phrase can be a noun, a pronoun or a group of words that does the function of a noun. A verb phrase consists of a helping verb or a main verb which may or may not be followed by other words.

Let’s split the subject and predicate parts of the first sentence.
The little child was crying for milk.

The headword of the noun phrase is a noun. The other words are used to give additional information about the headword. Thus the obligatory element in a noun phrase is the noun. Any other element is optional.

Determiner

Articles, possessives, and demonstratives which come before a noun are called determiners.

Articles: a, an, the
Possessives: my, our, your, his, her, their, its, Gandhiji’s, etc.
Demonstratives: this, that, these, those

Now, let’s have a look at the following sentence.
All the women in the camp who saw the child gave vent to cries of horror.
Identify the noun phrase and the verb phrase

Pre-determiners or pre-articles are those items which come before the determiner in an NP. Words and phrase like ‘half’, ‘half of’, ‘all’, ‘all of’ etc. are called pre-determiners.

a) The structure of the noun phrase of a sentence can be:

 A B Women noun The women article + noun All the women pre-determiner + article + noun All the women in the camp pre-determiner + article + noun + prepositional phrase All the women in the camp who saw the child pre-determiner + article + noun + prepositional phrase+ relative clause

b) Analyze the subject part of the other sentences in the passage given above in the same way and identify the constituents. Work with your partner and write down the ideas you have discussed.
The little child – article → adjective noun Mikali -noun
The poor boy – article → adjective → noun
All the women in the camp who saw the child – pre-determiner → article → noun → prepositional phrase → relative clause
A kind Chinese woman – article →> adjective → adjective → noun

1. A sentence has two parts.
2. The subject part is usually a noun phrase.
3. The noun phrase may consist of ……………..
4. The predicate is usually…..

1. A sentence has two parts.
2. The subject part is usually a noun phrase.
3. The noun phrase may consist of pre-determiner → determiner → adjective → noun → prepositional phrase → relative clause
4. The predicate is usually a verb phrase.

d) Let’s see what a verb phrase consists of
Look at the picture and identify the verb phrase in the sentences given below.

1. left the pool.
2. left the pool in the morning.
3. was waiting for its prey.
4. killed the duckling.
5. took rest at the pond.
6. died

Now, read the sentences again and identify the verb phrases along with its constituents.

Now, read the sentences again and identify the verb phrases along with its constituents.

 A B Verb phrase died Verb phrase + noun phrase left the pool Verb phrase + prep.phrase was waiting for its prey Verb phrase + noun phrase + prep.phrase left the pool in the morning Verb phrase + noun phrase killed the duckling Verb phrase + noun phrase + prep. phrase took rest at the pond

Make sentences of your own with the following constituents in the verb phrase given below.
1. VP —
2. VP + NP —
3. VP + NP + PREP. PHRASE —
4. ………………..
5. ……………….
6. ……………….

1. VP — Joe came.
2. VP + NP — Joe killed a rat.
3. VP+NP+Pre Phrase — Joe killed a snake with a stick.
4. VP — Rosy smiled.
5. VP + NP — Rosy read a book.
6. VP+NP+Pre Phrase — Rosy read a book in the morning.

## Kerala Syllabus 9th Standard Biology Solutions Chapter 5 Excretion to Maintain Homeostasis

You can Download Excretion to Maintain Homeostasis Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 2 Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 9th Standard Biology Solutions Chapter 5 Excretion to Maintain Homeostasis

### Excretion to Maintain Homeostasis Textual Questions and Answers

Excretion to Maintain Homeostasis Question 1.
How can make our external environment garbage-free?
We can made our external environment garbage free by processing reusing or recycling waste material.

Excretion to Maintain Homeostasis Question 2.
Different by-products are formed as a result of many …. in the cells.
Metabolic activities

The Main Excretory Product in Human Beings is Question 3.
What are the main excretory products in human beings?
Carbon dioxide, water, and nitrogenous compounds.

Homeostasis Bodybuilding Question 4.
………. carries excretory products to excretory compounds.
Blood

Synthesis of Urea in Liver Takes Place by Question 5.
Complete the flow chart of the waste materials formed inside the cells reach excretory organs?

a) Tissue fluid
b) Blood

Excretory Organs

Biology Class 9 Chapter 5 Question 6.
Name the organs that help to remove waste materials from blood and maintain homeostasis?
Liver, Lungs, skin, and kidney

HSSlive in Class 9 Question 7.
Complete the illustration.

Liver – The Waste Processing Unit

Question 8.
What is the function of liver?
Liver converts harmful substances entering the body and those produced inside the body into harmless substances. Synthesis of urea from ammonia is an example for this.

Question 9.
…………….. are formed by the breakdown of protein.
Amino acids

Question 10.
What are the uses of amino acids?
Amino acids are used for the synthesis of various substances like proteins, enzymes etc. which are used for bodybuilding.

Question 11.
What is the most harmful by-product formed by the metabolism of amino acid?
Ammonia

Question 12.
Prepare a note on the synthesis of urea.
Amino acids are formed by the breakdown of proteins. As a result of the metabolic activities of amino acids, several nitrogenous by-products are formed. The most harmful among these is ammonia. 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 13.
Write down the chemical equation of the synthesis of urea.
Ammonia + carbon dioxide + water → urea

Formation Of Sweat

Question 14.
How is sweat formed from blood?
Blood passes through the capillaries excess water and minerals enter the sweat glands. This is eliminated as sweat through the body surface.

Question 15.
What are the components of sweat?
Urea, salt, and water

Question 16.
Which is the largest organ in our body?
Skin

Question 17.
What do you mean by sweat gland?
Sweat gland is a long coiled tube that opens to the surface of the skin.

Question 18.
The lower portion of the sweat gland is rich in
Capillaries

Question 19.
……… helps in regulating our body temperature.
Sweating

Kidneys

Question 20.
What are the main functions of kidney?
Kidneys are vital organs which help in maintaining homeostasis by filtering waste products like urea, salts, vitamins, other harmful substances, etc. from blood. When blood passes through the kidneys, the waste materials present in it are filtered.

Question 21.
Prepare a short note on the position and size of the kidney?
Human beings possess a pair of kidneys, situated on both sides of the vertebral column adjoining the muscles in the abdominal cavity. They are bean-shaped and are about 11 cm long, 5 cm broad and 3 cm thick. Each kidney is covered by a strong but soft membrane.

Question 22.
Illustrate kidney and its parts.

Question 23.
Complete the illustration of kidneys and associated parts.

a) bean-shaped and are located in the abdominal cavity on either side of the vertebral column.
b) Renal artery
c) Renal vein
d) Ureters

Internal Structure of Kidney

Question 24.
Internal structure of kidney

Question 25.
Analyze illustration given below and prepare table including the parts and peculiarities of nephron.

 Parts Peculiarities Bowman’s capsule The double-walled cup­shaped structure at one end of the nephron. The space between the two walls is called capsular space. Afferent vessel The branch of renal artery which enters the Bow­mann’s capsule. Glomerulus The region where afferent vessel enters the Bow­mann’s capsule and splits into minute capillaries. Efferent vessel The blood vessel that comes out of Bowman’s capsule. Peritubular capillaries Blood capillaries seen around the renal tubules as the continuation of the efferent vessel. Renal tubule The long tubule which connects the Bowman’s capsule and the collecting duct. Collecting duct The part where renal tubules enter. Absorption of water takes place. Urine is collected and is carried to the pelvis.

Question 26.
……… are the structural and functional units of kidneys.
Nephrons

Question 27.
Where is Bowmann’s Capsule of nephrons distributed?
Cortex

Question 28.
What is capsular space?
It is the space between the double walls of the Bowman’s capsule.

Question 29.
Blood capillaries seen around the renal tubules as continuation of the efferent vessel are ………..
Peritubular capillaries

Question 30.
………… helps the ultrafiltration.
Glomerulus

Formation of urine

Question 31.
List out the process of formation of urine.

• Ultrafiltration
• Reabsorption and secretion
• Absorption of water

Question 32.
What are the characteristic that help in ultrafiltration?
This process is supported by the high pressure developed in the glomerulus, due to the difference in the diameter of afferent vessel and efferent vessel

Question 33.
How is urine formed?
When blood flows through the glomerulus, ultrafiltration takes place through its small pores. The glomerular filtrate formed as a result of this is collected in the capsular space. When glomerular filtrate flows through renal tubeless to the collecting duct, essential components are reabsorbed to the peritubular capillaries. The absorption of excess water from the glomerular filtrate takes place in the collecting duct. What is left behind is urine.

Question 34.
Prepare a table relating to the different components of the glomerular filtrate and urine.

 Components of glomerular filtrate Components of urine Water Water-96% Glucose Urea-2% Amino acids Sodium chloride Sodium, potassium Potassium chloride calcium ions, vitamins Calcium salts Urea, uric acid creatinine, etc Phosphate, Uric acid Creatinine etc. – 2%

Question 35.
Urine is temporarily stored in the

Question 36.
State whether true or false
Washing out of germs inside the urinary tract also
takes place during the process of micturition.
True

Question 37.
How does avoiding timely urination affect our body and list out the healthy habits to be followed?
Avoiding urination for a long time prevents the expulsion of bacteria that may be present in the urinary tract and urinary bladder. This causes infection in the inner membrane of the urinary bladder.
Females are more susceptible to urinary tract infections when compared to males.
1) Frequent urination
2) Drink plenty of water
3) Keep the personal hygiene

Question 38.
Prepare a flow chart on the role of kidneys in maintaining homeostasis

Kidney Diseases

 Disease Reason Symptoms Nephritis inflammation of kidneys due to infection or intoxication. Turbid and dark-colored mine, back pain, fever, oedema on face and ankle Kidney stone Deposition of crystals of calcium salts in kidney or urinary tract. Pain in the lower abdomen, blockage of mine, dizziness, vomiting Uremia Different types of kidney diseases, nephritis. diabetes. Irigli’s blood pressure. Anemia, loss ol body weight, dizziness suffocation, dianitoea production of urine stops gradually.

Haemodialysis

Question 39.
What do you mean by hemodialysis?
Haemodialysis is the process proposed by modern medicine for the removal of wastes from the blood when both the kidneys become non-functional. In this process, blood is pumped into an artificial kidney called haemodialyser and is purified.

Question 40.
Who designed the first artificial kidney?
William Johann Kolff in 1944.

Kidney Transplantation

When both kidneys of an individual get damaged completely a fully functioning kidney should be received from a donor to save life. Kidney of a healthy person who died in an accident or of a completely healthy person can be transplanted after considering the matching of blood groups and tissues.

Excretion In Other Organisms

Diversity in Excretion:

Question 41.
Prepare a table about the excretory organs and excretory products of different organisms?

 Organism Excretory product Excretory organ Amoeba Ammonia, excess water in the body No special excretory organ, contractile vacuoles function as excretory organs. Earthworm Urea, ammonia, water Special structures called nephridia collect excretory products from body cavity and eliminate through pores in the body surface. Insects Uric acid Malpighian tubules seen along with digestive tract. They separate excretory products and eliminates along with digestive wastes. Fishes Ammonia Kidneys filter the wastes and eliminate directly to water Frog Urea Nitrogenous wastes filtered by kidneys are excreted in the form of urine. Reptiles and birds Uric acid Kidneys filter waste products and eliminate along with digestive wastes.

Question 42.
What are the methods of the excretion in plants?
Stomata, hydathodes, formation of heartwood, abscision of leaves.

Question 43.
Illustration related to the excretion in plants.

Let Us Assess

Question 1.
Glucose, amino acids, etc. found in the glomerular filtrate are absent in urine. Why?
When glomerular filtrate flows through renal tubules to the collecting duct, essential components are reabsorbed to the peritubular capillaries. So glucose and amino acids are absent in urine.

Question 2.
The steps involved in the formation of urine are given below. Arrange them in the correct sequence.
1. Collects urine
2. Ultrafiltration takes place
3. Reabsorption of ions takes place towards this part from renal tubules.
4. Collects glomerular filtrate
5. Excess urea is secreted here from peritubular capillaries.

Question 3.
Alcoholism is a bad habit which should be avoided. Analyze this statement relating it to the health of liver.
The detoxification of alcohol in our body is done by liver cells. As a result liver cell become damaged.

Question 4.
Based on the similarities in major excretory materials, arrange the following organisms properly in the table given below
Frog, Amoeba, Human beings, Fish, Birds, Insects