PSEB 9th Class Science Notes Chapter 11 Work, Energy and Power

This PSEB 9th Class Science Notes Chapter 11 Work, Energy and Power will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 11 Work, Energy and Power

→ All living beings need food for their energy needs.

→ Machines too need fuels like petrol and diesel for energy.

→ In our daily life, we consider any useful physical or mental activity as work.

→ Work done by the force acting on a body is equal to the product of the magnitude of the force and the displacement in the direction of the force.

→ When a force acts in a direction opposite to the direction of displacement produced, then the work done is negative.

PSEB 9th Class Science Notes Chapter 11 Work, Energy and Power

→ When a force acts in the direction of displacement then the work done is positive.

→ For us, the sun is the major natural source of energy.

→ In addition to this, we can also get energy from nuclei of atoms, from the interior of earth, and tides.

→ If a body has the capacity to do work then it is said to possess energy.

→ The object which does work loses energy and the object on which work is done gains energy.

→ The sum total of potential and kinetic energy of the body is called mechanical energy.

→ The energy present in the body due to its motion is called it’s kinetic energy.

→ The kinetic energy of an object increases with the increase of its velocity.

→ When an object is raised to some height above the ground, the work done against gravitational force is stored in the object as potential energy.

PSEB 9th Class Science Notes Chapter 11 Work, Energy and Power

→ We can transform one form of energy into another form.

→ The rate of doing work or transformation of energy is called power.

→ The unit of energy is joule but this unit is small. Its bigger unit is kilowatt hour (kWh) 1 kWh = 3.6 × 106 J

→ HP Energy consumed in industries and commercial establishments are expressed in kilowatt-hour which is also called unit (B.O.T. unit)

→ In order to survive all living beings have to do some basic activities for which energy is required. This energy they get from the food they eat.

→ According to the law of conservation of energy, energy can neither be created nor destroyed but it can only be transformed from one form to another form total energy before and after transformation always remains constant.

→ Energy is present in different forms as for example kinetic energy, potential energy, heat energy, chemical energy, electric energy, and light energy.

→ From a scientific point of view for doing work two conditions must be satisfied:

  • some force must act on the object
  • the object must be displaced in the direction of the force.

→ The unit of power is the watt.

→ 1 kilowatt = 1000 watt.

→ 1 watt is the power of that agent which does 1 joule of work in 1 second.

→ Energy: The capacity of doing work is called energy.

→ Kinetic Energy: The capacity of the bodies to do work due to the motion present in them is called kinetic energy.

PSEB 9th Class Science Notes Chapter 11 Work, Energy and Power

→ Potential Energy: The capacity of an object to do work clue to change in its position or configuration is called its potential energy.

→ Law of Conservation of Energy: Energy can neither be created nor destroyed but it can be changed from one form into another form.

→ Joule: If one Newton (N) of force acts on a body of 1 kg and displaces it through 1 m then work done on the body is one joule.

→ Power: The rate at which energy is supplied or consumed is called power. The unit of power is the watt (W).
Power = \(\frac{\text { Energy Supplied }}{\text { Time }}\)

→ Watt: If a source supplies or consumes energy at the rate of 1 Joule (J) per second then the power of the source, is said to be one watt.

→ Work: If the force acting on a body displaces the body in the direction of force then work is said to be done by the force.
Work = Force × Displacement
i.e. W = F × S

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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 6 Lines and Angles Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

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

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.3 1

Answer:
Here, ∠ SPR and ∠ PQT are exterior angles.
Then, by theorem 6.8,
∠ SPR = ∠ PQR + ∠ PRQ and
∠ PQT = ∠ QPR + ∠ PRQ
∴ ∠ PQR + ∠ PRQ = 135° and
∠ QPR + ∠ PRQ = 110°
Adding these two equations,
∠ PQR + ∠ PRQ + ∠ QPR + ∠ PRQ = 135° + 110°
∴ 180° + ∠ PRQ = 245° (Theorem 6.7)
∴ ∠ PRQ = 245° – 180°
∴ ∠ PRQ = 65°

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

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

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.3 2

Answer:
In ∆ XYZ,
∠ X + ∠ XYZ + ∠ XZY = 180° (Theorem 6.7)
∴ 62° + 54° + ∠ XZY = 180°
∴ ∠ XZY = 180° – 62° – 54°
∴ ∠ XZY = 64°
YO and ZO are bisectors of ∠ XYZ and ∠ XZY respectively.
∴ ∠ OYZ = \(\frac{1}{2}\) ∠ XYZ = \(\frac{1}{2}\) × 54° = 27° and
∠ OZY = \(\frac{1}{2}\) ∠ XZY = \(\frac{1}{2}\) × 64° = 32°.
Now, in ∆ OYZ,
∠ OYZ + ∠ OZY + ∠ YOZ = 180° (Theorem 6.7)
∴ 27° + 32° + ∠ YOZ = 180°
∴ ∠ YOZ = 180° – 27° – 32°
∴ ∠ YOZ = 121°

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

Question 3.
In the given figure, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.
PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.3 3

Answer:
AB || DE and AE is transversal for them.
∴ ∠ AED = ∠ BAE (Alternate interior angles)
∴ ∠ CED = ∠ BAC (Point C lies on line AE)
∴ ∠ CED = 35° (Given : ∠ BAC = 35°)
In ∆ CDE, by theorem 6.8
∠ CDE + ∠ CED + ∠ DCE = 180°
∴ 53° + 35° + ∠ DCE = 180°
∴ ∠ DCE = 180° – 53° – 35°
∴ ∠ DCE = 92°

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

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

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.3 4

Answer:
In ∆ PRT,
∠ RPT + ∠ PRT + ∠ PTR = 180° (Theorem 6.7)
∴ 95° + 40° + ∠ PTR = 180°
∴ 135° + ∠ PTR = 180°
∴ ∠ PTR = 180°- 135°
∴ ∠ PTR = 45°
Lines PQ and RS intersect at point T.
∴ ∠ STQ = ∠ PTR (Vertically opposite angles)
∴ ∠ STQ = 45°
In ∆ STQ,
∠ TSQ + ∠ STQ + ∠ SQT = 180° (Theorem 6.7)
∴ 75° + 45° + ∠ SQT = 180°
∴ 120° + ∠ SQT = 180°
∴ ∠ SQT = 60°

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

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

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.3 5

Answer:
PQ || SR and QR is transversal for them.
∴ ∠ PQR = ∠ QRT (Alternate interior angles)
∴ ∠ PQR = 65° (Given : ∠ QRT = 65°)
∴ ∠ PQS + ∠ SQR = 65° (Adjacent angles)
∴ x + 28° = 65° (Given : ∠ SQR = 28°)
∴ x = 65° – 28°
∴ x = 37°
PQ ⊥ PS
∴ ∠ SPQ = 90°
In ∆ PSQ,
∠ SPQ + ∠ PQS + ∠ PSQ = 180° (Theorem 6.7)
∴ 90° + 37° + y = 180°
∴ 127° + y = 180°
∴ y = 180°- 127°
∴ y = 53°

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

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

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

Answer:
QT is the bisector of ∠ PQR and RT is the bisector of ∠ PRS.
∴ ∠ TQR = \(\frac{1}{2}\) ∠ PQR and ∠ TRS = \(\frac{1}{2}\) ∠ PRS ……………… (1)
∠ PRS is an exterior angle of ∆ PQR.
∴ ∠ PRS = ∠ QPR + ∠ PQR
∴ \(\frac{1}{2}\) ∠ PRS = \(\frac{1}{2}\) ∠ QPR + \(\frac{1}{2}\) ∠ PQR
∴ ∠ TRS = \(\frac{1}{2}\) ∠ QPR + ∠ TQR [By (1)] …………… (2)
In ∆ TQR, ∠ TRS is an exterior angle.
∴ ∠ TRS = ∠ QTR + ∠ TQR ……………. (3)
From (2) and (3), we get
∠ QTR + ∠ TQR = \(\frac{1}{2}\) ∠ QPR + ∠ TQR
∴ ∠ QTR = \(\frac{1}{2}\) ∠ QPR

PSEB 9th Class Science Notes Chapter 10 Gravitation

This PSEB 9th Class Science Notes Chapter 10 Gravitation will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 10 Gravitation

→ Everybody in this universe attracts every other body towards it with a force called Force of Gravitation.

→ The revolution of the moon around the earth and the falling down of a body projected upward is due to the force of gravitation.

→ Gravitation is a weak force unless it may not involve bodies having greater masses. The attractive force due to earth is called the force of gravity.

→ The law of gravitation is a universal law which means that this law applies to all big and small objects.

→ The magnitude of the force acting between the earth and the objects lying on the surface of the earth is given by formula F = \(\frac{\mathrm{GM} m}{d^{2}}\) where G = 6.673 × 10-11 N-m2/kg2

→ The acceleration produced in the body due to the force of gravity is called acceleration due to gravity. It is denoted by ‘g’.

PSEB 9th Class Science Notes Chapter 10 Gravitation

→ The value of ‘g’ is more on the poles than on the equatorial line.

→ The Force of gravity decreases with the increase of height above the surface of the earth.

→ The quantity of matter present in a body is called the mass of the body.

→ Mass is the measure of inertia of the body.

→ The mass of a body remains constant at all places.

→ The weight of a body is the force with which the body is attracted towards the centre of the earth.

→ The weight of an object depends upon the product of its mass (m) and the acceleration due to gravity (g). i.e. Weight (W) = mass (m) × acceleration due to gravity (g)

→ Force acting perpendicular to the surface of an object is called thrust.

→ Thrust acting per unit area is called pressure.
∴ Pressure = \(\frac {Thrust}{Area}\)

→ SI unit of pressure is Nm-2 (or N/m2) or pascal (Pa).

PSEB 9th Class Science Notes Chapter 10 Gravitation

→ The pressure exerted on an enclosed liquid is transmitted equally in all directions.

→ All bodies when immersed in liquid experience an upward force called upthrust or buoyant force.

→ The value of Buoyant force depends upon the density of the liquid.

→ If the weight of an immersed body is more than the upthrust then the body sinks in the liquid.

→ Those bodies float which has a density less than the density of the liquid in which they are immersed.

→ Those bodies, which have a density more than the density of the liquid in which it is immersed sink.

→ Archimedes’ principle states that when a solid body is wholly or partly immersed in a liquid then the body experiences an upward thrust due to liquid displaced by the immersed part of the body.

→ Relative density of a substance is defined as the ratio of the density of the substance to the density of the Same volume of water.
∴ Relative density = \(\frac {Density of substance}{Density of water}\)

→ Relative density has no unit.

→ According to Archimedes’ principle when a solid body is completely or partially immersed in a liquid and the weight of the body is equal to the weight of the liquid displaced by the immersed part of the body then the body floats.

→ Kepler gave the following three laws which govern the motion of the planets:

  • The planetary path of each planet is elliptical at the centre of which the sun is situated.
  • The line joining sun and the planet sweeps equal areas in equal intervals of time.
  • The cube of the average distance ‘r’ of a planet from the sun is inversely proportional to the square of the orbital period ‘T’ of that planet.

PSEB 9th Class Science Notes Chapter 10 Gravitation

→ Law of gravitation states that the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. This force always acts along the line joining the centres of two objects.

→ Force of Gravitation: It is that force with which two bodies attract each other but they are lying apart.

→ Acceleration due to Gravity: Acceleration produced in the bodies falling freely under the action of gravity is called acceleration due to gravity.

→ Force of Gravity: It is the force with which earth attracts all bodies towards its centre.

→ Weight: The force with which the earth attracts bodies towards it is called weight.

→ Mass: The quantity of matter contained in the body is called it’s mass.

→ Inertial Mass: This mass measures the resistance which is produced due to a change in the position of rest or motion.

→ Universal Gravitational Constant: It is the force that acts between two bodies each of unit mass and lying unit distance apart.

→ Kepler’s first law: All planets move around the sun in elliptical orbits.

→ Kepler’s second law: The line joining planet and sun sweeps out equal area in equal intervals of time.

→ Kepler’s third law: Cube of mean distance ‘r’ of the planet from the sun is proportional to the square of its orbital period.

PSEB 9th Class Science Notes Chapter 10 Gravitation

→ Thrust: The total force acting perpendicular to the surface of the body is called thrust.

→ Buoyant Force: The upward thrust applied by the liquid on the immersed body is called Buoyant force.

→ Pressure: Thrust acting per unit area of the surface in contact is called pressure.

→ Density: Mass per unit volume of the substance is called the density of the substance or it is termed as the ratio of the mass of a body to its volume.

  • Density = \(\frac {Mass of the body}{Volume of the body}\)
  • SI unit of density is kg/m3.

→ Relative Density: Relative density of a substance is defined as the ratio of the density of the substance to the density of the same volume of water. Since it is the ratio of two similar quantities, therefore it is only a number. It is also called the specific gravity of the substance.

→ Archimedes Principle: This principle states that whenever a solid body is immersed in a liquid either wholly or partly then it experiences an upward thrust due to the displaced liquid.

→ The universal law of Gravitation: The mutual force of attraction between every two bodies in this universe is directly proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance between them. This force acts along the line joining the centres of the two bodies.

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

Question 1.
Find the roots of the following quadratic equations if they exist, by the method of completing the square:
(i) 2x2 + 7x + 3
(ii) 2x2 + x – 4 = 0
(ili) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) Given quadratic equation is
2x2 – 7x + 3 = 0
Or 2x2 – 7x = -3
Or x2 – \(\frac{7}{2}\)x = –\(\frac{3}{2}\)
Or x2 – \(\frac{7}{2}\)x + (\(\frac{-7}{4}\))2 = \(\frac{-3}{2}+\left(\frac{-7}{4}\right)^{2}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-3}{2}+\frac{49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-24+49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}\)

Or x – \(\frac{7}{4}\) = \(\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}\)

Case I:
When x – \(\frac{7}{4}\) = \(\frac{5}{4}\)
Or x = \(\frac{5}{4}+\frac{7}{4}=\frac{5+7}{4}\)
Or x = \(\frac{12}{4}\) = 3

Case II:
When x – \(\frac{7}{4}\) = \(\frac{-5}{4}\)
Or x = \(\frac{-5}{4}+\frac{7}{4}=\frac{-5+7}{4}\)
Or x = \(\frac{2}{4}=\frac{1}{2}\)
Hence, roots of given quadratic equation is 3, \(\frac{1}{2}\).

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

(ii) Given Quadratic Equation is
2x2 + x – 4 = 0
Or 2x2 + x = 4
Or x2 + \(\frac{1}{2}\)x = \(\frac{4}{2}\)

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 1

Case I:
When x + \(\frac{1}{4}\) = \(\frac{-\sqrt{33}}{4}\)
Or x = \(-\frac{\sqrt{33}}{4}-\frac{1}{4}\)
Or x = \(\frac{-\sqrt{33}-1}{4}\)
Hence, roots of given quadratic equation are \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\).

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

(iii) Given quadratic equation is
4x2 + 4√3x + 3 = 0
Or 4x2 + 4√3x = -3
Or x2 + \(\frac{4 \sqrt{3}}{4}\)x = \(\frac{-3}{4}\)
Or x2 + √3x = \(\frac{-3}{4}\)
Or x2 + √3x + \(\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\left(\frac{\sqrt{3}}{2}\right)^{2}\)
Or \(\left(x+\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\frac{3}{4}\)
or (x + \(\frac{\sqrt{3}}{2}\))2 = 0
(x + \(\frac{\sqrt{3}}{2}\)) (x + \(\frac{\sqrt{3}}{2}\)) = 0
Either x + \(\frac{\sqrt{3}}{2}\) = 0
x = –\(\frac{\sqrt{3}}{2}\)
Or x + \(\frac{\sqrt{3}}{2}\) = 0
Or x = –\(\frac{\sqrt{3}}{2}\)
Hence, roots of given quadratic equation are –\(\frac{\sqrt{3}}{2}\) and –\(\frac{\sqrt{3}}{2}\).

(iv) Given quadratic equation is
2x2 + x + 4 = 0
2x2 + x = -4
x2 + \(\frac{1}{2}\)x = \(-\frac{4}{2}\)
Or x2 + \(\frac{1}{2}\)x + (\(\frac{1}{4}\))2 = -2 + (\(\frac{1}{4}\))2

Or \(\left(x+\frac{1}{4}\right)^{2}=-2+\frac{1}{16}\)
Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-32+1}{16}\)
Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-31}{16}<0\)

∴ square of any number cannot be negative. So, (x + \(\frac{1}{4}\))2 cannot be negative for any real x.
∴ There is no real x whith satisfied the given quadratic equation.
Hence, given quadratic equation has no real roots.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 2.
Find the roots of the quadratic equations given in Q. 1 by applying the quadratic formula. Which of the above two
methods do you prefer, and why?
Solution:
(i) Given quadratic equation is
2x2 – 7x + 3 = 0
Compare it with ax2 + bx + c = 0
a = 2, b = -7, c = 3
Now, b2 – 4ac = (-7)2 4 x 2 x 3
= 49 – 24
= 25 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-7) \pm \sqrt{25}}{2 \times 2}=\frac{7 \pm 5}{4}\)
= \(\frac{7+5}{4} \text { and } \frac{7-5}{4}\)
= \(\frac{12}{4} \text { and } \frac{2}{4}\)
= 3 and \(\frac{1}{2}\)
Hence, 3 and \(\frac{1}{2}\) are the roots of given quadratic equation.

(ii) Given quadratic equation is
2x2 + x – 4 = 0
Compare it with ax2 + bx + c = 0
∴ a = 2, b = 1, c = -4
Now,
b2 – 4ac = (1)2 – 4 × 2 × 4
= 1 + 32 = 33 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-1 \pm \sqrt{33}}{2 \times 2}=\frac{-1 \pm \sqrt{33}}{4}\)
= \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)
Hence, \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\) are the roots of given quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

(iii) Given quadratic equation is
4x2 + 4√3x + 3 = 0
Compare it with ax2 + bx + c = 0
a = 4, b = 4√3, c = 3
b2 – 4ac = (4√3)2 – 4 × 4 × (3)
= 48 – 48 = 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-4 \sqrt{3} \pm \sqrt{0}}{2 \times 4}\)

= \(-\frac{4 \sqrt{3}}{8}\), \(-\frac{4 \sqrt{3}}{8}\)

= –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\)

Hence, –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\) are the roots of given quadratic equation.

(iv) Given quadratic equation is 2x2 + x + 4 = 0
Compare it with ax2 + bx + c = 0
∴ a = 2, b = 1, c = 4
Now, b2 – 4ac = (1)2 – 4 × 2 × 4
= 1 – 32 = -31 < 0
But
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Since the square of a real number cannot be negative, therefore x will not have any real value.
Hence, there are no real roots for the given quadratic equation.

From above two questions, we used two methods to find the roots of the quadratic equations. It is very clear from above discussion that quadratic formula method is very convenient as compared to method of completing the square.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 3.
Find the roots of the following equations:

(i) x – \(\frac{1}{x}\) = 3, x ≠ 0
(ii) \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\), x ≠ -4, 7
Solution:
(i) Given Equation is
x – \(\frac{1}{x}\) = 3
Or \(\frac{x^{2}-1}{x}\) = 3
Or x2 – 1 = 3x
Or x2 – 3x – 1 = 0
Compare it with ax2 + bx + c = 0
∴ a = 1, b = -3, c = -1
Now, b2 – 4ac = (-3)2 – 4 . 1 . (-1)
= 9 + 4 = 13 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-3) \pm \sqrt{13}}{2 \times 1}\)
= \(\frac{3 \pm \sqrt{13}}{2 \times 1}\)
= \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\)
Hence, \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\) are the roots of given quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

(ii) Given equation is

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 2

-11 × 30 = 11 (x2 – 3x – 28)
Or -30 = x2 – 3x – 28
Or x2 – 3x – 28 + 30 = 0
Or x2 – 3x + 2 = 0
Compare it with ax2 + bx + c = 0
a = 1, b = – 3, c = 2
Now, b2 – 4ac = (-3)2 – 4 × 1 ×2
= 9 – 8 = 1 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-3) \pm \sqrt{1}}{2 \times 1}=\frac{3 \pm 1}{2}\)
= \(\frac{3+1}{2}\) and \(\frac{3+1}{2}\)
= \(\frac{4}{2}\) and\(\frac{2}{2}\) and 1
Hence, 2 and 1 are the roots of given quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). F1nd his present age.
Solution:
Let Rehman’s present age = x years
3 years ago Rehman’s age (x – 3) years
5 years from now Rehman’s age =(x + 5) years
According to question,
\(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

Or \(\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+5 x-3 x-15}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+2 x-15}=\frac{1}{3}\)
Or 6x + 6 = x2 + 2 -15
Or x2 + 2x – 15 – 6x – 6 = 0
Or x2 – 4x – 21 = 0, which is quadratic in x.
So compare it with ax2 + bx + c =0
a = 1, b = -4, c = -21
Now, b2 – 4ac = (- 4)2 4 × 1 × (-21)
= 16 + 84 = 100 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
x = \(\frac{-(-4) \pm \sqrt{100}}{2 \times 1}\)
= \(\frac{4 \pm 10}{2}\)
= \(\frac{4+10}{2}\) and \(\frac{4-10}{2}\)
\(\frac{14}{2}\) and \(\frac{-6}{2}\)
= 7 and -3
∵ age cannot be negative,
so, we reject x = – 3
∴ x = 7
Hence, Rehman’s present age = 7 years.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 5.
In a class test, the sum of Shefall’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less ¡n English, the product of their marks would have been 210. Find her marks In the two subjects.
Solution:
Let Shefali get marks in Mathematics = x
Shefali’s marks in English = 30 – x
According to 1st condition,
Shefali’s marks in Mathematics = x + 2
and Shefali’s marks in English = 30 – x – 3 = 27 – x
∴ Their product = (x + 2) (27 – x)
= 27x – x2 + 54 – 2x
= x2 + 25x + 54
According to 2nd condition,
-x2+ 25x+ 54 = 210
Or -x2 + 25x + 54 – 210 = 0
Or -x2 + 25x – 156 = 0
Or x2 – 25x+ 156 = o
Compare it with ax2 + bx + c = O
a = 1, b = -25, c = 156
Now, b2 – 4ac = (-25)2 – 4 × 1 × 156
= 625 – 624 = 1 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-25) \pm \sqrt{1}}{2 \times 1}\)
= \(\frac{25 \pm 1}{2}\)
= \(\frac{25+1}{2}\) and \(\frac{25-1}{2}\)
= \(\frac{26}{2}\) and \(\frac{24}{2}\)
= 13 and 12.

Case I:
When x = 13
then Shefaiis marks in Maths = 13
Shefali’s marks in English = 30 – 13 = 17.

Case II:
When x = 12
then Shefalis marks in Maths = 12
Shefali’s marks in English = 30 – 12 =18.
Hence, Shefalis marks in two subjects are 13 and 17 Or 12 and 18.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. if the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let shorter side of rectangular field = AD = x m

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 3

Longer side of rectangular field = AB = (x + 30) m
and diagonal of rectangular field = DB = (x + 60) m
In rectangle. the angle between the length and breadth is right angle.
∴ ∠DAB = 90°
Now, in right angled triangle DAB, using Pythagoras Theorem,
(DB)2 = (AD)2 + (AB)2
(x + 60)2 = (x)2 + (x + 30)2
Or x2 + 3600 + 120x = x2 + x2 + 900 + 60x
Or x2 + 3600 + 120x – 2x2 – 900 – 60x = 0
Or -x2 + 60x + 2700 = 0
Or x2 – 60x – 2700 = 0
Compare it with ax2 + bx + e = O
∴ a = 1, b = -60, c = -2700
and b2 – 4ac = (-60)2 – 4. 1 . (-2700)
= 3600 + 10800 = 14400 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-60) \pm \sqrt{14400}}{2 \times 1}\)

= \(\frac{60 \pm 120}{2}\)

= \(\frac{60+120}{2}\) and \(\frac{60-120}{2}\)
= \(\frac{180}{2}\) and \(\frac{-60}{2}\)
= 90 and – 30
∴ length of any side cannot be negative
So, we reject x = -30
∴ x = 90
Hence, shorter side of rectangular field = 90 m
Longer side of rectangular field = (90 + 30) m = 120 m.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find
the two numbers.
Solution:
Let larger number = x .
Smaller number = y
According to 1st condition,
x2 – y2 = 180 ……………(1)
According to 2nd condition,
y2 = 8x
From (1) and (2), we get
x2 – 8x = 180
Or x2 – 8x – 180 = 0
Compare it with ax2 + bx + c = 0
∴ a = -1, b = -8, c = -180
and b2 – 4ac = (-8)2 – 4 × 1 × (-180)
= 64 + 720 = 784 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-8) \pm \sqrt{784}}{2 \times 1}\)
= \(\frac{8 \pm 28}{2}\)
= \(\frac{8+28}{2}\) and \(\frac{8-28}{2}\)
= \(\frac{36}{2}\) and \(\frac{-20}{2}\)
= 18 and -10
When x = – 10 then from (2),
y2 = 8 (- 10) = – 80, which is impossible.
So, we reject x = – 10
When x = 18 then from (2).
y2 = 8(18) = 144
Or y = ±√144
Or y = ± 12
Hence, required numbers are 18 and 12 Or 18 and -12.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 8.
A train travels 360 km ¡t a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let constant speed of the train = x km/hour
Distance covered by the train = 360 km
Time taken by the train = \(=\frac{\text { distance }}{\text { speed }}\)
(∵ speed = \(\frac{\text { Distance }}{\text { Time }}\))
= \(\frac{360}{x}\)
Increased speed of the train = (x + 5) km/hour
∴ Time taken by the train with increased speed = \(\frac{360}{x+5}\) hour
According to question

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 4

Or 1800 = x2 + 5x
Or x2 + 5x – 1800 = 0
Compare it with, ax2 + bx + c = 0
a = 1, b = 5, c = – 1800
and b2 – 4ac = (5)2 4 × 1 × (- 1800)
= 25 + 7200 = 7225 > 0

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-5 \pm \sqrt{7225}}{2 \times 1}\)
= \(\frac{-5 \pm 85}{2}\)
= \(\frac{-5+85}{2}\) and \(\frac{-5-85}{2}\)
= \(\frac{80}{2}\) and \(\frac{-90}{2}\)
= 40 and – 45
∵ speed of any train cannot be negative.
So, we reject x = – 45
x = 40
Hence, speed of train = 40 km/hour.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 9.
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let time taken by larger tap to fill the tank = x hours.
Time taken by smaller tap to fill the tank = (x + 10) hours
In case of one hour:
Larger tap can fill the tank = \(\frac{1}{x}\)
Smaller tap can fill the tank = \(\frac{1}{x+10}\)
∴ Larger and smaller tap fill the tank = \(\frac{1}{x}\) + \(\frac{1}{x+10}\) ………….(1)
But, two taps together can fill the tank = 9\(\frac{3}{8}\)hour = \(\frac{75}{8}\) hour
Now, two taps together can fill the tank in one hour = \(\frac{8}{75}\) ……………..(2)
From (1) and (2), we get
\(\frac{1}{x}+\frac{1}{x+10}=\frac{8}{75} \)

Or \(\frac{x+10+x}{x(x+10)}=\frac{8}{75}\)

Or \(\frac{2 x+10}{x^{2}+10 x}=\frac{8}{75}\)

Or 75(2x + 10) = 8(x2 + 10x)
Or 150x + 750 = 8x2 + 80x
Or 8x2 + 80x – 150x – 750 = 0
Or 8x2 – 70x – 750 = 0
Or 4x2 – 35x – 375 = 0
Compare it with ax2 + bx + c = 0
∴ a = 4, b = -35, c = -375
and b2 – 4ac = (35)2 -4 × 4 × (-375)
= 1225 + 6000 = 7225 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-35) \pm \sqrt{7225}}{2 \times 4}\)

= \(\frac{35 \pm 85}{8}\)

= \(\frac{35+85}{8}\) and \(\frac{35-85}{8}\)

= \(\frac{120}{8}\) and \(\frac{-50}{8}\)

= 15 and \(\frac{-25}{4}\)

∵ time cannot be negative.
So,we reject x = \(\frac{-25}{4}\)
∴ x = 15
Hence, larger water tap fills the tank = 15 hours
and smaller water tap fills the tank = (15 + 10) hours = 25 hours.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop to intermediate stations), if the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Solution:
Let average speed of passenger train = x km/hour
Average speed of express train = (x+ 11) km/hour
Distance between Mysore and Bangalore = 132 km
Time taken by passenger train = \(\frac{132}{x}\) hour
[∵ Speed = \(=\frac{\text { Distance }}{\text { Time }}\) ]
Time taken by express train‚ = \(\frac{132}{x+11}\) hour
According to question,

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 5

Or 1452 = x2 + 11x
Or x2 + 11x – 1452 = 0
Compare it with ax2 + bx + c = 0
∴ a = 1, b = 11, c = -1452
and b2 – 4ac = (11)2 – 4 × 1 × (- 1452)
= 121 + 5808 = 5929 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-11 \pm \sqrt{5929}}{2 \times 1}\)

= \(\frac{-11 \pm 77}{2}\)

= \(\frac{-11+77}{2}\) and \(\frac{-11-77}{2}\)

= \(\frac{66}{2}\) and \(\frac{-88}{2}\) = 33 and -44

∵ speed of any train cannot be negative
∴ x = 33
Hence, speed of passenger train = 33 km/hour
and speed of express train = (33 + 11) km/hour = 44 km/hour.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 11.
Sum of the areas of two squares is 468 m2. If the difference of their perimeters
is 24 m, find the sides of the two squares.
Solution:
In case of larger square
Let length of each side of square = x m
Area of square = x2 m2
Perimeter of square = 4x m

In case of smaller square:

Let lenth of each side of square = y m
Area of square = y2 m2
Perimeter of square = 4y m
According to 1st condition,
x2 + y2 = 468 …………….(1)
According to 2nd condition,
4x – 4y = 24
Or 4(x – y) = 24
Or x – y = 6
x = 6 + y
From (1) and (2), we get
(6 + y)2 + y2 = 468
Or 36 + y2 + 12y + y2 = 468
Or 2y2 + 12y + 36 – 468 = 0
Or 2y2 + 12y – 432 = 0
Or y2 + 6y – 216 = 0
Compare it with ay2 + by + c = 0
∴ a = 1, b = 6, c = -216
and b2 – 4ac = (6)2 – 4 × 1 × (- 216) = 36 + 864 = 900 > 0
∴ y = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-6 \pm \sqrt{900}}{2 \times 1}\)

= \(\frac{-6 \pm 30}{2}\)

= \(\frac{-6+30}{2}\) and \(\frac{-6-30}{2}\)

= \(\frac{24}{2}\) and \(\frac{-36}{2}\) = 12 and -18

∵ length of square cannot be negative
So, we reject y = – 18
∴ y = 12
From (2), x = 6 + 12 = 18
Hence, sides of two squares are 12 m and 18 m.

PSEB 9th Class Science Notes Chapter 9 Force and Laws of Motion

This PSEB 9th Class Science Notes Chapter 9 Force and Laws of Motion will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 9 Force and Laws of Motion

→ Gallileo and Isac Newton put forth the scientific basis with regard to the motion of the objects.

→ The concept of force is based on the activity of pull, push or hit/kick.

→ By the application of force, the size and shape of an object can be changed.

→ Force is of two types:

  • Balanced force
  • Unbalanced force

→ An unbalanced force acting on an object produces motion in the object.

PSEB 9th Class Science Notes Chapter 9 Force and Laws of Motion

→ The frictional force acts against the direction of motion of an object.

→ The first law of motion is also called the Laws of Inertia.

→ The tendency of the body to remain in the state of rest or of uniform motion is called Inertia.

→ Everybody opposes the change in its state of motion.

→ The moment of inertia of the train is more than that of the cart so that it does not get displaced on being pushed. In other words, the heavier bodies have more inertia.

→ The inertia of an object is the measure of its mass.

→ The momentum ‘p’ of an object is measured by the product of its mass ‘m’ and velocity ‘v’. i.e. p = m × v

→ The momentum has both magnitude and velocity. Its direction is the same as that of velocity.

→ The S.I. unit of momentum is kg ms.

→ Force changes the momentum of an object.

→ To reduce the effect of friction the surface is either made plane or the surface is painted with a lubricant.

→ S.I. unit of force is the newton.

→ For the motion of the object due to the force acting on it, Newton gave three fundamental laws of motion which hold good for all types of motion.

→ According to the first law, everybody tends to remain in its state of rest or of uniform motion unless some external force acts on it.

PSEB 9th Class Science Notes Chapter 9 Force and Laws of Motion

→ According to the second law, the rate of change of momentum of a body is directly proportional to the unbalanced force acting on it and in its direction.

→ According to the third law, when one body applies force on another body, the second body also applies instantaneous force on the first body. These two forces are equal in magnitude but opposite in direction.

→ By applying force the motion can be given to a body at rest and the body in motion can be brought to rest or a change can be brought about in the same direction.

→ Force: It is an external cause that produces or tends to produce a change in its state of rest or of uniform motion.

→ One Newton Force: It is the force that produces an acceleration of 1 ms-2 in a body of mass 1 kg. It is denoted by ‘N’.

→ Balanced Forces: If a number of forces acting on an object do not produce any change in its state then these forces are called Balanced Forces. Their net result is zero force.

→ Unbalanced Forces: If the result of various forces acting on an object is not zero, then these forces are called unbalanced forces.

→ Frictional Force: It is the opposing force that comes into play when a body moves over the surface of another body.

→ Inertia: It is the property of the bodies due to which they cannot change their position of rest or of uniform motion unless some external force is applied.

→ The inertia of Rest: It is the property by virtue of which a body at rest will continue to remain at rest unless some external force is applied to bring about that change.

→ The inertia of Motion: It is the property of a body by virtue of which the body in motion, will continue to move with the same uniform speed unless some external force is applied to bring about that change.

→ Momentum: The product of mass and velocity possessed by a body is called its momentum.

→ Law of Conservation of Momentum: If no external force acts on a system of particles then the total momentum of the system remains conserved.

PSEB 9th Class Science Notes Chapter 9 Force and Laws of Motion

→ Newton’s Second Law of Motion: Tine rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.

→ Newton’s Third Law of Motion: Every action has an equal and opposite reaction.

→ Newton’s First Law of Motion: If a body is at rest, it will continue to remain at rest. And if a body is moving with a uniform velocity in a straight line it will continue to do so unless some external force is applied to bring about that change in its state.

PSEB 9th Class Science Notes Chapter 8 Motion

This PSEB 9th Class Science Notes Chapter 8 Motion will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 8 Motion

→ A body is said to be in motion when it changes its position with time.

→ At any instant, a body may appear to be in motion to one person but the same object may appear at rest to another person.

→ To describe the position of an object we are required to fix the directive point which is called the origin.

→ The motion of some objects can be controlled while that of some other objects remains uncontrolled and irregular.

PSEB 9th Class Science Notes Chapter 8 Motion

→ When an object moves along a straight line path then its motion is called linear motion.

→ Those quantities which can be described completely by their magnitude only are called physical quantities.

→ The shortest distance measured between initial and final position is called displacement of the body.

→ When the final position of the body in motion coincides with its initial position then the value of displacement becomes equal to zero.

→ The displacement of an object may be zero but its distance will not be zero.

→ An odometer is a device that measures the distance covered by automobiles.

→ The rate of change of motion of an object is called its speed. The unit of speed is ms-1.

→ The average speed of an object is obtained by dividing the distance covered by the body by the time taken to cover that distance.
∴ Average speed = Total distance covered/Time taken to cover that distance

→ The rate of change of motion of a body in a particular direction is called its velocity i.e. speed in a particular direction is called velocity.

→ Time interval is measured by digital wristwatch or stopwatch.

PSEB 9th Class Science Notes Chapter 8 Motion

→ Speed of sound in air = 340 m/s.

→ Speed of light in air = 3 × 108 m/s.

→ The rate of change of velocity of a body is called its acceleration

  • Acceleration = Change in velocity/Time taken
  • S.I. unit of acceleration is ms-2.

→ If the velocity of an object changes by the same rate then its mean (average) velocity, is obtained by taking the arithmetic mean of initial velocity and final velocity.
∴ Average Velocity = \(\frac{\text { Initial velocity + Final velocity }}{2}\)

→ If an object covers equal distances in equal intervals of time however small the interval maybe then its motion is called uniform motion.

→ If a moving object covers unequal distances in equal intervals of time then the motion of the object is called non-uniform motion.

→ Rest: A body is said to be at rest if it does not change its position with respect to its surroundings.

→ Motion: A body is said to be in motion if it changes its position with time with respect to its surroundings.

→ Distance: It is the length of the actual path travelled by a body between its initial position and final position.

→ Displacement: Tire change in position of an object in a given direction is known as displacement. It is measured by the shortest distance moved by a body from the initial position to the final position.

→ Uniform Motion: If an object covers equal distances in equal intervals of time however small the time intervals may be, then the motion of the object is said to be uniform motion.

PSEB 9th Class Science Notes Chapter 8 Motion

→ Non-Uniform Motion: If an object covers unequal distances in equal intervals of time then the object is said to be in non-uniform motion.

→ Speed: The distance travelled by a body in a unit of time
Speed = Distance traveled/Total time taken

→ Average Speed: It is the total distance travelled bv an object divided by the time taken to travel that distance.
Average speed = Total distance travelled/Total time taken

→ Uniform Velocity: If a body covers equal distances in equal intervals of time, however small these intervals maybe, in a particular direction its velocity is said to be uniform velocity.

→ Variable Velocity: If a body travels unequal distances in equal intervals of time or changes its direction or both then its velocity is said to be variable velocity.

→ Uniform Circular Motion: When a body moves along a circular path with uniform speed, its motion is called uniform circular motion. Tire motion is uniformly accelerated due to continuous change of direction.

→ Angular Velocity: It is the angle swept by the radius per unit time
ω = \(\frac{\text {θ}}{t}\) rad/s

→ Velocity: It is the displacement produced per unit time.
or
It is the distance travelled per unit time in a given direction.
Velocity = Displacement/Time

PSEB 9th Class Science Notes Chapter 8 Motion

→ Uniform Speed: When an object covers equal distances in equal intervals of time, however small these intervals may be, it is said to move with uniform speed.

→ Non-Uniform Speed: When an object covers unequal distances in equal intervals of time, it is said to move with non-uniform speed.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

This PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

→ There are about 10 million living organisms present on earth and only 1/3rd of these organisms have been identified.

→ All organisms differ from each other in their structure.

→ The classification of organisms is arranging organisms into groups or sets on the basis of similarities and differences which also exhibit their relationships.

→ Each organism is different from all others to a lesser or greater extent.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

→ The science of classification is called taxonomy.

→ Bacteria are microscopic of a few micrometres in size whereas blue whale and red Wood trees of California are of size 30 metres and 100 metres respectively.

→ Present-day diversity in living beings is the product of 3.5 billion years of organic evolution.

→ Carolus Linnaeus is the Father of Taxonomy.

→ Classification makes the study of a wide variety of organisms easy.

→ Classification helps us explore the diversity of life forms.

→ Linnaeus classified organisms on the basis of similarities and differences.

→ Binomial nomenclature was given by Linnaeus. Each organism is assigned two names, generic name, name of the genus, and specific name, of species.

→ In two-kingdom systems of classification, kingdom Plantae and kingdom Animalia were classified.

→ Plants are autotrophic, prepare their own food and plant cells have cell walls made up of cellulose. Divided into two groups crypto game and phanerogamae.

→ Animals lack chlorophyll and are heterotrophic in nutrition.

→ R. H. Whittaker (1969) proposed five kingdom systems of classification.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

→ The five kingdoms are Monera, Protista, Fungi, Plantae, and Animalia.

→ Monerans lack well-defined nuclei and membrane-bound cell organelles.

→ This group includes bacteria, blue-green algae (cyanobacteria), and Mycoplasma.

→ Protists include unicellular eukaryotic organisms. Some may have specialized structures for moving such as cilia, flagella, or pseudopodia.

→ Agar, diatoms, and protozoans are examples.

→ Plantae and Animalia are further divided into subdivisions on the basis of increasing complexity.

→ Charles Darwin (1959) described the idea of evolution in his book, ‘The Origin of Species’.

→ ‘Primitive’ or ‘lower’ organisms have ancient body designs.

→ Advanced or higher organisms acquired their particular designs relatively recently.

→ Complexity in design increases over evolutionary time.

→ Biodiversity refers to a variety of life forms found in a particular region.

→ The warm and humid tropical regions of the earth are rich in diversity of plant and animal life. The region is called the region of megadiversity.

→ Ernst Haeckel (1894), Robert Whittaker (1959), and Carl Woese (1977) have tried to classify all living organisms into broad categories, called Kingdoms.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

→ Aristotle classified animals according to their habitat (land or water).

→ Basis of classification

  • Nature of cell Prokaryotic or eukaryotic cell.
  • The number of cells Single-cell or multicellular.
  • Method of preparing food
  • Plants carry out photosynthesis thus autotrophs.
  • Fungi absorb food, thus absorptive.
  • Animals have an ingestive (holozoic) modes of feeding.

PSEB 9th Class Science Solutions Chapter 7 Diversity in Living Organisms 3

Classification of Kingdom Animalia:

→ Protozoa are included in Protista, include single-celled, aquatic, free-living or parasitic organisms.
Examples: Amoeba, Paramecium, Euglena.

→ Phylum Porifera era includes pore-bearing organisms called sponges which are the simplest multicellular animals.

→ Sponges are vase-like, rounded, sac-like, or branched. Euplectella, Sycon, Spongilla, Bath sponge are examples.

→ Phylum Cnidaria is a diploblastic radially symmetrical animal with two-layered body walls.
Examples: Hydra, Jellyfish, Sea anemone, and Corals.

→ In Phylum Platyhelminthes, flatworms have been included which are mostly parasitic.
Examples: Dugesia, Liver fluke, Tapeworm.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

→ Round or threadworms are included in Phylum Aschelminthes which are triploblastic, unsegmented, and of bilateral symmetry. Example: Ascaris.

→ Annelids are triploblastic, bilaterally symmetrically elongated with segmented bodies.
Examples: Earthworm, Leech, Nereis.

→ Animals with jointed legs have been placed in Phylum Arthropoda. These animals’ body is covered with chitinous cuticle.
Examples: It is the largest phylum and includes prawns, insects, spiders, scorpions, etc.

→ In Phylum Mollusca, the body is divided into three regions i.e. head, a dorsal visceral mass, and ventral foot. These are soft-bodied animals.
Examples: Pila, Unio.

→ Phylum Echinodermata includes starfishes, brittle stars, and sea urchins. These are spiny-skinned.
Examples: Starfish, Sea lily

→ In Phylum Hemichordata, the body is divided into three parts i.e. proboscis, collar, and trunk.
Examples: Balanoglossus.

→ Phylum Chordata represents the most advanced group of kingdom Animalia.

→ Protochordate group includes Herdmania and Amphioxus.

→ Chordates bear

  • a solid rod-like structure called the notochord, on the dorsal side above the gut
  • dorsal hollow nervous system
  • pharynx, perforated by gill slits

→ Systematics: It is the study of the diversity of organisms and all their comparative and evolutionary relationship.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

→ Taxonomy: It deals with the identification, nomenclature, and classification of different types of organisms.

→ Species: A group of living organisms of similar individuals capable of exchanging genes and interbreeding. The species is ranked below a genus.
Example: Homo sapiens.

→ Fertilization: Fusion of male gamete and female gamete is called fertilization.

→ Classification: It is the arrangement of organisms into groups on the basis of similarities and differences.

→ Binomial Nomenclature: Every organism is given a scientific name that has two parts, the first is the name of the genus (generic name) and the second is the name of the species (specific name).

→ Dicotyledonous: Plant having seeds with two cotyledons.

→ Ovule: A structure in the ovary of a seed plant that develops into a seed following fertilization.

→ Gametophyte: The haploid generation producing gametes in plants.

→ Annuals: A plant that completes its life cycle in one growing season.

→ Mycelium: A mass of filament or hyphae, composing the vegetative part of many fungi.

→ Hyphae: One of the filaments composing mycelium.

→ Saprophyte: A plant that lives on decaying organic matter.

→ Notochord: An elongated dorsal cord which is the primitive axial skeleton of chordates.

PSEB 9th Class Science Notes Chapter 7 Diversity in Living Organisms

→ Aerobe: An organism that can grow and live in the presence of oxygen.

→ Nocturnal: Active during night e.g. cockroach.

→ Polygamy: When one male lives in the company of many females e.g. Struthio.

→ Producers: The first trophic level in a food chain. Producers are those organisms that can prepare food from inorganic materials i.e. green plants.

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1

1. Draw a line, l, take a point p outside it. Through p, draw a line parallel to l using ruler and compass only.
Solution:
Steps of Construction :
Step 1. Draw a line l of any suitable length Mid a point ‘p’ outside l [see Fig. (i)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 1
Step 2. Take a point ‘q’ on l and join q to p [see Fig. (ii)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 2
Step 3. With q as centre and a convenient radius, draw an arc cutting l at E and pq at F [see Fig. (iii)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 3
Step 4. Now with p as a centre and the same radius as in step 3, draw an arc GH cutting pq at I [see Fig. (iv)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 4
Step 5. Place the pointed tip of the compasses at E and adjust the opening so that the pencil tip is at F [see Fig. (v)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 5

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

Step 6. With the same opening as in step 5 and with I as centre, draw an arc cutting the arc GH at J. [see Fig. (vi)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 6
Step 7. Now, join pand J to draw a line ‘m’ [see Fig. (vii)],
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 7
Note that : ∠Jpq and ∠pqE are alternate interior angles and ∠pqE = ∠qpJ
∴ m || l

2. Draw a line parallel to a line l at a distance of 3.5 cm from it.
Solution :
Steps of construction :
Step 1. Take a line ‘l’ and any point say O on it [see Fig. (i)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 8
Step 2. At O draw ∠AOB = 90°. [see Fig. (ii)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 9
Step 3. Place the pointed tip of the compasses at ‘0’ (zero) mark on ruler and adjust the opening so that the pencil tip is at 3.5 cm [see Fig. (iii)]
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 10
Step 4. With the same opening as in step 3 and with O as centre draw an arc cutting ray OB at X. [see Fig. (iv)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 11

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

Step 5. At X draw a line ‘m’ perpendicular to OB. In other words, draw ∠CXO = 90° [see Fig. (v)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 12
In this way, line m is parallel to l.
Note that. ∠AOX and ∠CXO are alternate angles and ∠AOX = ∠CXO (each = 90°).
∴ m || l.
Note. We may use any of three properties regarding the transversal OX and parallel lines l and m.

3. Let l be a line and P be a point not on l. Through P, draw a line ‘m’ parallel to l. Now, join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meets l at S. What shape do the two sets of parallel lines enclose ?
Solution:
Steps of Construction :
Step 1. Take a line ‘l’ and a point ‘P’ outside l. [see Fig . (i)]
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 13
Step 2. Take any point A on l and join P to A [see Fig. (ii)]
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 14
Step 3. With A as centre and convenient radius draw an arc cutting l at B and AP at C. [see Fig. (iii)]
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 15
Step 4. Now with P as centre and the same radius as in step 3, draw an arc DE cutting PA at F [see Fig. (iv)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 16
Step 5. Place the pointed tip of the compasses at B and adjust the opening so that the pencil tip is at C [see Fig. (v)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 17

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

Step 6. With the same opening as in step 5 and with F as centre, draw an arc cutting the arc DE at G [see Fig. (vi)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 18
Step 7. Now join PG to draw line ‘m’ [see Fig. (vii)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 19
Note that. ∠PAB and ∠APG are alternate interior angles and ∠PAB = ∠APG
∴ m || l
Step 8. Take any point Q on l. Join PQ [see Fig. (viii)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 20
Step 9. Take any other point R on m [see Fig. (ix)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 21
Step 10. With P as centre and convenient radius, draw an arc cutting line m at H and PQ at I [see Fig. (x)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 22
Step 11. Now with R as centre and the same radius as in step 10, draw an arc JK [see Fig. (xi)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 23

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

Step 12. Place the pointed tip of compasses at H and adjust the opening so that the pencil tip is at I.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 24
Step 13. With the same opening as in step 12 and with R as centre, draw an arc cutting the arc JK at L [see Fig. (xii)].
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 25
Step 14. Now join RL to draw a line parallel to PQ. Let this meet l at S. [see Fig. (xiv)]
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 26
Note that. ∠RPQ and ∠LRP are alternate interior angles
and ∠RPQ = ∠LRP
∴ RS || PQ.
Now we have
PR || QS
[∵ m || l and PR is part of m and QS is part of line l]
and PQ || RS
∴ PQSR is a parallelogram.

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

4.

Question (i).
How many parallel lines can be drawn, passing through a point not lying on the given line ?
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(c) 1

Question (ii).
Which of the following is used to draw a line parallel to a given line ?
(a) A protractor
(b) A ruler
(c) A compasses
(d) A ruler and compasses.
Answer:
(d) A ruler and compasses.

PSEB 9th Class Science Notes Chapter 6 Tissues

This PSEB 9th Class Science Notes Chapter 6 Tissues will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 6 Tissues

→ A cell is the structural and functional unit of life, while the group of cells coordinating to perform a specific function is called tissue.

→ The scientific study of tissues is called histology.

→ A cluster of cells called a tissue is arranged and designed so as to give the highest possible efficiency as tissues have provided division of labor in multicellular organisms.

→ A Group of tissues is called an organ.

→ A Group of organs constitute an organ system.

PSEB 9th Class Science Notes Chapter 6 Tissues

→ Due to improved organisation, higher efficiency, multicellular organisms have higher survival value.

→ Plant Tissues:

  • Most of the plant cells are dead which provides mechanical strength and needs less maintenance.
  • In plants, there is some tissue that continue dividing.
  • On the basis of the power of division of cells, plant tissues are of two types: Meristematic tissues and permanent tissues.
  • Meristematic tissue has the power of division throughout life, so helps in the growth of the plant but in some specific regions.
  • Meristematic tissue is of three types: apical, lateral, and intercalary.
  • Meristem is of two types depending upon the power of division: primary meristem and secondary meristem.
  • Simple Permanent tissue has lost the division power and is of three types: Parenchyma, Collenchyma, and Sclerenchyma.
  • Tissue may be simple or complex.
  • Xylem and phloem are complex tissues.

→ Animal Tissues:

  • On the basis of their functions, animal tissues are of four types: Epithelial, connective, muscular, and nervous tissue.
    PSEB 9th Class Science Notes Chapter 6 Tissues
  • The muscles of the heart show rhythmic contraction and relaxation throughout life.
  • The functional combination of nerve and muscle tissue is fundamental to most animals.
  • This combination enables the animals to move rapidly in response to stimuli.

→ Tissue: A group of similar or dissimilar cells alongwith intercellular substances coordinating to perform a specific function. Blood, phloem, and muscles are examples of tissues.

→ Meristematic tissue: The immature cells which are in a state of division and growth with no intercellular spaces.

→ Meristem: A group of cells capable of dividing to form new cells.

→ Companion cells: These are narrow, elongated, and nucleated cells that are connected to sieve tube cells.

PSEB 9th Class Science Notes Chapter 6 Tissues

→ Phloem Parenchyma: Parenchyma tissue associated with phloem.

→ Tendon: A band of white fibres surrounded by connective tissue sheath which joins muscles to bone.

→ Sarcolemma: Surface covering of striated muscle fibres.

→ Sarcoplasm: Cytoplasm of a muscle fibre.

→ Sarcomere: A structural and functional unit of muscle fibre.

→ Chlorenchyma: The type of parenchyma cells having chloroplast and carrying out photosynthesis.

→ Actin: It is a type of protein present in muscle fibres.

→ Bone marrow: A hemopoietic tissue inside the marrow cavity of the bones.

→ Cartilage: An elastic skeletal tissue that acts as a shock absorber.

→ Neurilemma: It is a covering around nerve fibres formed of Schwann cells.

→ Ligament: A band of yellow fibres surrounded by connective tissue sheath which joins bone to bone.

→ Osteoblasts and Osteocytes: They are bone-forming and bone cells respectively.

→ Schwann Cells: Cells around nerve fibres that form neurilemma.

→ Blood platelets: Blood corpuscles that help in blood clotting at an injury.

PSEB 9th Class Science Notes Chapter 6 Tissues

→ Axon: An efferent nerve process of a neuron.

→ I-band: It is a thin band of multiple fibres formed of actin protein.

→ Epithelial tissue: An animal covering and protective tissue.

→ Histology: Microscopic study of tissues.

→ Xylem: A compound tissue formed of tracheids, vessels, parenchyma, and fibres, and helps in the conduction of water and minerals in the plants.

→ Phloem: A compound tissue formed of sieve tube cells, companion cells, parenchyma, and fibres, and helps in the conduction of food in the plants.

→ Tracheids: These are elongated dead cells with large cavities and possess highly lignified cell walls.

→ Sieve tubes: These are tubular cells with perforated walls.

→ Vessels: These are composed of many cells joined end to end with their perforated walls to give a tube-like appearance.

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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 6 Lines and Angles Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2

Question 1.
In the given figure, find the values of x and y and then show that AB || C

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 1
Answer:
Ray QA stands on line PS.
∴ ∠ PQA and ∠ AQR form a linear pair.

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 2
∴ ∠ PQA + ∠ AQR = 180° [Linear pair axiom]
∴ 50° + x = 180°
∴ x = 130° ……………. (1)
Lines PS and CD intersect at R.
∴ ∠ CRS and ∠ QRD are vertically opposite angles.
∴ ∠ QRD = ∠ CRS
∴ y = 130° ……………. (2)
From (1) and (2),
x = y.
But, these angles are alternate interior angles formed by transversal PS for lines AB and CD and they are equal.
Hence, AB || CD.

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

Question 2.
In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 3
Answer:
AB || CD and CD || EF.
∴ AB || EF (Lines parallel to the same line)
∴ x = z (Alternate interior angles)
Now, AB || CD.
∴ x + y = 180° (Interior angles on the same side of transversal)
Now, x = z and x + y = 180°
∴ z + y = 180°
Moreover, y : z = 3 : 7
Sum of the ratios 3 + 7 = 10
Then, z = \(\frac{7}{10}\) × 180° = 126°
Now, x = z = 126°

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

Question 3.
In the given figure, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 4

Answer:
Here, AB || CD and GE is transversal for them.
∴ ∠ AGE and ∠ GED are equal alternate interior angles.
∴ ∠ GED = ∠ AGE
∴ ∠ AGE = 126° (Given : ∠ GED= 126°)
EF ⊥ CD
∴ ∠ FED = 90°
∠ GEF + ∠ FED = ∠ GED (Adjacent angles)
∴ ∠ GEF + 90° = 126°
∴ ∠ GEF = 126° – 90°
∴ ∠ GEF = 36°
Ray GE stands on line AF.
∴ ∠ AGE + ∠ FGE = 180° (Angles of linear pair)
∴ 126° + ∠ FGE = 180°
∴ ∠ FGE =180°- 126°
∴ ∠ FGE = 54°

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

Question 4.
In the given figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.
[Hint: Draw a line parallel to ST through point R.]

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 5

Answer:

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 6
Draw line RU parallel to line ST.
PQ || ST and ST || RU
∴ PQ || RU
ST || RU and SR is a transversal for them.
∴ ∠ TSR + ∠ SRU = 180° (Interior angles on the same . side of the transversal)
∴ 130° + ∠ SRU = 180°
∴ ∠ SRU = 50° ………………. (1)
PQ || RU and QR is a transversal for them.
∴ ∠ PQR = ∠ QRU (Alternate interior angles)
∴ 110° = ∠ QRU
∴ ∠ QRU =110° ……………… (2)
Now, ∠ QRS and ∠ SRU are adjacent angles.
∴ ∠ QRS + ∠ SRU = ∠ QRU
∴ ∠ QRS + 50° = 110° [by (1) and (2)]
∴ ∠ QRS = 60°

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

Question 5.
In the given figure, if AB || CD, ∠ APQ = 50° and ∠ P RD = 127°, find x and y.

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 7

Answer:
Here, AB || CD and PQ is transversal for them.
∴ ∠ APQ = ∠ PQR (Alternate interior angles)
∴ 50° = x (Given: ∠ APQ = 50°)
∴ x = 50°
Again, AB || CD and PR is transversal for them.
∴ ∠ APR = ∠ PRD (Alternate interior angles) )
∴ ∠ APR = 127° (Given: ∠ PRD = 127°)
∴ ∠ APQ + ∠ QPR = 127°
(∵ ∠ APQ and ∠ QPR are adjacent angles and their non-common arms form ∠ APR)
∴ 50° + y = 127°
∴ y = 127°- 50°
∴ y = 77°

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

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

PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 8

Answer:

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

Construction: Draw ray BE perpendicular to line PQ and ray CF perpendicular to line RS.
We know that for plane mirrors, angle of incidence = angle of reflection.
Hence, at point B,
∠ ABE = ∠ EBC …………. (1)
and ’ at point C,
∠ BCF = ∠ FCD ………….. (2)
Now, PQ || RS, BE ⊥ PQ and CF ⊥ RS.
Hence, BE || CE
∴ ∠ EBC = ∠ BCF (Alternate interior angles) …………… (3)
From (1), (2) and (3), we get
∠ ABE = ∠ EBC = ∠ BCF = ∠ FCD
∴ ∠ ABE + ∠ EBC = ∠ BCF + ∠ FCD
∴ ∠ ABC = ∠ BCD [Adjacent angles: ∠ ABE and ∠ EBC, ∠ BCF and ∠ FCD]
But ∠ ABC and ∠ BCD are alternate interior angles formed by transversal BC for lines AB and CD and they are equal.
∴ AB || CD