Bhagya

Students should go through these JAC Class 9 Maths Notes Chapter 8 Quadrilaterals will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 8 Quadrilaterals

Quadrilateral
A quadrilateral is a closed figure obtained by joining four points (with no three points collinear) in an order.
→ Since, ‘quad’ means ‘four’ and ‘lateral’ is for ‘sides therefore quadrilateral means a figure bounded by four sides’
→ Every quadrilateral has:
(A) Four vertices
(B) Four sides
(C) Four angles and
(D) Two diagonals.
→ A diagonal is a line segment obtained on joining the opposite vertices.

Sum of the Angles of a Quadrilateral:

Consider a quadrilateral ABCD as shown in figure. Join A and C to get the diagonal AC which divides the quadrilateral ABCD into two triangles ABC and ADC.
We know the sum of the angles of each triangle is 180°
∴ In ΔABC; ∠CAB + ∠B + ∠BCA = 180°
and In ΔADC; ∠DAC + ∠D + ∠DCA = 180°
On adding, we get:
(∠CAB + ∠DAC) + ∠B + ∠D + (∠BCA + ∠DCA) = 180° + 180°
⇒ ∠A + ∠B + ∠D + ∠C = 360°
Thus, the sum of the angles of a quadrilateral is 360°.

Types of Quadrilaterals:
→ Trapezium: It is a quadrilateral in which one pair of opposite sides are parallel and one pair is unparallel. In the quadrilateral ABCD, drawn alongside, sides AB and DC are parallel and AD and BC are unparallel therefore it is a trapezium

→ Parallelogram: It is a quadrilateral in which both the pairs of opposite sides are equal and parallel. The figure shows a quadrilateral ABCD in which AB is parallel and equal to DC and AD is parallel and equal to BC, therefore ABCD is a parallelogram.

Here, (A) ∠A = ∠C and ∠B = ∠D
(B) AB = CD and AD = BC
(C) AB || CD and AD || BC
→ Rectangle: It is a parallelogram whose each angle is 90°.
(a) ∠A + ∠B = 90° + 90° = 180°
⇒ AD || BC, also AD = BC.
(b) ∠B + ∠C = 90° + 90° = 180°
⇒ AB || DC, also AB = DC.
(c) Diagonals AC and BD are equal.

Rectangle ABCD is also a parallelogram.
→ Rhombus: It is a also parallelogram whose all the sides are equal and diagonals are perpendicular to each other. The figure shows a parallelogram ABCD in which AB = BC = CD = DA; AC ⊥ BD.; therefore it is a rhombus.

→ Square: It is a parallelogram whose all the sides are equal and each angle is 90°. Also, diagonals are equal and perpendicular to each other. The figure shows a parallelogram ABCD in which AB = BC = CD = DA, ∠A = ∠B = ∠C = ∠D = 90°, AC ⊥ BD and AC = BD, therefore ABCD is a square.

→ Kite: It is not parallelogram in which two pairs of adjacent sides are equal The figure shows a quadrilateral ABCD in which adjacent sides AB and AD are equal i.e. AB = AD and also the other pair of adjacent sides are equal i.e., BC = CD; therefore it is a kite or kite shaped figure.

Remarks:

• Square, rectangle and rhombus are all parallelograms.
• Kite and trapezium are not parallelograms.
• A square is a rectangle.
• A square is a rhombus.
• A parallelogram is a trapezium.

Parallelogram Theorems
A parallelogram is a quadrilateral in which both the pairs of opposite sides are equal and parallel.

Theorem 1.
A diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Proof:
Given: A parallelogram ABCD.
To Prove: A diagonal divides the parallelogram
into two congruent triangles i.e., if diagonal AC is drawn then ΔABC ≅ ΔCDA and if diagonal BD is drawn
then ΔABD ≅ ΔCDB.
Construction: Join A and C
Proof: Since, ABCD is a parallelogram
AB || DC and AD || BC
In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
And, AC = AC [Common side]
∴ ΔABC ≅ ΔCDA [By ASA]
Similarly, we can prove that
ΔABD ≅ ΔCDB

Theorem 2.
In a parallelogram, opposite sides are equal.

Proof:
Given: A parallelogram ABCD in which
AB || DC and AD || BC.
To Prove: Opposite sides are equal i.e.. AB = DC and AD = BC
Construction: Join A and C
Proof: In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By ASA]
⇒ AB = DC and AD = BC [By CPCT]
Hence, proved.

Theorem 3.
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Proof:
Given: A quadrilateral ABCD in which AB = DC and AD = BC.
To Prove: ABCD is a parallelogram ie, AB || DC and AD || BC
Construction: Join A and C
Proof: In ΔABC and ΔCDA
AB = DC [Given]
AD = BC [Given]
And AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By SSS]
⇒ ∠1 = ∠3 [By CPCT]
And ∠2 = ∠4 [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
∴ AB || DC and AD || BC
⇒ ABCD is a parallelogram.
Hence, proved.

Theorem 4.
In a parallelogram, opposite angles are equal.

Proof:
Given: A parallelogram ABCD in which AB || DC and AD || BC.
To Prove: Opposite angles are equal i.e. ∠A = ∠C and ∠B = ∠D
Construction: Draw diagonal AC.
Proof: In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By ASA]
⇒ ∠B = ∠D [By CPCT]
Similarly, we can prove that
∠A = ∠C Hence, proved.

Theorem 5.
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof:
Given: A quadrilateral ABCD in which opposite angles are equal. i.e., ∠A = ∠C and ∠B = ∠D
To prove: ABCD is a parallelogram i.e.,
AB || DC and AD || BC
Proof: Since the sum of the angles of quadrilateral is 360°
⇒ ∠A + ∠B + ∠C+ ∠D = 360°
⇒ ∠A + ∠D + ∠A + ∠D = 360°
[∵ ∠A = ∠C and ∠B = ∠D]
⇒ 2∠A + 2∠D = 360°
⇒ ∠A + ∠D = 180°
[∵ The sum of interior angles on the same side of transversal AB is 180°]
⇒ AB || DC
Similarly, ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠B + ∠A + ∠B = 360°
[∵ ∠A = ∠C and ∠B = ∠D]
⇒ 2∠A + 2∠B = 360°
⇒ ∠A + ∠B = 180°
[∵ The sum of interior angles on the same side of transversal AB is 180°]
∴ AD || BC
So, AB || DC and AD || BC
⇒ ABCD is a parallelogram.
Hence, proved.

Theorem 6.
The diagonal of a parallelogram bisect each other.

Proof:
Given: A parallelogram ABCD. Its diagonals AC and BD intersect each other at point O..
To Prove: Diagonals AC and BD bisect each other i.e., OA = OC and OB = OD.
Proof: In ΔAOB and ΔCOD
∵ AB || DC and BD is a transversal.
∴ ∠ABO = ∠CDO [Alternate angles]
∵ AB || DC and AC is a transversal line.
∴ ∠BAO = ∠DCO [Alternate angles]
And, AB = DC
⇒ ΔAOB ≅ ΔCOD [By ASA]
⇒ OA = OC and OB = OD [By CPCT]
Hence, proved.

Theorem 7.
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Proof:
Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at point O.
i.e., OA = OC and OB = OD
To prove: ABCD is a parallelogram i.e..
AB || DC and AD || BC.
Proof: In ΔAOB and ΔCOD
OA = OC [Given]
OB = OD [Given]
And, ∠AOB = ∠COD [Vertically opposite angles]
⇒ ΔAOB ≅ ΔCOD [By SAS]
⇒ ∠1 = ∠2 [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
∴ AB is parallel to DC ie., AB || DC Similarly,
ΔAOD ≅ ΔCOB [By SAS]
⇒ ∠3 = ∠4
But these are also alternate angles
⇒ AD || BC
AB || DC and AD || BC
⇒ ABCD is parallelogram.
Hence, proved.

Theorem 8.
A quadrilateral is a parallelogram, if a pair of opposite sides is equal and parallel.

Proof:
Given: A quadrilateral ABCD in which AB = DC and AB || DC.
To Prove: ABCD is a parallelogram, i.e. AB || DC and AD || BC.
Construction: Join A and C.
Proof: Since AB is parallel to DC and AC is transversal
∠BAC = ∠DCA [Alternate angles]
AB = DC [Given]
And AC = AC [Common]
⇒ ΔBAC ≅ ΔDCA [By SAS]
⇒ ∠BCA = ∠DAC [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
⇒ AD || BC
Now, AB || DC (given) and AD || BC
[Proved above]
⇒ ABCD is a parallelogram
Hence, proved.

Remarks:
In order to prove that given quadrilateral is parallelogram, we can prove any one of the following.

• Opposite angles of the quadrilateral are equal, or
• Diagonals of the quadrilateral bisect each other, or
• A pair of opposite sides is parallel and is of equal length, or
• Opposite sides are equal.
• Every diagonal divides the parallelogram into two congruent triangles.

Mid-Point Theorem
Statement: In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it.

Given: A triangle ABC in which P is the mid-point of side AB and Q is the mid-point of side AC.
To Prove: PQ is parallel to BC and is half of it
i.e., PQ || BC and PQ = \(\frac{1}{2}\)BC
Construction: Produce PQ upto point such that PQ = QR. Join R and C.
Proof: In ΔAPQ and ΔCRQ
PQ = QR [By construction]
AQ = QC [Given]
And, ∠AQP = ∠CQR [Vertically opposite angles]
⇒ ΔAPQ ≅ ΔCRQ [By SAS]
⇒ AP = CR [By CPCT]
And, ∠APQ = ∠CRQ [By CPCT]
But, ∠APQ and ∠CRQ are alternate angles and we know, whenever the alternate angles are equal, the lines are parallel.
⇒ AP || CR
⇒ AB || CR
⇒ BP || CR
Given, P is mid-point of AB
⇒ AP = BP
⇒ CR = BP [As, AP = CR]
Now, BP = CR and BP || CR
⇒ BCRP is a parallelogram.
[When any pair of opposite sides are equal and parallel, the quadrilateral is a parallelogram]
BCRP is a parallelogram and opposite sides of a parallelogram are equal and parallel.
∴ PR = BC and PR || BC
Since, PQ = QR
⇒ PQ = \(\frac{1}{2}\)PR = \(\frac{1}{2}\)BC [AS, PR = BC]
Also, PQ || BC [As, PR || BC]
∴ PQ || BC and PQ = \(\frac{1}{2}\)BC
Hence, proved.

Converse of the Mid-Point Theorem
Statement: The line drawn through the midpoint of one side of a triangle parallel to the another side bisects the third side.

Given: A triangle ABC in which is the midpoint of side AB and PQ is parallel to BC.
To prove: PQ bisects the third side AC i.e., AQ = QC.
Construction: Through C, draw CR parallel to BA, which meets PQ produced at point R.
Proof: Since, PQ || BC i.e., PR || BC [Given]
CR || BA i.e., CR || BP [By construction]
∴ Opposite sides of quadrilateral PBCR are parallel.
⇒ PBCR is a parallelogram
⇒ BP = CR
Also, BP = AP [As Pis mid-point of AB]
∴ CR = AP (As CR = BP)
Now, AB || CR and AC is transversal, ∠PAQ = ∠ROQ [Alternate angles]
Also, AB || CR and PR is transversal, ∠APQ = ∠CRQ [Alternate angles]
In ΔAPQ and ΔCRQ
CR = AP, ∠PAQ = ∠RCQ and ∠APQ = ∠CRQ
⇒ ΔAPQ ≅ ΔCRO [By ASA]
⇒ AQ = QC Hence, proved.

Students should go through these JAC Class 9 Maths Notes Chapter 5 Introduction to Euclid’s Geometry will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 5 Introduction to Euclid’s Geometry

Introduction:
The credit for introducing geometrical concepts goes to the distinguished Greek mathematician ‘Euclid’ who is known as the “Father of Geometry” and the word geometry comes from the Geek words ‘geo’ which means “Earth’ and ‘Metreon’ which means ‘measure’.

Basic Concepts In Geometry:
A point, a ‘line’ and a plane are the basic concepts to be used in geometry.
→ Axioms:
The statement that is taken to be true without proof, to serve as a premise for further reasoning and arguments, are called axioms.

Euclid’s Definitions:

• A point is that which has no part.
• A line is breadthless length.
• The ends of a line segment are points.
• A straight line is that which has length only.
• A surface is that which has length and breadth only.
• The edges of surface are lines.
• A plane surface is a surface which lies evenly with the straight lines on itself.

Euclid’s Five Postulates:
→ A straight line may be drawn from any one point to any other point.
→ A terminated line or a line segment can be produced infinitely.

→ A circle can be drawn with any centre and of any radius.
→ All right angles are equal to one another.
→ If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced infinitely, meet on that side on which the sum of angles is less than two right angles.

Important Axioms:
→ A line is the collection of infinite number of points.
→ Through a given point, infinite lines can be drawn.

→ Given two distinct points, there is one and only one line that contains both the points.
→ If P is a point not lying on a line l, then one and only one line can be drawn through P which is parallel to l.

→ Two distinct lines cannot have more than one point in common.

→ Two lines which are both parallel to the same line, are parallel to each other. i.e. IF l || n, m || n ⇒ l || m.

Some Important Definitions:
→ Collinear points: Three or more points are said to be collinear if there is a line which contains all of them.

→ Concurrent lines: Three or more lines are said to be concurrent if there is a point which lies on all of them.

→ Intersecting lines: Two lines are intersecting if they have a common point. The common point is called the “point of intersection”.

→ Parallel lines: Two lines I and m in a plane are said to be parallel lines if they do not have a common point.

→ Line segment: Given two points A and B on a line l, the connected part (segment) of the line with end points at A and B is called the line segment AB.

→ Interior point of a line segment: A point R is called an interior point of a line segment PQ if R lies between Pand O but Ris neither P nor Q.

→ Congruence of line segment: Two line segments AB and CD are congruent if trace copy of one can be superposed on the other so as to cover it completely and exactly in this case we write AB ≅ CD. In other words we can say two lines are congruent if their lengths are same.
→ Distance between two points: The distance between two points P and Q is the length of the line segment PO.
→ Ray: Directed line segment is called a ray. If AB is a ray, then it is denoted by \(\overrightarrow{\mathrm{AB}}\). Point A is called initial point of ray.

→ Opposite rays: Two rays AB and AC are said to be opposite rays if they are collinear and point A is the only common point of the two rays and A lies in between B and C.

Theorem 1.
If l, m, n are lines in the same plane such that l intersects m and n || m, then l also intersects n.
Answer:
Given: Three lines l, m, n in the same plane such that intersects m and n || m.
To prove: Lines land n are intersecting lines.

Proof: Let l and n be non intersecting lines. Then. l || n. But, n || m [Given]
∴ l || n and n || m
⇒ l || m
⇒ l and m are non-intersecting lines.
This is a contradiction to the hypothesis that I and m are intersecting lines. So our supposition is wrong.
Hence, l intersects line n.

Theorem 2.
If lines AB, AC, AD and AE are parallel to a line l, then points A, B, C, D and E are collinear.
Answer:
Given: Lines AB, AC, AD and AE are parallel to a line l.
To prove: A, B, C, D, E are collinear.
Proof: Since AB, AC, AD and AE are all parallel to a line l. Therefore point A is outside l and lines AB, AC, AD, AE are drawn through A and each line is parallel to l.
But by parallel lines axiom, one and only one line can be drawn through the point A outside a line l and parallel to it.
This is possible only when A, B, C, D and E all lie on the same line. Hence, A, B, C, D and E are collinear.

JAC Jharkhand Board Class 9th Maths Solutions in Hindi & English Medium

JAC Board Class 9th Maths Solutions in English Medium

JAC Class 9 Maths Chapter 1 Number Systems

• Chapter 1 Number Systems Ex 1.1
• Chapter 1 Number Systems Ex 1.2
• Chapter 1 Number Systems Ex 1.3
• Chapter 1 Number Systems Ex 1.4
• Chapter 1 Number Systems Ex 1.5
• Chapter 1 Number Systems Ex 1.6

JAC Class 9 Maths Chapter 2 Polynomials

• Chapter 2 Polynomials Ex 2.1
• Chapter 2 Polynomials Ex 2.2
• Chapter 2 Polynomials Ex 2.3
• Chapter 2 Polynomials Ex 2.4
• Chapter 2 Polynomials Ex 2.5

JAC Class 9 Maths Chapter 3 Coordinate Geometry

• Chapter 3 Coordinate Geometry Ex 3.1
• Chapter 3 Coordinate Geometry Ex 3.2
• Chapter 3 Coordinate Geometry Ex 3.3

JAC Class 9 Maths Chapter 4 Linear Equations in Two Variables

• Chapter 4 Linear Equations in Two Variables Ex 4.1
• Chapter 4 Linear Equations in Two Variables Ex 4.2
• Chapter 4 Linear Equations in Two Variables Ex 4.3
• Chapter 4 Linear Equations in Two Variables Ex 4.4

JAC Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry

• Chapter 5 Introduction to Euclid’s Geometry Ex 5.1
• Chapter 5 Introduction to Euclid’s Geometry Ex 5.2

JAC Class 9 Maths Chapter 6 Lines and Angles

• Chapter 6 Lines and Angles Ex 6.1
• Chapter 6 Lines and Angles Ex 6.2
• Chapter 6 Lines and Angles Ex 6.3

JAC Class 9 Maths Chapter 7 Triangles

• Chapter 7 Triangles Ex 7.1
• Chapter 7 Triangles Ex 7.2
• Chapter 7 Triangles Ex 7.3
• Chapter 7 Triangles Ex 7.4
• Chapter 7 Triangles Ex 7.5

JAC Class 9 Maths Chapter 8 Quadrilaterals

• Chapter 8 Quadrilaterals Ex 8.1
• Chapter 8 Quadrilaterals Ex 8.2

JAC Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

• Chapter 9 Areas of Parallelograms and Triangles Ex 9.1
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.2
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.3
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

JAC Class 9 Maths Chapter 10 Circles

• Chapter 10 Circles Ex 10.1
• Chapter 10 Circles Ex 10.2
• Chapter 10 Circles Ex 10.3
• Chapter 10 Circles Ex 10.4
• Chapter 10 Circles Ex 10.5
• Chapter 10 Circles Ex 10.6

JAC Class 9 Maths Chapter 11 Constructions

• Chapter 11 Constructions Ex 11.1
• Chapter 11 Constructions Ex 11.2

JAC Class 9 Maths Chapter 12 Heron’s Formula

• Chapter 12 Heron’s Formula Ex 12.1
• Chapter 12 Heron’s Formula Ex 12.2

JAC Class 9 Maths Chapter 13 Surface Areas and Volumes

• Chapter 13 Surface Areas and Volumes Ex 13.1
• Chapter 13 Surface Areas and Volumes Ex 13.2
• Chapter 13 Surface Areas and Volumes Ex 13.3
• Chapter 13 Surface Areas and Volumes Ex 13.4
• Chapter 13 Surface Areas and Volumes Ex 13.5
• Chapter 13 Surface Areas and Volumes Ex 13.6
• Chapter 13 Surface Areas and Volumes Ex 13.7
• Chapter 13 Surface Areas and Volumes Ex 13.8
• Chapter 13 Surface Areas and Volumes Ex 13.9

JAC Class 9 Maths Chapter 14 Statistics

• Chapter 14 Statistics Ex 14.1
• Chapter 14 Statistics Ex 14.2
• Chapter 14 Statistics Ex 14.3
• Chapter 14 Statistics Ex 14.4

JAC Class 9 Maths Chapter 15 Probability

• Chapter 15 Probability Ex 15.1

JAC Board Class 9th Maths Solutions in Hindi Medium

JAC Class 9 Maths Chapter 1 संख्या पद्धति

• Chapter 1 संख्या पद्धति Ex 1.1
• Chapter 1 संख्या पद्धति Ex 1.2
• Chapter 1 संख्या पद्धति Ex 1.3
• Chapter 1 संख्या पद्धति Ex 1.4
• Chapter 1 संख्या पद्धति Ex 1.5
• Chapter 1 संख्या पद्धति Ex 1.6

JAC Class 9 Maths Chapter 2 बहुपद

• Chapter 2 बहुपद Ex 2.1
• Chapter 2 बहुपद Ex 2.2
• Chapter 2 बहुपद Ex 2.3
• Chapter 2 बहुपद Ex 2.4
• Chapter 2 बहुपद Ex 2.5

JAC Class 9 Maths Chapter 3 निर्देशांक ज्यामिति

• Chapter 3 निर्देशांक ज्यामिति Ex 3.1
• Chapter 3 निर्देशांक ज्यामिति Ex 3.2
• Chapter 3 निर्देशांक ज्यामिति Ex 3.3

JAC Class 9 Maths Chapter 4 दो चरों वाले रैखिक समीकरण

• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.1
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.2
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.3
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.4

JAC Class 9 Maths Chapter 5 युक्लिड के ज्यामिति का परिचय

• Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.1
• Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.2

JAC Class 9 Maths Chapter 6 रेखाएँ और कोण

• Chapter 6 रेखाएँ और कोण Ex 6.1
• Chapter 6 रेखाएँ और कोण Ex 6.2
• Chapter 6 रेखाएँ और कोण Ex 6.3

JAC Class 9 Maths Chapter 7 त्रिभुज

• Chapter 7 त्रिभुज Ex 7.1
• Chapter 7 त्रिभुज Ex 7.2
• Chapter 7 त्रिभुज Ex 7.3
• Chapter 7 त्रिभुज Ex 7.4
• Chapter 7 त्रिभुज Ex 7.5

JAC Class 9 Maths Chapter 8 चतुर्भुज

• Chapter 8 चतुर्भुज Ex 8.1
• Chapter 8 चतुर्भुज Ex 8.2

JAC Class 9 Maths Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल

• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.1
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.2
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.3
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.4

JAC Class 9 Maths Chapter 10 वृत्त

• Chapter 10 वृत्त Ex 10.1
• Chapter 10 वृत्त Ex 10.2
• Chapter 10 वृत्त Ex 10.3
• Chapter 10 वृत्त Ex 10.4
• Chapter 10 वृत्त Ex 10.5
• Chapter 10 वृत्त Ex 10.6

JAC Class 9 Maths Chapter 11 रचनाएँ

• Chapter 11 रचनाएँ Ex 11.1
• Chapter 11 रचनाएँ Ex 11.2

JAC Class 9 Maths Chapter 12 हीरोन का सूत्र

• Chapter 12 हीरोन का सूत्र Ex 12.1
• Chapter 12 हीरोन का सूत्र Ex 12.2

JAC Class 9 Maths Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.1
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.2
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.3
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.4
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.5
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.6
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.7
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.8
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.9

JAC Class 9 Maths Chapter 14 सांख्यिकी

• Chapter 14 सांख्यिकी Ex 14.1
• Chapter 14 सांख्यिकी Ex 14.2
• Chapter 14 सांख्यिकी Ex 14.3
• Chapter 14 सांख्यिकी Ex 14.4

JAC Class 9 Maths Chapter 15 प्रायिकता

• Chapter 15 प्रायिकता Ex 15.1

Students should go through these JAC Class 9 Maths Notes Chapter 7 Triangles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 7 Triangles

Triangle
A plane figure bounded by three lines in a plane is called a triangle. Every triangle has three sides and three angels. If ABC is any triangle then AB, BC and CA are three sides and ∠A, ∠B and ∠C are three angles.

Types of Triangles
→ On the basis of sides we have three types of triangles:

• Scalene triangle – A triangle whose no two sides are equal is called a scalene triangle.
• Isosceles triangle – A triangle having two sides equal is called an isosceles triangle.
• Equilateral triangle – A triangle in which all sides are equal is called an equilateral triangle.

→ On the basis of angles we have three types of triangles:

• Right triangle – A triangle in which any one angle is a right angle (= 90°) is called right triangle.
• Acute triangle – A triangle in which all angles are acute (0° >angle >90°) is called an acute triangle.
• Obtuse (90° < angle < 180° ) triangle – A triangle in which any one angle is obtuse is called an obtuse triangle.

Congruent Figures
The figures are called congruent if they have same shape and same size. In other words, two figures are called congruent if they are having equal length, width and height.

In the above figures {Fig. (i) and Fig. (ii)} both are equal in length, width and height, so these are congruent figures.

Congruent Triangles
Two triangles are congruent if and only if one of them can be made to superimposed on the other, so as to cover it exactly.

If two triangles ΔABC and ΔDEF are congruent then there exist a one to one correspondence between their vertices and sides. i.e. we get following six equalities.
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB = DE, BC = EF, AC = DF.
If ΔABC and ΔDEF are congruent under one to one correspondence A ↔ D, B ↔ E, C ↔ F then we write ΔABC ≅ ΔDEF We cannot write it as ΔABC ≅ ΔDFE or ΔABC ≅ ΔEDF or in other forms because ΔABC ≅ ΔDFE have following one-one correspondence A ↔ D, B ↔ F, C ↔ E.
Hence, we can say that ‘two triangles are congruent if and only if there exists a oneone correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal.

Sufficient Conditions for Congruence of two Triangles
→ SAS Congruence Criterion:

Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle.

→ ASA Congruence Criterion:

Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

→ AAS Congruence Criterion:
If any two angles and a non included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

→ SSS Congruence Criterion:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

→ RHS Congruence Criterion:
Two right angled triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.

→ Congruence Relation in the Set of all Triangles:
By the definition of congruence of two triangles, we have following results.

• Every triangle is congruent to itself i.e., ΔABC ≅ ΔABC
• If ΔABC ≅ ΔDEF then ΔDEF ≅ ΔABC
• If ΔABC ≅ ΔDEF and ΔDEF ≅ ΔΡQR then ΔΑΒC ≅ ΔΡQR

NOTE: If two triangles are congruent then their corresponding sides and angles are also congruent by CPCT (corresponding parts of congruent triangles are also congruent).

Theorem 1.
Angles opposite to equal sides of an isosceles triangle are equal.

Given:
ΔABC in which AB = AC
To Prove: ∠B = ∠C
Construction: We draw the bisector AD of ∠A which meets BC in D.
Proof: In ΔABD and ΔACD, we have
AB = AC [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
And, AD = AD [Common side]
∴ By SAS criterion of congruence, we have
ΔΑΒD ≅ ΔΑCD
⇒ ∠B = ∠C [by CPCT]
Hence, proved.

Theorem 2.
If two angles of a triangle are equal, then sides opposite to them are also equal.
Given: ΔABC in which ∠B = ∠C
To Prove: AB = AC
Construction: We draw the bisector of ∠A

which meets BC in D.
Proof: In ΔABD and ΔACD, we have
∠B = ∠C [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
AD = AD [Common side]
∴ By AAS criterion of congruence, we get
ΔΑΒD ≅ ΔΑCD
⇒ AB = AC [By CPCT] Hence, proved.

Theorem 3.
If the bisector of the vertical angle bisects the base of the triangle, then the triangle is isosceles.

Given: ΔABC in which AD is the bisector of ∠A meeting BC in D such that BD = CD
To Prove: ΔABC is an isosceles triangle.
Construction: We produce AD to E such that AD = DE and join EC
Proof: In ΔADB and ΔEDC, we have
AD = DE [By construction]
∠ADB = ∠CDE [Vertically opposite angles]
BD = DC [Given]
∴ By SAS criterion of congruence, we get
ΔADR ≅ ΔEDC ⇒ AB = EC ……(i)
And, ∠BAD = ∠CED [By CPCT]
But, ∠BAD = ∠CAD
∴ ∠CAD = ∠CED
⇒ AC = EC [Sides opposite to equal angles are equal]
⇒ AC = AB [By eq. (i)] Hence, proved

Some Inequality Relations In A Triangle
→ If two sides of a triangle are unequal, then the longer side has greater angle opposite to it, i.e., if in any ΔABC, AB > AC then ∠C > ∠B.
→ In a triangle the greater angle has the longer side opposite to it, ie, if in any ΔABC, ∠A > ∠B then BC > AC.
→ The sum of any two sides of a triangle is always greater than the third side, i.e., in any ΔABC, AB + BC > AC, BC + CA > AB and AC + AB > BC.
→ Of all the line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

P is any point not lying on line l, PM ⊥ l then PM < PN.
→ The difference of any two sides of a triangle is less than the third side, i.e., in any ΔABC, AB – BC < AC, BC – CA < AB and AC – AB < BC.

Students should go through these JAC Class 9 Maths Notes Chapter 3 Coordinate Geometry will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 3 Coordinate Geometry

Co-Ordinate System:
In two dimensional coordinate geometry, we generally use two types of coordinate systems.

• Cartesian or Rectangular coordinate system.
• Polar coordinate system.

In cartesian coordinate system we represent any point by ordered pair (x, y) where x and y are called x and y coordinate of that point respectively.
In polar coordinate system we represent any point by ordered pair (r, θ) where is called radius vector and ‘θ’ is called vectorial angle of that point, which will be studied in higher classes.

Cartesian Coordinate System:
→ Rectangular Coordinate Axes:
Let XX’ and YY’ are two lines such that XX’ is horizontal and YY’ is vertical lines in the same plane and they intersect each other at O. This intersecting point is called origin Now choose a convenient unit of length and starting from origin as zero, mark off a number scale on the horizontal line XX’, positive to the right of origin O and negative to the left of origin O. Also mark off the same scale on the vertical line YY’, positive upwards and negative downwards of the origin. The line XX’ is called X-axis and the line YY’ is known as Y-axis and the two lines taken together are called the coordinate axes.

→ Quadrants:
The coordinates axes XX’ and YY’ divide the plane of graph paper into four parts XY, X’Y, X’Y’ and XY’. These four parts are called the quadrants. The parts XY, X’Y, X’Y’ and XY’ are known as the first second, third and fourth quadrants respectively.

→ Cartesian Coordinates of a Point:
Let -axis and y-axis be the coordinate axes and P be any point in the plane. To find the position of P with respect of x-axis and y-axis, we draw two perpendicular line segment from P on both coordinate axes.

Let PM and PN be the perpendiculars on x-axis and y-axis resepectively. The length of the line segment OM is called the x-coordinate or abscissa of point P. Similarly the length of line segment ON is called the y-coordinate or ordinate of point P.

Let OM = x and ON = y. The position of the point P in the plane with respect to the coordinate axes is represented by the ordered pair (x, y). The ordered pair (x, y) is called the coordinates of point P. “Thus, for a given point, the abscissa and ordinate are the distances of the given point from y-axis and x-axis respectively”.

The above system of coordinating of ordered pair (x, y) with every point in plane is called the Rectangular or Cartesian coordinate system.

Cartesian coordinate system

→ Convention of Signs:
As discussed earlier that regions XOY, Χ’ΟΥ, Χ’ΟΥ’ and ΧΟΥ’ are known as the first second, third and fourth quadrants respectively. The ray OX is taken as positive X-axis, OX’ as negative x-axis, OY as positive y-axis and OY as negative y-axis. Thus we have.
In first quadrant: x > 0, y > 0
In second quadrant: x < 0, y > 0
In third quadrant: x < 0, y < 0
In fourth quadrant: x > 0, y < 0

→ Points on Axis:
If point P lies on x-axis then clearly its distance from x-axis will be zero, therefore we can say that its ordinate will be zero. In general, if any point lies on x-axis then its y-coordinate will be zero. Similarly if any point Q lies on y-axis, then its distance from y-axis will be zero therefore we can say its x-coordinate will be zero. In general, if any point lies on y-axis then its x-coordinate will be zero.

→ Plotting of Points:
In order to plot the points in a plane, we may use the following algorithm.
Step I: Draw two mutually perpendicular lines on the graph paper, one horizontal and other vertical.
Step II: Mark their intersection point as O (origin).
Step III: Choose a suitable scale on X-axis and Y-axis and mark the points on both the axes.
Step IV: Obtain the coordinates of the point which is to be plotted. Let the point be P(a, b). To plot this point start from the origin and |a| units move along OX, OX’ according as ‘a’ is positive or negative respectively. Suppose we arrive at point M. From point M move vertically upward or downward |b| units according as ‘b’ is positive or negative respectively The point where we arrive finally is the required point P(a, b).

Distance Between Two Points:
→ If there are two points A (x₁, y₁) and B(x₂, y₂) on the XY plane, the distance between them is given by
AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Students should go through these JAC Class 9 Maths Notes Chapter 14 Statistics will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 14 Statistics

Introduction:
The branch of science known as Statistics has been used in India from ancient times. Statistics deals with collection of numerical facts ie, data, their classification and tabulation and their interpretation. In statistics we shall try to study, in detail about collection, classification and tabulation of such data.
→ Importance of Data: Expressing facts with the helps of data is of great importance in our day-to-day life. For example, instead of saying that India has a large population it is more appropriate to say that the population of India, based on the census of 2001 is more than one billion.

→ Collection of Data: On the basis of methods of collection, data can be divided into two categories:
(i) Primary data: Data which are collected for the first time by the statistical investigator or with help of his workers is called primary data. For example if an investigator wants to study the condition of the workers working in a factory then for this he collects some data like their monthly income, expenditure, number of brothers, sisters, etc.

(ii) Secondary data: Data already collected by a person or a society and may be available in published or unpublished form is known as secondary data. Secondary should be carefully used. Such data is generally obtained from the following two sources.

• Published sources
• Unpublished sources

→ Classification of Data: When the data is compiled the same form and order in which it is collected, it is known as Raw Data. It is also called Crude Data. For example, the marks obtained by 20 students of class X in English out of 10 marks are as follow:
7. 4, 9, 5, 8, 9, 6, 7, 9, 2, 0 3, 7, 6, 2, 1, 9, 8, 3, 8
(i) Geographical basis: Here, the data is classified on the basis of place or region. For example, the production of food grains of different states is shown in the following table:

S. No.	State	Production (in Tons)
1	Andhra Pradesh	9690
2	Bihar	8074
3	Haryana	10065
4	Punjab	17065
5	Uttar Pradesh	28095

(ii) Chronological classification data’s classification is based on hour, day, week, month or year, then it is called chronological classification. For example, the population of India in different years is shown in following table:

S.No	Year	Production (in Crores)
1	1951	46.1
2	1961	53.9
3	1971	61.8
4	1981	68.5
5	1991	88.4
6	2001	100.01

(iii) Qualitative basis: When the data is classified into different groups on the basis of their descriptive qualities and properties such a classification is known as descriptive or qualitative classification. Since the attributes cannot be measured directly they are counted on the basis of presence or absence of qualities. For example intelligence, literacy, unemployment, honesty etc. The following table shows classification on the basis of sex and employment.

Population (in lacs)
Gender → Position of Employment ↓	Male	Female
Employed	16.2	13.7
Unemployed	26.4	24.8
Total	42.6	38.5

(iv) Quantitative basis: If facts are such that they can be measured physically e.g, marks obtained, height, weight, age, income, expenditure etc, such facts are known as variable values. If such facts are kept into classes then it is called classification according to quantitative or class intervals.

Marks obtained	10 – 20	20 – 30	30 – 40	40 – 50
No. of students	7	9	15	6

Definitions:
→ Variate: The numerical quantity whose value varies is called a variate, generally a variate is represented by x. There are two types of variate.
(i) Discrete variate: Its magnitude is fixed. For example, the number of teachers in different branches of a institute are 30, 35, 40 etc.
(ii) Continuous variate: Its magnitude is not fixed. It is expressed in groups like 10 – 20, 20 – 30, … etc.
→ Range: The difference between the maximum and the minimum values of the variable x is called range.
→ Class frequency: in each class, the number of times a data is repeated is known as its class frequency.
→ Class limits: The lowest and the highest value of the class are known as lower and upper limits respectively of that class.
→ Classmark: The average of the lower and the upper limits of a class is called the mid value or the class mark of that class. It is generally denoted by x.
If x is the mid value and his the class size, then the class limits are \(\left(x-\frac{h}{2}, x+\frac{h}{2}\right)\).

Example:
The mid values of a distribution are 54, 64, 74, 84 and 94. Find the class size and class limits.
Solution:
The class size is the difference of two consecutive class marks, therefore class size (h) = 64 – 54 = 10.
Here the mid values are given and the class size is 10. So, class limits are:

Therefore, class limits are 49 – 59, 59 – 69, 69 – 79, 79 – 89, and 89 – 99.

Frequency Distribution:
The marks scored by 30 students of IX class of a school in the first test of Mathematics out of 50 marks are as follows:

The number of times a mark is repeated is called its frequency. It is denoted by f.

Above type of frequency distribution is called ungrouped frequency distribution. Although this representation of data is shorter than representation of raw data, but from the angle of comparison and analysis it is quite bit. So to reduce the frequency distribution, it can be classified into groups in following ways and it is called grouped frequency distribution.

Class	Frequency
1 – 10	8
11 – 20	2
21 – 30	12
31 – 40	5
41 – 50	3

(a) Kinds of Frequency Distribution: Statistical methods like comparison, decision taken etc. depends on frequency distribution. Frequency distribution are of three types.
(i) Individual frequency distribution: Here each item or original price of unit is written separately. In this category, frequency of each variable is one.

Example:
Total marks obtained by 10 students in a class.

Marks obtained	S. No.
46	1
18	2
79	3
12	4
97	5
80	6
5	7
27	8
67	9
54	10

(ii) Discrete frequency distribution: When number of terms is large and variable are discrete, i.e., variate can accept some particular values only under finite limits and is repeated then it is called discrete frequency distribution. For example, the wages of employees and their numbers is shown in following table.

Monthly wages	No. of employees
4000	10
6000	8
8000	5
11000	7
20000	2
25000	1

The above table shows ungrouped frequency distribution and same facts can be written in grouped frequency as follows:

Monthly wages	No. of employees
0 – 10,000	23
11,000 – 20,000	9
21,000 – 30,000	1

Note: If variable is repeated in individual distribution then it can be converted into discrete frequency distribution.
(iii) Continuous frequency distribution:
When number of terms is large and variate is continuous, i.e. variate can accept all values under finite limits and they are repeated then it is called continuous frequency distribution. For example age of students in a school is shown in the following table:

Age (in years)	Class	No. of students
Less than 5 years	0 – 51	72
Between 5 and 10 years	5 – 10	103
Between 10 and 15 years	10 – 15	50
Between 15 and 20 years	15 – 20	25

Classes can be made mainly by two methods:
(i) Exclusive series: In this method upper limit of the previous class and lower limit of the next class is same. In this method the value of upper limit in a class is not considered in the same class, it is considered in the next class.

(ii) Inclusive series: In this method value of upper and lower limit are both contained in same class. In this method the upper limit of class and lower limit of next class are not same. Some time the value is not a whole number, it is a fraction or in decimals and lies in between the two intervals then in such situation the class interval can be constructed as follows:

A		B
Class	Frequency	Class	Frequency
0 – 9	4	0 – 9.5	4
10 – 19	7	9.5 – 19.5	7
20 – 29	6	19.5 – 29.5	6
30 – 39	3	29.5 – 39.5	3
40 – 49	3	39.5 – 49.5	3

Cumulative Frequency:
→ Discrete frequency distribution:
From the table of discrete frequency distribution, it can be identified that number of employees whose monthly income is 4000 or how many employees of monthly income 11000 are there. But if we want to know how many employees whose monthly income is upto 11000, then we should add 10, 8, 5 and 7 i.e., number of employees whose monthly income is upto 11000 is 10 + 8 + 5 + 7 = 30. Here we add all previous frequency and get cumulative frequency. It will be more clear from the following table:

Income	Frequency (f)	Cumulative frequency (cf)	Explanation
4000	10	10	10 = 10
6000	8	18	10 + 8
8000	5	23	18 + 5
11000	7	30	23 + 7
20000	2	32	30 + 2
25000	1	33	32 + 1

→ Continuous frequency distribution: In (a) part, we obtained cumulative frequency for discrete series. Similarly, cumulative frequency table can be made from continuous frequency distribution also.
For example, for table:

Monthly income	No. of employees	Cumulative	Explanation
Variate (x)	Frequency (F)	Frequency (cf)
0 – 5	72	72	72 = 72
5 – 10	103	175	72 + 103 = 175
10 – 15	50	225	175 + 50 = 225
15 – 20	25	250	225 + 25 = 250

Graphical Representation Of Data:
(i) Bar graphs
(ii) Histograms
(iii) Frequency polygons
(iv) Frequency curves
(v) Cumulative frequency curves or Ogives
(vi) Pie Diagrams

(i) Bar Graphs. A bar graph is a graph that present categorical data with rectangular bars with heights or lengths proportional to the values that they represent.

Example:
A family with monthly income of ₹ 20,000 had planned the following expenditure per month under various heads: Draw bar graph for the data giyen below:

Monthly income	No. of employees	Cumulative	Explanation
Variate (x)	Frequency (F)	Frequency (cf)
0 – 5	72	72	72 = 72
5 – 10	103	175	72 + 103 = 175
10 – 15	50	225	175 + 50 = 225
15 – 20	25	250	225 + 25 = 250

Solution:

To draw a bar graph, class intervals are marked along x-axis on a suitable scale. Frequencies are marked along y-axis on a suitable scale, such that the areas of drawn rectangles are proportional to corresponding frequencies.

(ii) Histogram: Histogram is rectangular representation of grouped and continuous frequency distribution in which class intervals are taken as base and height of rectangles are proportional to corresponding frequencies.

Now we shall study construction of histo grams related with four different kinds of frequency distributions.

• When frequency distribution is grouped and continuous and class intervals are also equal.
• When frequency distribution is grouped and continuous but class interval are not equal.
• When frequency distribution is grouped but not continuous.
• When frequency distribution is ungrouped and middle points of the distribution are given.

Now we try to make the above facts clear with some examples.

Example:
Draw a histogram of the following frequency distribution.

Class (Age in year)	0 – 5	5 – 10	10 – 15	15 – 20
No. of students	72	103	50	25

Solution:
Here frequency distribution is grouped and continuous and class intervals are also equal. So mark the class intervals on the x-axis i.e., age in year (scale 1 cm = 5 year). Mark frequency i.e., number of students (scale 1 cm = 25 students) on the y-axis.

Example:
The weekly wages of workers of a factory are given in the following table. Draw histogram for it.

Weekly wages	1000 – 2000	2000 – 2500	2500 – 3000	3000 – 5000	5000 – 5500
No. of worker	26	30	20	16	1

Solution:
Here frequency distribution is grouped and continuous but class intervals are not same. Under such circumstances the following method is used to find heights of rectangle so that heights are proportional to frequencies the least.
(i) Write the least class size (h), here h = 500.
(ii) Redefine the frequencies of classes by using following formula.
Redefined frequency of class = \(\frac{\mathrm{h}}{\text { class size }}\) × frequency of class interval.
So, here the redefined frequency table is obtained as follows:

Weekly wages (in Rs.)	No. of workers	Redefined frequency of workers
1000 – 2000	26	500/1000 × 26 = 13
2000 – 2500	30	500/500 × 30 = 30
2500 – 3000	20	500/500 × 20 = 20
3000 – 5000	16	500/2000 × 16 = 4
5000 – 5500	1	500/500 × 1 = 1

Now mark class interval on x-axis (scale 1 cm = 500) and no of workers on y-axis (scale 1 cm = 5). On the basis of redefined frequency distribution, construct rectangles A, B, C, D and E.

This is the required histogram of the given frequency distribution.

(a) Difference between Bar Graph and Histogram

• In histogram there is no gap in between consecutive rectangles as in bar graph.
• The width of the bar is significant in histogram. In bar graph, width is not important at all.
• In histogram the areas of rectangles are proportional to the frequency, however if the class size of the classes are equal then heights of the rectangle are proportional to the frequencies.

(iii) Frequency polygon: A frequency polygon is also a form of a graphical representation of frequency distribution Frequency polygon can be constructed in two ways:

• With the help of histogram.
• Without the help of histogram.

Following procedure is useful to draw a frequency polygon with the help of histogram.

• Construct the histogram for the given frequency distribution.
• Find the middle point of each upper horizontal line of the rectangle.
• Join these middle points of the successive rectangles by straight lines.
• Join the middle point of the initial rectangle with the middle point of the previous expected class interval on the x-axis.
• Join the middle point of the last rectangle with the middle point of the next expected class interval on the x-axis.

Example:
For the following frequency distribution, draw a histogram and construct a frequency polygon with it.

Class	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Frequency	8	12	17	9	4

Solution:
The given frequency distribution is grouped and continuous, so we construct a histogram by the method given earlier Join the middle points P, Q, R, S, T of upper horizontal line of each rectangles A, B, C, D, E by straight lines.

Example:
Draw a frequency polygon of the following frequency distribution table.

Marks obtained	Frequency (No. of students)
0 – 10	8
10 – 20	10
20 – 30	6
30 – 40	7
40 – 50	9
50 – 60	8
60 – 70	8
70 – 80	6
80 – 90	3
90 – 100	4

Solution:
Given frequency distribution is grouped and continuous. So we construct a histogram by using earlier method. Join the middle points P, Q, R, S, T, U, V, W, X, Y of upper horizontal lines of each rectangle A, B, C, D, E, F, G, H, I, J by straight line in successions.

Example:
Draw a frequency polygon of the following frequency distribution.

Age (in years)	Frequency
0 – 10	15
10 – 20	12
20 – 30	10
30 – 40	4
40 – 50	10
50 – 60	4

Solution:
Here frequency distribution is grouped and continuous so here we obtain following table on the basis of class.

Age (in years)	Classmark	Frequency
0 – 10	5	15
10 – 20	15	12
20 – 30	25	10
30 – 40	35	4
40 – 50	45	10
50 – 60	55	14

Now taking suitable scale on graph mark the points (5, 15), (15, 12), (25, 10) (35, 4), (45, 11), (55, 14).

Measures Of Central Tendency:
The commonly used measure of central tendency are:
(a) Mean,
(b) Median,
(c) Mode

(a) Mean: The mean of a number of observations is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol \(\overline{\mathrm{x}}\), read as x bar.
(i) Properties of mean:
→ If a constant real number ‘a’ is added to each of the observations, then new mean will be \(\overline{\mathrm{x}}\) + a.
→ If a constant real number ‘a’ is subtracted from each of the observations, then new mean will be \(\overline{\mathrm{x}}\) – a.
→ If a constant real number ‘a’ is multiplied with each of the observations, then new mean will be \(\overline{\mathrm{x}}\).
→ If each of the observation is divided by a constant no ‘a’ then new mean will be \(\frac{\overline{\mathrm{x}}}{\mathrm{a}}\).

(ii) Mean of ungrouped data: If x₁, x₂, x₃, ….., x_n are n values (or observations) then A.M. (Arithmetic mean) is

i.e. product of mean and no. of items gives sum of observations.

Example:
Find the mean of the factors of 10.
Solution:
Factors of 10 are 1, 2, 5 and 10.
\(\overline{\mathrm{x}}\) = \(\frac{1+2+5+10}{4}=\frac{18}{4}\) = 4.5

Example:
If the mean of 6, 4, 7, P and 10 is 8, find P.
Solution:
8 = \(\frac{6+4+7+P+10}{5}\)
⇒ P = 13

(iii) Method for Mean of ungrouped frequency distribution:

xi	fi	fixi
x1	f1	f1x1
x2	f2	f2x2
x3	f3	f3x3
·	·	·
·	·	·
xn	fn	fnxn
	Σfi	Σfixi

Then, mean \(\overline{\mathrm{x}}\) = \(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\)

(iv) Method for Mean of grouped frequency distribution:
Example:
Direct Method: For finding mean

Marks	No. of students (fi)	Mid values (xi)	fixi
10 – 20	6	15	9
20 – 30	8	25	200
30 – 40	13	35	455
40 – 50	7	45	315
50 – 60	3	55	165
60 – 70	2	65	130
70 – 80	1	75	75
	Σfi = 40		Σfixi = 1430

\(\overline{\mathrm{x}}\) = \(\frac{\sum f_i x_i}{\sum f_i}=\frac{1430}{40}\) = 35.75

(v) Combined Mean:
\(\overline{\mathrm{x}}\) = \(\frac{\mathrm{n}_1 \overline{\mathrm{x}}_1+\mathrm{n}_2 \overline{\mathrm{x}}_2+\ldots \ldots .}{\mathrm{n}_1+\mathrm{n}_2+\ldots \ldots}\)

(vi) Uses of Arithmetic Mean

• It is used for calculating average marks obtained by a student.
• It is extensively used in practical statistics.
• It is used to obtain estimates.
• It is used by businessmen to find out profit per unit article, output per machine, average monthly income and expenditure etc.

(b) Median: Median of a distribution is the value of the variable which divides the distribution into two equal parts.
(i) Median of ungrouped data:

• Arrange the data in ascending or descending order.
  Count the no. of observations (Let there be ‘n’ observations)
  If n is odd then median = value of \(\left(\frac{\mathrm{n}+1}{2}\right)^{\mathrm{th}}\) observation.
  If n is even then median = value of mean of \(\left(\frac{n}{2}\right)^{\text {th }}\) observation and \(\left(\frac{\mathrm{n}}{2}+1\right)^{\mathrm{th}}\) observation.

Example:
Find the median of the following values:
37, 31, 42, 43, 16, 25, 39, 45, 32
Solution:
Arranging the data in ascending order, we have
25, 31, 32, 37, 39, 42, 43, 45, 46
Here the number of observations, n
= 9 (odd)
∴ Median
= Value of \(\left(\frac{9+1}{2}\right)^{\text {th }}\) observation
= Value of 5th observation = 39.

Example:
The median of the observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.
Solution:
Here, the number of observations, n = 10.
Since n is even, therefore
Median

(ii) Uses of Median:
(A) Median is the only average to be used while dealing with qualitative data which cannot be measured quantitatively but can be arranged in ascending or descending order of magnitude.
(B) It is used for determining the typical value in problems concerning wages, distribution of wealth etc.

(c) Mode:
(i) Mode of ungrouped data (By inspection only): Arrange the data in an array and then count the frequencies of each variate. The variate having maximum frequency is the mode.

Example: Find the mode of the following array of an individual series of scores 7, 10, 12, 12, 12, 11, 13, 13, 17.

Number	7	10	11	12	13	17
Frequency	1	1	1	3	2	1

∴ Mode is 12
(ii) Uses of Mode: Mode is the average to be used to find the ideal size, e.g., in business forecasting, in manufacture of ready-made garments, shoes etc.

Empirical Relation Between Mode, Median And Mean:
Mode = 3 Median – 2 Mean
Range:
The range is the difference between the highest and lowest scores of a distribution. It is the simplest measure of dispersion. It gives a rough idea of dispersion. This measure is useful for ungrouped data.
(a) Coefficient of the Range:
If R and h are the lowest and highest scores in a distribution then the coefficient of the Range = \(\frac{\mathrm{h}-\mathrm{R}}{\mathrm{h}+\mathrm{R}}\)

Example: Find the range of the following distribution: 1, 3, 4, 7, 9, 10, 12, 13, 14, 16 and 19.
Solution:
R = 1, h = 19
∴ Range = h – R = 19 – 1 = 18.

Students should go through these JAC Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles

Polygonal Region

Polygon region can be expressed as the union of a finite number of triangular regions in a plane such that if two of these intersect, their intersection is either a point or a line segment. It is the shaded portion including its sides as shown in the figure.

Parallelogram

(a) Base and Altitude of a Parallelogram:
→ Base: Any side of parallelogram can be called its base.
→ Altitude: The perpendicular to the base from the opposite side is called the altitude of the parallelogram corresponding to the given base.
In the given Figure
→ DL is the altitude of ||^gm ABCD corresponding to the base AB.
→ DM is the altitude of ||^gm ABCD, corresponding to the base BC.

Theorem 1.
A diagonal of parallelogram divides it into two triangles of equal area.

Proof:
Given: A parallelogram ABCD whose one of the diagonals is BD.
To prove: ar(ΔABD) = ar(ΔCDB).
Proof: In ΔABD and ΔCDB;
AB = DC [Opp sides of a ||^gm]
AD = BC [Opp. sides of a ||^gm]
BD = BD [Common side]
∴ ΔΑΒD ≅ ΔCDB [By SSS]
ar(ΔABD) = ar (ΔCDB) [Areas of two congruent triangles are equal]
Hence, proved.

Theorem 2.
Parallelograms on the same base or equal base and between the same parallels are equal in area.

Proof:
Given: Two Parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and FC.
To Prove: ar(||^gm ABCD) = ar(||^gm ABEF)
Proof: In ΔADF and ΔBCE, we have
AD = BC [Opposite sides of a ||^gm]
and AF = BE [Opposite sides of a ||^gm]
AD || BC (Opposite sides of a parallelogram)
⇒ ∠ADF = ∠BCE (Alternate interior angles)
∴ ΔADF ≅ ΔBCE [By AAS]
∴ ar(ΔADF) = ar(ΔBCE) …..(i)
[Congruent triangles have equal area]
∴ ar (||^gm ABCD) = ar(ABED) + ar(ΔBCE)
= ar (ABED) + ar(ΔADF) [Using (1)]
= ar(||^gm ABEF).
Hence, ar(||^gm ABCD) = ar(||^gm ABEF).
Hence, proved.

Theorem 3.
The area of parallelogram is the product of its base and the corresponding altitude.

Proof:
Given: A ||^gm ABCD in which AB is the base and AL is the corresponding height.
To prove: Area (||^gm ABCD) = AB × AL.
Construction: Draw BM ⊥ DC so that rectangle ABML is formed.
Proof: ||^gm ABCD and rectangle ABML are on the same base AB and between the same parallel lines AB and LC.
∴ ar(||^gm ABCD) = ar(rectangle ABML)
= AB × AL
∴ area of a ||^gm = base × height.
Hence, proved.

Area Of A Triangle
Theorem 4.
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Proof:
Given: Two triangles ABC and PCB on the same base BC and between the same parallel lines BC and AP.
To prove: ar(ΔABC) = ar(ΔPBC)
Construction: Through B, draw BD || CA intersecting PA produced in D and through C, draw CQ || BP, intersecting line AP produced in Q.
Proof: We have, BD || CA (By construction) And, BC || DA [Given]
∴ Quad. BCAD is a parallelogram.
Similarly, Quad. BCQP is a parallelogram.
Now, parallelogram BOQP and BCAD are on the same base BC, and between the same parallels.
∴ ar (||^gm BCQP) = ar (||^gm BCAD)….(i)
We know that the diagonals of a parallelogram divides it into two triangles of equal area.
∴ ar(ΔPBC) = \(\frac{1}{2}\)(||^gm BCQP) …..(ii)
And ar (ΔABC) = \(\frac{1}{2}\)(||^gm BCAD)….(iii)
Now, ar (||^gm BCQP) = ar(||^gm BCAD) [From (i)]
\(\frac{1}{2}\)ar(||^gm BCAD) = \(\frac{1}{2}\)ar(||^gm BCQP)
Hence, \(\frac{1}{2}\)ar(ABC) = ar(APBC) (Using (ii) and (iii) Hence, proved.

Theorem 5.
The area of a trapezium is half the product of its height and the sum of the parallel sides.

Proof:
Given: Trapezium ABCD in which AB || DC, AL ⊥ DC, CN ⊥ AB and AL = CN = h (say). AB = a, DC = b.
To prove: ar(trap. ABCD) = \(\frac{1}{2}\)h × (a + b).
Construction: Join AC.
Proof: AC is a diagonal of quad. ABCD.
∴ ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD)
= \(\frac{1}{2}\)h × a+\(\frac{1}{2}\)h × b= \(\frac{1}{2}\)h(a + b).
Hence, proved.

Theorem 6.
Triangles having equal areas and having one side of the triangle equal to corresponding side of the other, have their corresponding altitudes equal.

Proof:
Given: Two triangles ABC and PQR such that
(i) ar(ΔABC) = ar(ΔPQR) and
(ii) AB = PQ.
CN and RT and the altitudes corresponding to AB and PQ respectively of the two triangles.
To prove: CN = RT.
Proof: In ΔABC, CN is the altitude corresponding to the side AB
ar(ΔABC) = \(\frac{1}{2}\)AB × CN ……(i)
Similarly, ar(ΔPQR) = \(\frac{1}{2}\)PQ × RT ……(ii)
Since, ar(ΔABC) = ar(ΔPQR) [Given]
∴ \(\frac{1}{2}\)AB × CN = PQ × RT
Also, AB = PQ [Given]
∴ CN = RT Hence, proved.

Students should go through these JAC Class 9 Maths Notes Chapter 15 Probability will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 15 Probability

Probability:
Theory of probability deals with measurement of uncertainty of the occurrence of some event or incident in terms of percentage or ratio.
→ Sample Space: Set of all possible outcomes.
→ Trial: Trial is an action which results in one of several outcomes.
→ An experiment: An experiment is any kind the of activity such as throwing a die, tossing a coin, drawing a card. The different possibilities which can occur during an experiment. e.g. on throwing a dice, 1 dot, 2 dots, 3 dots, 4 dots, 5 dots, 6 dots can occur.
→ An event: Getting a ‘six’ in a throw of dice, getting a head, in a toss of a coin.
→ A random experiment: is an experiement that can be repeated numerous times under the same conditions.
→ Equally likely outcomes: The outcomes of a sample space are called equally likely if all of them have same chance of occurring.
→ Probability of an event A: Written as P(A) in a random experiment and is defined as –
P(A) = \(\frac{\text { Number of outcomes in favour of A }}{\text { Total number of possible outcomes }}\)

Important Properties:
(i) 0 ≤ P(A) ≤ 1
(ii) P (not happening of A) + P (happening of A) = 1
or, P(\(\bar{A}\)) + P(A) = 1
∴ P(\(\bar{A}\)) = 1 – P(A)
Probability of the happening of A = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}\)
Probability of not happening of A (failing of A) = \(\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}}\)
where event A can happen in m ways and fail in n ways. All these ways being equally likely to occur.

Problems of Die:
A die is thrown once. The probability of
→ Getting an even number in the throwing of a die: the total number of outcomes is 6. Let A be the event of getting an even number then there are three even numbers 2, 4, 6.
∴ Number of favourable outcomes = 3.
P(A) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{3}{6}=\frac{1}{2}\)

→ Getting an odd number: Total no. of outcomes = 6,
favourable outcomes = 3 i.e. {1, 3, 5}
∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\)
→ Getting a natural number P(A) = \(\frac{6}{6}\) = 1
→ Getting a number which is multiple of 2 and 3 = \(\frac{1}{6}\)
→ Getting a number ≥3 i.e. {3, 4, 5, 6},
P(A) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
→ Getting a number 5 or 6, P(A) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
→ Getting a number ≤5 i.e. {1, 2, 3, 4, 5},
P(A) = \(\frac{5}{6}\)

Problems Concerning Drawing a Card:

• A pack of 52 cards
• Face cards (King, Queen, Jack)

JAC Board Class 9th Science Important Questions Chapter 11 Work and Energy

Multiple Choice Questions

Question 1.
The gravitational potential energy of an object is due to
(a) it’s mass
(b) its acceleration due to gravity
(c) its height above the earth’s surface
(d) all of the above
Answer:
(d) all of the above

Question 2.
The unit of work is the joule. The other physical quantity that has the same unit is
(a) power
(b) velocity
(c) energy
(d) force
Answer:
(c) energy

Question 3.
If the velocity of a body is doubled, its kinetic energy
(a) gets doubled
(b) becomes half
(c) does not change
(d) becomes 4 times
Answer:
(d) becomes 4 times

Question 4.
A student carries a bag weighing 5 kg from the ground floor to his class on the first floor that is 2 m high. The work done by the boy is
(a) 1J
(b) 10J
(c) 100J
(d) 1000J
Answer:
(c) 100J

Question 5.
How much time will be required to perform 520 J of work at the rate of 20 W?
(a) 24s
(b) 16s
(c) 20s
(d) 26s
Answer:
(d) 26s

Question 6.
A body of mass 2 kg is dropped from a height of lm. Its kinetic energy as it touches the ground is
(a) 19.6N
(b) 19.6J
(c) 9.8m
(d) 9.8J
Answer:
(b) 19.6J

Question 7.
The unit of power is
(a) watt per second
(b) joule
(c) kilojoule
(d) joule per second
Answer:
(d) joule per second

Question 8.
A coolie carries a load of 500 N to a distance of 100 m. The work done by him is
(a) 5 Nm
(b) 50,000 Nm
(c) 0 Nm
(d) 1/5 N m
Answer:
(c) 0 Nm

Question 9.
Power is a measure of the
(a) rate of change of momentum
(b) force which produces motion
(c) change of energy
(d) rate of change of energy
Answer:
(d) rate of change of energy

Question 10.
If the speed of an object is doubled, its kinetic energy is
(a) doubled
(b) quadrupled
(c) halved
(d) tripled
Answer:
(b) quadrupled

Question 11.
Which of the following is not correct?
(a) Energy is the ability of doing work
(b) Work can be expressed as F × s
(c) Unit of power is joule
(d) Power is the amount of work done per unit of time
Answer:
(d) Power is the amount of work done per unit of time

Question 12.
kW h is the unit of
(a) acceleration
(b) work
(c) power
(d) energy
Answer:
(c) power

Question 13.
Considering air resistance negligible, the sum of potential and kinetic energies of a free falling body would be
(a) zero
(b) increasing
(c) decreasing
(d) fixed
Answer:
(d) fixed

Question 14.
Two bodies of masses m] and m2 have equal kinetic energies. If P₁ and P₂ are their respective momenta, the ratio of P₁ to P₂ is
(a) m₁ : m₂
(b) m₂ : m₁
(c) \(\sqrt{\mathrm{m}_{1}} : \sqrt{\mathrm{m}_{2}}\)
(d) \(m_{1}^{2} : m_{2}^{2}\)
Answer:
(c) \(\sqrt{\mathrm{m}_{1}} : \sqrt{\mathrm{m}_{2}}\)

Question 15.
A light and a heavy body have equal momenta Which one has greater kinetic energy?
(a) The lighter body
(b) The heavier body
(c) Both have same KE
(d) None of these
Answer:
(b) The heavier body

Analysing & Evaluating Questions

Question 16.
A car is accelerated on a levelled road and attains a velocity four times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of intial
(d) becomes 16 times that of initial
Answer:
(a) does not change

Question 17.
An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) kinetic energy
Answer:
(a) acceleration

Question 18.
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g = 10 m^-2)
(a) 6 × 10³J
(b) 6J
(c) 0.6J
(d) zero
Answer:
(d) zero

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: Work, the product of force and displacement, is a vector quantity.
Reason: Product of two vector quantities is always a vector quantity.
Answer:
(D) Both the statements are false.

2. Assertion: When a book is moved from a table to the top of an almirah, its potential energy increases.
Reason: Higher the height of a body from the ground level, higher is its potential energy.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: A person carrying a suitcase on his head is not doing any work. Reason: The force applied by the person is acting in the downward direction.
Answer:
(C) The assertion is true but the reason is false.

4. Assertion: The work done by the force of gravity on the moon revolving around the earth is zero.
Reason: The gravitational force and the displacement of moon are at right angles to each other.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion: Work done on an object can be positive, negative or zero.
Reason: Work done is a scalar quantity.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
Define work.
Answer:
Work is said to be done when a force applied on a body produces a displacement of the body. It is given by W = F x s where ‘F’ is the force applied and ‘s’ is the displacement caused.

Question 2.
State reason why work is a scalar quantity.
Answer:
Work is the product of force (F) and displacement (s). Since both F and s are vector quantities and the dot product of vector quantities produces a scalar quantity, therefore, work is a scalar quantity.

Question 3.
Name two kinds of potential energies.
Answer:
Gravitational potential energy and elastic potential energy.

Question 4.
If the work done is 20J and displacement is 2m, find the force applied.
Answer:
Given, W = 20J and s = 2m.
W = F × s
20 = F × 2
F= 10 N

Question 5.
Name the energy stored in a rubber band when it is stretched?
Answer:
On stretching a rubber band, potential energy is stored in it.

Question 6.
State the law of conservation of energy.
Answer:
It states that energy can neither be created nor destroyed. It can only change its form.

Question 7.
When a book is lifted from a table, against which force is the work done?
Answer:
Work is done against the force of gravity.

Question 8.
Define the commercial unit of energy.
Answer:
The commercial unit of energy is kW h (kilowatt hour). 1 kW h is the energy used in one hour at the rate of 1000 J per second.
1 kWh = 1 kW × 1 h = 3.6 × 10⁶j

Question 9.
A light and a heavy body have equal kinetic energies. Which one is moving faster?
Answer:
The lighter body is moving faster because for the same kinetic energy, velocity is inversely proportional to the mass.
Analysing & Evaluating Question uestions

Question 10.
Two objects of same mass are placed at positions A and B as shown in the figure. Both of them are raised to the position C. In which case, the potential energy is more?
Answer:
The object at A will gain more potential energy than the object at B. But the final potential energy of both A and B will be equal when raised to position C.

Question 11.
A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of the itwo kinetic energies?
Answer:
K.E. °c (velocity)²
When the velocity is tripled, K.E. increases by a factor of 9 ( = 3²)
Thus, the ratio of the two kinetic energies is 1 : 9.

Question 12.
Can any object have momentum even if its mechanical energy is zero? Explain.
Answer:
No. Zero mechanical energy means no potential energy and no kinetic energy. Zero kinetic energy means, zero velocity. As a result, momentum is also zero (as P = mv).

Short Answer Type Questions

Question 1.
Calculate the work done by a man in rotating a wheel of an amusement slide in a fair 40 times in 1 minute?
Answer:
The man is rotating the wheel of an amusement slide by just standing at a place. This concludes that the wheel is not undergoing any displacement. Since displacement is zero, therefore, work done is zero.

Question 2.
Define positive work done and negative work done.
Answer:

1. Positive work (done: Work done is positive when the displacement occurs in the direction of force.
2. Negative work done: Work done is negative when the displacement occurs opposite to the direction of force.

Question 3.
In which of the following cases, work is said to be done?
1. When we push a table and the table is displaced.
2. When a person holds a book in his hand and keeps it stationary.
3. When a wire is twisted.

Answer:

1. When we push a table and the force applied by us is large enough to move it from its original position, then work is said to be done.
2. When a person holds a book in his hand and keeps it stationary, there occurs no movement of the book. In this case, though a force is constantly being applied, there is no displacement and hence work done is zero.
3. When a wire is twisted, the shape of the wire changes which concludes that work is done as there occurred changes in the configuration of the wire.

Question 4.
What will be the nature of work done when the force acting on a body retards its motion? Justify your answer by quoting examples.
Answer:
When force retards the motion of a body, the motion is stopped, i.e., a force opposite to the direction of the motion is applied. Thus, a negative work is done by the force.
For example:

1. In tug of war, the work done by the losing team is negative.
2. When a ball is thrown up in the air, the gravitational force acting downwards upon the ball does negative work on the ball.

Question 5.
What is gravitational potential energy?
Answer:
The gravitational potential energy of an object at a point above the ground
is defined as the work done in raising an object from the ground to that point against gravity.
G_PE = mgh

Question 6.
Differentiate between potential energy and kinetic energy.
Answer:

Potential Energy	Kinetic Energy
(a) Energy possessed by a body due to its position, shape or configuration.	(a) Energy possessed by a body due to its motion.
(b) P.E. = mgh where, m = mass, g = acceleration due to gravity, h = height.	(b) K.E. = \(\frac{1}{2}\) mv2 where, m = mass, v = velocity.

Question 7.
State two situations where energy is supplied but no work is done.
Answer:
(a) A person pushing a heavy rock is using all the energy but if the rock does not move, no work is done, (b) A person standing with heavy load on his head is spending energy in doing this, but no work is done.

Question 8.
How are work and energy related to each other?
Answer:
An object having a capability to do work is said to possess energy. The object which does work loses energy and the object on which work is done, gains energy. The unit of both energy and work is joule.

Question 9.
On what factor does the gravitational potential energy depend?
Answer:
Gravitational potential energy depends on the height of object from the ground level zero level we choose. Example: A ball tossed from the second floor of a building will attain a height, say h, from its roof, but from the first floor its height will be h ‘where h > h’.
Hence, the potential energy of the ball on the first floor level is less as compared to that on the second floor.

Question 10.
Write the form of energy possessed by the body in the following situations:
(a) A coconut falling from tree
(b) An object raised to a certain height
(c) Blowing wind
(d) A child driving a bicycle on the road
Answer:
(a) Kinetic energy + Potential energy
(b) Potential energy
(c) Kinetic energy
(d) Kinetic energy

Question 11.
What is energy? Give the unit of energy. Name the different forms of energy.
Answer:
Energy of a body is defined as its capacity or ability to do work. When a body is capable of doing more work, it is said to possess more energy.
The SI unit of energy is joule (J).
Energy has many forms: potential energy, kinetic energy, heat energy, chemical energy, electrical energy, light energy, solar energy, etc.

Question 12.
Derive an equation for kinetic energy of an object?
Answer:
The kinetic energy of a body can be determined by calculating the amount of work required to set the body into motion with the velocity ‘v’ from its state of rest. Suppose,
m = mass of the body
u = 0 = initial velocity of the body
F = force applied on the body
a = acceleration produced in the body in
the direction of force
v = final velocity of the body
s = distance covered by the body
As v² – u² = 2as
= v² – 0² = 2as
a= \(\frac{v^{2}}{2 s}\)
As the force and displacement are in the same direction, the work done on the body is
W = Fs = mas = m \( \frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\)s = \(\frac{1}{2}\) mv²
This work done appears as the kinetic energy of the body.
∴ KE = \(\frac{1}{2}\) mv²

Question 13.
Derive an equation for potential energy?
Answer:
Let the work done on the object against gravity be W.
Work done, W = force × displacement
Work done, W = mg × h
Work done, W = mgh
Since work done on the object is equal to mgh, an energy equal to mgh units is gained by the object. This is the potential energy (E_p) of the object.
E_p = mgh

Question 14.
An electric heater of 1000 W is used for 2 hours a day. What is the cost of using it for a month of 28 days, if 1 unit costs ₹ 3.00?
Answer:
Here, P = 1000W = lkW
Total time, t = 2 × 28 hours = 56 hours
Total energy consumed = P × t
= 1 kW × 56 h = 56 kW h
Cost of 1 kWh = ₹ 3.00
Cost of 56 kWh = 3 × 56 = ₹ 168.

Analysing & Evaluating Questions

Question 15.
Two identical pointed objects made from iron and wood are allowed to fall on a heap of sand from the same height. The iron object penetrates more in sand than the wooden object. Which of the objects has more potential energy?
Answer:
Of the two identical objects, the one made from iron will have greater mass. So when it falls from a height, it will possess greater kinetic energy as compared to the wooden object. As a result,
(a) the iron object will penetrate more in sand.
(b) the iron object will have more potential energy.

Question 16.
The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high can he jump on the planet A?
Answer:
The weight of a person on planet A is about half that on the earth. This means, the acceleration due to gravity of the planet A is half that of the earth. So, Height of the jump on the surface of planet A = \(\frac{0.4 \mathrm{~m}}{1 / 2}\) = 0.8 m

Long Answer Type Questions

Question 1.
How is work done measured when a body moves in a direction inclined to the direction of the applied force?
Answer:
In the figure, a force F pulls a block making angle 0 with the horizontal surface. Under this force, suppose the block moves from position A to B after covering a distance ‘s’.

Let, F₁ = Component of force in the direction of displacement ‘s’
Then, \(\frac{F_{1}}{F}\) = cos θ or F₁ = F cos θ
Work done = Component of force in the direction of displacement × displacement
W = F₁ × s = F cos θ x s
W = Fs cos θ
Special cases:
(a) When θ = 0°,
cos θ = 1 and W = Fs
Thus, work done is maximum when the displacement of the body is along the direction of the force.

(b) When θ = 90°,
cos θ = 0 and W = 0
Thus, work done is zero when the displacement of the body is perpendicular to the direction of force.

(c) When θ = 180°,
cos θ = -1 and W = – Fs
Thus, work done is negative when displacement is opposite to the direction of force.

(d) When s = 0, W = 0 Thus, work done on a stationary body is zero.

Question 2.
What is meant by potential energy of a body? Give some examples.
Answer:
The energy possessed by a body by the virtue of its position, shape or configuration is called its potential energy.

1. Examples of P.E. due to position:
   • Water stored in a dam at a height has potential energy.
   • A stone lying on the roof of the building has potential energy.
2. Examples of P.E. due to shape:
   • In a toy car, the wound spring possesses potential energy. As the spring is released, its potential energy changes into kinetic energy which moves the toy car.
   • Energy possessed by a stretched rubber band is potential energy.
3. Example of P.E. due to configuration:
   • A stretched bow possesses potential energy. As soon as it is released, it shoots the arrow in the forward direction with a large velocity.
   • The potential energy of the stretched bow gets converted into the kinetic energy of the arrow.
     

Question 3.
A 100 W electric bulb is lighted for 2 hours every day and five 40 W tubes are lighted for 4 hours every day. Calculate:
(a) the energy consumed for 60 days and
(b) the cost of electricity consumed at the rate of ₹3 per kW h.
Answer:
(a) Energy consumed by a 100 W bulb each day = 100 W × 2 h
= 200 Wh = \(\frac{200}{1000}\)= 0.2 kW h
Energy consumed by five 40 W tubes each day = 5 × 40 W × 4h
= 800 Wh = \(\frac{800}{1000}\) = 0.8 kW h
Total energy consumed each day = 0.2 + 0.8 = 1.0 kWh
Total energy consumed in 60 days = 1.0 x 60 = 60 kWh

(b) Cost of 1 kW h = ₹ 3
Cost of 60 kW h = 3 × 60 = 180

Question 4.
Answer the following:
(a) List any three situations in your daily life where you can say that work has been done.
(b) How much work is done in increasing the velocity of a car from 15 km/h to 30 km/h if the mass of the car is 1000 kg?
Answer:
(a) Three situations where work is done are:

1. Pushing a pebble lying on the ground. The pebble moves through some distance. Here, we apply a force and the pebble gets displaced. So, we have done work on the pebble.
2. We apply a force to lift a book through a height. The book rises up. We have done work in moving up the book.
3. A bullock is pulling a cart and the cart moves. There is a force on the cart and the cart has moved. The bullock has done work on the cart.

(b) According to the question,
u = 15 km/h = 4 m/s
v = 30 km/h = 8 m/s
Mass = 1000 kg
W = ?
W = K.E. = \(\frac{1}{2}\) m(v² – u²)
\(\frac{1}{2}\) × 1000 × ((8)² – (4)²)
= \(\frac{1}{2}\) × 1000 × (64 – 16)
= \(\frac{1}{2}\) × 1000 × (675) = 24,000J
Hence, work done is 24,000J.

Analysing & Evaluating Questions

Question 5.
A boy is moving on a straight road against a frictional force of 5N. After traveling a distance of 1.5 km he forgot the correct path at a roundabout of radius 100 m. However, he moves on that circular path for one and half cycle and then he moves forward up to 2.0 km. Calculate the work done by him.
Answer:
Work done by the boy while moving on a straight road
W = F × s
W = 5N × 1.5 km
= 5 kg m s^-2 × 1500 m = 7500J
Work done during moving around circular path
= 5N × (2 × 100 m) = 1000J
Work done during moving further by
2.0 km = 5N × (2 × 1000 m) = 10,000J
Total work done by the boy = 7500J + 1000J + 10,000J
= 18500J

Activity 1

• Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from a height of about 25 cm and observe the results.
• Repeat this activity from heights of 50 cm, lm and 1.5 m.
• Ensure that all the depressions are distinctly visible.
• Mark the depressions to indicate the height from which the ball was dropped.
• Compare their depths.

Observations

• The ball that falls from the height of 1.5 m creates the deepest depression.
• The ball that falls from the height of 50 cm creates the shallowest depression.
• Larger the height from which the ball is dropped, larger is the kinetic energy gained by the ball on reaching the ground and more is its capability of doing work.

Activity 2

1. Set up the apparatus as shown in the figure.
2. Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
3. Place a known mass on the pan so that the trolley starts moving.
4. The trolley moves forward and hits the wooden block.
5. Fix a stop on the table in such a manner that the trolley stops after hitting the block. The block gets displaced.
6. Note down the displacement of the block.
7. Repeat this activity by increasing the mass on the pan. Observe, in which case is the displacement more.
   

Observations

1. The force of gravity pulls the mass in the pan in the downward direction. This force gets transferred to the trolley through the string.
2. The trolley moves and hits the block with a force.
3. The larger the mass in the pan, the larger is the force with which the trolley hits the block.
4. Consequently, larger will be the displacement and larger will be the work done. The moving trolley possess energy and hence does work on the block.

Activity 3

• Take a slinky as shown below.
• Ask a friend to hold one of its ends. You hold the other end and move away from your friend. Now, you release the slinky and observe.
  

Observations

• When released, the slinky regains its original length. The slinky has acquired potential energy due to the work done on it during stretching. On releasing, potential energy is converted into kinetic energy.
• The slinky will also acquire energy when it is compressed.

Value Based Questions

Question 1.
Apoorva saw few planter pots kept on the balcony sill of fourth floor in her building. She makes an effort and keeps all the planter pots down the sill.
1. What type of energy is present in the pot kept on the balcony sill of fourth floor?
2. If the pot falls from the fourth floor, what type of energy will be seen in the falling pot?
3. What value of Apoorva is reflected in the above act?
Answer:
1. The pot on the sill possesses potential energy.
2. The falling pot possesses both the kinetic energy and the potential energy.
3. Apporva showed the value of moral responsibility and awareness.

Question 2.
Siddharth saw a lady labourer who carried bricks on her head from one point of the construction site to the other end which was some 500 m away. He prepares a trolley for the labourer to carry the bricks to make her work easier.
1. In carrying the bricks from point A to point B on the head by the lady labourer in the construction site, is any work done by the labourer?
2. By pulling the trolley of bricks from point A to point B, is any work done?
3. What value of Siddharth is seen in the above act?
Answer:
1. In carrying the bricks from point A to point B on head by the lady, no work is said to be done.
2. By pulling the trolley of bricks, work is said to be done.
3. Siddharth showed kindness, general awareness and sympathy.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 10 Gravitation

Multiple Choice Questions 

Question 1.
The mass of a body on moon is 40 kg. What is its weight on the earth?
(a) 240 kg
(b) 392 N
(c) 240 N
(c) 400 kg
Answer:
(b) 392 N

Question 2.
The gravitational force between two objects is
(a) attractive at large distance only
(b) attractive at small distance only
(c) attractive at all distances
(d) attractive at large distance but repulsive at small distance
Answer:
(c) attractive at all distances

Question 3.
A body of mass 1 kg is attracted by the earth with a force equal to
(a) 9.8 N
(b) 1 N
(c) 6.67 × 10¹¹ N
(d) 9.8 m/s
Answer:
(a) 9.8 N

Question 4.
The earth attracts a body with a force of 10 N. With what force does that the body attract the earth?
(a) 10N
(b) 1 N
(c) \(\frac{1}{10 \mathrm{~N}}\)
(d) 2 N
Answer:
(a) 10N

Question 5.
The SI unit of G is
(a) Nm² kg^-1
(b) Nm² kg^-2
(c) Nm² kg
(d) N^-1 m² kg^-2
Answer:
(b) Nm² kg^-2

Question 6.
The gravitational force of attraction between two bodies of 1 kg each and 1 m apart is
(a) 6.67 × 10^-11 N
(b) 6.67 × 10^-8 N
(c) 6.67 × 10¹¹ N
(d) 6.67 × 10⁸ N
Answer:
(a) 6.67 × 10^-11 N

Question 7.
The value of acceleration due to gravity near the earth’s surface is
(a) 8.9 m^-2
(b) 9.8 ms^-2
(c) 8.9 cms^-2
(d) 9.8 cms^-2
Answer:
(b) 9.8 ms^-2

Question 8.
Consider an elevator moving downwards with an acceleration a The force exerted by a passenger of mass m on the floor of the elevator is
(a) ma
(b) ma – mg
(c) mg – ma
(d) mg + ma
Answer:
(c) mg – ma

Question 9.
If the earth suddenly shrinks to half of its present size, the value of acceleration due to gravity will
(a) become twice
(b) remain unchanged
(c) become half
(d) become four times
Answer:
(d) become four times

Question 10.
The weight of a body would not be zero
(a) at the centre of the earth
(b) during a free fall of an elevator
(c) in interplanetary space
(d) on a frictionless surface
Answer:
(d) on a frictionless surface

Question 11.
Newton’s law of gravitation
(a) can be verified in a laboratory
(b) cannot be verified but is true
(c) is valid only on earth
(d) is valid only in the solar system
Answer:
(a) can be verified in a laboratory

Question 12.
10 kg weight is equal to
(a) 9.8 N
(b) 98 N
(c) 980 N
(d) \(\frac{1}{9.8}\) N
Answer:
(b) 98 N

Question 13.
The weight of a body cannot be expressed in
(a) kg wt
(b) N
(c) dyne
(d) kg
Answer:
(d) kg

Question 14.
A stone is dropped from a cliff. Its speed after it has fallen 100m is
(a) 9.8 m/s
(b) 44.2 m/s
(c) 19.8 m/s
(d) 98 m/s
Answer:
(b) 44.2 m/s

Question 15.
At which of the following locations, the value of g is the largest?
(a) On top of the Mount Everest
(b) On top of Qutub Minar
(c) At a place on the equator
(d) A camp site in Antarctica
Answer:
(d) A camp site in Antarctica

Analysing & Evaluating Questions

Question 16.
An apple falls from a tree because of gravitational attraction between the earth and apple. If Ft is the magnitude of force exerted by the earth on the apple and F2 is the magnitude of force exerted by the apple on the earth, then
(a) F₁ is very much greater than F₂
(b) F₂ is very much greater than F₁
(c) F₁ is only a little greater than F₂
(d) F₁ and F₂ are equal
Answer:
(d) F₁ and F₂ are equal

Question 17.
An object is put one by one in three liquids having differei dnsities. The object floats with\(\frac{1}{9}\), \(\frac{2}{11}\) and \(\frac{3}{7}\) part of its volume outside dit liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following is correct?
(a) d₁ > d₂ > d₃
(b) d₁ >d₂ <d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃
Answer:
(d) d₁ < d₂ < d₃

Question 18.
An object weighs 10N in air. When immersed fully in water, it weighs only 8N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12
Answer:
(a) 2 N

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: The law of gravitation is not applicable on the two bodies lying on the surface of the earth.
Reason: Law of gravitation is applied to celestial bodies only.
Answer:
(D) Both the statements are false.

2. Assertion: Mass of a body remains same at all the places.
Reason: Mass of a body is independent of acceleration due to gravity.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: Value of acceleration due to gravity is greater at the poles than at the equator.
Reason: Distance between pole and the centre of the earth is less than the distance between the equator and the centre of the earth.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

4. Assertion: One can jump higher on the surface of the moon than on the earth. Reason: The value of g (acceleration due to gravity) on the moon is greater than that on the earth.
Answer:
(C) The assertion is true but the reason is false.

5. Assertion: Fluids exert an upthrust on the objects immersed in them.
Reason: The upthrust is equal to the weight of the liquid displaced.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
Define the term ‘gravitation’.
Answer:
Every object in this universe attracts every other object with a force known as ‘force of gravitation’. Gravitation is the force of attraction between any two bodies in the universe.

Question 2.
Give the SI unit and value of G.
Answer:
SI unit of G = \(\frac{\mathrm{Nm}^{2}}{\mathrm{~kg} \cdot \mathrm{kg}}\) = N m² kg^-2
Its value = 6.67 × 10^-11 N m² kg^-2.

Question 3.
g = GM/R², what do the symbols in this formula denote?
Answer:
g = Acceleration due to gravity
G = Gravitational constant
M = Mass of the earth
R = Radius of the earth

Question 4.
Name the scientist who first determined the value of G experimentally.
Answer:
Henry Cavendish first determined the value of G experimentally, in the year 1778, using a sensitive balance.

Question 5.
Why is G called a universal gravitational constant’?
Answer:
The value of constant G is same for any pair of objects in the universe. Also, its value does not depend on the nature of the intervening medium. That is why constant G is called ‘universal’, whether the bodies are big or small, or whether they are celestial or terrestrial.

Question 6.
State any two properties of gravitational force.
Answer:

1. It is always attractive in nature.
2. It does not depend on the nature of the medium between the two bodies.

Question 7.
Which force is responsible for the stability of our universe?
Answer:
The force of gravitation.

Question 8.
How is ‘g’ on the surface of the earth related to ‘G’?
Answer:
g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\) = \(\frac{1.67 \times\left(1.74 \times 10^{6}\right)^{2}}{6.67 \times 10^{-11}}\)

where M = mass of the earth,
R = radius of the earth.

Question 9.
How is weight of an object related to its mass?
Answer:
Weight (w) = mass (m) × acceleration due to gravity (g).

Question 10.
What is the SI unit of pressure?
Answer:
The SI unit of pressure = N/m² = Pascal.

Question 11.
Define thrust. What is the SI unit of thrust?
Answer:
The net force exerted by a body in a particular direction is called thrust. The SI unit of thrust is newton.

Question 12.
Why does a truck or a motorbike have much wider tyres?
Answer:
A truck or a motorbike has much wider types so that the pressure exerted by it can be distributed to more surface area of the road and avoid the wear and tear of tyres.

Question 13.
In what sense does the moon fall towards the earth? Why does it not actually fall on the earth’s surface?
Answer:
At each point of its orbit, the moon falls towards the earth instead of going straight. It does not fall on the earth because it is moving in a circular orbit. The moon is kept in its circular orbit by the centripetal force provided by the force of attraction of the earth.

Question 14.
The earth attracts an apple from the tree and it falls but the earth does not appear to move towards the apple. Why?
Answer:
The mass of the earth is extremely large as compared to that of the apple. So the acceleration produced is very small as compared to that in the apple. Hence, the motion of the earth towards the apple is not noticeable.

Question 15.
What do you mean by the term ‘free fall’?
Answer:
The motion of a body under the influence of the force of gravity alone is called a ‘free fall’.

Question 16.
Suppose a planet exists whose mass and radius, both are half that of the earth. Calculate the acceleration due to gravity on the surface of this planet.
Answer:
On the surface of the earth, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
On the surface of the planet, \(g^{\prime}=\frac{G\left(\frac{M}{2}\right)}{\left(\frac{R}{2}\right)^{2}}=\frac{2 G M}{R^{2}}=2 g\)

Question 17.
Define mass of a body. What is its SI unit?
Answer:
The mass of a body represents the quantity of matter contained in the body. It gives a measure of inertia of the body. The greater the mass of a body, the greater is its inertia. The SI unit of mass is kilogram (kg).

Question 18.
Define one kilogram – weight. How many newtons are there in 1kg wt?
Answer:
One kilogram – weight (kg – wt)
= 1kg × 9.8 m/s² = 9.8 N.

Question 19.
State Archimedes’ principle.
Answer:
This principle states that when a body is immersed fully or partially in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it.

Analysing & Evaluating Questions

Question 20.
A beam balance is used for measuring the mass of a body, whereas a spring balance gives the weight of a body. What is measured by a digital balance?
Answer:
Digital balance measures the weight of a body.

Question 21.
Under what conditions do the two hollow balls – one of aluminium and the other of iron, experience equal upthrust when placed in water?
Answer:
The two balls will experience equal upthrust in water when their volumes inside the water are equal.

Question 22.
The density of liquid B is greater than that of liquid A. A hydrometer is placed one by one in the two liquids. In which liquid will the hydrometer sink to the greater depth?
Answer:
The hydrometer will sink deeper in liquid A.

Short Answer Type Questions

Question 1.
During free – fall of an elevator, what is the weight of a body inside it? Give reasons for your answer.
Answer:
During free – fall of an elevator the apparent weight of the object becomes zero. This is because both the body and the elevator are in free fall, and have a downward acceleration of ‘g’. In this situation normal reaction on the body becomes zero. Since normal reaction is responsible for the sensation of weight in a body, the apparent weight of the body in this case becomes zero.

Question 2.
What is relative density?
Answer:
The relative density of a substance is the ratio of its density to that of the water at 4°C. Relative density Density of the substance Density of water at 4°C

Question 3.
Differentiate between ‘g’ and ‘G’ in tabular form.
Answer:

Acceleration due to gravity ‘g’	Universal gravitational constant ‘G’
(a) It is the acceleration acquired by a body due to earth’s gravitational pull on it.	(a) It is numerically equal to the force of attraction between two masses of 1kg each separated by a distance of 1 m.
(b) ‘g’ is not a universal constant. It is different at different places on the surface of the earth. Its value varies from one celestial body to another.	(b) ‘G’ is a universal constant and its value is same, i.e., 6.67 × 10-11 N m2 kg-2 everywhere in the universe.
(c) It is a vector quantity.	(c) It is a scalar quantity.

Question 4.
Why is it easier to swim in sea water than in river water?
Answer:
The density of sea water is high due to dissolved salts in it as compared to the density of river water. Hence, the buoyant force applied on the swimmer by the sea water is high which helps in floating and makes swimming simpler.

Question 5.
A ship made of iron does not sink but an iron nail sinks in water. Why?
Answer:
The iron nail sinks because of its high density and less buoyant force acting on it due to lesser surface area. Whereas, the surface area of a ship is greater and thus experiences a higher buoyant force. Due to this fact, a ship floats but an iron nail sinks.

Question 6.
A stone and feather are thrown from a tower. Both the objects should reach the ground at the same time but it does not happen. Give reasons.
Answer:
As per the motion of objects due to gravitational pull of the earth, both the bodies are acted upon by the same force of the earth but the stone will fall first and then the feather. Feather being lighter, experiences greater air resistance, so it will reach later.

Question 7.
The relative density of gold is 18.3. The density of water is 10³ kg/m³? What is the density of gold in S.I units?
Answer:
The relative density of gold is 18.3.
Relative density of gold = \(\frac{Density of gold}{Density of water}\)
That is, density of gold = Relative density of gold × Density of water = 18.3 × 10³ kg/m³ = 18300 kg/m³.

Question 8.
The acceleration due to gravity at the moon’s surface is 1.67 m/s². If the radius of the moon is 1.74 × 106m, calculate the mass of the moon.
Answer:
Here, G = 6.67 × 10^-11 Nm²/kg²
g = 1.67 m/s²
R= 1.74 × 10⁶ m.
Mass of the moon is
M = \(\frac{\mathrm{gR}^{2}}{\mathrm{G}}\) = \(\frac{1.67 \times\left(1.74 \times 10^{6}\right)^{2}}{6.67 \times 10^{-11}}\)
= 7.58 × 10²² kg.

Question 9.
Derive the formula for the gravitational force acting between two objects.
Answer:
Let two objects A and B of masses ‘M’ and’m’, lie at a distance d from each other as shown in the figure.

Let the force of attraction between two objects be F.
According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses.
F ∝ Mm … (1)
And the force between two objects is inversely proportional to the square of the distance between them,
F ∝ \(\frac{1}{\mathrm{~d}^{2}}\) …………(2)
F ∝ \(\frac{\mathrm{Mm}}{\mathrm{d}^{2}}\)
where G is universal gravitational constant

Question 10.
Find the value of ‘g’, acceleration due to gravity.
Answer:
g = \( \frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
G = 6.67 × 10^-11 Nm²/kg² (constant)
M = 6 × 10²⁴ kg (mass of the earth)
= \( \frac{6.7 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2} \times 6 \times 10^{24} \mathrm{~kg}}{\left(6.4 \times 10^{6} \mathrm{~m}\right)^{2}}\)
= 9.8 m/s²

Question 11.
Camels can walk easily on desert sand but we are not comfortable walking on the sand. Why do you think so?
Answer:
The surface area of a camel’s feet is broad and large. Thus, the pressure exerted is low. However, when we walk, our legs sink because the pressure exerted by our body is not equally distributed but is directed towards the legs.

Question 12.
Define lactometer and hydrometer.
Answer:
Lactometer is a device used to find the purity of milk. Hydrometer is a device used to find the density of a liquid.

Question 13.
Two wood pieces of same size and mass are dipped in two beakers containing water and oil. One wood floats on water, but the other one sinks in oil. Why?
Answer:
The wood floats on water because the density of wood is lower than the density of water, and the other wood sinks in the oil because the density of wood is more than that of the oil.

Question 14.
What are fluids? Why is Archimedes’ principle applicable only for fluids? Give the applications of Archimedes’ principle.
Answer:
Liquids are the substances which can flow. Archimedes principle is based on the upward force exerted by fluids on any object immersed in them. Hence, it is applicable only for fluids. Applications of Archimedes’ principle are as follows:

1. It is used to design ships and submarines.
2. To determine the purity of milk using lactometers which are designed based on the Archimedes’ principle.
3. To make hydrometers which are used to determine the density of liquids.

Question 15.
The volume of 60g of a substance is 10cm². If the density of water is 1g/cm², find out whether the substance will float or sink in water.
Answer:
Mass = 60g
Volume = 10 cm³
The density of water is 1g/cm². As the density of the given substance is more than the density of water, the substance will get submerged in water.

Question 16.
What is meant by buoyancy?
Answer:
When a body is submerged in a fluid, the fluid exerts an upward force on the submerged body. This upward force is equal to the weight of the fluid displaced by the submerged body and is called the buoyant force. In other words, it is the force exerted by the fluid when an object is submerged in it.

Analysing & Evaluating Questions

Question 17.
Identical packets are dropped from two aeroplanes, one over the equator and the other over the north pole, both from the height h. Assuming that all conditions are identical, explain if those packets take same time to reach the surface of earth or not. Justify your answer.
Answer:
No. The two packets will take different time intervals to reach the earth’s surface. This is because the acceleration due to gravity (g) is greater at the poles than that at the equator. The packet will fall slowly at the equator than that at the north pole.

Question 18.
You must have seen two types of balances. A grocer has a balance with two platforms or pan, and a needle moving on a scale. Some junk dealers (kabadiwalas) may be using a spring balance to weigh used newspapers. Suppose, the two balances give the same measure for a given body on the earth. Now, you take the balances to the moon. Would the two balances give the same measure? Explain your answer.
Answer:
No. The two balances will not give the same measure on the moon. This is because, the spring balance measures the weight, and the grocer’s balance measures the mass. The weight of a body depends upon the acceleration due to gravity at the place of measurement. The acceleration due to gravity on the moon is not equal to that on the earth. Therefore, spring balance will give different readings relative to the grocer’s balance on the moon.

Long Answer Type Questions

Question 1.
What is centripetal force? From where does the moon get the centripetal force required for its motion around the earth?
Answer:

1. Centripetal force When a body moves along a circular path with a uniform speed, its direction of motion changes at every point. The change in direction involves change in velocity or acceleration.
2. The force that provides this acceleration and keeps the body moving along the circular path, acts towards the centre. This force is called centripetal (centre seeking) force.
3. Therefore, a force which is required to make a body move along a circular path with uniform speed is called centripetal force. Centripetal force always acts along the radius and towards the centre of the circular path.
   
4. The figure given here shows that it is the centripetal force which continuously deflects a particle from its straight line path to make it move along a circle.
5. Example: The moon needs a centripetal force for its circular motion around the earth. This centripetal force is provided by the gravitational attraction exerted by the earth on the moon.

Question 2.
Show that the weight of an object on the moon is 1/6,h its weight on the earth. Given that the mass of the earth is 100 times the mass of the moon and its radius is 4 times that of the moon.
Answer:
Suppose, mass of the object = m
mass of the earth = M_e
mass of the moon = M_m
radius of the earth = R_e
radius of the moon = R_m
Then, M_e = 100 Mm and Re = 4Rm
Weight of an object of mass m on the earth is
We = the force with which the earth
attracts the object = G \(\frac{\mathrm{M}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}^{2}}\)
Weight of the object on the moon is
W_m = The force with which the moon
attracts the object = \( \mathrm{G} \frac{\mathrm{M}_{\mathrm{m}} \mathrm{m}}{\mathrm{R}_{\mathrm{m}}^{2}}\)
\(\frac{W_{m}}{W_{e}}\) = \(\frac{\left(G \frac{M_{m} m}{R_{m}^{2}}\right)}{\left(G \frac{M_{m} m}{R_{e}^{2}}\right)}\)
= \(\frac{M_{m}}{M_{e}} \times \frac{\left(R_{e}\right)^{2}}{\left(R_{m}\right)^{2}}\)
= \(\frac{\mathrm{M}_{\mathrm{m}}}{100 \mathrm{M}_{\mathrm{m}}} \times \frac{4 \mathrm{R}_{\mathrm{m}}^{2}}{\mathrm{R}_{\mathrm{m}}^{2}}\) = \(\frac{16}{100} \approx \frac{1}{6}\)
Thus, the weight of an object on the moon is about one – sixth of its weight on the earth.

Question 3.
Briefly explain how can Archimedes’ principle be verified experimentally.
Answer:
According to Archimedes’ principle, when a body is immersed in a liquid, completely or partially, it loses its weight. The loss in weight is equal to the weight of liquid displaced by the body. The loss in weight of a body is due to the presence of upthrust which is equal to the weight of liquid displaced.
Thus, a Loss in weight = Weight of body in air – Weight of body immersed in water =
W W₁ = upthrust in water on the body = Weight of liquid displaced.

As shown in figure, measure the weight of a solid, say a metallic ball, in air using spring balance. Weigh the empty beakers using spring balance. Set the spring balance, overflow can with tap water and beaker. Now, allow the bob to immerse completely in water in overflow can. Note down the new position of the pointer of the spring balance.

This will give you the weight of the brass bob in tap water. It is found to be less than the weight of bob in air. Weigh the beaker containing displaced water which is collected from the overflow can while immersing the bob in it completely. It is observed that the loss in weight of the bob is equal to the weight of the water collected in the beaker, i.e., weight of the water displaced.

Question 4.
With the help of an activity, prove that the force acting on a lesser area exerts a larger pressure.
Answer:
Consider a block of wood kept on a tabletop. The mass of the wooden block is 5kg. Its dimensions are 30 cm × 20 cm × 10 cm.
Now, we have to find the pressure applied by the wooden block on the tabletop by keeping it vertically and horizontally.

The mass of the wooden block = 5kg
Weight of the wooden block applies a thrust on the tabletop.
Thrust = F = mg = 5kg × 9.8 m/s² = 49 N
(Case a) : When the wooden box is kept vertically with sides 20 cm × 10 cm,
Area of a side = length × breadth = 20 cm × 10 cm = 200 cm² = 0.02m²
Pressure = \(\frac{Thrust}{Area}\) = \(\frac{49 \mathrm{~N}}{0.02 \mathrm{~m}^{2}}\)
= 2450 N/m²

(Case b): When the block is kept horizontally with side 30 cm × 20 cm, Area = length × breadth = 30 cm × 20 cm = 600 cm² = 0.06 m²

Pressure = \(\frac{Thrust}{Area}\) = [/latex] \frac{49 \mathrm{~N}}{0.02 \mathrm{~m}^{2}}[/latex]
= 816.7 N/m²
The pressure exerted by the box in case (a) in more as compared to the pressure exerted in case (b). The area is reduced and the pressure exerted is more.
This shows that pressure ∝= \( \frac{1}{\text { Area }}\)
Hence, pressure will be larger if the area is reduced.

Analysing & Evaluating Questions

Question 5.
(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case it will experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water.

(b) A ball of weight 4 kg and density 4000 kg m³ is completely immersed in water of density 10³ kg m³. Find the force of buoyancy on it. (Given g = 10 ms^-2)
Answer:
(a) The cube will experience greater buoyant force in salt solution. When the side of the cube is reduced to 4 cm, it will experience lesser buoyant force as compared to the first case.
This is because the volume of water displaced by the smaller cube is lesser than that displaced by the original cube,

(b) Volume of water displaced by the ball = Volume of the ball = \(\frac{Mass}{Density}\)
= \(\frac{4 \mathrm{~kg}}{4000 \mathrm{~kg} \mathrm{~m}^{-3}}\) = 10^-3 m³
Weight of water displaced by the ball = Volume of water × Density × g
= 10^-3 m³ × 10³ kg m³ × 10 ms^-2 = 10 N
So, Buoyant force acting on the ball = Weight of water displaced = 10 N

Activity 1

• Take a piece of thread.
• Tie a small stone at one end. Hold the other end of the thread and whirl it round.
• Note the motion of the stone.
• Now release the thread.
• Again, note the direction of motion of the stone.
• Record your observations and draw conclusion about the direction of the stone.
  
• If thread is released when stone is here, stone goes straight towards A not towards B Top view A stone describing a circular path

Observations

• The motion gets accelerated and stone moves in a circular path.
• When the thread is released, the stone makes a tangent to the circle and falls down.

ACTIVITY 2

• Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building.
• Observe whether both of them reach the ground simultaneously or not.

Observations

• We see that the paper reaches the ground a little later than the stone. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the paper is more than the resistance offered to the stone.
• If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.

ACTIVITY 3

• Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water and observe.
• Push the bottle into the water and observe again. Try to push it further down.
• Now, release the bottle. Record your observations.
• Does the force due to the gravitational attraction of the earth act on this bottle? If so, why doesn’t the bottle stay immersed in water after it is released? How can you immerse the bottle in water?

Gravity

Observations

• The plastic bottle floats on water. When it is pushed deeper, the upward force on the bottle increases.
• When released, the bottle bounces back to the surface because the upthrust on the bottle was larger than the downward pull of gravity
• Take a piece of stone and tie it to one end of a rubber string or a spring balance.

ACTIVITY-4

• Take a piece of stone and tie it to one end of a rubber string or a spring balance.
  Thread
  
• Suspend the stone by holding the balance or the string as shown in the figure (A).
• Note the elongation of the string or the reading on the spring balance due to the weight of the stone.
• Now, slowly dip the stone in the water in a container as shown in Figure (B).
• Observe what happens to the elongation of the string or the reading on the balance.

Observations

• In fig. A, the elongation of the string is more.
• In fig. B, when the stone is dipped in water, the length of the string is reduced.
• The length of the string in case (B) decreases due to the upward force exerted by water on the stone called buoyant force.

Value Based Questions

Question 1.
Priya had a bad experience during the take off of the plane when she boarded it for the first time. Her friend assisted and helped her during the landing of plane. She told her to fasten the seat belt and involved her in gossip. Priya faced less problems while landing of the plane.
1. Why do we tie our seat belts in moving cars or during take off of the plane?
2. What is free fall?
3. What value of Priya’s friend is seen in this set?
Answer:

1. We tie our seat belts to remain intact on the seat so that our body does not fall forward.
2. When an object falls towards the earth under the gravitational force alone, we say that the object is in free fall.
3. Priya’s friend showed the value of concerned, sympathetic, responsible and caring friend.

Question 2.
A goldsmith measured the purity of gold by using a special measuring device. He told the customer that there was impurity present in gold ornament that he wanted to buy and was not 22 carat but 18 carat jewellery.
1. How can we find the purity of gold?
2. What is the unit of relative density?
3. What value of goldsmith is reflected in this act?
Answer:

1. The purity of gold can be obtained by knowing the density of the gold.
2. Relative density does not have any unit.
3. Goldsmith showed the value of honesty and trustworthiness.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 9 Force and Laws of Motion

Multiple Choice Questions

Question 1.
The SI unit of force is
(a) kg m/s
(b) newton
(c) dyne
(d) kg wt
Answer:
(b) newton

Question 2.
The combined effect of mass and velocity is taken into account by a physical quantity called
(a) torque
(b) momentum
(c) moment
(d) force
Answer:
(b) momentum

Question 3.
Quantitative definition of force is given by
(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) Newton’s law of gravitation
Answer:
(b) Newton’s second law of motion

Question 4.
Momentum gives a measure of
(a) mass
(b) weight
(c) velocity
(d) motion
Answer:
(d) motion

Question 5.
An athlete runs some distance before taking a long jump because
(a) he gains energy to take himself through long distance
(b) it helps him to apply a larger force
(c) by running, action and reaction forces increase
(d) by running, the athlete gives himself larger inertia of motion
Answer:
(d) by running, the athlete gives himself larger inertia of motion

Question 6.
Rocket works on the principle of conservation of
(a) velocity
(b) energy
(c) linear momentum
(d) mass
Answer:
(c) linear momentum

Question 7.
The rate of change of momentum of an object is proportional to the
(a) mass of the body
(b) velocity of the body
(c) net force applied on the body
(d) none of these
Answer:
(c) net force applied on the body

Question 8.
A force of 50 N moves a body. Which of the following is correct?
(a) Frictional force exerted on the body is less than 50N
(b) Frictional force exerted on the body is more than 50N
(c) None of these
(d) Both (a) and (b)
Answer:
(a) Frictional force exerted on the body is less than 50N

Question 9.
If a football and a stone have same mass, then both will have
(a) same inertia
(b) same momentum
(c) different inertia
(d) different momentum
Answer:
(a) same inertia

Question 10.
The rate of change of momentum w. r. t time is measured in
(a) kg m
(b) kg
(c) kg ms^-1
(d) kg ms^-1
Answer:
(d) kg ms^-1

Question 11.
A block of mass M is pulled with a force F along a smooth horizontal surface with a rope of mass m. The acceleration of the block will be
(a) F/M
(b) F/m
(c) F/(M + m)
(d)F/(M – m)
Answer:
(c) F/(M + m)

Question 12.
Action and reaction forces act on two
(a) different objects
(b) same objects
(c) either (a) or (b)
(d) none of these
Answer:
(a) different objects

Question 13.
A man is standing in a boat in still water. If he tries to walk towards the shore, the boat will
(a) move away from the shore
(b) remain stationary
(c) sink
(d) move towards the shore
Answer:
(a) move away from the shore

Question 14.
When an object undergoes acceleration
(a) its speed always increases
(b) its velocity always increases
(c) it always falls towards the earth
(d) a force always acts on it
Answer:
(d) a force always acts on it

Question 15.
On applying a constant force to a body, it moves with uniform
(a) momentum
(b) speed
(c) acceleration
(d) velocity
Answer:
(c) acceleration

Analysing & Evaluating Questions

Question 16.
A goalkeeper in the game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to
(a) exert larger force on the ball
(b) reduce the force exerted by the ball on hands
(c) increase the rate of change of momentum
(d) decrease the rate of change of momentum
Answer:
(b) reduce the force exerted by the ball on hands

Question 17.
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is
(a) accelerated
(b) uniform
(c) retarded
(d) along circular tracks
Answer:
(a) accelerated

Question 18.
An object of mass 2kg is sliding with a constant velocity of 4 ms^-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N
Answer:
(b) 0 N

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both statements are false.
1. Assertion: A brick has more inertia than a hollow wooden block of the same shape and size.
Reason: Heavier the body, more is the inertia.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

2. Assertion: A body falling towards the earth also pulls the earth towards itself.
Reason: The forces always occur in pairs.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: When a bus starts, the person inside it falls forward.
Reason: The bus pushes the person forward.
Answer:
(D) Both statements are false.

4. Assertion: The third law of motion defines acceleration.
Reason: The third law of motion states that the net force experienced by a body is proportional to the rate of change in momentum of the body.
Answer:
(D) Both statements are false.

5. Assertion: Action and reaction forces are always equal and opposite.
Reason: Action and reaction forces cancel each other.
Answer:
(C) The assertion is true but the reason is false.

Very Short Answer Type Questions

Question 1.
Define force.
Answer:
Force may be defined as a push or a pull which changes or tends to change the state of rest or of uniform motion or direction of motion of a body.

Question 2.
Define one Newton.
Answer:
One newton is the force which produces an acceleration of m/s² in an object of mass 1kg.
1N = 1kg m/s²

Question 3.
State the various effects of force.
Answer:
A force applied on an object can do five things:
(a) It can change the speed of the object.
(b) It can change the direction of motion of the object.
(c) It can change the shape and size of the object.
(d) It can set an object at rest in motion.
(e) It can bring a moving object to rest.

Question 4.
Define resultant force.
Answer:
The resultant force or resultant of several forces acting simultaneously on a body is that single force which produces the same effect on a body as all these forces produce together.

Question 5.
What is frictional force?
Answer:
The force that always opposes the motion of an object is called force of friction.

Question 6.
What is inertia?
Answer:
The natural tendency of an object to resist a change in their state of rest or of uniform motion along a straight line is called inertia.

Question 7.
State Newton’s first law of motion.
Answer:
An object remains in its state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.

Question 8.
Place a water – filled tumbler on a tray. Hold the tray and turn around as fast as you can. We observe that the water spills. Why?
Answer:
The water spills when the tray turns around fastly, because initially, the water filled in the tumbler placed on the tray was in a state of rest. While turning, we apply force on the tray which starts moving but the water remains at rest due to its inertia, and hence, it spills out of the tumbler.

Question 9.
State Newton’s second law of motion.
Answer:
The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Question 10.
State Newton’s third law of motion.
Answer:
To every action, there is an equal and opposite reaction and they act on two different bodies.

Question 11.
Why are shockers used in cars, scooters and motorcycles?
Answer:
Due to the shockers, the time interval of the jerk increases. As the rate of change of momentum will be smaller, comparatively less force acts on the passengers during the jerks.

Question 12.
Why is glass or chinaware packed with straw?
Answer:
The straw paper between the chinaware increases the time of experiencing the jerk during transportation. Hence, they strike against each other with a less force and are less likely to be damaged.

Question 13.
A bird hit the windscreen of a fast moving car and fell on the bonnet. Which of the two, the car or the bird, suffers greater change in momentum?
Answer:
By the law of conservation of momentum, both the car and the bird suffered equal and opposite change in momentum.

Question 14.
What decides the rate of change of momentum of an object?
Answer:
The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

Question 15.
The diagram shows a moving truck. Forces A, B, C and D are acting on the truck. Name the type of each of these forces acting on the truck.

Answer:
The forces A, B, C and D acting on the truck are:
A driving force
B normal reaction
C frictional force
D weight gravitational force.

Analysing & Evaluating Questions

Question 16.
Two identical bullets are fired, one by a light rifle and another by a heavy rifle, with the same force. Which rifle will hurt the shoulder more and why?
Answer:
The velocity of recoil will be higher for ligher rifle. So lighter rifle will hurt the shoulder more.

Question 17.
When a fast moving carrom striker hits the carrom coin at the bottom of a pile, ¡t moves out without disturbing
the pile. Which kind of inertia of the pile is responsible for it?
Answer:
Inertia of rest of the pile is responsible for it.

Question 18.
There are three solid balls made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have the highest inertia?
Answer:
Steel ball will have the highest inertia This is because the density of steel is higher than aluminium and wood As a result, the steel ball will have the highest mass and hence highest inertia

Short Answer Type Questions

Question 1.
Differentiate between balanced and unbalanced forces.
Answer:

Balanced forces	Balanced forces
1. The resultant of balanced forces is zero.	1. The resultant of balanced forces is zero.
2. These forces donot produce anyacceleration in the body.	2. These forces donot produce anyacceleration in the body.
3. Example: A book kept on the table (the downward force of gravity on the book is balanced by the normal reaction by the table on the book in upward direction).	3. Example: When the book that is kept on the table is pushed to the left side (the book moves as the push towards left is greater than the frictional force acting towards right).

Question 2.
Give examples to show that balanced forces can change the shape of an object.
Answer:
Consider a spring attached to a rigid support at one of its ends, as shown in the figure.

If we pull the free ends of the spring, it gets elongated Thus, on applying force, a spring expands. Similarly, if we hold a rubber ball between our palms and push the two palms against each other, we find that the ball is no longer round but oblong. The force exerted on the ball changes its shape, as shown in the figure. In both cases, the two forces, being equal and opposite, balance each other but change the shape of the object.

Question 3.
A football player kicks the ball which travels in the air for a while and lands on the ground where the ball travels on the ground for a short distance and then stops. What may be the reason behind it?
Answer:
The reason is frictional force. There is always a contact between the ball and the ground The opposing force always acts against the motion of the ball, thereby stopping the ball.

Question 4.
A table is moved across the floor with a constant velocity where the horizontal force acting on it is 400N. What will be the frictional force that will be acting on the table?
Answer:
According to Newton’s third law of motion, for every action there will be an equal and opposite reaction. The net force acting on the table is zero because the table moves with constant velocity in forward direction. Hence, an equal amount of frictional force must act in opposite direction as that of the table, i.e., F = – 400N.

Question 5.
When the car, we are moving in, makes a sharp turn at a very high speed, we tend to get thrown to one side. Explain the statement.
Answer:
The reason behind this is the law of inertia While moving in a car along a straight line, we tend to maintain our state of motion. But when the car takes a sharp turn, our bodies resist the change in direction and are thrown to the opposite side due to inertia.

Question 6.
Athletes always have a special posture by resting their right foot on a solid support, why?
Answer:
During the race, athletes have to run the heats and they rest their right foot on the solid support before the start, so that this support can give them a lot of force during the start of the race. They push the support backwards and get an equal and opposite forward push to get a very good start.

Question 7.
The safety belts in the car help in preventing accidents. Justify the statement.
Yes, safety belts help in preventing accidents. When a car is moving with a high speed, our body tends to be in movement due to inertia of motion in the forward direction. So when there is a sudden  collision, serious injuries can happen. However, seat belts exert a force on our body to slow down the forward motion and hence prevent injuries.

Question 8.
A karate player breaks the pile of tiles with a single blow. Give reason.
Answer:
A karate player strikes the pile of tiles by applying a very large velocity. The overall momentum of fast moving hand is reduced to zero in a very short interval. So this increases the rate of change of momentum which increases the force on the tiles and they eventually break.

Question 9.
Athletes are made to fall on a sand bed while performing a high jump. Give reason.
Answer:
During the high jump event, an athlete is made to fall on the sand bed because it increases the time to attain the rest
position. We know that, F = \(\frac{m(v-u)}{t}\)
So, force and time are inversely proportional to each other. If there is an increase in time, there is a decrease in rate of change of momentum and therefore the force or the impact is reduced This prevents injuries to the feet of athlete.

Question 10.
Explain the law of conservation of momentum.
Answer:
According to the law of conservation of momentum, the total momentum of the objects before collision is equal to the total momentum after collision, provided there is no external unbalanced force acting on them.

Question 11.
A bullet of mass 20g moving with a speed of 500 ms^-2 strikes a wooden block of mass 1kg and gets embedded in it. Find the speed with which the block moves along with the bullet.
Answer:
Let the final velocity of the block along with the bullet embedded in it be v.
For bullet, m₁ = 20g = 0.02 kg,
U₁ = 500 m/s, v₁ = v
For block, m₁ = 1kg, u₂= 0, v₂ = v
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ + m₂u₂= (m₁ + m₂) v
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{m_{1}+m_{2}}\)
= \(\frac{(0.02 \times 500)+(1 \times 0)}{(0.02+1)}\)
=\(\frac{10}{1.02}\) = \(\frac{1000}{102}\)
= 9.8 m/s.

Analysing & Evaluating Questions

Question 12.
Two balls of the same size but of different materials, rubber and iron, are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed?

Answer:
The balls will start rolling in the forward direction. No. The two balls will move with different speed This is because the inertia of the two balls is not the same.

Question 13.
Two forces F₁ = 40N and F₂ = 60N are acting on an object as shown in the figure.

1. What is the net force acting on the object?
2. What is the direction of the net force acting on the object?
3. How much extra force is acting on the object if the object is not moving due to the application of these two forces? Name that force. Where does that force act and what is the direction of that force?
Answer:

1. Net force on the object = 60N – 40N = 20N
2. Net force of 20N acts in the direction of F₂ (i.e., towards left)
3. A force equal to 20N is acting on the object.
   This force is the frictional force. This force acts in the direction of F₁

Long Answer Type Questions

Question 1.
Describe Galileo’s experiments with inclined planes and state their conclusion.
Answer:
By observing the motion of objects on inclined planes, Galileo deduced that objects move with constant speed when no force acts on them.

1. In the first experiment, Galileo studied the motion of marbles on an inclined plane. He observed that when a marble rolls down an inclined plane, its velocity increases, as shown in the figure A.
2. Here the marble falls under the unbalanced force of gravity. The velocity of the marble decreases when it rolls up the inclined plane (against the force of gravity), as shown in the figure B.
3. From these observations, Galileo argued that the velocity of a marble rolling on a flat horizontal surface should remain constant. To test his idea, Galileo used double inclined planes.
4. He observed that when the inclinations of the planes on both sides were equal, the marble rolled down from one plane from a certain height will climb up to the same height on the other plane.
   
   side plane were gradually decreased, the marble would travel larger and larger distances to reach the same height.
5. Ultimately, if the right side plane were made horizontal, the marble would continue to travel forever to reach the same height from which it was released No unbalanced force acts on the marble in this case.
6. The above experiment suggests that an unbalanced force (external force) is required to change the motion of an object while no unbalanced force is needed to keep an object moving with a constant velocity.
7. In actual practice, the bodies stop due to the force of friction which always acts opposite to the direction of motion.

Question 2.
Mathematically explain the concept of conservation of momentum during collision between two bodies.
Answer:
Let us consider two bodies, viz., A and B, moving with initial velocities u_A and u_B respectively, such that u_A > u_B.
Let mass of body A be m_A and that of B be m_B.
Before collision
For body A, momentum = m_Au_A
For bodv B. momentum = m_Bu_B

After collision, the bodies A and B move with velocities, say, v_A and v_B respectively. Let the collision occur for a time, say, t.
After collision
For body A, momentum = m_Av_A
For body B, momentum = m_Bv_B
The rate of change of momentum for the body A = \(\frac{m_{A}\left(v_{A}-u_{A}\right)}{t}\)
The rate of change of momentum for the body A = \(\frac{m_{B}\left(v_{B}-u_{B}\right)}{t}\)

According to Newton’s third law of motion, The force F_AB exerted by ball A on ball B and the force F_BA exerted by the ball B on ball A must be equal and opposite to each other. Therefore
\(\left\{\frac{m_{A}\left(v_{A}-u_{A}\right)}{t}=-\left(\frac{m_{B}\left(v_{B}-u_{B}\right)}{t}\right)\right\}\)
m_Au_A + m_Bu_B= m_Av_A + m_Bv_B
Here, (m_Au_A + m_Bu_A) is the total momentum of the two bodies, A and B, before collision and (m_Av_A+ m_Bv_B) is the total momentum of the two bodies, A and B, after collision. So, the total momentum of the two bodies remains constant during collision, when no external force acts on them.

Question 3.
Give some examples of Newton’s third law of motion.
Answer:
Newton’s third law of motion states that: For every action, there is always an equal and opposite reaction acting on two different bodies.
Examples:

1. The person who is swimming, pushes the water backward (action), while water pushes the person forward (reaction).
2. When a bullet is fired from a gun, the gun exerts a force on the bullet (action), the bullet exerts an equal and opposite force on the gun (reaction),
3. When a rower jumps out of the boat in the forward direction (action), the boat moves backward (reaction).

Question 4.
Define the term ‘inertia’. Explain its types and state some of its examples in daily life.
Answer:

1. The tendency of a body to resist change in its state of rest or of uniform motion is called inertia.
   • Inertia of rest: The ability of a body to resist any change in its state of rest.
   • Inertia of motion: The ability of a body to resist any change in its state of uniform motion.
   • Inertia of direction: The ability of a body to resist any change in its direction of motion.
2. Examples of application of inertia in our day to day life are as follows:
   • When a carpet is beaten with a stick, the dust falls down.
   • When the car applies a sudden brake, we tend to fall forward.
   • When a bus suddenly takes a turn, we fall in the direction opposite to the turn.

Question 5.
Mathematically explain the concept of Newton’s second law of motion and also explain its importance in sports.
Answer:
Newton’s second law of motion: The rate of change of momentum of an object is directly proportional to the applied unbalanced force and it acts in the direction of force.

Example: Let us consider an object of mass (m), moving along a straight line with an initial velocity (u). Suppose, it attains a final velocity (v) in time (t).
Initial momentum of the object
= p₁ = mu
Final momentum of the object
= p₂ = mv
The change in momentum
= p₂ – p₁ = (mv) – (mu) = m (v – u)
The rate of change in momentum
= \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)
Therefore,
The applied force, F is
F ∝ \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)
F = \(\frac{\mathrm{mk}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\) …………..(i)
(where k is proportionality constant)
From the first equation of motion,
v = u + at
Substituting (ii) in (i),
So, F = kma
The unit of force is so chosen that the value of k becomes one.
Hence, F = kma or
F = ma
The second law is related to sports in the following examples:
(a) In a cricket match, a fielder who is trying to attempt a catch, generally pulls his hands along the moving ball because it consumes time to bring the momentum of the ball to zero. This decreases the rate change of mometum, and hence the force. So, when the time is increased, the overall force reduces, thus saving the hand from getting injured.

(b) In a high jump athletic event, the athletes are made to fall on a sand bed to increase the time of the athlete’s fall to stop after making the jump.
This decreases the rate of change of momentum and hence the force,

(c) A karate expert breaks a slab of ice with a single blow as he moves his hand very fast and then stops it in a very short time. Since the rate of change of momentum increases, the force also increases and the ice slab breaks.

Analysing & Evaluating Questions

Question 6.
A test tube containing little water is corked and suspended from two strings in a slanting position. Afterwards, the test tube is heated Answer the following:

1. What happens to the cork?
2. In which direction does the test tube move?
3. Compare the velocities of the cork and of the recoiling test tube.
Answer:

1. On heating, the water inside the test tube vaporises and the cork is blown out.
2. The test tube moves in the direction opposite to the direction in which the cork has blown out.
3. Cork blows out with greater velocity than the recoiling test tube.

Activity 1

• Make a pile of similar carom coins on a table, as shown in the figure.
• Attempt a sharp and strong horizontal hit at the bottom of the pile using another carom coin or striker.

Observations

• If the hit is strong enough the bottom coin moves out quickly.
• Once the lowermost coin is removed, the inertia of the other coins makes them fall vertically on the table.

Activity 2

• Set a five – rupee coin on a stiff card covering an empty glass tumbler standing on a table as shown in the figure.
• Give the card a sharp horizontal flick with a finger.

Observations

• If we do it fast, the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia.
• The inertia of the coin tries to maintain its state of rest even when the card is flown away.
• The force applied on the card due to flicking changes the inertia of the card but the coin resists this change and stays at rest, i.e, due to the inertia and gravity, the coin falls down into the tumbler.

Activity 3

• Request two children to stand on two separate carts as shown in the figure.
  
• Give them a bag full of sand or some other heavy object.
• Ask them to play a game of catch with the bag and observe.
• You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.

Observations

• In this case, each of them receives an instantaneous reaction as a result of throwing the sand bag.
• This activity explains Newton’s third law of motion, i.e., a force is exerted in forward direction in throwing the bag full of sand and the person who is throwing it, gets pushed backward.

Activity 4

• Take a big rubber balloon and inflate it fully.
• Tie its neck using a thread Also, using adhesive tape, fix a straw on the surface of this balloon.
• Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.
• Ask your friend to hold the other end of the thread or fix it on a wall at some distance.
• The arrangement is shown in the figure.
• Now, remove the thread tied on the neck of the balloon and let the air escape from the mouth of the balloon.
• Observe the direction in which the straw moves.

Observations.

• When the air escapes out from the balloon, the straw moves in the direction opposite to the direction in which the air moved out of the balloon. This activity explains the law of conservation of momentum.
• This activity is also based on Newton’s third law of motion, i.e., for every action, there is always an equal and opposite reaction between two bodies.

Value Based Questions

Question 1.
Class IX students were playing cricket with cork ball in the school campus. Abhishek, senior student, told them about the accidents that can occur due to cork ball in the campus and also advised them to bring a soft cosco ball to play the game.
1. Why is it safe to play with a soft ball and not with a hard cork ball?
2. A player pulls his hands backwards while catching the ball shot at high speed Why?
3. What value of Abhishek is seen in this act?
Answer:

1. The soft ball has less inertia as compared to the heavy ball and it would not hurt the players.
2. By pulling the hand backwards, it reduces the force exerted by the ball
   on the hand.
3. Abhishek showed the value of being responsible and helpful by nature.

Question 2.
Rahul saw his karate expert friend breaking a brick. He tried to break the brick but his friend stopped him from doing so and told him that it would hurt him as one needs a lot of practice in doing so.
1. How can a karate expert break the brick without any injury to his hand?
2. What is Newton’s third law of motion?
3. What value of Rahul’s friend is seen in the above case?
Answer:

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

JAC Class 9 Science Important Questions