Bhagya

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 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 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 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.

Students should go through these JAC Class 9 Maths Notes Chapter 4 Linear Equations in Two Variables will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 4 Linear Equations in Two Variables

Linear Equation In One Variable:
An equation of the form ax + b = 0, where a and bare real numbers with a = 0 and ‘x’ is a variable, is called a linear equation in one variable.
Here ‘a’ is called coefficient of x and ‘b’ is called a constant term. i.e. 3x + 5 = 0, 7x – 2 = 0 etc.

Linear Equation In Two Variables:
An equation of the form ax + by + c = 0 where a, b, c are real numbers and a and b both are not to be zero, and x, y are variables, is called a linear equation in two variables, here ‘a’ is called coefficient of x, ‘b’ is called coefficient of y and ‘c’ is called constant term.
Any pair of values of x and y which satisfies the equation ax + by + c = 0, is called a solution of it.

Graph Of A Linear Equation:
→ In order to draw the graph of a linear equation in one variable we may follow the following algorithm.
Step I: Obtain the linear equation.
Step II: If the equation is of the form ax = b, a ≠ 0, then plot the point \(\left(\frac{\mathrm{b}}{\mathrm{a}}, 0\right)\) and one more point \(\left(\frac{b}{a}, \alpha\right)\) where α is any real number. If the equation is of the form ay = b, a ≠ 0, then plot the point \(\left(0, \frac{b}{a}\right)\) and \(\left(\beta, \frac{b}{a}\right)\) where β is any real number.
Step III: Join the points plotted in step II to obtain the required line. Note: If equation is in the form ax = b then we get a line parallel to y-axis and if equation is in form ay = b then we get a line parallel to x-axis.

→ In order to draw the graph of a linear equation ax + by + c = 0, we may follow the following algorithm.
Step I: Obtain the linear equation ax + by + c = 0.
Step II: Express y in terms of x i.e. y = \(-\frac{a}{b} x-\frac{c}{b}\) or x in terms of y i.e. x = \(-\frac{b}{a} y-\frac{c}{a}\).
Step III: Put any two or three values for x or y and calculate the corresponding values of y or x respectively from the expression obtained in Step II. Let us suppose that we get the points as (α₁, β₁),(α₂, β₂), (α₃, β₃).
Step IV: Plot the points (α₁, β₁), (α₂, β₂), (α₃, β₃) on graph paper.
Step V: Join the points marked in step IV. The line obtained is the graph of the equation ax + by + c = 0.

Different Forms Of A Line:
→ Slope of a Line: If a line makes an angle θ with positive direction of x-axis, then tangent of this angle is called the slope of a line, it is denoted by m i.e. m = tan θ.
→ Slope-intercept form is y = mx + c where m is the slope of line and c is intercept made by line with y-axis.
→ The equation of a line passing through origin is y = mx. When c = 0, the line always passes through origin.

→ Intercept form of line is \(\frac{x}{a}+\frac{y}{b}=1\), where a and b are the intercepts on positive direction of x-axis and y-axis respectively made by the line.

Solution Of Linear Equation In One Variable:
Let ax + b = 0 be the equation then ax = – b
⇒ x= \(-\frac{b}{a}\) is a solution.

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 2 Polynomials will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 2 Polynomials

Polynomials:
An algebraic expression f(x) of the form f(x) = a₀ + a₁x + a₂x² + …… + a_nxⁿ, where a₀, a₁, a₂ ……, a_n are real numbers and all the index of x’ are nonnegative integers is called a polynomial in x.
→ Degree of a Polynomial: Highest Index of x in algebraic expression is called the degree of the polynomial, here a₀, a₁x, a₂x² ….. a_nxⁿ, are called the terms of the polynomial and a₀, a₁, a₂, …… a_n are called various coefficients of the polynomial f(x).
Note: A polynomial in x is said to be in standard form when the terms are written either in increasing order or decreasing order of the indices of x in various terms.

→ Different Types of Polynomials: Generally, we divide the polynomials in the following categories.
→ Based on degrees:
There are four types of polynomials based on degrees. These are listed below:

• Linear Polynomials: A polynomial of degree one is called a linear polynomial. The general form of linear polynomial is ax + b, where a and b are any real constant and a ≠ 0.
• Quadratic Polynomials: A polynomial of degree two is called a quadratic polynomial. The general form of a quadratic polynomial is ax² + bx + c, where a ≠ 0, a, b, c ∈ R.
• Cubic Polynomials: A polynomial of degree three is called a cubic polynomial. The general form of a cubic polynomial is ax³ + bx² + cx + d, where a ≠ 0 and a, b, c, d ∈ R.
• Biquadratic (or quadric) Polynomials: A polynomial of degree four is called a biquadratic (quadric) polynomial. The general form of a biquadratic polynomial is ax⁴ + bx³ + cx² + dx + e, where a ≠ 0 and a, b, c, d, e are real numbers.

Note: A polynomial of degree five or more than five does not have any particular name. Such a polynomial usually called a polynomial of degree five or six or ….etc.

→ Based on number of terms:
There are three types of polynomials based on number of terms. These are as follow:

• Monomial: A polynomial is said to be monomial if it has only one term. e.g. x, 9x², 5x³ all are monomials.
• Binomial: A polynomial is said to be binomial if it contains only two terms e.g. 2x² + 3x, \(\sqrt{3}\)x + 5x³, -8x³ + 3, all are binomials.
• Trinomial: A polynomial is said to be a trinomial if it contains only three terms.e.g. 3x³ – 8x + \(\frac{1}{2}\), \(\sqrt{7}\) x¹⁰ + 8x⁴ – 3x², 5 – 7x + 8x⁹, are all trinomials.

Note: A polynomial having four or more than four terms does not have particular name. These are simply called polynomials.

→ Zero degree polynomial: Any non-zero number (constant) is regarded as polynomial of degree zero or zero degree polynomial. i.e. f(x) = a. where a ≠ 0 is a zero degree polynomial, since we can write f(x) = a, as f(x) = ax⁰.

→ Zero polynomial: A polynomial whose all coefficients are zero is called as zero polynomial i.e. f(x) = 0, we cannot determine the degree of zero polynomial.

Algebraic Identities:
An identity is an equality which is true for all values of the variables.
Some important identities are:
(i) (a + b)² = a² + 2ab + b²
(ii) (a – b)² = a² – 2ab + b²
(iii) a² – b² = (a + b)(a – b)
(iv) a³ + b³ = (a + b)(a² – ab + b²)
(v) a³ – b³ = (a – b)(a² + ab + b²)
(vi) (a + b)³ = a³ + b³ + 3ab (a + b)
(vii) (a – b)³ = a³ – b³ – 3ab (a – b)
(viii) a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² – ab + b²)
(ix) a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ac)

Special case: if a + b + c = 0 then a³ + b³ + c³ = 3abc.
Other Important Identities
(i) a² + b² = (a + b)² – 2ab,
if a + b and ab are given
(ii) a² + b² = (a – b)² + 2ab
if a – b and ab are given
(iii) a + b = \(\sqrt{(a-b)^2+4 a b}\)
if a – b and ab are given
(iv) a – b = \(\sqrt{(a+b)^2-4 a b}\)
if a + b and ab are given

Factors Of A Polynomial:
→ If a polynomial f(x) can be written as a product of two or more other polynomials f₁(x), f₂(x), f₃(x)…. then each of the polynomials f₁(x), f₂(x), f₃(x)….. is called a factor of polynomial f(x). The method of finding the factors of a polynomial is called factorisation.

Zeroes Of A Polynomial:
→ A real number α is a zero of polynomial f(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + ….. +a₁x + a₀, if f(α) = 0. i.e. a_nαⁿ + a_n-1α^n-1 + a_n-2α^n-2+ ….. + a₁α + a₀ = 0.
For example x = 3 is a zero of the polynomial f(x) = x³ – 6x² + 11x – 6, because f(3) = (3)³ – 6(3)² + 11(3) – 6 = 27 – 54 + 33 – 6 = 0.
but x = -2 is not a zero of the above mentioned polynomial,
∵ f(-2) = (-2)³ – 6(-2)² + 11(-2) – 6
f(-2) = -8 – 24 – 22 – 6
f(-2) = -60 ≠ 0.

→ Value of a Polynomial: The value of a polynomial f(x) at x = a is obtained by substituting x a in the given polynomial and is denoted by f(a). Eg if f(x) = 2x³ – 13x² + 17x + 12 then its value at x = 1 is
f(1) = 2(1)³ – 13(1)² + 17(1) + 12
= 2 – 13 + 17 + 12 = 18.

Remainder Theorem:
Let ‘p(x)’ be any polynomial of degree greater than or equal to one and ‘a’ be any real number and if p(x) is divided by (x – a). then the remainder is equal to p(a). Let q(x) be the quotient and r(x) be the remainder when p(x) is divided by (x – a), then
Dividend = Divisor × Quotient + Remainder
∴ p(x) = (x – a) × q(x) + [r(x) or r], where r(x) = 0 or degree of r(x) < degree of (x – a). But (x – 2) is a polynomial of degree 1 and a polynomial of degree less than 1 is a constant. Therefore, either r(x) = 0 or r(x) = Constant. Let r(x) = r, then p(x) = (x – a)q(x) + r.
Putting x = a in above equation, p(a)
p(a) = (a – a)q(a) + r = 0 × q(a) + r
p(a) = 0 + r
⇒ p(a) = r
This shows that the remainder is p(a) when p(x) is divided by (x – a).
Remark: If a polynomial p(x) is divided by (x + a),(ax – b), (ax + b), (b – ax) then the remainder is the value of p(x) at x.
= \(-a, \frac{b}{a},-\frac{b}{a}, \frac{b}{a} \text { i.e. } p(-a)\)
\(p\left(\frac{b}{a}\right), p\left(-\frac{b}{a}\right), p\left(\frac{b}{a}\right)\) respectively.

Factor Theorem:
Let ‘p(x)’ be a polynomial of degree greater than or equal to 1 and ‘a’ be a real number such that p(a) = 0, then (x – a) is a factor of p(x). Conversely, if(x – a) is a factor of p(x). then p(a) = 0.

Factorisation Of A Quadratic Polynomial:
→ For factorisation of a quadratic expression ax² + bx + c where a ≠ 0, there are two methods.
→ By Method of Completion of Square:
In the form ax² + bx + c where a ≠ 0, firstly we take ‘a’ common in the whole expression then factorise by converting the expression \(a\left\{x^2+\frac{b}{a} x+\frac{c}{a}\right\}\) as the difference of two squares, which is

→ By Splitting the Middle Term:
→ x² + lx + m = x² + (a + b)x + ab
Where l = a + b and m = ab, such that a and b are real numbers
= x² + ax + bx + ab
= x (x + a) + b (x + a)
= (x + a) (x + b)
Method: We express l as the sum of two such numbers whose product is m.

→ ax² + bx + c = prx² + (ps + qr)x + qs
where b = ps + qr, a = pr, c = qs
so that (ps) (gr) (pr) (qs) = ac
∴ prx² + (ps + qr)x + qs
= prx² + psx + qrx + qs
= px (rx + s) + q(rx + s)
= (px + q) (rx + x)
Method: We express b as the sum of two such numbers whose product is ac.

→ Integral Root Theorem:
If f(x) is a polynomial with integral coefficient and the leading coefficient is 1, then any integral root of f(x) is a factor of the constant term. Thus if f(x) = x³ – 6x² + 11x – 6 has an Integral root, then it is one of the factors of 6 which are ±1, ±2, ±3, ±6.
Now in fact,
f(1) = (1)³ – 6(1)² + 11(1) – 6 = 1 – 6 + 11 – 6 = 0
f(2) = (2)³ – 6(2)² + 11(2) – 6
= 8 – 24 + 22 – 6 = 0
f(3) = (3)³ – 6(3)² + 11(3) – 6
27 – 54 + 33 – 6 = 0
Therefore Integral roots of f(x) are 1, 2, 3.

→ Rational Root Theorem:
Let \(\frac{b}{c}\) be a rational fraction in lowest terms. If \(\frac{b}{c}\) is a rational root of the polynomial f(x) = a_nxⁿ + a_n-1x^n-1 +…+ a₁x + a₀, an ≠ 0 with integral coefficients, then b is a factor of constant term a₀, and C is a factor of the leading coefficient a_n.
For example: If \(\frac{b}{c}\) is a rational root of the polynomial f(x) = 6x³ + 5x² – 3x – 2, then the values of b are limited to the factors of -2, which are ±1, ±2 and the values of care limited to the factors of 6, which are ±1, ±2, ±3, ±6. Hence, the possible rational roots of f(x) are ±1, ±2, \(\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{2}{3}\). In fact -1 is an integral root and \(\frac{2}{3}\), –\(\frac{1}{2}\) are the rational roots of f(x) = 6x³ + 5x² – 3x – 2.
Note: (i) n^th degree polynomial can have at most n real roots.
→ Finding a zero of polynomial f(x) means solving the polynomial equation f(x) = 0. It follows from the above discussion that if f(x) = ax + b, a ≠ 0 is a linear polynomial, then it has only one zero given by
f(x) = 0 i.e. f(x) = ax + b = 0
⇒ ax = -b
⇒ x = –\(\frac{b}{a}\)
Thus, x = –\(\frac{b}{a}\) is the only zero of f(x) = ax + b.
→ If a polynomial of degree n has more than n zeros then all the coefficients of powers of x including constant term of polynomial are zero.

JAC Jharkhand Board Class 9th Sanskrit Solutions शेमुषी भाग 1

• Chapter 1 भारतीवसन्तगीतिः
• Chapter 2 स्वर्णकाकः
• Chapter 3 गोदोहनम्
• Chapter 4 कल्पतरुः
• Chapter 5 सूक्तिमौक्तिकम्
• Chapter 6 भ्रान्तो बालः
• Chapter 7 प्रत्यभिज्ञानम्
• Chapter 8 लौहतुला
• Chapter 9 सिकतासेतुः
• Chapter 10 जटायोः शौर्यम्
• Chapter 11 पर्यावरणम्
• Chapter 12 वाङ्मनः प्राणस्वरूपम्

JAC Class 9 Sanskrit अपठितावबोधनम्

• अपठित अनुच्छेदाः

JAC Class 9 Sanskrit व्याकरणम्

• उच्चारणस्थानानि
• सन्धि-प्रकरणम्
• समास प्रकरणम्
• कारक प्रकरणम्
• उपसर्ग प्रकरणम्
• प्रत्यय प्रकरणम्
• संज्ञा-शब्दरूप-प्रकरणम्
• सर्वनाम शब्दरूप प्रकरणम्
• संख्याज्ञानम्
• धातुरूप-प्रकरणम्

JAC Class 9 Sanskrit रचनात्मक कार्यम्

• पत्र-लेखनम्
• संकेताधारितलघुकथा, चित्रवर्णनम्
• संकेताधारित अनुच्छेदलेखनम्
• अनुवाद-प्रकरणम्

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

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