Category: Class 9

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.2

Question 1.
Construct a triangle ABC in which BC = 7 cm, B = ∠75° and AB + AC = 13 cm.
Solution:
BC = 7 cm
∠B = 15°
AB + BC = 13 cm.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° AB – AC = 3.5 cm.
Solution:
BC = 8 cm
∠B = 45°
AB – AC = 3.5 cm.

Question 3.
Construct a triangle PQR in which QR = 6 cm. ∠Q = 60° and PR – PQ = 2 cm.
Solution:
QR = 6 cm
∠Q =60°
PR – PQ = 2 cm

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:
XY + YZ + ZX = 11 cm
∠Y = 30°
∠Z =90°

1. Draw a line segment BC =11 cm.
2. At B construct an angle of 30° and at C, draw angle of 90°.
3. Bisect these angles. Let the bisectors of these angles intersect atX.
4. Draw perpendicular bisectors AC of BX to intersect BC at Y and DZ of XC to intersect BC at Z.
5. Join XY and XZ. XYZ is the required D.

Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Solution:
Steps of construction:

1. Draw \(\overline { BC } \) = 12 cm.
2. Construct ∠CBY = 90°.
3. From \(\overline { BY } \), cut off BX = 18 cm.
4. Join CX.
5. Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.
6. JoinAC.

Thus, ABC is the required triangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.4

Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
Given
Let O and O₁ be the centre of bigger and smaller circle respectively
OA = OB = 5 cm
O₁A – O₁B = 3 cm
OO₁ = 4 cm

To find: AB.
Construction:
Join OA, OB, O₁A and O₁B join AB also.
In ∆OAO₁ and ∆OBO₁
OA = OB (Radii of a circle)
O₁A = O₁B (Radii of a circle)
OO₁ = OO₁ (Common)
so ∆OAO₁ = ∆OBO₁ (By SSS)
and so ∠1 = ∠2 (By CPCT)
In ∆OCA and ∆OCB,
OA = OB (Radii of a circle)
∠1 = ∠2 (Proved)
OC = OC (Common)
∆OCA = ∆OCB (By SAS)
so AC = BC (By CPCT)
and ∠ACO = ∠BCO (By CPCT)
∠ACO + ∠BCO = 180° (LPA’s)
⇒ ∠ACO + ∠BCO = 180°
2∠ACO = 180°
∠ACO = 90°
ar (OAO₁) = \(\frac{1}{2}\) x OO₁ x AC
= \(\frac{1}{2}\) x 4 x AC = 2ACcm² …..(i)
In ∆QAO₁, a = 5 cm, bc = 4 cm, c = 3 cm
12
s = \(\frac{5+4+3}{2}\) = \(\frac{12}{2}\) = 6 cm
s – a = 6 – 5 = 1 cm
s – b = 6 – 4 = 2 cm
s – c = b – 3 = 3 cm
ar(OAO₁) = \(\sqrt{s(s-a)(s -b)(s- c)}\).
= \(\sqrt{6x1x2x3}\)
= 6 cm² …..(ii)
From (i) and (ii), we get
2AC =6
AC = 3 cm
Now, AB = 2AC (∴ AC = BC)
= 2 x 3 = 6 cm.

Method II. By Construction:
Geometrically, AB is the diameter of the circle of radius 3 cm as it passes through centre O₁
AB = 2 x 3 = 6 cm.

Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segment of the other chord.
Solution:
Given
C (O, r) is a circle in which AB and CD are two equal chords which intersect at P.
To prove:
CP = BP and AP = DP.
Construction:
Draw OE and OF perpendiculars on AB and CD respectively. Join OP.
Proof:
In ∆OPF and ∆OPE,
OP = OP (Common)
OE = OF (∴ AB = CD)
∠F = ∠E (Each 90°)
∆OPF = ∆OPE (By RHS)
and so PE = PF …..(1) (By CPCT)
AB = CD (Given)

\(\frac{1}{2}\) AB = \(\frac{1}{2}\) CD
BE = CF
and AE = DF …..(2)
Adding (1) and (2), we get,
PE + AE = PF + DF
∴ AP = DP
Subtracting (1) and (2) we get,
BE – PE = CF – PF
∴ BP = CP

Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Given
AB and CD are two equal chords of a circle which intersect at E.
To prove:
∠1 = ∠2
Construction:
Draw OL ⊥ AB and OM ⊥ CD. Join OE.

Proof:
In ∆OLE and ∆OME,
OE = OE
OL = OM (∴ AB = CD)
∠L = ∠M (Each 90°)
∆OLE = ∆OME (By RHS)
and so∠1 = ∠2 (By CPCT)

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD. (see Fig. adjacent)

Solution:
Given:
C (O, r) and C (O, r) are two concentric circles.
To prove: AB = CD

Construction: Draw OP ⊥ AD.
Proof:
In circle I, AD is the chord and OP ⊥ AD.
AP = DP …(1)
In circle II, BC is the Chord and OP L BC.
∴ BP = CP …(2)
Subtracting (1) and (2), we get
AP – BP = DP – CP
AB = CD

Question 5.
Three girls Reshma. Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Given:
OR = OM = 5 m and SR = SM = 6 m.
To find: MR.

Constrution:
Join OR, OM and OS. Draw ON ⊥ SR. In AORS and AOMS,
OS = OS (Common)
RS = MS (Given)
OR = OM (Given)
∆ORS = ∆OMS (By SSS)
and ∠1 = ∠2 (By CPCT)
SP = SP (Common)
SR = SM (Given)
∠1 = ∠2 (Proved)
∆SPR = ∆SPM (By SAS)
and so PR = PM (By CPCT)
and ∠3 = ∠4 (By CPCT)
∠3 + ∠4 = 180° (LPA’s)
2∠3 = 180°
∠3 = \(\frac{180^{\circ}}{2}\) = 90°
ar (∆OSR) = \(\frac{1}{2}\) x OS x PR …(i)
= \(\frac{1}{2}\) x 5 x PR
ON ⊥ SR
RN = \(\frac{1}{2}\) SR
(Perpendicular drawn from the centre of a circle to a chord bisects the chord)
= \(\frac{6}{2}\) = 3m
In ∆ONR ON² = \(\sqrt{O R^{2}-N R^{2}}\) (Using Pythagoras Theorem)
= \(\sqrt{5^{2}-3^{2}}\) = \(\sqrt{4^{2}}\) = 4m
ar (∆OSR) = \(\frac{1}{2}\) x SR x ON
= \(\frac{1}{2}\) x 6 x \(\frac{1}{2}\) x 4 = 12m² …..(ii)
From (i) and (ii), we get
PR = \(\frac{2×12}{2}\) = 4.8m
MR = 2 PR
= 2 x 4.8
= 9.6 m

Question 6.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Solution:
Given: OS = OA = 20 m and AS = SD = AD
To find: AS, SD and AD.
Construction:
Draw AE ⊥ SD. Join OS.
Let AS = SD = AD = 2x (say)

In equilateral ∠ASD, AE ⊥ SD
⇒ E is the mid-point of SD
SE = \(\frac{2x}{2}\) = x
In ∆AES, AS² = AE² + SE²
(2x)² = AE² + x²
4x² – x² = AE²
AE = \(\sqrt{3x^{2}}\) = \(\sqrt{3}\)
OE = AE – AO
= (\(\sqrt{3}\) – 20)m
In ∆OES, OS² = OE² + SE²
(20)² = [(\(\sqrt{3}\)x) 20]² + x²
400 = (\(\sqrt{3}\)x)² – 2 x \(\sqrt{3}\)x × 20 + (20)² + x²
= 3x² + 400 – 40\(\sqrt{3}\)x + x²
400 – 400 = 4x² – 40\(\sqrt{3}\)x
0 = 4x² – 40\(\sqrt{3}\)x
40\(\sqrt{3}\)x = 4x²
40\(\frac { 40\sqrt { 3 } }{ 4 } \) = x
x = 10\(\sqrt{3}\)m .
2x = 2 x 10\(\sqrt{3}\) = 20\(\sqrt{3}\)m
AS = SD = AD = 20\(\sqrt{3}\)m.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

Question 1.
In Fig. given below E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE).

Solution:
Given.
E is any point on median AD of ∆ABC.
To prove
ar (ABE) = ar (ACE)

Proof:
In ∆ABC, AD is the median
ar (ABD) = ar(ACD) ….(1)
In ∆EBC, ED is the median
∴ ar (BDE) = ar (CDE) ….(2)
Subtracting (2) from (1), we get
ar (ABD) – ar (BDE) = ar (ACD) – ar (CDE)
ar (ABE) = ar (ACE)

Question 2.
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = \(\frac{1}{4}\) ar (ABC).

Solution:
Given
E is the mid point of – median AD of ∆ABC.
To prove:
ar (BED) = \(\frac{1}{4}\) ar (ABC)
Proof:
In ∆ABC, AD is the media
ar (ABD) = \(\frac{1}{2}\) ar (ABC) …..(1)
In ∆ABD, BE is the median

ar (BED) = \(\frac{1}{2}\) ar (ABD)
= \(\frac{1}{2}\) [\(\frac{1}{2}\) ar (ABC)]
= \(\frac{1}{4}\) ar (ABC)

Question 3.
Show that the diagonals of parallelogram divide it into four triangles of equal area.
Solution:
Given.
ABCD is a parallelogram.
To prove:
ar (AOB) = ar (BOC) = ar (COD) = ar (AOD)

Proof:
In ∆ABC, OA is the median
∴ ar (AOB) = ar (AOD) …..(1)
In AABC, BO is the median
∴ ar (AOB) = ar (BOC) …..(2)
In ABCD, CO is the median
ar (BOC) = ar (COD) …(3)
From (1), (2) and (3), we get
ar (AOB) ar (BOC) = ar (COD) = ar (AOD)

Question 4.
In Fig. given below, ABC and ABC are two triangles on the same base AB. If line – segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

Solution:
Given.
∆ABC and ∆ABD have a common base AB.
OC = OD
To prove
ar (ABC) = ar (ABD)

Proof:
As OC = OD,
O is the mid-point of CD
In ∆ACD, O is the median
ar (AOC) = ar(AOD)
In ABCD, BO is the median
ar (BOC) = ar (BOD)
Adding (1) and (2), we get
ar (AOC) + (BOC) = ar (AOD) + ar (BOD)
ar (ABC) = ar (ABC)

Question 5.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a AABC. Show that

1. BDEF is a parallelogram.
2. ar (DEF) = \(\frac{1}{2}\) ar (ABC)
3. ar (BDEF) = \(\frac{1}{4}\) ar (ABC).

Solution:
D, E, F are the mid points of sides BC, CA and AB of ∆ABC.
To prove:

1. BDEF is a parallelogram.
2. ar (DEF) = \(\frac{1}{4}\) ar (ABC)
3. ar (BDEF) = \(\frac{1}{2}\) ar (ABC)

Proof:
In ∆ABC, F is the mid-point of AB, E is the mid point of AC.
∴ FE ∥ BC and FE = – BC (By MPT)
⇒ FE ∥ BD and FE =BD

1. BDEF is a ∥^gm
Similarly CDFE and AEDF are ∥^gm
In parallelogram BDEF, DF is the diagonal
∴ ax (BDF) = ar (DEF)
[In a ∥^gm diagonal divides it into 2∆s of equal areas]
Similarly, In ∥^gm CDFE, DE is a diagoilal …(1)
ar (CDE) = (DEF) …(2)
In ∥^gm AEDF, FE is a diagonal
ar (AEF) = ar (DEF) …(3)
From (1), (2) and (3), we get
ar (BDF) = ar (CDF) = ar (AEF) = (DEF) …(4)

2. ar (ABC) = ar (AEF) + ar (ADF) + ar (CDE) + ar (DEF)
ar (ABC) = 4 ar (DEF) [Using (4)]
ar (DEF) = \(\frac{1}{4}\) ar (ABC)

3. ar (BDEF) = ar (BDF) + ar (DEF)
ar (BDEF) = 2ar (DEF) [∴ ar (BDF) = ar (DEF)]
= 2 x \(\frac{1}{4}\) ar (ABC) = \(\frac{1}{2}\) ar (ABC)

Question 6.
In the Fig. diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB – CD, then show that:

1. ar (DOC) = ar (AOB)
2. ar DCB = ar (ACB)
3. DA ∥ CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC.]

Solution:
Given
ABCD is a quadrilateral in which OB = OD and AB = CD.
To prove:

1. ar (DOC) = ar (AOB)
2. ar (DCB) = ar (ACB)
3. DA ∥ CB or ABCD is a ∥^gm

Construction:
Draw DE and BF perpendicular from point D and B on AC.
Proof:
1. In ∆OED and ∆OFB
∠1 = ∠2 (V.O.A.’s)
OD = OB (given)
∠E = ∠F (each 90°)
∆ OED = ∆ OFB (by AAS)
and so DE = BF (by CPCT)
In ∆DEC and ∆BFA, DE = BF (proved)
∠E = ∠F (each 90°)
DC = BA (given)
∆DEC = ∆BFA (by RHS)
and so ∠3 = ∠4 (by CPCT)
ar (OED) = ar (OFB) [∴ AOED = AOFB] …(1)
ar (DEC) = ar (BFB) [∴ ADEC = ABEA] …..(2)
Adding (1) and (2), we get
ar (OED) + ar (DEC) = ar (OFB) = ar (BFA)
ar (OCD) = ar (OAB) …..(3)

2. Adding ar (OBC) on both sides of equation (3)
ar (OCD) + (OBC) = ar (OAB) + ar (OBC)
ar (DCB) = ar (ACB)

3. ∆DCB and ∆ACB have the same base BC and have equal area
∴ they will lie between the same parallels BC and AD
and so BC ∥ AD
∠3 and ∠4 are A.I.A’s and are equal
∴ AB ∥ DC
In quadrilateral ABCD, AB ∥ DC and AB = DC (given)
ABCD is a parallelogram.

Question 7.
D and E are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBQ. Prove that DE ∥ BC.
Solution:
Given
ar (DBC) = ar (EBC)
To prove:
DE ∥ BC

Proof:
∆ DBC and ∆EBC have the same base BC and
ar (DBC) = ar (EBC)
∴ They will lie between the same parallel lines DE and BC.
and so DE ∥ BC

Question 8.
XY is a line parallel to side BC of a triangle ABC. If BE ∥ AC and CF ∥ AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).
Solution:
Given
XY ∥ BC, BE ∥ AC and CF ∥ AB.
To prove
ar (ABE) = ar (ACF)
Proof:
In quadrilateral ACBE, AE ∥ CB (∴ BC ∥ XY)
and AC ∥ EB
ACBE is a ∥^gm

Similarly ABCF is a parallelogram.
∥^gm ACBE and ABCF have the same base BC and are between the same parallels BC and AY.
∴ ar (ACBE) = ar (ABCF)
ar (ABE) + ar (ABC) = ar (ABC) + ar (ACF)
∴ ar (ABE) = ar (ACF)

Question 9.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. below). Show that ar (ABCD) = ar (PBQR).
[Hint: Joint AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Solution:
Given
ABCD and BPRQ are parallelograms. CP ∥ AQ
To prove:
ar (ABCD) = ar (PBQR)
Construcion:
Join AC and PQ
Proof:
∆ACQ and ∆APQ lie on the same base AQ and are between the same parallels AQ and CP.
∴ ar (ACQ) = ar (APQ)
⇒ ar (ABQ) + ar (ABC) = ar (ABQ) + ar (BQP)
∴ ar (ABC) = ar (BQP) …..(i)

Multiplying (1) by 2 on both sides
2 ar (ABC) = 2 ar (BQP)
∴ ar (ABCD) = ar (PBQR)
[∴ ABCD and PBQR are parallelogram]

Question 10.
Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at C. Prove that ar (AOD) = ar (BOC).
Solution:
Given
ABCD is a trapezium in which AB ∥ DC.
To prove:
ar (AOD) = ar (BOC).

Proof:
∆ADC and ∆BCD lie on the same base DC and between the same parallels AB and CD.
∴ ar (ADC) = ar (BCD)
Subtracting ar (DOC) from both sides
ar (ADC) – ar (DOC) = ar (BCD) – ar (DOC)
ar (AOD) = ar (BOC)

Question 11.
In Fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:

1. ar (ACB) = ar (ACF)
2. ar (AEDF) = ar (ABCDE).

Solution:
Given
ABCDE is a pentagon.
To prove:
1. ar (ACB) = ar (ACF)

2. ar (AEDF) = ar (ABCDE)
Proof:
∆ACB and ∆ACF lie on the same base AC and between the same parallels AC and BE.
ar (ACB) = ar (ACF)

3. Adding ar (AEDC) on both sides
ar (ACB) + ar (AEDC) = ar (ACF) + ar (AEDC)
ar (ABCDE) = ar (AEDF)

Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Solution:
Given.
ABCD is a quadrilateral.
Construction:
Join AC. Draw DE ∥ CA which intersect BA produced at E.

Proof:
∆ADC and ∆ACE both lie on the same base AC and between the same parallels AC and DE.
ar (ADC) = ar (ACE)
Adding ar (ABC) on both sides
ar (ADC) + ar (ABC) = ar (ACE) + ar (ABC)
ar (ABCD) = ar (EBC)

Question 13.
ABCD is a trapezium with AB ∥ DC. Aline parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
Solution:
Given
ABCD is trapezium with AB ∥ DC and AC ∥ XY.
To prove:
ar (ADX) = ar (ACY).
Construction:
Join CX.
Proof:
∆ADX and ∆ACX both lie on the same base AX and between the same parallel AN and DC.

∴ ar (ADX) = ar (ACX) …(1)
∆ACX and ∆ACT both lie on the same base AC and between the same parallels AC and AT.
ar (ADY) = ar (ACX) …(2)
From (1) and (2), we get
ar (ADX) = ar(ACT)

Question 14.
In Fig. below AP ∥ BQ ∥ CR. Prove that ar (AQC) = ar (PBR).

Solution:
Given:
AP∥BQ∥CR.
To prove:
ar (AQC) = ar (PBR)
Proof:
∆ABQ and ∆PBQ both lie on the same base BQ and between the same parallels AP and BQ.
∴ ar (ABQ) = ar (PBQ) …(1)
∆BCQ and ∆BRQ both lie on the same base BQ and between the same parallels BQ and CR.

∴ ar (BCQ) = ar (BRQ) …(2)
Adding (1) and (2), we get
ar (ABQ) tar (BCQ) = ar (PBQ) + ar (BRQ)
∴ ar (AQC) = ar (PBR)

Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
Solution:
Given:
ar (AOD) = ar (BOC)
To prove:
ABCD is a trapezium.

Proof:
ar ∠AOD = ar (BOC) (given)
Adding ar (AOB) on both sides
ar (AOD) + ar (AOB) = ar (BOC) + ar (AOB)
ar (ABD) = ar (ABC)
∆ABD and ∆ABC both lie on same base AB and have equal area.
∴ they will lie between the same parallels.
⇒ AB ∥ DC
∴ ABCD is a trapezium.

Question 16.
In Fig. below, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
Given
ar (DRC) = ar (DPC)
ar (BDP) = ar (ARC)
To prove:
ABCD and DCPR are trapezium.

Proof:
∆DRC and ∆DPC both lie on the same base DC and have equal area .•. They will lie between the same parallels.
⇒ DC ∥ BP
and so DCPR is a trapezium.
ar (BDP) = ar (ARC) [given] …(1)
ar (DPC) = ar (DRC) [given] …(2)
Subtracting (1) and (2), we get
ar (BDP) – ar (DPC) = ar (ARC) – (DRC)
ar (BDC) = ar (ADC)
∆ADC and ∆BDC both lie on same base DC and have equal area.
∴ they will lie between the same parallels
⇒ AB ∥ DC
and so ABCD is a trapezium.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 12 Sound

Sound Intext Questions

Sound Intext Questions Page No. 162

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
When an object vibrates, it allows the particles of the medium around it to vibrate. It exerts force on the adjacent particles and continue oscillating in all directions and one of it, hit our ear’s medium which creates sound. This process continues in the medium till the sound reaches your ear.

Sound Intext Questions Page No. 163

Question 1.
Explain how sound is produced by your school bell.
Answer:
When the bell rings, it continues to move forward and backward which creates vibration and simultaneously a series of compressions and rarefactions which produce a very loud sound.

Question 2.
Why are sound waves called mechanical waves?
Answer:
Sound waves needs medium to propagate therefore, they are called mechanical waves. Sound cannot travel in the absence of a medium. Sound waves are propagated through a medium because of the interaction of the particles present in that medium.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, because sound waves needs a medium through which they can propagate. And since there is no medium on the moon due to absence of atmosphere, one cannot hear any sound produced by his / her friend on the moon.

Sound Intext Questions Page No. 166

Question 1.
Which wave property determines.

1. loudness
2. pitch?

Answer:

1. Amplitude determines loudness of a sound wave.
2. Frequency determines pitch of a sound wave.

Question 2.
Guess which sound has a higher pitch: guitar or car horn?
Answer:
Guitar has a higher pitch than car horn, because sound produced by the strings of guitar has higher frequency and since pitch is proportional to frequency, pitch of guitar will be higher.

Question 3.
What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:

1. Wavelength: Wave length is the length between two consecutive peaks it is represented by Greek letter λ (lambda). Louder sound has shorter wavelength and softer sound has longer wavelength.
2. Frequency: The number of sound waves produced in unit time is called the frequency of sound waves. Frequency is measured in seconds. Frequency is denoted by Greek letter v (nu). The SI unit of frequency is ‘hertz’.
3. Amplitude: Amplitude of a wave is magnitude of maximum disturbance on either side of the normal position or mean value in a medium.

Question 4.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
Speed, wavelength, and frequency of a sound wave are related as follows:
Speed (ν) = Wavelength [λ] × Frequency (v)
⇒ ν = λ × v

Question 5.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
Given,
Frequency of the sound wave, v = 220 Hz
Speed of the sound wave, u = 440 ms^-1
Putting the equations,
Speed = Wavelength (λ) × Frequency (v)
ν = λ × v
∴ λ = \(\frac { ν }{ v  }\) = \(\frac { 440 }{ 220 }\) = 2 m
Hence, Wavelength = 2 m.

Question 6.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Answer:
Given,
Frequency = 500 Hz
Distance = 450 m
Since, T = \(\frac { 1 }{ Frequency }\)
= \(\frac { 1 }{ 500 }\) = 0.002 sec.
Hence, person will listen the sound after every 0.002 sec.

Question 7.
Distinguish between loudness and intensity of sound.
Answer:
Intensity drives loudness of a sound. These both qualities of sound are proportional to each other. The amount of sound passing through a unit area per second represents intensity of a sound wave. While loudness is the response of the ear to the sound (amount received to pinna). The loudness of a sound is defined by its amplitude. The amplitude of a sound decides its intensity, which in turn is perceived by the ear as loudness.

Sound Intext Questions Page No. 167

Question 1.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
Sound travels the fastest in solids medium (here Iron) then in liquids (water) and it is the slowest in gases (air). Therefore, for a given temperature, sound travels as follows (decreasing order): ⇒ Iron > water >air.

Sound Intext Questions Page No. 168

Question 1.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms^-1?
Answer:
Given,
Speed, v = 342 ms^-1
Time, t = 3 s
Since, Distance = v × t
= 342 × 3 = 1026 m
Condition,
In fixed time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source = \(\frac { 1026 }{ 2 }\) m = 513 m.

Sound Intext Questions Page No. 169

Question 1.
Why are the ceilings of concert halls curved?
Answer:
Ceilings of concert halls are curved to:

1. Enhance loudness and echo of sound created.
2. Sound after reflection (from the walls) spreads uniformly in all directions.

Sound Intext Questions Page No. 170

Question 1.
What is the audible range of the average human ear?
Answer:
The audible range of an average human ear is between 20 Hz to 20,000 Hz.

Question 2.
What is the range of frequencies associated with

1. Infrasound?
2. Ultrasound?

Answer:

1. Infrasound: frequencies less than 20 Hz.
2. Ultrasound: frequencies more than 20,000 Hz.

Sound Intext Questions Page No. 172

Question 1.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Answer:
Given,
Time taken by the sonar pulse to return, t = 1.02s
Speed of sound in salt water, v = 1531 ms^-1
Distance of the cliff from the submarine = Speed of sound × Time taken
= 1531 × 1.02 m = 1561.62 m
Distance travelled by the sonar pulse during its transmission and reception in water
= 2 × Actual distance = 2d
Actual distance,
d = Distance of the cliff from the submarine / 2 = \(\frac { 1561 }{ 2 }\) = 780.31 m.

Sound NCERT Textbook Exercises

Question 1.
What is sound and how is it produced?
Answer:
Sound is a form of energy which is received at our ear pinna and gives the sensation of hearing. It is a vibration which propagates in air and developed by vibrating objects.

Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
When vibration is produced by a body, it moves in backward and forward direction till the energy lasts. During forward movement it creates a region of high pressure. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure. This region is known as a rarefaction. This is shown in figure given below.

Here,
C = compressions
R = rarefaction.

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Answer:
Arrange the instrument according to the picture given below:

• Take an electric bell and an air tight jar with glass bell and connect it to a vacuum pump.
• Suspend the bell inside the jar, and press the switch of the bell.

Method: Now pump out the air from the glass jar. The sound of the bell will become fainter and after some time, the sound will not be heard. This shows that sound needs a material medium to travel.

Question 4.
Why is sound wave called a ‘longitudinal wave’?
Answer:
Sound wave is called longitudinal wave because the air particles vibrates parallel to the direction of propagation of sound wave,it is produced by compressions and rarefactions in the air.

Question 5.
Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The quality of pitch and loudness of sound enables us to identify our friend by his voice.

Question 6.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer:
Speed difference is the main reason of this happening. The speed of sound (344 m/s) is very less than the speed of light (3 × 10⁸ m/s). A flash is seen before we hear a thunder because sound of thunder takes more time to reach the Earth as compared to light.

Question 7.
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms^-1.
Answer:
We know,
Speed (ν) = Wavelength × Frequency v
ν = λ × v
Given,
Speed of sound in air ν = 344 m/s (Given)

(i) For, v = 20 Hz
λ₁ = \(\frac { ν }{ v }\)
= \(\frac { 344 }{ 20 }\) = 17.2 m

(ii) For, v = 20000 Hz
12 = \(\frac { ν }{ v }\)
= \(\frac { 344 }{ 20000 }\) = 0.172 m
Hence, for humans, the wavelength range for hearing is 0. 0172 m to 17.2 m.

Question 8.
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of time taken by the sound wave in air and in aluminium to reach the second child.
Answer:
Velocity of sound in air= 346 m/s
Velocity of sound waves in aluminium 6420 m/s
Let length of rod be 1.
Time taken for sound wave in air, t₁ = \(\frac { 1 }{ Velocity }\) in air.
Time taken for sound wave in aluminium, t₂ = \(\frac { 1 }{ Velocity }\) in aluminium.
Therefore,

= \(\frac { 6420 }{ 346 }\) = 18.55.

Question 9.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Frequency =100 Hz. (Given)
This means the source of sound vibrates 100 times in one second.
Therefore, number of vibrations in 1 minute, i.e. in 60 seconds = 100 × 60 = 6000 times.

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
Yes, sound waves also follow the same laws of reflection as light wave do because:

1. Angle of incidence of sound is always equal to that of angle of reflection of sound waves.
2. The direction in which sound is incident, the direction in which it is reflected and normal all lie in the some plane.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of
sound production remains the same. Do you hear echo sound on a hotter day?
Answer:
An echo is heard when the time for the reflected sound is just 0.1 s
Time taken = \(\frac { Total Distance }{ Velocity }\)
On a hotter day, due to lighter medium the velocity of sound is more then a colder day. Hence, sound wave will move faster and if time taken by echo is less than 0.1 sec it will not be heard.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of reflection of sound waves are:

1. Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.
2. Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflection of sound.

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms^-2 and speed of sound = 340 ms^-1.
Answer:
Height of the tower, s = 500 m
Velocity of sound, v = 340 ms^-1
Acceleration due to gravity, g = 10 ms^-2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower t₁,
According to the second equation of motion:
s = ut₁ + \(\frac { 1 }{ 2 }\) gt₁²
500 = 0 × t₁ + \(\frac { 1 }{ 2 }\) × 10 t₁²
t₁² = 500
t₁ = 10 s
Now, time taken by the sound to reach the top from the base of the tower,  t₂ = \(\frac { 500 }{ 340 }\) = 1.47 s.
Therefore, the splash is heard at the top after time, t Where, t = t₁ + t₂ = 10 + 1.47 = 11.47 s.

Question 14.
A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
Speed of sound, v = 339 ms^-1
Wavelength of sound, λ = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
ν = λ × v
∴ v = \(\frac { ν }{ λ }\)
= \(\frac { 339 }{ 0.015 }\)
= 22600 Hz.
The frequency range of audible sound for humans lies between 20 Hz to 20,000 Hz.
Since, the frequency of the given sound is more than 20,000 Hz, it is not audible.

Question 15.
What is reverberation? How can it be reduced?
Answer:
The repeated multiple reflections of sound in any big enclosed space is known as reverberation. The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials, such as fibre board, loose woollens, etc.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
The effect produced in the brain by the sound of different frequencies is called loudness of sound. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Question 17.
Explain how bats use ultrasound to catch a prey.
Answer:
Bats produce high – pitched ultrasonic squeaks. These high – pitched squeaks are reflected by objects such as preys and returned to the bat’s ear. This allows a bat to know the distance of his prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

Question 19.
Explain the working and application of a SONAR.
Answer:
Sonar stands for Sound Navigation And Ranging. It is a device used to measure the depth, direction and speed of under – water objects such as submarines and ship wrecks with the help of ultrasounds and is also used to measure the depth of seas and oceans. An ultrasonic sound is produced and transmitted by the transducer (it is a device that produces ultrasonic sound) of the Sonar, which travels through sea water.

The echo produced by the reflection of this ultrasonic sound is detected and recorded by the detector, which is converted into electrical signals. The distance (d) of the under – water object is calculated from the time (t) taken by the echo to return with speed (v) is given by 2d = v × t.

Question 20.
A SONAR device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer:
Given,
Time taken to hear the echo, t = 5 s
Distance of the object from the submarine, d = 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water = 2d
Using formula:
Velocity of sound in water,
v = 2 \(\frac { d }{ t }\)
= 2 × \(\frac { 3625 }{ 5 }\) = 1450 ms^-1.

Question 21.
Explain how defects in a metal block can be detected using ultrasound.
Answer:
Defects in metal blocks do not allow ultrasound to pass through them and they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound.

Question 22.
Explain how the human ear works.
Answer:
The human ear consists of three parts – the outer ear, middle ear and inner ear.

1. Outer ear: It is also termed ‘pinna’. It collects the sound from the surrounding and directs it towards auditory canal.
2. Middle ear: It is at the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane to vibrate. These vibrations are amplified by three small bones hammer, anvil and stirrup.
3. Inner ear: When vibration reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.
   

Sound Additional Questions

Sound Multiple Choice Questions

Question 1.
Which of the following waves have no requirement of any medium to propagate?
(a) sound
(b) radio
(c) light waves
(d) none of above.
Answer:
(c) light waves

Question 2.
What kinds of waves are produced by sound?
(a) longitudinal only
(b) transverse waves only
(c) electromagnetic Waves
(d) both Longitudinal and Transversal.
Answer:
(a) longitudinal only

Question 3.
When a wave propagate, it transfers ___________ .
(a) energy only
(b) matter only
(c) both energy and matter
(d) none of these.
Answer:
(a) energy only

Question 4.
Note is a sound ___________ .
(a) of mixture of several frequencies
(b) of mixture of two frequencies only
(c) of a single frequency
(d) always unpleasant to listen.
Answer:
(a) of mixture of several frequencies

Question 5.
A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case:
(a) sound will be louder but pitch will not be different.
(b) sound will be louder and pitch will also be higher.
(c) sound will be louder but pitch will be lower.
(d) both loudness and pitch will remain unaffected.
Answer:
(d) both loudness and pitch will remain unaffected.

Question 6.
Earthquake produces which kind of sound before the main shock wave begins ___________ .
(a) ultrasound
(b) Infrasound
(c) audible sound
(d) none of the above.
Answer:
(b) Infrasound

Question 7.
Infrasound can be heard by ___________ .
(a) dog
(b) bat
(c) rhinoceros
(d) human beings.
Answer:
(c) rhinoceros

Question 8.
Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting ___________ .
(a) intensity of sound only.
(b) amplitude of sound only.
(c) frequency of the sitar string with the frequency of other musical instruments.
(d) loudness of sound.
Answer:
(c) frequency of the sitar string with the frequency of other musical instruments.

Question 9.
What we term to the number of oscillations completed in one second?
(a) time period
(b) velocity
(c) frequency
(d) wavelength.
Answer:
(c) frequency

Question 10.
Sound waves which can be propagate in solids are ___________ .
(a) Longitudinal only.
(b) Transverse only.
(c) Either longitudinal or transverse.
(d) Non mechanical waves only.
Answer:
(c) Either longitudinal or transverse.

Question 11.
In SONAR, we use ___________ .
(a) ultrasonic waves
(b) infrasonic waves
(c) radio waves.
(d) audible sound waves.
Answer:
(a) ultrasonic waves

Question 12.
Sound travels in air if ___________ .
(a) particles of medium travel from one place to another.
(b) there is no moisture in the atmosphere.
(c) disturbance moves.
(d) both particles as well as disturbance travel from one place to another.
Answer:
(c) disturbance moves.

Question 13.
When we change feeble sound to loud sound we increases ___________ .
(a) Frequency
(b) Amplitude
(c) Velocity
(d) Wavelength.
Answer:
(b) Amplitude

Question 14.
In the curve (Figure) half the wavelength is ___________ .
image
(a) AB
(b) BD
(c) DE
(d) AE
Answer:
(a) AB

Question 15.
The time taken to complete an osoillation ___________ .
(a) time period
(b) velocity
(c) frequency
(d) wavelength.
Answer:
(a) time period

Question 16.
The period of a vibrating body of frequency 100 Hz is ___________ .
(a) 100 seconds
(b) 10 seconds
(c) 0.1 second
(d) 0.01 second.
Answer:
(a) 100 seconds

Question 17.
Which of the following is SI unit of amplitude?
(a) metre
(b) s^-1
(c) metre / second
(d) Hertz.
Answer:
(a) metre

Question 18.
The quantity λ is known as ___________ .
(a) wave velocity
(b) frequency
(c) wavelength
(d) wave number.
Answer:
(c) wavelength

Question 19.
Sound wave of which of the following frequency is an ultrasonic sound?
(a) 30 Hz
(b) 300 Hz
(c) 3000 Hz
(d) 30,000 Hz.
Answer:
(d) 30,000 Hz.

Question 20.
In which of the following, speed of the sound is maximum?
(a) air
(b) water
(c) steel
(d) kerosene.
Answer:
(c) steel

Sound Very Short Answer Type Questions

Question 1.
How does sound travels in gases and liquids – as longitudinal or as transverse waves?
Answer:
As longitudinal waves.

Question 2.
Give examples of longitudinal waves.
Answer:
Sound waves.

Question 3.
What is speed of sound in air?
Answer:
At 0° C, it is 331 m/s. At 20°C, it is 341 m/s.

Question 4.
What is reverberation?
Answer:
The repeated reflection that results in the persistence of sound is called reverberation.

Question 5.
Among solids, liquids and gases sound travels faster in which medium?
Answer:
Sound travels the fastest in solids.

Question 6.
What is one Hz?
Answer:
Hz is the unit of frequency, called as Hertz. One Hertz is equal to one cycle per second.

Question 7.
What is ‘note’ of second?
Answer:
The sound produced due to a mixture of several frequency is called a note, it is pleasant to listen to.

Question 8.
What is pitch?
Answer:
The way our brain interprets the frequency of an emitted sound is called the pitch.

Sound Short Answer Type Questions

Question 1.
What is Sound? Why it is important for us?
Answer:
Sound is a longitudinal mechanical wave. Sound has great importance in our daily life. It gives us a sensation of hearing. It makes it possible to communicate with other persons through speech.

Question 2.
What is a mechanical wave?
Answer:
A mechanical wave is a disturbance that moves through a medium when the particles of the medium set neighbouring particles into motion. The particles vibrating in turn do not move forward but the disturbance is carried forward.

Question 3.
What is a longitudinal wave?
Answer:
If the particles of a medium vibrate in a direction, parallel to or along the direction of propagation of wave, it is called longitudinal wave.

Question 4.
What type of waves can travel in vacuum? Give example(s).
Answer:
Electromagnetic waves can travel in vacuum. Sun light, x – rays are examples of electromagnetic waves.

Question 5.
Suppose you and your friend are on the Moon. Will you be able to hear any sound produced by your friend?
Answer:
No, we will not hear any sound on the Moon. The Moon does not have any atmosphere, since sound is a mechanical wave and requires a medium to travel.

Question 6.
What are the factors, speed of sound wave depends upon?
Answer:
Speed of the sound depends on the following factors:

1. Inertial property of the medium (to store kinetic energy).
2. Elastic property of the medium (to store potential energy).
3. Temperature of the medium.
4. Density of the medium.
5. Humidity present in the medium (in air / gases).

Question 7.
Three persons A, B and C are made to hear a sound travelling through different mediums as given below:

Who will hear the sound first ? Why ?
Answer:
Person A will hear the sound first because sound travels the faster in solids than in liquids and gases.

Question 8.
If 20 waves are produced per second, what is the frequency in hertz?
Answer:
Number of waves per second is known as frequency.
∴ Frequency (v) = 20 Hz.

Question 9.
What is echo?
Answer:
The sound waves produced bounce back or gets reflected from the mountains or buildings and come to our ears, this reflected sound is known as Echo. To hear echo, the barrier reflecting the sound should be atleast at a distance of 17 metres.

Question 10.
What is infrasonic? Give an example.
Answer:
Sound having frequency less than 20 Hz is known as infrasonic sound or infrasonic. Waves produced during earthquake are infrasonic.

Sound Long Answer Type Questions

Question 1.
Establish the relationship between speed of sound, its wave length and frequency. If velocity of sound in air is 340 m s^-1, calculate:

1. wavelength when frequency is 256 Hz.
2. frequency when wavelength is 0.85 m.

Answer:
Derivation of formula ν = v λ

1. 340 = 256 λ ⇒ λ = 1.33 m.
2. 340 = v (0.85) ⇒ v = 400 Hz.

Question 2.
A girl is sitting in the middle of a park of dimension 12 m × 12 m. On the left side of it there is a building adjoining the park and on right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain answer.
Answer:
If the time gap between the original sound and reflected sound received by the listener is around 0.1 s, only then the echo can be heard. The minimum distance travelled by the reflected sound wave for the distinctly listening the echo, distance = velocity of sound × time interval;
= 344 × 0.1 = 34.4 m.
But, in this case the distance travelled by the sound reflected from the building and then reaching the girl will be (6 + 6 = 12 m),  which is much smaller than the required distance. Therefore, no echo can be heard.

Sound Higher Order Thinking Skills (HOTS)

Question 1.
A key of piano is struck gently and then struck again but much harder this time. What will happen in the second case?
Answer:
In second case, the loudness will increase as this will increase the amplitude of vibration of string. Pitch and frequency will also increase in force or tension in the string.

Question 2.
At a hill station, a child could hear the echo of his voice after 0.2 s. But, when he went to the same place in the afternoon, he could not hear echo at all. What could be the reason?
Answer:
In afternoon, the temperature rises, therefore the speed of sound also increased. The reflected sound will take very less time to travel back and no echo is heard.

Sound Value Based Question

Question 1.
It is not advisable to construct houses near airports, inspite of that many new residential apartments are constructed near airports. Rajesh / Sumit files RTI and also complains the municipal office about the same.

1. Why one should not reside near airport?
2. Name other two places where there is noise pollution.
3. What value of Rajesh is reflected in this act?

Answer:

1. The landing and taking off of the air – planes causes lot of noise-pollution which may lead to deafness, high blood pressure and other health problems.
2. The other two places where there is noise-pollution are residing near the heavy traffic routes and railway stations or lines.
3. Rajesh shows participating citizen and moral responsibility values.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
Let us draw different pairs of circles as shown below:

We gave,
In figure           Maximum number of common points
(i)             –                     nil
(ii)            –                    one
(iii)           –                    two
Thus, two circles can have at the most two points in common.

Question 2.
Suppose you are given a circle. Give a construction to find its center.
Solution:
Steps of construction:

1. Mark any three points A, B and C on the circle.
   
2. Join AB and BC.
3. Draw the perpendicular bisector of AS. Let it be PQ.
4. Draw the perpendicular bisector of BC. Let it be RS.
5. PQ and RS intersect at point O.
6. O is the center of given circle.

Question 3.
If two circles intersect at two points, prove that their centers lie on the perpendicular bisector of the common chord.
Solution:
Given:
Circles C(O, r) and C(D, r₁) intersect at A and B.
To prove:
OD is perpendicular bisector of AS.

Construction:
Join OA, AD, OB and BD.
Proof:
In ∆AOD and ∆BOD,
OA = OB
DA = DB (Radii of the circle)
OD = OD (Common)
∆AOD ≅ ∆BOD (By SXS)
and ∠1 = ∠2 (By CPCT)
In ∆OAC and ∆OBC,
OA = OB (Radii of the circle)
OC =OC (Common)
∠1 = ∠2 (Proved)
∆AOC = ∆OBC (By SAS)
So ∠3 = ∠4 (By CPCT)
and AC = BC (By CPCT)
∠3 + ∠4 = 180.°
2∠3 = 180°
∠3 = \(\frac{180^{\circ}}{2}\) =90° (UA’s)
OD is perpendicular bisector of AB.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.2

Question 1.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centers.
Solution:
We have a circles having its centers at O and O’ two equal chords AB and CD such that they subtend ∠AOB and ∠COD respectively at their centers, i.e. at O and O’.

We have to prove that
∠AOB = ∠CO’D
Now, in ∆AOB and ∆CO’D, we have
AO = CO’ [Radii of the same circle]
BO = DO’ [Radii of the same circle]
AB = CD [Given]
∆AOB = ∆CO’D [SSS criterion]
Their corresponding parts are equal.
∴ ∠AOB = ∠CO’D

Question 2.
Prove that if chords of congruent circles subtend equal angles at their centers, then the chords are equal.
Solution:
We have circles having their centers at O and O’, and its two chords AB and CD such that
∠AOB = ∠CO’D
we have to prove that
AB = CD
In ∆AOB and ∆COO’D, we have:
AO = CO’ [Radii of the same circle]
BO = DO’ [Radii of the same circle]
∠AOB = ∠CO’D [Given]
∆AOB = ∆CO’D [SAS criterion]
Their corresponding parts are equal, i.e.,
AB = CD.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.1

Question 1.
Fill in the blanks:

1. The center of a circle lies in …………of the circle.
2. A point, whose distance from the center of a circle is greater then its radius lies in …………..of the circle.
3. The longest chord of a circle is a ………… of the circle.
4. An arc is a ………….. when its ends are the ends of a diameter.
5. Segment of a circle is the region between an arc and ……….. of the circle.
6. A circle divides the plane, on which it lies, in ……….. parts.

Solution:

1. interior
2. exterior
3. diameter
4. semicircle
5. the chord
6. three.

Question 2.
Write True or False. Give reasons for your answers.

1. Line segment joining the center to any point on the circle is a radius of the circle.
2. A circle has only finite number of equal chords.
3. If a circle is divided into three equal arcs, each is a major arc.
4. A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
5. Sector is the region between the chord and its corresponding arc.
6. A circle is a plane figure.

Solution:

1. True [∵ All points on the circle are equidistant from the center]
2. False [∵ A circle can have an infinite number of equal chords.]
3. False [∵ Each part will be less than a semicircle.]
4. True [∵ Diameter = 2 x Radius]
5. False [∵ The region between the chord and its corresponding arc is a segment.]
6. True [∵ A circle is drawn on a plane.]

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

Question 1.
In fig. given below, ABCD is a parallelogram, AE ⊥ DCand CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Solution:
ar (ABCD) = AE x DC ….(i)
ar (ABCD) = AD x CF ….(ii)
From (i) and (ii), we get
AE x DC = AD x CF
⇒ AE x AB = AD x CF (∵ AB = DC)

8 x 16 = AD x 10
\(\frac{8×16}{10}\) = AD
\(\frac{128}{10}\) = AD
AD = 12.8 cm.

Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) \(\frac{1}{2}\) = ar (ABCD)
Solution:
Given
ABCD is a ∥^gm E, F, G, H, are the mid-points of AB, BC, CD and DA respectively.

To prove:
ar (EFGH) = \(\frac{1}{2}\) ar (ABCD)
Construction: Join GE.
Proof:
ABCD is a ∥gm
∴ AB ∥ DC and AB = DC
⇒ \(\frac{1}{2}\)AB ∥ \(\frac{1}{2}\)DC and \(\frac{1}{2}\) AB = \(\frac{1}{2}\) DC
AE ∥ DG and AE = DG
∴ AEGD is a parallelogram.
Similarly BCGE is also a parallelogram.
∥gm AEGD and AEGH both lie on the same base EG and are between the same parallels AD and EG.
ar (EGH) = \(\frac{1}{2}\) ar {AEGD) …(1)
Similarly ar {GEF) = \(\frac{1}{2}\) ar {BCGE) …(2)
Adding (1) and (2), we get
ar (EGH) + ar (GEF) = ar (AEGD) + \(\frac{1}{2}\) ar {BCGE)
ar (EFGH) ar {EFGH) = \(\frac{1}{2}\) [ar (AEGD) + ar (BCGE)]

Question 3.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Solution:
Given:
ABCD is a ∥^gm
To prove:
ar (APB) = ar (BQC)..
Proof:
∆APB and ∥^gm ABCD lie on the same base BC and are between the same parallels AB and DC.

ar(APB) = \(\frac{1}{2}\) ar (ABCD)
∆BQC and ∥^gm ABCD lie on the same base BC and are between the same parallels BC and AD.
ar (BQC) = \(\frac{1}{2}\) ar (ABCD) …(2)
From (1) and (2), we get
ar (APB) = ar (BQC)

Question 4.
In Fig. given below, P is a point in the interior of a parallelogram ABCD, Show that

1. ar (APB) + ar (PCD) = \(\frac{1}{2}\) ar (ABCD)
2. ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint: Thorough P, draw a line parallel to AB.]
Solution:
Given:
P is any point in the interior of ∥^gm ABCD.
To prove:

1. ar (APB) + ar (PCD) \(\frac{1}{2}\) ar (ABCD)
2. ar (APD) + ar (PBC) ar (APB) + ar (PCD)

1. Construction:
Draw a line EF passing through point P parallel to AB.
Proof:
AB ∥EF (By construction) …(1)
AB ∥ DC (∵ ABCD is a ∥gm) …(2)
EF ∥ DC [From (1) and (2)]

In quadrilateral ABFE, AB ∥ EF and AE ∥ BF (∵ AD ∥ BC)
∴ ABFE is a ∥gm
∆ APB and ∥gm ABEF lie on same base base AB and are between the same parallels AB and EF.
ar (APB) = \(\frac{1}{2}\) ar (ABFE) …. (A)
Similarly, ar (PCD) = \(\frac{1}{2}\) ar (DCFE) …(B)
Adding (A) and (B), we get
ar (APB) + ar (PCD) = \(\frac{1}{2}\) ar (ABFE) + \(\frac{1}{2}\) (DCFE)
= \(\frac{1}{2}\) [ar (ABFE) + ar (DCFE)]
= \(\frac{1}{2}\) ar (ABCD) …(3)

2. Construction:
Draw a line GH passing through point P parallel to AD.
Proof:
AD ∥ GH (by construction) …(4)
AD ∥ BC (∵ ABCD is a ∥^gm) …(5)
GH ∥ BC [From (4) and (5)]
In quadrilateral ADHG,
AD ∥ GH (by construction)
and AG ∥ DH (∵ AB\\DC)
∴ ADHG is a parallelogram.
∆APD and ∥gm ADHG lie on the same base AD and lie betweeiwthe same parallels AD and GH.
ar(APD) = \(\frac{1}{2}\) ar (ADHG) …(6)
Similarly, ar (BCP) = \(\frac{1}{2}\) ar (BCHG) …(7)
Adding (6) and (7), we get
ar (APD) + ar (BCP) = \(\frac{1}{2}\) ar (ADHG) + \(\frac{1}{2}\) ar (BCHG)
= \(\frac{1}{2}\) ar (ADHG) + ar (BCHG)
= \(\frac{1}{2}\) [ar (ADHG) + ar (BCHG)]
= \(\frac{1}{2}\) ar (ABCD) …(8)
From (3) and (8), we get
ar (APB) + ar (PCD) = ar (APD) + ar (BCP)

Question 5.
In Fig. given below PQRS and ABRS are parallelograms and X is any point on side BR. Show that

1. ar (PQRS) = ar (ABRS)
2. ar (AZS) = \(\frac{1}{2}\) ar (PQRS)

Solution:
Given
PQRS and ABRS are H801 and X is any point on side BR.
To prove:

1. ar (PQRS) = ar (ABRS)
2. ar (AXS) \(\frac{1}{2}\) ar (PQRS)

Proof:
1. ∥^gm PQRS and ABRS lie on the same base SR and are on the same parallels SR and PB.
∴ ar (PQRS) = ar (ABRS) …(i)

2. D AXS and ∥^gm ABRS lie on the same base AS and are on the same parallels AS and BR.
ar (∆ AXS) = \(\frac{1}{2}\) ar (ABRS) …(ii)

From (i) and (ii), we get
ar (AXS) = \(\frac{1}{2}\) ar (PQRS)

Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Solution:
The field will be divided in three parts. The shapes of three parts are triangles.
Given:
PQRS is a ∥^gm
To prove:
ar (APQ) = ar (APS) + ar (AQR)
Construction:
Draw a line AB ∥ SP.
Proof:
∆APQ and ∥^gm PQRS are on the same base PQ and are between the same parallels PQ and SR.

ar (PQRS) = \(\frac{1}{2}\) ar (PQRS)
⇒ 2ar (APQ) = ar (PQRS)
2ar (APQ) = ar (APS) + ar (APQ) + ar (AQR)
2ar (APQ) – ar (APQ) = ar (APQ) ar (AQR)
ar (APQ) = ar (APS) + ar (AQR)

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 13 Why do we Fall Ill

Why do we Fall Ill Intext Questions

Why do we Fall Ill Intext Questions Page No. 178

Question 1.
State any two conditions essential for good health.
Answer:
Two conditions essential for good health are:

1. Balanced diet and good nutrition.
2. Good social surrounding and economic wellness.

Question 2.
State any two conditions essential for being free of disease.
Answer:
Two conditions essential for being disease – free are:

1. Personal and community hygiene.
2. Proper diet and economic growth.

Question 3.
Are the conditions essential for maintaining good health and being free of diseases same or different? Why?
Answer:
Yes, to some extent they are the similar, because if the conditions that are essential for good health are maintained, then the chances of getting a disease will be minimized. But at the same time, we can say that they are different because being health or good health means physical, mental and social well-being while being disease-free means not suffering from a particular disease.

Why do we Fall Ill Intext Questions Page No. 180

Question 1.
List any three reasons why you would think that you are sick and ought to see a doctor. If only one of these symptoms were present, would you still go to the doctor? Why or why not?
Answer:
Common symptoms which indicate sickness are:

1. Headache
2. Cough
3. Dysentery

If only one of these symptoms is present, we usually do not visit a doctor. This is because such symptoms do not have much effect on our general health and ability to work. However, if a person is experiencing these symptoms for quite sometime, then he needs to visit a doctor for proper treatment.

Question 2.
In which of the following case do you think the long – term effects on your health are likely to be most unpleasant?

1. If you get jaundice
2. If you get lice
3. If you get acne.
   Why?

Answer:
Jaundice is a disease that can cause long – term effects on our health. It is a chronic disease that lasts for a long period of time. Jaundice does not spread rapidly, but it develops slowly over a period of time.

Why do we Fall Ill Intext Questions Page No. 187

Question 1.
Why are we normally advised to take bland and nourishing food when we are sick?
Answer:
When we are sick, the normal body functions get disturbed. In such a situation, food that is easily digestible and contains adequate nutrients are required for the speedy recovery. Thus, bland and nourishing food is given during sickness.

Question 2.
What are the different means by which infectious diseases spread?
Answer:
The different modes of transmission of infectious diseases are:

1. Through Air: Certain disease – causing micro – organisms are expelled in air by coughing, sneezing, talking etc. These micro-organisms can travel through dust particles or water droplets in air to reach other people. For example, tuberculosis, pneumonia etc. spread through air.
2. Through Water: Sometimes causal micro – organisms get mixed with drinking water and spread water borne diseases. Cholera, for example is a water borne disease.
3. Through Sexual Contact: Sexual act between two people can lead to the transfer of diseases such as syphilis, gonorrhoea, AIDS etc.
4. Through Vectors: Certain diseases spread by animals called vectors. For example, mosquitoes spread malaria.

Question 3.
What precautions can you take in your school to reduce the incidence of infectious diseases?
Answer:
Precautions to reduce incidence of infectious diseases are:

1. Staying away from the infected person.
2. Covering mouth or nose while coughing or sneezing to prevent the spread of disease.
3. Drinking safe water.
4. Keeping the school environment clean to prevent multiplication vectors.

Question 4.
What is immunisation?
Answer:
Immunisation is defined as protection of the body from communicable diseases by administration of some agent that mimics the pathogen.

Question 5.
What are the immunisation programmes available at the nearest health centre in your locality? Which of these diseases are the major health problems in your area?
Answer:
The immunisation programmes available at the nearest health centre are DPT (Diphtheria, Pertusis, and Tetanus), polio vaccine, hepatitis B, MMR (Measles, Mumps, and Rubella), jaundice, typhoid etc. Of all these diseases, jaundice and typhoid are major health problems.

Why do we Fall Ill NCERT Textbook Exercises

Question 1.
How many times did you fall ill in the last one year? What were the illnesses?

1. Think of one change you could make in your habits in order to avoid any of/most of the above illnesses.
2. Think of one change you would wish for in your surroundings in order to avoid any of / most of the above illnesses.

Answer:
The illness was 2 – 3 times common – cold, occurred in a year:

1. One change I would make in my habits in order to avoid the above illness is that I would take proper diet rich in vitamin C and would avoid too cold food.
2. The surroundings should be neat and clean.

Question 2.
A doctor / nurse / health – worker is exposed to more sick people than others in the community. Find out how she / he avoids getting sick herself / himself.
Answer:
The following precautions must be taken by a doctor / nurse / health – worker:

1. Wearing a mask when in contact with a diseased person.
2. Keeping yourself covered while moving around an infected place.
3. Drinking safe water.
4. Eating healthy and nutritious food.
5. Ensuring proper cleanliness and personal hygiene.

Question 3.
Conduct a survey in your neighbourhood to find out what the three most common diseases are. Suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.
Answer:
Common diseases are dysentry, malaria, viral fever / jaundice. Preventive Measures:

1. Cleanliness of surrounding neighbourhood i.e, removal of garbage from the streets / house and storage in a covered place till it is disposed off. Proper cleaning of drains and proper disposal of sewer water.
2. Removal of breeding places of mosquitoes like ditches with stagnant water.
3. Ensure supply of safe drinking water.

Or
Periodical programmes to educate people about prevention of disease.

Question 4.
A baby is not able to tell her / his caretakers that she / he is sick. What would help us to find out

1. that the baby is sick?
2. what is the sickness?

Answer:

1. The baby is sick can be determined by his / her behavioural changes such as constant crying of baby, improper intake of food, frequent mood changes etc.
2. The sickness is determined by symptoms or indications that can be seen in the baby. The symptoms include vomiting, fever, loose motion, paleness in the body etc.

Question 5.
Under which of the following conditions is a person most likely to fall sick?
(a) when she is recovering from malaria.
(b) when she has recovered from malaria and is taking care of someone suffering from chicken – pox.
(c) when she is on a four – day fast after recovering from malaria and is taking care of someone suffering from chicken – pox. Why?
Answer:
(c) A person is more likely to fall sick when she is on a four day fast after recovering from malaria and is taking care of someone who is suffering from chicken pox. This is because she is fasting during recovery and her immune system is so weak that it is not able to protect its own body from any foreign infection. If she is taking care of someone suffering from chicken pox, then she has more chances of getting infected from chicken pox virus and will get sick again with this disease.

Question 6.
Under which of the following conditions are you most likely to fall sick?
(a) when you are taking examinations.
(b) when you have travelled by bus and train for two days.
(c) when your friend is suffering from measles. Why?
Answer:
(c) You are more likely to fall sick when your friend is suffering from measles. This is because measles is highly contagious and can easily spread through respiration i. e., through air. Thus, if your friend is suffering from measles, stay away from him otherwise you might easily get infected with the disease.

Why do we Fall Ill Additional Questions

Why do we Fall Ill Multiple Choice Questions

Question 1.
Community health is related to:
(a) Social health
(b) Physical health
(c) Mental health
(d) All.
Answer:
(d) All.

Question 2.
Every disease have particular:
(a) Body weight
(b) temperature
(c) Sign
(d) Appearance change.
Answer:
(c) Sign

Question 3.
Condition for good community health is by maintaining:
(a) Cleanliness
(b) Pollution less atmosphere
(c) Economic growth
(d) All.
Answer:
(d) All.

Question 4.
Malaria is caused by:
(a) Vibrio cholerae
(b) Plasmodium vivax
(c) Andies
(d) H. pylori.
Answer:
(b) Plasmodium vivax

Question 5.
Vibrio cholerae is a:
(a) Agent
(b) Host
(c) Vector
(d) None.
Answer:
(a) Agent

Question 6.
Which one of the following is water born infectious disease?
(a) Common cold
(b) Peptic ulcers
(c) Cholera
(d) AIDS.
Answer:
(c) Cholera

Question 7.
Which one among the following is a viral disease?
(a) AIDS
(b) Cholera
(c) TB
(d) Typhoid.
Answer:
(a) AIDS

Question 8.
Which one among the following is a chronic disease?
(a) Cold
(b) Malaria
(c) Elephantiasis
(d) None.
Answer:
(c) Elephantiasis

Question 9.
In malaria disease, which one of the following is vector?
(a) Plasmodium vivax
(b) Mosquito
(c) Human
(d) Dog.
Answer:
(b) Mosquito

Question 10.
Which one of the following is mechanical cause of disease?
(a) Injury
(b) Microbes
(c) Heat
(d) All.
Answer:
(a) Injury

Question 11.
HIV – AIDS is transmitted by:
(a) Microbes
(b) Sexual contact
(c) Water
(d) All.
Answer:
(b) Sexual contact

Question 12.
Main factors which can help in public health hygiene:
(a) Availability of safe drinking water.
(b) Availability of clean environment.
(c) Spacious condition for living.
(d) All.
Answer:
(d) All.

Question 13.
Penicillin is a / an:
(a) Antigen
(b) Antibody
(c) Pathogen
(d) Host.
Answer:
(a) Antigen

Why do we Fall Ill Very Short Answer Type Questions

Question 1.
Write symptoms of a disease.
Answer:
Discomfort pain and tiredness is symptoms of a disease.

Question 2.
Is mental discomfort a disease?
Answer:
Yes.

Question 3.
When a part of body is hurt, can it affect all body? Why?
Answer:
Yes, because all body parts are connected internally.

Question 4.
Write name of two diseases caused by virus.
Answer:
AIDS, common cold.

Question 5.
Write name of two diseases caused by worms.
Answer:
Elephantiasis, intestine infection.

Question 6.
Write name of two deficiency diseases.
Answer:
Beri – Beri, Rickets.

Question 7.
What is a vector?
Answer:
Organism which transfer a disease is called a vector.

Question 8.
What is a host?
Answer:
Organism which bear disease and their causative agents is called host.

Question 9.
Name one genetical disorder.
Answer:
Haemophilia.

Question 10.
How malaria and dengue are spread?
Answer:
By species of mosquito.

Question 11.
Name two communicable diseases.
Answer:

1. AIDS
2. cold.

Question 12.
What become vector for a communicable disease?
Answer:
Host itself.

Question 13.
What kind of disease is AIDS?
Answer:
AIDS is a fatal viral disease.

Question 14.
Name two air born diseases.
Answer:

1. Influenza
2. pneumonia.

Question 15.
Name a disease against which vaccines are available.
Answer:
Typhoid.

Why do we Fall Ill Short Answer Type Questions

Question 1.
Define congenital diseases. Give some examples of such disease.
Answer:
Diseases or abnormalities present since birth are called congenital diseases. It is due to gene mutations (genetic factor) or environmental factors.
Examples:

• Colour blindness
• Cleft lip or palate.

Question 2.
Name the agent and vector which cause rabies.
Answer:
The causative agent of rabies is rabies virus (RV). The animals included humans can be infected and all of them can spread the disease.
Causes:

• Dogs
• rats
• cats
• monkeys
• squirrels
• cattle
• wolves
• racoons
• bears etc. can spread this disease.

Question 3.
What are acquired diseases?
Answer:
The diseases which develop after birth are called acquired diseases. It is categorized into

• Communicable Diseases(Infectious Diseases): malaria, influenza etc. and
• Non – Communicable Diseases: diabetes, scurvy, high blood pressure.

Question 4.
Write some common signs and symptoms of a disease if a brain is affected.
Answer:
Headache, fits, vomitting, unconsciousness etc.

Question 5.
What are infectious agents?
Answer:
The microorganisms which spread the disease from one person to other are called infectious agents.

Question 6.
Define ‘germ theory of disease’?
Answer:
Louis Pasteur proposed ‘germ theory of disease’ also called pathogenic theory of medicine. He stated that micro – organisms are the causes of many diseases.

Question 7.
What happen when kidneys of a person do not filter urine properly?
Answer:
If the kidneys of a person do not filter urine properly then the toxic substances which disturbs metabolism could be fatal.

Question 8.
Is skin, hairs, saliva form the first line of defence against diseases? If yes, then how?
Answer:
Yes, skin, nose hairs, saliva act as the first line of defence.

• Skin: It prevents the entry of microbes.
• Hairs: These prevent the entry of dust particles and germs.
• Saliva: It washes away bacterial growth from teeth and mouth.

Question 9.
What is immunity?
Answer:
Immunity refers to ability of the body to be resistant towards injury, poison or harmful pathogens. It is part of the defence reaction in the body. There are two types of immunity – natural or innate immunity and acquired or specific immunity.

Question 10.
Why are kids and elderly people more vulnerable to cold / flu?
Answer:
Kids are more vulnerable to cold and flu because their immune systems haven’t fully developed. Elderly people also more prone to catch a cold because of their poor health.

Question 11.
Define antibiotic. Explain how it is able to control bacterial infections but not viral infections.
Answer:
Antibiotics (anti means against and biotic means living) are types of medications which slows and destroy the growth of bacteria. They commonly block biochemical pathways important for bacteria. Many bacteria, for example, make a cell – wall to protect themselves. The antibiotic penicillin blocks the bacterial processes that build the cell wall. Thus, bacteria fail to build their cell – wall, stop multiplying and die out.

Question 12.
Name infectious diseases that are spread through air.
Answer:
Infectious diseases that spread through air are – sneezing, coughing, open pneumonia, tuberculosis, measles, SARS, chicken – pox.

Question 13.
How infectious diseases are spread through water? Give two examples.
Answer:
Open defecation, consuming contaminated water, inadequate hand washing causes infectious diseases through water. Water – borne diseases are cholera, typhoid etc.

Question 14.
Give few examples of direct and indirect contact diseases.
Answer:

1. Direct Contact Diseases: Examples of diseases spread by direct contact are common cold, tuberculosis, chicken pox, warts etc. Sexual Transmitted Diseases (STD) are Syphilis, AIDS, gonorrhoea.
2. Indirect Contact Diseases: Examples of diseases spread by indirect contact are Flu, TB, Chickenpox, urinal infections etc.

Why do we Fall Ill Long Answer Type Questions

Question 1.
Becoming exposed to or infected with an infectious microbe does not necessarily mean developing noticeable disease. Explain.
Answer:
Infected with a microbe does not mean developing a disease because an infectious microbe is able to cause a disease only. If the immune system of the person is weak but a person with strong immune system normally fights off microbes. We have cells which are specialised to kill the pathogenic microbes. These cells are active when infecting microbes enter the body and if they are successful in removing the pathogen, we remain disease – free. So, even if we are exposed to infectious microbes, the person will not catch the disease.

Question 2.
Give any four factors necessary for a healthy person.
Answer:
Factors necessary for a healthy person are as follows:

1. A clean environment with proper public health services.
2. Personal hygiene prevents infectious diseases.
3. A proper balanced diet and sufficient nourishment are necessary for good immune system of our body.
4. Immunisation / vaccination against severe diseases.

Question 3.
Why is AIDS considered to be a ‘Syndrome’ and not a disease?
Answer:
AIDS is considered a syndrome and not a disease because AIDS causing virus – HIV comes into the body via the sexual organs or any other means like blood transfusion and spread to lymph nodes all over the body. The virus damages the immune system of the body and the body can no longer fight off many minor infectious. Instead, every small cold can become pneumonia, or minor get infection can become severe diarrhoea. The effect of disease becomes very severe and complex, at times killing the person suffering from AIDS. Hence, there is no specific disease symptoms for AIDS but it results in a complex disease and symptoms.

Question 4.
(a) The signs and symptoms of a disease depend upon the tissues or organ the microbe targets. Explain giving any two examples.
(b) How does the immune system work against microbes?
(c) Name two diseases against which vaccines are available.
Answer:
(a) Disease – causing microbes enter the body through different means. These disease – causing micro – organisms are tissues specific or we can say that they target a specific organ.  If these microbes enter the body via the nose, the lungs are target and the symptoms are:

1. cough
2. breathlessness.

This is seen in the bacteria causing tuberculosis. If brain is targetted by these microbes, the symptoms will be headache, vomitting, fits or unconsciousness. This is seen in the bacteria causing meningitis.

(b) Immunisation is most important and effective way to raise resistance against the disease. Immunity may be two types:

1. Active immunity
2. Passive Immunity (artificial).

In active immunity, the body responds against disease – causing microbes and produces antibodies. The antibodies attack organ and kill the disease – causing microbes. In passive immunity, prepared antibodies are used against disease – causing microbes.

(c) We have vaccine against polio and tuberculosis.

Why do we Fall Ill Higher Order Thinking Skills (HOTS)

Question 1.
Why we cannot destroy virus?
Answer:
Virus attacks our body and consume proteins and RNA only. They do not have any metabolic pathway while other pathogens have their metabolic pathways. Our medicines stops the metabolic activity of pathogens and hence they get destroyed, but it is not possible in case of virus.

Question 2.
How vaccine stops disease while it is, itself a low dose of similar causative agent?
Answer:
Our body has a very strong immune system and it develops antigen to subside the effect of any new high quantity antibody entering in the body. Presence of chemical in small harmless quantity initiate development of antigen in body and when actually agent enters in the body, antigen fights easily with it.

Why do we Fall Ill Value Based Questions

Question 1.
Anya got vaccination, she told her friend Sakshi. Sakshi got scared and doubted if it was safe as it contains weakened pathogens which cause diseases. Later, Anya explained her the concept of vaccination and immunity. Answer the following questions:

1. What did Anya explain Sakshi?
2. What values did Anya possess?

Answer:

1. Anya explained that vaccine or weakened pathogen are not harmful to the body. Rather, they boost the person immunity against the same pathogen when it is actually encountered by the body is attacked by it leading to its destruction.
2. Anya possess the values of helpfulness and intelligence. She is practical.

Question 2.
Akash found in a blood test that his brother was HIV positive. His brother, due to this was expelled from the job. Only after intervention of NGO, he got it back.

1. Which disease is his brother suffering from?
2. Write its modes of transmission.
3. Write your views about the NGO and the organization where his brother worked.

Answer:

1. AIDS (caused by HIV).
2. • Mother to child (during pregnancy).
   • Using infected syringes.
   • Blood transfusion.
   • Sexual contact with infected person.
3. NGO has taken correct step and has helped Akash in fighting for injustice. While, organisation was wrong in its step as it does not spread by touching or by working with the infected person.

MP Board Class 9th Science Solutions

MP Board Class 9th Science Solutions Chapter 14 Natural Resources

Natural Resources Intext Questions

Natural Resources Intext Questions Page No. 193

Question 1.
How is our atmosphere different from the atmospheres on Venus and Mars?
Answer:
Earth’s atmosphere has mixture of nitrogen (79%), oxygen (20%), and a small fraction of carbon dioxide, water vapours and other gases which makes the existence of life possible on Earth while the atmospheres on Venus and Mars mainly has carbon dioxide. Approximately, 95% to 97% is carbon dioxide is present on these planets which do not support existence of life.

Question 2.
How does the atmosphere act as a blanket?
Answer:
Atmosphere is a mixture of various gases along with vapour and submerged particles. It surrounds the earth and separates it from outer space. As a blanket stop interaction between body covered inside and atmosphere outside, in similar way atmosphere stops interaction between earth and space.
The most important isolations done by atmosphere are:

1. It keeps the average temperature of the Earth fairly constant during day time and even during the course of whole year.
2. It prevents a sudden increase in the temperature during day time.
3. It slows down the escape of heat from the surface of the Earth into outer space during night time.

Question 3.
What causes winds?
Answer:
Atmospheric pressure and temperature difference generates wind. As earth surface is not similar everywhere even at some places it is occupied by huge water bodies, lava emitting volcanoes, moist forest etc., a difference in temperature and pressure arises and to maintain equilibrium wave of pressure moves which we feel or notice as wind.

Question 4.
How are clouds formed?
Answer:
During day time, a large amount of water evaporates from various water bodies and of earth surface and through biological activities such as transpiration and respiration and mix up into the air. This causes the air in the atmosphere to heat up and since air is a bad conductor of heat, when this heated air rises, it expands and cools, which results in the condensation and formation of water droplets. The presence of dust and other suspended particles in air also facilitates cloud formation. The gathering of water droplets leads to the formation of clouds.

Question 5.
List any three human activities that you think would lead to air pollution.
Answer:
Three human activities leading to air pollution are:

1. Smoke from industrial and manufacturing activities etc.
2. Burning of fossil fuels for household and commercial purpose.
3. Spreading farming chemicals such as crop dusting, insect / pest killers, fertilizer dust etc.

Natural Resources Intext Questions Page No. 194

Question 1.
Why do organisms need water?
Answer:
Water is the major component of living organism some of the living organism has up to 90% of water of their body weight. Water is a wonderful liquid, it helps in performing most of the functions of our body like digestion, cell sap and oilier content formation, transportation of substance from one place to another inside the body etc.

Question 2.
What is the major source of fresh water in the city’town/ village where you live?
Answer:
The major source of fresh water near me is river.

Question 3.
Do you know of any activity which may be polluting this water source?
Answer:
Dumps from industrial activities, household works for various purpose in the fresh water sources are main activities which causes pollution. Air pollution which generates acid rain and spread submerged dust is also responsible for polluting water resources.

Natural Resources Intext Questions Page No. 196

Question 1.
How is soil formed?
Answer:
The process of soil formation is termed ‘paedogenesis’. Soil is formed mainly by the weathering of rocks through various physical chemical and biological processes with the help of various factors such as the sun, water, wind, and living organisms. It is made up of mineral particles, organic materials, air, water and living organisms. During day time, the rocks are heated up by solar rays. This causes the rocks to expand. During night time, these rocks cool down and contracts. It causes weathering of rock and formation of soil.

1. Water: It helps in breaking of rocks in two ways:
   • It goes into the cracks and crevices formed in the rocks. When this water freezes, its volume increases. As a result, the size of the cracks also increases. This helps in the weathering of rocks.
   • Water moving at fast speed carries big and small particles of rock downstream. These rocks rub against each other, resulting in breaking down of rocks. These smaller particles are carried away by running water and deposited down its path.
2. Wind: Strong winds carry away rocks, which causes rubbing of rocks. This results in the breaking down of rocks into smaller and smaller particles.
3. Living Organism: Some living organisms like lichens help in the formation of soil. Lichens grow on rock surfaces and converts them into powdery form and make soil layer. In the same way, the plants like moss also helps in the making of fine soil particles.

Question 2.
What is soil erosion?
Answer:
The blowing away or washing away of land surface by wind or water is known as soil erosion.

Question 3.
What are the methods of preventing or reducing soil erosion?
Answer:
The methods of preventing or reducing soil erosion are:

1. Plantation of tress and plants.
2. Prevention of deforestation.
3. Prevent excessive grazing.

Natural Resources Intext Questions Page No. 201

Question 1.
What are the different states in which water is found during the water cycle?
Answer:
Water is found in three different states during the water cycle:

1. Solid (snow, ice).
2. Liquid water (ground water, river water, etc.)
3. Gaseous state (water vapours).

Question 2.
Name two biologically important compounds that contain both oxygen and nitrogen.
Answer:
Two biologically important compounds that contain both oxygen and nitrogen are:

1. Amino acids.
2. Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA).

Question 3.
List any three human activities which would lead to an increase in the carbon dioxide content of air.
Answer:
Three Human activities are:

1. Burning of fuels in various processes like heating, cooking, transportation and industry.
2. Human induced forest fires.
3. The process of deforestation includes the cutting down of trees. This decreases the uptake of carbon dioxide for photosynthesis. Eventually, the content of carbon dioxide increases.

Question 4.
What is the greenhouse effect?
Answer:
Some gases like carbon dioxide, methane, nitrous oxide prevent the escape of heat from the Earth’s surface by trapping it. This increases the average temperature of the Earth. This is called the green house effect.

Question 5.
What are the two forms of oxygen found in the atmosphere?
Answer:
The two forms of oxygen found in the atmosphere are:

1. Diatomic molecular form with chemical formula O₂.
2. Triatomic molecular form with chemical formula O₃ known as ozone.

Natural Resources NCERT Textbook Exercise

Question 1.
Why is the atmosphere essential for life?
Answer:
The atmosphere is essential for life because it maintains an appropriate climate for the sustenance of life by carrying out the following activities:

1. Atmosphere keeps the average temperature of the earth fairly constant during day time.
2. It prevents a sudden increase in temperature during day time.
3. It also slows down the escape of heat from the surface of the earth into outer space during night time.

Question 2.
Why is water essential for life?
Answer:
Water is essential for life because of the following reasons:

1. Most biological reactions occur when substances are dissolved in water. Thus, all cellular processes need water as a medium to take place.
2. Transportation of biological substances needs water as a medium.

Question 3.
How are living organisms dependent on the soil?
Or
Are organisms that live in water totally independent of soil as a resource?
Answer:
Almost all living organisms are dependent on soil. Some depends directly, while some depends indirectly. Plants need soil for getting support as well as nutrients to prepare their food. On the other hand, organisms depend on plants for food and other substances that are essential for life. Herbivores depend directly upon plants and carnivores depend upon animals, which in turn depend upon plants for food. This makes them depend on soil indirectly.

Organisms that live in water are not totally independent of soil as a resource. These organisms depend on aquatic plants for food and other substances. These aquatic plants in turn require minerals for their sustenance. These minerals are carried to water bodies from soil by rivers, rain water etc. Without the supply of minerals from the soil to the water bodies, it is impossible to imagine aquatic life.

Question 4.
You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather?
Answer:
The meteorological department of the government collects data on the elements of weather such as maximum and minimum temperatures, maximum and minimum humidity, rainfall, wind speed etc. They are able to study these elements using various instruments. The maximum and minimum temperature of a day is measured by a thermometer known as the maximum – minimum thermometer. Rain fall is measured by an instrument known as the rain gauge. Wind speed is measured by anemometers. There are various instruments used to measure humidity.

Question 5.
We know that many human activities lead to increasing levels of pollution of the air, water-bodies and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution?
Answer:
Yes. Isolating human activities to specific areas would help in reducing levels of pollution. For example, setting up of industries in isolated regions will control pollution to some extent. The pollution caused by these industries will not contaminate water resources, agriculture land, fertile land, etc.

Question 6.
Write a note on how forests influence the quality of our air, soil and water resources.
Answer:
Forests influence the quality of our air, soil, and water resources in various ways. Some of them are:

1. Forests balance the percentages of carbon dioxide and oxygen in the atmosphere. The increasing amount of carbon dioxide caused by human activities is balanced by a larger intake of carbon dioxide by plants during
   the process of photosynthesis. Simultaneously, a large amount of oxygen is released.
2. Forests prevent soil erosion. Roots of plants bind the soil tightly in a way that the surface of the soil cannot be eroded away by wind, water etc.
3. Forests help in the replenishment of water resources. During the process of transpiration, a huge amount of water vapour goes into the air and condenses to form clouds. These clouds cause rainfall that recharge water bodies.

Natural Resources Additional Questions

Natural Resources Multiple Choice Questions

Question 1.
The atmosphere of the earth is heated by radiations which are mainly ___________ .
(a) radiated by the sun
(b) re – radiated by land
(c) re – radiated by water
(d) re – radiated by land and water.
Answer:
(d) re – radiated by land and water.

Question 2.
If there were no atmosphere around the earth, the temperature of the earth will ___________ .
(a) increase.
(b) go on decreasing.
(c) increase during day and decrease during night.
(d) be unaffected.
Answer:
(c) increase during day and decrease during night.

Question 3.
What would happen, if all the oxygen present in the environment is converted to ozone?
(a) We will be protected more.
(b) It will become poisonous and kill living forms.
(c) Ozone is not stable, hence it will be toxic.
(d) It will help harmful sun radiations to reach earth and damage many life forms.
Answer:
(b) It will become poisonous and kill living forms.

Question 4.
One of the following factors does not lead to soil formation in nature ___________ .
(a) the sun
(b) water
(c) wind
(d) polythene bags.
Answer:
(d) polythene bags.

Question 5.
The two forms of oxygen found in the atmosphere are ___________ .
(a) water and ozone
(b) water and oxygen
(c) ozone and oxygen
(d) water and carbon – dioxide.
Answer:
(c) ozone and oxygen

Question 6.
The process of nitrogen – fixation by bacteria does not take place in the presence of ___________ .
(a) molecular form of hydrogen
(b) elemental form of oxygen
(c) water
(d) elemental form of nitrogen.
Answer:
(b) elemental form of oxygen

Question 7.
Rainfall patterns depend on ___________ .
(a) the underground water table.
(b) the number of water bodies in an area.
(c) the density pattern of human population in an area.
(d) the prevailing season in an area.
Answer:
(b) the number of water bodies in an area.

Question 8.
Among the given options, which one is not correct for the use of large amount of fertilisers and pesticides?
(a) They are eco – friendly.
(,b) They turn the fields barren after sometime.
(c) They adversely affect the useful component from the soil.
(d) They destroy the soil fertility.
Answer:
(a) They are eco – friendly.

Question 9.
The nitrogen molecules present in air can be converted into nitrates and nitrites by ___________ .
(a) a biological process of nitrogen fixing bacteria present in soil.
(b) a biological process of carbon fixing factor present in soil.
(c) any of the industries manufacturing nitrogenous compounds.
(d) the plants used as cereal crops in field.
Answer:
(a) a biological process of nitrogen fixing bacteria present in soil.

Question 10.
One of the following processes is not a step involved in the water – cycle operating in nature ___________ .
(a) evaporation
(b) transpiration
(c) precipitation
(d) photosynthesis.
Answer:
(d) photosynthesis.

Question 11.
The term “water – pollution” can be defined in several ways. Which of the following statements does not give the correct definition?
(a) The addition of undesirable substances to water – bodies.
(b) The removal of desirable substances from water – bodies.
(c) A change in pressure of the water bodies.
(d) A change in temperature of the water bodies.
Answer:
(c) A change in pressure of the water bodies.

Question 12.
Which of the following is not a green house gas?
(a) Methane
(b) Carbon dioxide
(c) Carbon monoxide
(d) Ammonia.
Answer:
(d) Ammonia.

Question 13.
Which step is not involved in the carbon – cycle?
(a) Photosynthesis
(b) Transpiration
(c) Respiration
(d) Burning of fossil fuels.
Answer:
(b) Transpiration

Question 14.
‘Ozone – hole’ means
(a) a large sized hole in the ozone layer.
(b) thinning of the ozone layer.
(c) small holes scattered in the ozone layer.
(d) thickening of ozone in the ozone layer.
Answer:
(b) thinning of the ozone layer.

Question 15.
Ozone – layer is getting depleted because of ___________ .
(a) excessive use of automobiles.
(b) excessive formation of industrial units.
(c) excessive use of man – made compounds containing both fluorine and chlorine.
(d) excessive deforestation.
Answer:
(c) excessive use of man – made compounds containing both fluorine and chlorine.

Question 16.
Which of the following is a recently originated problem of environment?
(a) Ozone layer depletion.
(b) Green house effect.
(c) Global warming.
(d) All of the above.
Answer:
(d) All of the above.

Question 17.
When we breathe in air, nitrogen also goes inside along with oxygen. What is the fete of this nitrogen?
(a) It moves along with oxygen into the cells.
(b) It comes out with the CO₂ during exhalation.
(c) It is absorbed only by the nasal cells.
(d) Nitrogen concentration is already more in the cells so it is not at all absorbed.
Answer:
(b) It comes out with the CO₂ during exhalation.

Question 18.
Top – soil contains the following ___________ .
(a) Humus and living organisms only.
(b) Humus and soil particles only.
(c) Humus, living organisms and plants.
(d) Humus, living organisms and soil particles.
Answer:
(d) Humus, living organisms and soil particles.

Question 19.
Choose the correct sequences ___________ .
(a) CO₂ in atmosphere → decomposers → organic carbon in animals → organic carbon in plants.
(b) CO₂  in atmosphere → organic carbon in plants → organic carbon in animals → inorganic carbon in soil.
(c) Inorganic carbonates in water → organic carbon in plants → organic carbon in animals → scavengers.
(d) Organic carbon in animals → decomposers → CO₂ in atmosphere → organic carbon in plants.
Answer:
(b) CO₂  in atmosphere → organic carbon in plants → organic carbon in animals → inorganic carbon in soil.

Question 20.
Major source of mineral in soil is the ___________ .
(a) parent rock from which soil is formed.
(b) plants.
(c) animals
(d) bacteria.
Answer:
(a) parent rock from which soil is formed.

Natural Resources Very Short Answer Type Questions

Question 1.
Name the factor responsible for rainfall patterns in India.
Answer:
North – East or South – West Monsoons.

Question 2.
Name any two organisms. Which play important role in nitrogen – fixation.
Answer:
Rhizobium and Nitrosomonas.

Question 3.
State the process and the form by which carbon enters in the living system.
Answer:
Respiration.

Question 4.
Write any one difference between oxygen and ozone.
Answer:
Oxygen is diatomic (O₂) whereas ozone is triatomic (O₃).

Question 5.
What is the temperature range on the surface of moon?
Answer:
-190°C to 110°C.

Natural Resources Short Answer Type Questions

Question 1.
Rivers from land, add minerals to sea water. Discuss, how?
Answer:
Water is wonderful solvent. As water flows over the rocks which containing soluble minerals and these mineral get dissolved in the water. Thus, Rivers from land, add minerals to sea water.

Question 2.
How can we prevent the loss of top soil?
Answer:
Loss of top soil can be prevented by stopping the falling of trees and excessive grazing by animals, growing more the vegetation cover.

Question 3.
How is the life of organisms living in water affected when water gets polluted?
Answer:
When water gets polluted, oxygen does not dissolves in water properly, which effects the aquatic organisms adversely. Also, due to undesirable chemicals and waste discharge from factories and households cause diseases to the aquatic organisms.

Question 4.
During summer, if you go near the lake, you feel relief from the heat, why?
Answer:
Due to evaporation of water, air nearby water bodies become cooler.

Question 5.
In coastal area, wind current moves from the sea towards the land during day; but during night it moves from land to the sea. Discuss the reason.
Answer:
Air above the land gets heated quickly during day and starts rising. This creates a region of low pressure as a result air over sea rushes into this area of low pressure. This movement of air from one region to the other creates winds. During night, as water cools down slowly, the air above water is warmer than the air on land. So, air moves from land to sea creating winds.

Question 6.
Following are a few organisms: (a) lichen (b) mosses (c) mango tree (d) cactus.
Which among the above can grow on stones; and also help in formation of soil? Write the mode of their action for making soil.
Answer:
Lichens and Mosses: (a) and (b).
Lichens and mosses release materials which break down the stones causes formation of soil.

Question 7.
Soil formation is done by both abiotic and biotic factors. List the names of these factors by classifying them as abiotic and biotic?
Answer:

1. Sun, water and wind are abiotic factors that forms soil.
2. Lichens, mosses and trees are biotic factors which form soil.

Question 8.
All the living organisms are basically made up of C, N, S, P, H and O. How do they enter the living forms? Discuss.
Answer:
In 1952, Stanley Miller along with Harold C. Urey designed an experiment to see how complex organic molecules might have formed under the conditions of early Earth. From long experimental method he developed organic matter from non living things. And, concluded that on specific environmental conditions non living things grow up to living form.

Question 9.
Why does the percentage of gases like oxygen, nitrogen and carbon dioxide remain almost the same in the atmosphere?
Answer:
The percentage of gases like oxygen, nitrogen and carbon dioxide remain almost the same in the atmosphere because cycling of these gases maintains consistency.

Question 10.
Why does moon have very cold and very hot temperature variations e.g., from -190°C to 110°C even though it is at the same distance from the sun as the earth is?
Answer:
No atmosphere exists on the moon thus, moon have very cold and very hot temperature variations e.g., from -190°C to 110°C even though it is at the same distance from the sun as the earth.

Question 11.
Why do people love to fly kites near the seashore?
Answer:
People love to fly kites near the seashore because air gets heated quickly during day and creates wind, which helps them to fly kites easily.

Question 12.
Why does Mathura refinery pose problems to the Taj Mahal?
Answer:
Polluted environment of Mathura due to refinery and other infrastructural facilities, industries etc. release toxic gases (like oxides of sulphur) which causes acid rain and corrode the marbles of Taj Mahal.

Question 13.
Why does water need conservation even though large oceans surround the land masses?
Answer:
Marine water is not useful for human and plant life directly. Hence fresh water need conservation. Uneven distribution of limited fresh water resources need conservation to ft ever rising demands.

Question 14.
There is mass mortality of fishes in a pond. What ma) the reasons?
Answer:

1. Thermal pollution
2. Addition of poisonous (mercury) compounds in water
3. Due to blockage of gills with any pollutant.

Natural Resources Long Answer Type Questions

Question 1.
How do fossil fuels cause air pollution?
Answer:
Most of air pollution of our environment is caused by the burning of fossil fuels, such as coal, oil natural gas and gasoline to produce electricity and power out vehicles. Carbon dioxide (CO₂) is a good indicator of how much fossil fuel is burnt and how much of other pollutants are emitted as a result. When fossil fuels are burnt, they release primarily nitrogen oxides into the atmosphere, which leads to the formation of smog and simultaneously acid rain.

The most common nitrogen – related compounds emitted into the air by human activities are collectively referred to as nitrogen oxides. Ammonia is also a harmful nitrogen compound emitted to the air, agricultural activities also contribute in air pollution but fossil fuels are most common reason for its increase day by day.

Question 2.
What are the causes of water pollution? Discuss how you can contribute in reducing water pollution.
Answer:
Domestic and industrial waste dump to water – bodies are main reason of water pollution. Industries produce huge amount of waste which contains toxic chemicals and pollutants which can cause air pollution and damage us and our environment. They contain pollutants such as lead, mercury, sulphur, asbestos, nitrates and many harmful chemicals.

Prevention of water pollution:
Stop dumping waste in water – bodies is the only way to get clear water resources. Do not throw chemicals, oils, paints and medicines down the sink drain or the toilet. If you use chemicals and pesticides for your gardens and farms, be mindful not to overuse pesticides and fertilizers. This will reduce run offs of the chemical into nearby water sources.

Natural Resources Higher Order Thinking Skills (HOTS)

Question 1.
If decomposers are removed completely from the Earth, what will be the result? Give reasons.
Answer:
Decomposers are the organisms that feed on dead and decaying matter present on the Earth. These organisms decompose the dead plants and animals returning the complex nutrients present in them in available form to the nature.

Question 2.
Why is nitrogen – fixing bacteria found only in the roots of leguminous plants?
Answer:
The roots of leguminous plants contains root nodules. These nodules provide shelter to nitrogen fixing bacteria like Rhizobium.

Natural Resources Value Based Questions

Question 1.
Manisha lived near industrial area in Delhi. During winters . she saw that the smoke has reduced the visibility in the area. Manisha requested the industries to fit in chimney the precipitators that would not release smoke particulates in air.

1. What is smog?
2. Name two disease caused due to air pollution.
3. What values are displayed by Manisha?

Answer:

1. When smoke is added to fog, smog is formed.
2. Air – pollution causes respiratory problem and allergies.
3. Manisha showed the value of self-awareness, and responsible behaviour.

Question 2.
Radha saw a child sleeping in a car parked with closed doors and glass rolled up in the open area on a sunny day near the market. She immediately raised an alarm and with the help of police, she got the window glass rolled down.

1. Why was it not safe to keep the doors with window glass rolled Up for a child inside the car?
2. Name two gases that can lead to above effect.
3. What values are exhibited by Radha?

Answer:

1. It was not safe for the child in the car with locked doors and window rolled up, the sunlight would result in the green house effect in the car. This would increase the temperature in the car and also result in the increase in CO₂ level which would lead to suffocation.
2. Carbon dioxide gas and methane gas can lead to green house effect.
3. Radha displayed the value of aware citizen and responsible behaviour.

MP Board Class 9th Science Solutions