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Karnataka Board Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.1

Question 1.
Which of the following figures lie on the same base and between the same • parallels. In such a case, write the common base and the two parallels.

Solution:
(i) ∆PDC and quadrilateral ABCD are on DC. They are in between AD || DC.

(ii) Parallelogram PQRS is on base SR. Trapezium SRMN is on base SR. But they are not in S between SR || PQ.

(iii) Parallelogram PQRS and ∆TRQ are on the base RQ and in between PS || RQ.

(iv) Parallelogram ABCD is on base BC and in between AD || BC.

∆RQP is on base PQ and in between AD || BC.
∴ Parallelogrm ABCD and ∆RQP are not on same base. But they are in between AD || DC.

(v) Here,
AD || DC,
AD || BC,
AP || DQ.
AD || PQ.

Parallelogrm ABCD and Parallelogram APQD are on same base AD and in between AD || BC.

(vi) Here, PQ || SR, PS || AD || BC || QR .
These are on the same base and in between a pair of parallel lines.

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.1.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.1

Question 1.
In quadrilateral ACBD, AC = AD, and AB bisect ∠A. Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?
Solution:
Data: In a quadrilateral ABCD,
AC = AD and AB bisect ∠A.

To Prove: ∆ABC ≅ ∆ABD
Proof: In ∆ABC and ∆ABD
AC = AD (Data)
∠CAB = ∠DAB .
∠A bisected
AB is common.
Here Side, angle, side rule is there.
∴ ∆ABC ≅ ∆ABD.

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(1) ∆ABD ≅ ∆BAC
(2) BD = AC
(3) ∠ABD = ∠BAC A
Solution:
Data: ABCD is a quadrilateral.
AD = BC and ∠DAB = ∠CBA.

To Prove:
(1) ∆ABD ≅ ∆BAC
(2) BD = AC
(3) ∠ABD = ∠BAC
Proof: (i) In ∆ABD and ∆BAC
AD = BC (Data)
∠DAB = ∠CBA (Data)
AB is common SAS postulate.
∴ ∆ABD ≅ ∆BAC.

(ii) ∆ABD ≅ ∆BAC.
∵ Corresponding sides are equal.
∴ BD = AC.

(iii) ∆ABD ≅ ∆BAC. (Proved)
Equal sides of Adjacent angles are equal.
As we have AD = BC,
The adjacent angle for AD is ABD
The adjacent angle for BC is BAC
∴∠ABD = ∠BAC.

Question 3.
AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Solution:
Data: AD and BC are equal perpendiculars to a line segment AB.

To Prove: CD Bisects AB.
Proof: In ∆CBO and ∆DAO,
BC = AD
∠CBO = ∠DAO = 90°
∠BOC = ∠AOD (Vertically opposite angles)
SAS postulate,
∴ ∆CBO = ∆DAO ∴ OA = OB
∴ CD bisects AB at ‘O’.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Solution:
Data: l and m are two parallel lines intersected by another pair of parallel lines p and q.
To Prove: ∆ABC ≅ ∆CDA
Proof: In AABC and ACDA,
∠ACB = ∠DAC [∵ Alternate angles.]
∠BAC = ∠ACD
AC is common.
A.S.A. postulate.
∆ABC ≅ ∆CDA

Question 5.
Line 1 is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that :

(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Solution:
Data: Line l is the bisector of an angle ∠A and B is any point on l.
BP and BQ are perpendiculars from B to the arms of ∠A.
To Prove:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Proof: In ∆APB and ∆AQB
∠APB = ∠AQB = 90°
∠PAB = ∠QAB (∵ l bisects ∠A).
AB is common.
∴ ∆APB ≅ ∆AQB (ASA postulate)
∴ BP = BQ.
B is equidistant from the arms of ∠A.

Question 6.
In AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
Data: AC = AE, AB = AD and ∠BAD = ∠EAC
To Prove : BC = DE
Proof: In ∆ABC and ∆EAD,
AB = AD (Data)
AC = AE (Data)
∆BAC = ∆EAD
(∵∠BAD + ∠DAC = AEAC + ∠DAC and ∠DAC = ∠DAC.)
Side – Angle – Side (SAS) Postulate
∴ ∆ABC = ∆EAD
∴ BC = DE.

Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that

(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
Solution:
Data: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD= ∠ABE and ∠EPA = ∠DPB.
To Prove :
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Proof : (i) In ∆DAP and ∆EBP,
AP = BP (∵ P is the mid-point of AB)
∠BAD = ∠ABE (Data)
∠APD = ∠BPE
∵ (∠EPA = ∠DPB Adding ∠EPD to both sides)
∠EPA + ∠EPD = ∠DPB + ∠EPD
∴ ∠APD = -BPE.
Now, Angle, Side, Angle postulate.
∴ ∆DAP ≅ ∆EBP.

(ii) As it is ∆DAP ≅ ∆EBP,
Three sides and three angles are equal to each other.
∴ AD = BE.

Question 8.
In the right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that :

(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle.
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac{1}{2}\) AB.
Solution:
Data: In a right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB.
C is joined to M and produced to a point D such that DM = CM.
Point D is joined to point B.
To Prove:
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle.
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac{1}{2}\) AB.
Proof: (i) In ∆AMC and ∆BMD,
BM = AM (∵ M is the mid-point of AB)
DM = CM (Data)
∠BMD = ∠AMC (Vertically opposite angles)
Now, Side, Angle, Side postulate.
∴ ∆AMC ≅ ∆BMD

(ii) ∆AMC ≅ ∆BMD (Proved)
Adding AMBC to both sides,
∆BMD + ∆MBC = ∆AMC + ∆MBC
∆DBC ≅ ∆ACB.
∴ AB Hypotenuse = DC Hypotenuse
∠ACB = ∠DBC = 90°
∴ ∠DBC is a right angle.

(iii) In ∆DBC and ∆ACB,
DB = AC
∵ ∆DMB ≅ DMC (Proved)
∠DBC = ∠ACB (Proved)
BC is Common.
Side, angle, side postulate.
∴ ∠DBC ≅ ∆ACB.

(iv) ∆DBC ≅ ∆ACB (proved)
Hypotenuse AB = Hypotenuse DC
\(\frac{1}{2}\) AB = \(\frac{1}{2}\) DC
\(\frac{1}{2}\) AB = CM
∴ CM = \(\frac{1}{2}\)AB.

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.5.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.5

Question 1.
Use suitable identities to find the following products :
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) \(\left(\mathrm{y}^{2}+\frac{3}{2}\right)\left(\mathrm{y}^{2}-\frac{3}{2}\right)\)
(v) (3 – 2x) (3 + 2x)
Solution:
(i) (x + 4) (x + 10)
Suitable identity is (x + a) (x + b) = x² + (a + b)x + ab
(x + 4) (x + 10) = x² + (4 + 10)x + (4 × 10)
= x² + 14x + 40

(ii) (x + 8) (x- 10)
Suitable identity is (x + a) (x + b) = x² + (a + b)x + ab
(x + 8) (x – 10)= (x)² + (8 – 10)x + (8 × -10)
= x² + (-2)x + (-80)
= x² – 2x – 80

(iii) (3x + 4) (3x – 5)
Suitable identity is (x + a) (x + b) = x² + (a + b)x + ab
(3x + 4) (3x – 5) = (3x)² + (+4 – 5)3x + (4 x -5)
= 9x² + (-1)3x + (-20)
= 9x² – 3x – 20

(iv) \(\left(\mathrm{y}^{2}+\frac{3}{2}\right)\left(\mathrm{y}^{2}-\frac{3}{2}\right)\)
Suitable identity is (a + b) (a – b) = a² – b²
\(\begin{aligned}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) &=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2} \\ &=y^{4}-\frac{9}{4} \end{aligned}\)

(v) (3 – 2x) (3 + 2x)
Suitable identity is (a + b) (a – b) = a² – b²
(3 – 2x) (3 + 2x) = (3)² – (2x)²
= 9 – 4x².

Question 2.
Evaluate the following products without mulitplying directly :
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) 103 × 107
= (100 + 3) (100 – 3)
As per Identity,
(a + b)(a – b) = a² – b²
(100 + 3)(100 – 3) = (100)² – (3)²
= 10000 – 9
= 9991

(ii) 95 × 96
= (100 – 5) (100 + 6)
As per Identity.
(x + a) (x + b) = x² + (a + b)x + ab
(100 – 5) (100 + 6) = (100)² + (-5 + 6)100 + (-5 × 6)
= 10000 + (1)100 + (-30)
= 10000 + 100 – 30
= 10100 – 30
= 10070

(iii) 104 × 96
= (100 + 4) (100 – 4)
As per Identity,
(a + b) (a – b) = a² – b²
(100 + 4) (100 – 4) = (100)² – (4)²
= 10000 – 16
= 9984

Question 3.
Factorise the following using appropriate identities :
(i) 9x² + 6xy + y²
(ii) 4y² – 4y + 1
(iii) \(x^{2}-\frac{y^{2}}{100}\)
Solution:
(i) 9x² + 6xy + y²
= (3x)² + 2(3x)(y) + (y)²
= (3x + y)² [∵ x² + 2xy + y² = (x + y)²]

(ii) 4y² – 4y + 1
= (2y)² – 2(2y)(1) + (1)²
= (2y – 1)² [∵ x² – 2xy + y² = (x – y)²]

(iii) \(x^{2}-\frac{y^{2}}{100}\)

Question 4.
Expand each of the following, using suitable identities :
(i) (x + 2y + 4z)²
(ii) (2x – y + z)²
(iii) (-2x + 3y + 2z)²
(iv) (3a – 7b – c)²
(v) (-2x + 5y – 3z)²
(vi) \(\left(\frac{1}{4} a-\frac{1}{2} b+1\right)^{2}\)
Solution:
(i) (x + 2y + 4z)²
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(x + 2y + 4z)² =(x)² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx

(ii) (2x – y + z)²
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx (2x – y + z)²= (2x)² + (-y)² + (z)² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx.

(iii) (-2x + 3y + 2z)²
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx (-2x + 3y + 2z)²
= (-2x)² + (3y)²2 + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x² + 9x² + 4z² – 12xy + 12yz – 8zx

(iv) (3a – 7b – c)²
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca (3a – 7b – c)² =(3a)² + (-7b)² + (-c)² + 2(3a)(-7b)
+ 2(-7b)(-c) + 2(-c)(3a) = 9a² + 49b² + c² – 42ab – 14bc – 6ca.

(v) (-2x + 5y – 3z)²
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx (-2x + 5y – 3z)² = (-2x)² + (5y)² + (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x) = 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

(vi) \(\left(\frac{1}{4} a-\frac{1}{2} b+1\right)^{2}\)
(a + b + c)² = a² + b² + c² +2ab + 2bc + 2ca

Question 5.
Factorise :
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz- 8xz
Solution:
(i) 4x² + 9y² + 16z²+ 12xy – 24yz – 16xz
= (2x)²+ (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
= (2x + 3y – 4z)².

(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz- 8xz

Question 6.
Write the following cubes in expanded form:
(i) (2x + 1)³
(ii) (2a – 3b)³
(iii) \(\left[\frac{3}{2} x+1\right]^{3}\)
(iv) \(\left[\mathrm{x}-\frac{2}{3} \mathrm{y}\right]^{3}\)
Solution:
(i) (2x + 1)³
As per Identity,
(a + b)³ = a³ + b³ + 3ab(a + b)
(2x + 1)³ = (2x)³ + (1)³ + 3(2x)(1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 1 + 12x² + 6x
(2x + 1)³ = 8x³ + 12x² + 6x + 1.

(ii) (2a – 3b)³
As per Identity,
(a – b)³ = a³ – b³ – 3ab(a – b)
(2a – 3b)³= (2a)³– (3b)³ – 3(2a)(3b)(2a-3b)
= 8a³ – 27b³ – 18ab(2a – 3b)
(2a – 3b)³= 8a³ – 27b³ – 36a²b + 54ab²

(iii) \(\left[\frac{3}{2} x+1\right]^{3}\)
As per Identity,
(a + b)³ = a³ – b³ – 3ab(a + b)

(iv) \(\left[\mathrm{x}-\frac{2}{3} \mathrm{y}\right]^{3}\)
As per Identity,
(a – b)³ = a³ – b³ – 3ab(a – b)

Question 7.
Evaluate the following using suitable identities :
(i) (99)³
(ii) (102)³
(iii) (998)³
Solution:
(i) (99)³ = (100 – 1)³
As per Identity,
(a – b)³ = a³ – b³ – 3ab(a – b)
(100 – 1)³ = (100)³ – (1)³ – 3(100)(1)(100- 1)
(99)³ = 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 970299
∴ (99)³ = 970299

(ii) (102)³ = (100 + 2)³
As per Identity,
(a + b)³ = a³ + b³ + 3ab(a + b)
(100 + 2)³ = (100)³ + (2)³ + 3( 100)(2)( 100 + 2)
= 1000000 + 8 + 600( 100 + 2)
= 1000000 + 8 + 60000 + 1200
(100 + 2)³ = 1061208
∴ (102)³= 1061208

(iii) (998)³ = (1000 – 2)³
As per Identity,
(a – b)³ = a³ – b³ – 3ab(a – b)
(1000 – 2)³ = (1000)³ – (2)3 – 3( 1000) (2)(1000 – 2)
= 1000000000 – 8 – 6000(1000 – 1)
= 1000000000 – 8 – 6000000 + 6000
= 994005992
∴ (998)³ = 994005992

Question 8.
Factorise each of the following :
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27- 125a³– 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108ab²
(v) \(27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\)
Solution:
(i) 8a³ + b³ + 12a²b + 6ab²
= (2a)³ + (b)³ + 3(2a)(b)(2a + b)
= (2a + b)³ ∵ a³ + b³ + 3ab(a + b) = (a + b)³

(ii) 8a³ – b³ – 12ab² + 6ab³
= (2a)³ + (-b)³ + 3(2a)(-b)(2a – b)
= (2a – b)³ ∵ a³ – b³ – 3ab(a – b) = (a – b)³

(iii) 27 – 125a³ – 135a + 225a²
OR 125a³ – 225a² + 135a – 27
= (5a)³ + (-3)³ + 3(5a)(-3)(5a – 3)
= (5a – 3)³ ∵ a³ – b³ – 3ab(a – b) = (a – b)³

(iv) 64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ + (-3b)³ + 3(4a)(-3b)(4a – 3b)
= (4a-3b)³ ∵ a³ – b³ – 3ab(a – b) = (a – b)³

Question 9.
Verify :
(i) x³ + y³ = (x + y) (x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
(i) x³ + y³ = (x + y)(x² – xy + y²)
L.H.S. = x³ + y³
R.H.S. = (x + y)(x² – xy + y²)
= x(x² – xy + y²) + y(x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
R.H.S. = x³ + y³
∴ L.H.S. = R.H.S.
∴ x³ + y³ = (x + y)(x² – xy + y²)

(ii) x³ – y³ = (x – y)(x² + xy + y²)
L.H.S. = x³ – y³
R.H.S. = (x – y)(x² + xy + y²)
= x(x² + xy + y²) – y(x² + xy + y²)
= x³ + x²y + xy² – x²y – xy³ – y³
R.H.S. = x³ – y³
∴L.H.S. = R.H.S.
∴ x³ – y³ = (x – y)(x² + xy + y²)

Question 10.
Factorise each of the following :
(i) 27y³ + 125z³
(ii) 64m³ – 343n³
[Hint: See Question 9).
Solution:
(i) 27y³ + 125z³
= (3y)³ + (5z)³
= (3y + 5z) {(3y)² – (3y)(5z) + (5z)²}
= (3y + 5z) (9y² – 15yz + 25z²)

(ii) 64m³ – 343n³
= (4m)³ – (7n)³
= (4m – 7n) {(4m)² + (4m) (7n) + (7n)²}
= (4m – 7n) (17m² + 28mn + 49n²)

Question 11.
Factorise : 27x³ + y³ + z³ – 9xyz
Solution:
27x³ + y³ + z³ – 9xyz
= (3x)³ + (y)3³ + (z)³ – 3(3x)(y)(z)
= (3x + y + z) {(3x)² + (y)² + (z)² – (3x)(y) – (y)(z) – (z)(3x)}
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3zx)

Question 12.
Verify that x³ + y³ + z³ – 3xyz = \(\frac{1}{2}\) (x + y + z)[(x – y)² + (y – z)² + (z – x)²]
Solution:
L.H.S.= x³ + y³+ z³ – 3xyz
R.H.S. = \(\frac{1}{2}\) (x+y+z) [(x – y)² + (y – z)² + (z – x)²]
= \(\frac{1}{2}\) (x+y+z) [x² 2xy + y²+ y² – 2yz + z² + z² -2zx + x²]
= \(\frac{1}{2}\) (x+y+z) [2x²+ 2y²+ 2z²– 2xy – 2yz – 2zx]
= \(\frac{1}{2}\) (x+y+z) × (x² + y² + z² – xy – yz – zx)
= (x + y + z)(x² + y² + z² – xy – yz – zx)
R.H.S.= x³ + y³ + z³ – 3xyz
∴ L.H.S. = R.H.S.
∴ x³ + y³ + z³ – 3xyz = \(\frac{1}{2}\) (x + y + z)[(x – y)² + (y – z)² + (z – x)³]

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Solution:
x + y + z = 0
x + y = -z
by cubing on both sides,
(x + y)³ = (-z)3
x³+ y³ + 3xy(x + y) = -z³
x³ + y³ + 3xy (-z) = -z³
(∵ x + y =-z data given)
x³ + y³ – 3xyz = -z³
∴ x³ + y³ + z³ = 3xyz.

Question 14.
Without actually calculating the cubes, find the value of each of the following:
(i) (-12)³ + (7)³ + (5)³
(ii) (28)³ + (-15)³ + (-13)³
Solution:
(i) (-12)³ + (7)³ + (5)³
Let -12 = x, 7 = y, 5 = z, then
x³ + y³ + z³ = 3xyz.
∴ (-12)³ + (7)³ + (5)³
= 3(-12)(7)(5)
= -36 × 35
= -1260.

(ii) (28)³ + (-15)³ + (-13)³
Let x = 28, y = -15, z = -13, then
x³ + y³ + z³ = 3xyz
∴ (28)³ + (-15)³ + (-13)³
= 3(28)(- 15)(-13)
= 84 × 195
= 16380

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :
(i) Area: 25a² – 35a +12
(ii) Area : 35y² + 13y – 12
Solution:
(i) Area: 25a² – 35a + 12
= 25a² – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
= (5a -4) (5a – 3)
length × breadth = Area of rectangle
(5a – 4) (5a – 3) = 25a² – 35a + 12
∴ Length = (5a – 4)
Breadth = (5a – 3)

(ii) Area : 35y² + 13y – 12
Area of Rectangle – Length × Breadth
35y² + 13y – 12
35y² + 28y – 15y
7y(5y + 4) – 3(5y + 4)
(7y – 3) (5y + 4)
∴ Length = (7y – 3)
Breadth = (5y + 4)

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below :
(i) Volume: 3x² – 12x
(ii) Volume: 12ky² + 8ky – 20k
Solution:
(i) length × breadth × height = Volume of Cuboid
= 3x² – 12x
= 3x² – 12x
= 3x(x – 4)
= 3 × x × (x – 4)
l × b × h
∴ Length = 3
Breadth = x
Height = (x – 4)

(ii) length × breadth × height = Volume of Cuboid
= 12ky² + 8ky – 20k
12ky² + 8ky – 20k
= 4k(3y² + 2y – 5)
= 4k (3y² + 5y – 3y – 5)
= 4k (y(3y + 5) – 1 (3y + 5))
= 4k × (3y + 5) × (y – 1)
= l × b × h
∴ Length = 4k
Breadth = (3y + 5)
Height = (y – 1)

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.5, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Science Chapter 11 Work and Energy

KSEEB Solutions Class 9 Science Chapter 11 Intext Questions

Question 1.
A force of 7N acts on an object. The displacement is, say 8m, in the direction of the force. Let us take it that they force acts on the object through the displacement. What is the work done in this case?
Answer:
Work done on an object = 7N × 8m = 56 Nm or 56 J.

Question 2.
When do we say that work is done?
Answer:
Two conditions need to be satisfied for work to be done:

1. a force should act on an object and
2. the object must be displaced.

Question 3.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
Let a constant force F act on an object. Let the object be displaced through a distance S in the direction of the force. Let W be work done. We define work to be equal to the product of the force and displacement.
Work done = force × displacement W = F × S.

Question 4.
Define 1 J of work.
Answer:
1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

Question 5.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?
Answer:
Force exerted, F = 140N.
Displacement, S = 15m
Work done in ploughing the field
W = F × S
= 140 × 15
= 2100J
= 2.1 × 10³ J.

Question 6.
What is the kinetic energy of an object?
Answer:
Objects in motion possess energy, we call this energy kinetic energy.

Question 7.
Write an expression for the kinetic energy of an object.
Answer:
KE = \(\frac{1}{2}\) mv².

Question 8.
The kinetic energy of an object of mass, m moving with a velocity of 5 ms_-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
Kinetic energy, k = \(\frac{1}{2}\) mv².
Where m = mass of the object
V = velocity of the object.
Here mass (m) is the same in both cases (∵ object is same)
\(\frac{\mathbf{k}_{1}}{\mathbf{k}_{2}}=\left(\frac{\mathbf{v}_{1}}{\mathbf{v}_{2}}\right)^{2}\)
Initial kinetic energy k₁ = 25J
Initial velocity V₂ = 5 ms^-1
New kinetic Energy K₂ = ?
New velocity v₂ = 3v₁ = 3 × 5 = 10 ms^-1

∴ When velocity is increased three times its K.E. is 225J.

Question 9.
What is power?
Answer:
Power is defined as the rate of doing work or the rate of transfer of energy w
P = \(\frac{w}{t}\)

Question 10.
Define 1 watt of power.
Answer:
Power is 1 W when the rate of consumption of energy is 1 JS^-1.

Question 11.
A lamp consumes 1000 J of electrical energy in 10s. What is its power?
Answer:
Power = 1000 J Work w = 1000J
Time = 10 s
Power of lamp, P = w/t
= 1000/10 = 100W
∴ Power of a lamp = 100 KW.

Question 12.
Define average power.
Answer:
We obtain average power by dividing the total energy consumed by the total time taken.

KSEEB Solutions for Class 9 Science Chapter 11 Textbook Exercises

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’

1. Suma is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Food grains are getting dried in the sun.
7. A sailboat is moving due to wind energy.

Answer:

1. Work is being done by Seema because she displaces the water by applying the force.
2. No work is being done by the donkey because the direction of force i.e the load is vertically downward and displacement is along the horizontal. If displacement and force are perpendiculars then no work is done.
3. Work is done because the windmill is lifting the water i.e., it is changing the position of water.
4. No work is done because there are no force and displacement.
5. Work is done because the engine is changing the position of the train.
6. No work’ is done because there is no force and no displacement.
7. Work is done because the force acting on the boat is moving it.

Question 2.
An object was thrown at a certain angle to the ground moves in a curved path and falls back to the ground the initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Work done by the force of gravity, W = mgh.
Where h = difference in height of initial and final positions of the object.
According to the question, the initial and final positions of the object lie in the same horizontal line. So h = 0.
∴ Work done W = mg × 0 = 0

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
In the case given in the question, the battery has chemical energy which is converted into energy. Electric energy provided to the bulb further converted into light energy.

Question 4.
A certain force acting on a 20 kg mass changes its velocity from 5 ms_-1 to 2 ms_-1. Calculate the work done by the force.
Answer:
Mass, m = 20 kg.
Initial velocity, u = 5 ms^-1
Final velocity v = 2ms^-1
Work done by the force = change in kinetic energy
= Final kinetic energy – Initial kinetic energy

Question 5.
A mass of 10kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravitational force W = mgh
Where h = Difference in the heights of initial and final positions of the object.
Here both the initial and final positions are on the same horizontal line.
So there is no difference in height i.e., h = 0.
∴ work done W = mg × 0 = 0.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
The total mechanical energy remains constant / as P.E. of the freely falling object decreases. Its kinetic energy and the kinetic energy increases on account of an increases its velocity) the law of conservation of energy is not violated.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:

1. Muscular energy into kinetic energy
2. the kinetic energy of the rotation of the which into kinetic energy of the bicycle.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
When we push a huge rock and fall to move it. The energy spent in doing so is absorbed by the.rock. This energy is converted into potential energy of the configuration of the rock which results in its deformation. However t his deformation is not visible on account of the huge size of the rock.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is this in Joules?
Answer:
Energy consumed w = 250 units
= 250 kwh
= 250 × 1000 W × 3600 S
= 250 × 1000 J/S × 3600 S
= 9 × 10⁸ J.

Question 10.
An object of mass 40 kg is raised to a height of 5m above the ground what is its potential energy? If the object is allowed to fall, find its kinetic energy when it is halfway down.
Answer:
Mass m = 40 kg.
height h = 5 m.
Potential energy PE = mgh = 40 × 9.8 × 5 = 1960J.
KE at half way down = PE at halfway down = \(\mathrm{mg} \frac{\mathrm{h}}{2}\)
= 40 × 9.8 × \(\frac{5}{2}\) = 980 J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:

The satellite round the earth moves in a circular orbit. Here the force of gravity acts towards the centre of the earth and displacement of the satellite is along the tangent of the circular path that means therefore and displacement are perpendicular to each other.
So, work done, W = F.S cosθ
= F × S cos 90°
= F × S × 0
= 0
That is, no work is done by the force of gravity.

Question 12.
Can there be displacement of an object in the absence of any force acting on it? think. Discuss this question with your friends and teacher.
Answer:
If an object moves with a constant velocity (i.e, there is no acceleration) then no force acts on it. As the object is moving ie it is displaced from one position to another position.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
The person has no movement ie his displacement is zero. So the person had done no work (∵ work is done only when the object is displaced).

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer:
Time = 10 h = 10 × 60 min
= 10 × 60 × 60 S

Energy = power × time
= 1500 × 10 × 60 × 60
= 5.4 × 10⁷ J.

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate why does the bob eventually comes to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
When the pendulum oscillates in the air, the air friction opposes its motion. So some part of the kinetic energy of the pendulum is used to overcome this friction. With the passage of time, the kinetic energy of the pendulum goes on decreasing and finally becomes zero. The kinetic energy of the pendulum is transferred to the atmosphere. So energy is being transferred ie it is converted into one form to another. So here is no violation of the law of conservation of energy.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer:
The work done on the object to bring the object to rest
= change in kinetic energy
= Final kinetic energy – Initial Kinetic energy
Here final kinetic energy is zero because the object is brought to rest.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at the velocity of 60km/h?
Answer:
Mass, m = 1500 kg.
Initial velocity, u = 60 kmh^-1
= 60 × \(\frac{5}{18}\) = 16.67 ms^-1
Final velocity, v = 0
(∵ the car comes to rest)
Work done to stop the car = change in kinetic energy
= Final kinetic energy – Initial Kinetic energy

Question 18.
In each of the following a force, F is acting on an object of mass M. The direction of displacement is from west to east shown by the longer arrow. Observe the diagram carefully and state whether the work done by the force is negative, positive or zero.

Answer:
Cas i) The force and displacement are perpendicular to each other. So S = 90°
Work done = FScos θ°
= FScos 90°
= FS × 0 = 0
(∵ cos90° = 0)
Cas ii) The force and displacement are in the same direction so θ = 0°
Work done = FScos θ°
= FScos 90°
= FS × 1 = FS
(∵ cos θ° = 1)
That is the work done is positive.
Cas iii) The force and displacement in opposite direction θ = 180°
Workdone = FScos θ°
= FScos 180°
= FS × -1 = FS
(∵ cos 180° = -1)
That is workdone is negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Yes, I agree with Soni, the acceleration of an object can be zero even when several forces are acting on it if the resultant of all the forces acting is zero.

Question 20.
Find the energy in kWh consumed in 10 hours by four devices of power 500 w each.
Answer:
Total power p = 500w × 4 = 2000w.
Time, t = 10h
Energy = p × t.
= 2000w × 10h
= 2kwh × 10h = 20kwh.
∴ Energy = 20kwh.

Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
P.E. of the configuration of the body and the ground (the body may be deformed and the ground may at the place of collision).
This process energy in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy charges is called dissipation of energy.

KSEEB Solutions for Class 9 Science Chapter 11 Additional Questions and Answer

Fill in the blanks:
Question 1.
Work done = ___________
Answer:
force, displacement

Question 2.
An object having the capability to do work is said to possess __________
Answer:
energy.

Question 3.
__________ is the energy possessed by an object due to its motion.
Answer:
Kinetic energy

Question 4.
P.E
Answer:
mgh.

Question 5.
Larger unit of energy is called
Answer:
kilojoule.

Answer the following questions:

Question 1.
Can there be displacement of an object in the absence of any force acting on it?
Answer :
In the absence of any force on the object i.e., F = 0, ma, (as F = ma) since m ≠ 0 and a = 0. In such a case the object is either at rest or in a state of uniform motion in a straight line in the latter case there is a displacement . of the object without any force acting on it.

Question 2.
How does a bullet pierce a target?
Answer :
A bullet moves with large velocity and as such possesses a lot of kinetic energy the work in piercing the target is derived from the kinetic energy of the bullet.

Question 3.
Why do some engines require fuels like petrol and diesel?
Answer :
Internal combustion heat engines use the chemical energy of fossil fuel (petrol and diesel) for their operation. These engines first convert the chemical energy of the fuels into heat energy. Which is later on converted into mechanical energy.

Question 4.
Calculate the work done by a body. By the force of 5 N makes to move through a distance of 12 m.
Answer :
W = F x s = 5 x 12 = 60 joules

Question 5.
When force 6 N applied on a wall, the wall remains in the same position, calculate the work done.
Answer :
W = F x s
W – 6 x s = 6 x 0 = 0 Joules No work is done on the body

Question 6.
Electricity is the most convenient form of energy. Why?
Answer :
It can be converted into other forms of energy easily.
It can be produced by different means. It is ecofriendly

Question 7.
Calculate the power experienced by a source that can do work of 50 joules in 5 seconds.
Answer :
Power = \($\frac{\text { Workdone }}{\text { Time taken }}$\)
\(=\frac{50}{5} \frac{\text { Joules }}{\text { Second }}\) = 10 J/s = 10 watts

KSEEB Solutions for Class 9 Science

 

Karnataka Board Class 9 Science Chapter 12 Sound

KSEEB Solutions Class 9 Science Chapter 12 Intext Questions

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
Sound is produced by vibrating objects. When an object vibrates, it sets the particles of the medium around it vibrating. The particles do not travel all the way from the vibrating object in the ear. A particle of the medium in contact with the vibrating objects is first displaced from its equilibrium position.

It then exerts a force on the adjacent particle. As a result of which the adjacent particle. As a result of which the adjacent particle get displaced from its position at rest. After displacing the particle comes back to its original position. This process continues in the medium in the sound reaches our ears.

Question 2.
Explain how sound is produced by your school bell?
Answer:
The bell produces the sound when it is struck by air hammer when the bell is struck by a hammer it starts vibrating. Since the vibrating objects produce sound. So the bell produces sound.

Question 3.
Why are sound waves called mechanical waves?
Answer:
Sound waves are called mechanical waves because they are produced by the motion of particle of a medium. Hence sound waves are called mechanical waves.

Question 4.
Suppose you and your Mend are on the moon. Will you be able to hear any sound produced by your Mend?
Answer:
I will not be able to hear any sound produced by my friend because the sound waves require some material medium like air to travel. There is no atmosphere or air on the moon. So the sound produced by my friend will not reach me and I will not be able to hear.

Question 5.
Which wave property determines
(a) loudness
(b) pitch?
Answer:
(a) The loudness or softness of a sound is determined by its amplitude.
(b) Pitch of the sound is determined by its frequency.

Question 6.
Guess which sound has a higher pitch: guitar or car horn?
Answer:
Guitar has a higher pitch

Question 7.
What are the wavelength, frequency, time period, and amplitude of a sound wave?
Answer:

1. Wavelength: The distance travelled by the wave during one complete oscillation is called wavelength of the wave.
2. Frequency: The number of oscillations made by the wave in one second is called the frequency of the wave.
3. Time period: The time taken by the wave to complete one oscillation is called the amplitude of the wave.
4. amplitude: The maximum displacement of the wave from its mean or equilibrium position is called the amplitude of the wave.

Question 8.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
The relation between wavelength (λ), frequency (ν) and speed of the wave (v) is
υ = λv.

Question 9.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440ms-1 in a given medium.
Answer:
Frequency v = 220 Hz,
Speed u = 440 ms^-1 Wavelength, l = ?
From the relation υ = vλ
440 = 220A.
∴ λ = \(\frac{440}{220}\) = 2.0 m

Question 10.
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:
The time interval between two successive compressions or rarefactions is equal to the time period of the wave.
∴ Required time interval = time period
= \(\frac{1}{\text { Frequency }}=\frac{1}{500}\) = 0.002 S
= 2 × 10^-3 S = 2ms.

Question 11.
Distinguish between loudness and intensity of sound.
Answer:

Loudness	Intensity
Loudness is a measure of the response of the ear to the sound.	The amount of sound energy passing each second through unit area is called the intensity of sound.

Question 12.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
In these three media, sound travel in iron fastest at a particular temperature.

Question 13.
An echo returned in 3 S. What is the distance of the reflecting surface from the source, given that the speed of sound is 342ms?
Answer:
Echo returns in 3S. Therefore time taken by sound to travel from the source to the reflecting surface
t = \(\frac{3}{2}=\) = 1.5 S
Speed v = 342ms^-1
∴ Distance of reflecting surface from the source = Speed × Time
= 342 × 1.5 = 513 m

Question 14.
Why are the ceilings of concert halls curved?
Answer:
Generally, the ceilings of concert halls, are curved so that sound after reflection reaches all corners of the hall.

Question 15.
What is the audible range of the average human ear?
Answer:
The audible range of sound for human beings extends from about 20Hz to 20000Hz.

Question 16.
What is the range of frequencies associated with
(a) Infrasound
(b) Ultrasound
Answer:
(a) Infrasound: Sounds of frequencies below 20 Hz are called infrasonic sound or infrasound.
(b) Ultrasound: Frequencies higher than 20KHz are called ultrasound or ultrasonic sound.

Question 17.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02s. If the speed of sound in saltwater is 1531 m/s how far away is the cliff?
Answer:
Speed of sound in seawater, v = 1531 m/s.
Distance travelled ultrasonic sound = 2 × depth sea = 2d.
Here d means depth of sea
2d = Speed of sound × time
= 1531 m/s × 1.02
= 1561.62 m
d = 1561.62/2 = 780.81 m.
Cliff is 780.81 m away from the sea.

KSEEB Solutions for Class 9 Science Chapter 12 Textbook Exercises

Question 1.
What is sound and how is it produced?
Answer:
Sound is a form of energy – the sound is produced due to the vibration of different objects.

Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of the sound.
Answer:
Air is the most common medium through which sound travels. When a vibrating object moves forward, it pushes and compresses the air in front of

it creating a region of high pressure. This region is called a compression (c) as shown in Fig. This compression starts to move away from the vibrating object when the vibrating object moves backward, it creates a region of low pressure called rarefaction (R) as shown in Fig. As the object moves back and forth rapidly, a series of compressions and rarefactions is created in the air.

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Answer:

Experiment: Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar. The bell jar is connected to a vacuum pump as shown in figure. If we press the switch we will be able to hear the bell. Now start the vacuum pump. When the air in the jar is pumped out gradually, the sound becomes painter, although the same current is passing through the bell. After sometime when less air is left inside the bell jar we will near a feeble sound. If the air is removed completely we can’t hear the bell. From this experiment, we conclude that sound needs a material medium for its propagation.

Question 4.
Why is a sound wave called a longitudinal wave?
Answer:
Sound waves are called longitudinal waves because the particles do not move from one place to another but they simply oscillate back and forth about their position of rest.

Question 5.
Which characteristic of the sound helps you to identify your Mend by his voice while sitting with others in a dark room?
Answer:
The quality or timbre of a sound is the characteristic of the sound which helps to us to identify our friend by his voice while sitting with others in a dark room.

Question 6.
Plash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?
Answer:
Thunder is heard a few seconds later after the flash because the speed of light in the atmosphere for air is 3 × 10⁸ ms^-2 which is very high, as compared to the speed of sound which is only 330 ms^-1. So the sound of thunder reaches us later than the flash.

Question 7.
A person has a hearing range from 20Hz to 20KHz. 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^-2).
Answer:
The relation between speed (v), wavelength (X) and frequency (v) of a wave, we have

Question 8.
Two children are at opposite ends of an alluminium rod. One strikes the end of the rod with a stone. Find the ratio or times taken by the sound wave in air and in alluminium to reach the second child?
Answer:
Let l = length of the rod.
Time taken by the sound to travel through the alluminium rod

Question 9.
The frequency of a source of sound is 100Hz. How many times does it vibrate in a minute?
Answer:
Frequency = 100Hz.
From the definition of frequency, we can say that Number of oscillations in 1 min or 60 Sec. = 100 × 60
= 6000 = 6 × 10³.

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
Yes. Sound waves follow the same laws of reflection as light does. The directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface at the point of incidence and three are in the same 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:
The time taken by echo to be heard
t = \(\frac{2 \mathrm{d}}{v}\)
where d = distance between the reflecting surface and source of the sound and v = speed of sound in air. As we know the speed of sound increases with an increase in temperature. So on a hotter day speed of sound will be higher, so the time after which echo is heard will decrease. If the time taken by the reflected sound is less than 0.15 after the production of the original sound, then echo is not heard.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
(1) Megaphone (2) Hearing aid are two practical applications of reflection of sound waves.

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= 10ms^-2 and speed of sound = 340ms^-1)
Answer:
Time after which splash is heard = Time taken by the stone to reach the pond + Time taken by the splash sound to reach the top of tower.
(i) For the time taken by the stone to reach the pond.
Here u = 0 (∵ stone is dropped from rest)
From equation of motion against gravity.

(ii) Time taken by splash sound to reach the top of lower

∴ Time after which splash is heard = 10 + 1.47 = 11.47 S.

Question 14.
A sound wave travels at a speed of 339ms^-1. If its wavelength is 1.5 cm. What is the frequency of the wave? Will it be audible?
Answer:
Speed = 339 ms^-1
Wavelength λ = 1.5 cm. = 1.5 × 10²

= 22600 Hz.
This frequency is greater than 20000 Hz. So it will not be audible. For an audible range of the human ear is 20Hz to 20000 Hz.

Question 15.
What is reverberation? How can it be reduced?
Answer:
The phenomenon of propagation of original sound due to the multiple reflections of sound waves even after the source of sound and stops producing sound is called reverberation. It can be reduced by covering the roofs and walls of the hall by absorbing materials.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
Loudness of a sound is a subjective quantity. It causes an unpleasant effect on air. Loudness depends on

1. the amplitude of the vibrating body and
2. The sensitivity of the human ear.

Question 17.
Explain how bats use ultrasound to catch a prey.
Answer:
Bats emit ultrasonic waves and search their food. Ultrasonic waves having higher pitch reflects from prey and reaches ears of the Bat. Nature of reflection indicates to Bat regarding position and type of prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to high frequency, the particles of dust, grease, and dirt get detached and drop out. The objects thus get thoroughly cleaned.

Question 19.
Explain the working and application of a sonar.
Answer:
Working: Sonar consists of a transmitter and a detector and is installed in a boat or a ship as shown in Fig. The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector.

The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the soundwave can be calculated by knowing the speed of sound in water and the time interval between the transmission and reception of the ultrasound. Let the time interval between transmission and reception of ultrasound signal be t and the speed of sound through seawater be v. The total distance, 2d travelled by the ultrasound is then, 2d = vxt.

Applications: The sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer:
Time taken by submarine to sends out a signal and receives an echo,
t = \(\frac{5}{2}\) = 2.5s.
distance of the object from the submarine = 3625m.
∴ Speed of sound in water,

Question 21.
Explain how defects in a metal block can be detected using ultrasound.
Answer:

The cracks or holes inside the metal blocks, which are invisible from outside reduces the strength of the structure Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect.

Question 22.
Explain how the human ear works.
Answer:
Working on the human ear. The outer ear is called ‘pinna’. It collects the sound from the surroundings. The collected sound passes through the auditory canal. A the end of the auditory canal there is a thin membrane called the eardrum or tympanic membrane. When compression of the medium reaches the eardrum the pressure on the outside of the membrane increases and forces the eardrum inward.

Similarly, the eardrum moves outward when a rarefaction reaches it. In this way, the eardrum vibrates. The vibrations are amplified several times by three bones (the hammer, anvil, and stirrup) in the middle ear. The middle ear transmits the amplified pressure variations received from the sound wave to the inner ear. In the inner ear, the pressure variations are turned into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve and the brain interprets them as sound.

KSEEB Solutions for Class 9 Science Chapter 12 Additional Questions and Answer

Answer the following questions:
Question 1.
What is a wave?
Answer:
A wave is a disturbance that moves through a medium -when the particles of the medium set neighbouring particles into motion.

Question 2.
What type are sound waves?
Answer:
Sound waves are mechanical waves (longitudinal waves).

Question 3.
What is S.I. unit of a sound wave?
Answer:
Hertz (Hz).

Question 4.
What do you mean by tone?
Answer:
A sound of a single frequency is called a tone.

Question 5.
What is the intensity of sound?
Answer:
The amount of sound energy passing each second through unit area is called the intensity of sound.1

Question 6.
Expand SONAR.
Answer:
SOUND NAVIGATION AND RANGING.

KSEEB Solutions for Class 9 Science

 

Karnataka Board Class 9 Science Chapter 13 Why Do We Fall ill

KSEEB Solutions Class 9 Science Chapter 13 Intext Questions

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

1. Public cleanliness
2. Good quality of food and water are two conditions which are essential for good health.

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

1. By doing exercises and walking, our body is healthy
2. We should consume balanced food and follow health tips.

These are the conditions which are essential for being free of diseases.

Question 3.
Are the answers to the above questions necessarily the same or different? Why?
Answer:
Answer of Both questions is alike. Both are related to maintaining good health.

Question 4.
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:
Three reasons why we would think that wear sick and ought to see a doctor are:

1. There is a chance of getting sick due to microbial activity.
2. Due to the deficiency of vitamins, minerals (nutrients) we get diseases.
3. Due to genetic differences also we get diseases.

Question 5.
In which of the following case do you think the long-term effects on your health are likely to be most unpleasant?
i) if you get jaundice
ii) if you get lice.
iii) if you get acne. Why?
Answer:
In these three cases, if we get jaundice is most unpleasant. Because jaundice virus is present in liver cells. Liver is the largest and important organ of our body. If it is damaged, leads to more danger to our body.

Question 6.
Why are we normally advised to take bland and nourishing food when we are sick?
Answer:
When we are sick, we have a weakness. If we consume hard food, it is difficult to digest it. Hence we should consume bland food. Nourishing food is also essential as they provide nutrients for us. Therefore we should take bland and nourishing food when we are sick.

Question 7.
What are the different means by which infectious diseases are spread?
Answer:

1. Many pathogenic microbes spread from an infected person to others by different modes. These are called infectious diseases.
2. Some of the microbes spread through air. These are called airborne infections. Eg: Common cold, pneumonia, and tuberculosis etc.
3. Some of the diseases spread through water. If we drink polluted water which contains germs we get diseases. This is called water-borne infections. Eg: Cholera, Typhoid, Jaundice etc.
4. Some of the diseases spread through sexual contact. Eg: Syphilis or AIDS.
5. Some of the diseases spread through animals bite. Eg: Rabies.

Question 8.
What precautions can you take in your school to reduce the incidence of infectious diseases?
Answer:

1. We should maintain the school environment clean.
2. There should be a good drinking facility.
3. There must be a Lavatory system.
4. Children should be provided with vaccination periodically. Nutritious food should be provided to children.

Question 9.
What is an immunisation?
Answer:
We have cells that specialise in killing infecting microbes. These cells go into action each time infecting microbes enter the body. If they are successful, we do not actually come down with any disease. The immune cells manage to kill off the infection long before it assumes major proportions. As we noted earlier, if the number of the infecting microbes is controlled, the manifestations of the disease will be minor. This is called the immunisation.

Question 10.
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:
Various vaccines are supplied from the health centre in our locality.
Eg: There are vaccines which prevent diseases like tetanus. Diphtheria, whooping cough, measles, polio, and other diseases.
Some of the diseases which cause major health problems in our area are cholera, Gastro enterprise, jaundice, measles etc.

KSEEB Solutions for Class 9 Science Chapter 13 Textbook Exercises

Question 1.
How many times did you fall ill in the last one year? What were the illnesses?
a) Think of one change you could make in your habits in order to avoid any of/most of the above illnesses.
b) Think of one change you would wish for in your surroundings in order to avoid any of / most of the above illnesses.
Answer:
In the last one year, I had illness for two times. I was suffering from cold, dysentery and skin disease.
(a) As per the doctor’s advice we should take injections and medicines and nutritious food.
(b) Our surrounding should be clean, neat and tidy. We should drink filtered water, we must follow health tips such as exercises and walking etc.

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:
Doctor. Nurse and health worker will take precautionary measures to avoid getting sick herself/himself. When they suffer from sick they must take injection and medicines concerning to diseases along with nutritious food.

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 cold, cholera, cough are the three diseases, common to all people, we should prevent as follows:

1. Public hygiene is one basic key to the prevention of infectious diseases.
2. We can prevent such diseases by our immunity power.
3. We should take proper vaccinations and food meant for different diseases.

Question 4.
A baby is not able to tell her/his caretakers that she/he is sick. What would help us to find out.
a) that the baby is sick?
b) what is the sickness?
Answer:
(a) Yes, the bay is sick.
(b) If the baby is suffering from dysentery and omitting we can find out it is suffering from cholera. By knowing the symptoms, we can find out diseases.

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 chickenpox.
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:
Among three conditions in the condition (c) a person most likely to fall sick because After fasting of four days, and taking care of someone suffering from chicken-pox. In this case, she becomes weak and she may fall sick.

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:
Under these conditions, in condition (b) we get sick because in two days travelling because of cold air and sneezing of co-travelers in the bus or train. Because of overcrowded and lack of ventilation. Common cold, pneumonia and T.B. are the diseases we may get by contact with others.

KSEEB Solutions for Class 9 Science Chapter 13 Additional Questions and Answer

Question 1.
What do you mean by health?
Answer:
Health is therefore a state of being well enough to function well physically, mentally, and socially.

Question 2.
How do doctors diagnose diseases?
Answer:
With the help of information from the Laboratory, doctors are diagnosing the diseases accurately.

Question 3.
What do you mean by acute diseases?
Answer:
Some diseases last for only very short periods of time and these are called acute diseases.

Question 4.
What are Chronic Diseases?
Answer:
Some ailments can last for a long time, even as much as a lifetime and are called chronic diseases.

Question 5.
How do we get Blood Pressure?
Answer:
Because of overweight and lack of exercise, we get B.P.

Question 6.
How do we get skin infections?
Answer:
From fungus, we may get skin infections.

KSEEB Solutions for Class 9 Science

 

Karnataka Board Class 9 Science Chapter 14 Natural Resources

KSEEB Solutions Class 9 Science Chapter 14 Intext Questions

Question 1.
How is our atmosphere different from the atmospheres on Venus and Mars?
Answer:
Both Venus and Mars have 95.97% of the carbon dioxide. But in our earth 0.02% of carbon dioxide is present. Our earth has 21% of oxygen which is necessary for all living beings. In this way, Earth’s atmosphere is different from the atmosphere on Venus and Mars.

Question 2.
How does the atmosphere act as a blanket?
Answer:
Atmosphere is covering the earth, like a blanket, we know that air is a bad conductor of heat. The atmosphere keeps the average temperature of the earth fairly steady during the day and even during the course of the whole year. The atmosphere prevents the sudden increase in temperature during the daylight hours. And during the night, it slows down the escape of heat into outer space.

Question 3.
What causes winds?
Answer:
In the coastal regions, during the day, the air above the land gets heated and starts rising. As this air rises, a region of low pressure is created and air over the sea moves into this area of low pressure. The movement of air from one region to the other creates winds,

Question 4.
How are clouds formed?
Answer:
When water bodies are heated during the day, a large amount of water evaporates and goes into the air-some amount of water vapour also get into the atmosphere because of various biological activities. This air also gets heated. The hot air rises up carrying the water vapour with it. As the air rises, it expands and cools. The cooling causes the water vapour in the air to condense in the form of tiny droplets. This condensation of water is facilitated if some particles could act as the nucleus for these drops to form around. Normally dust and other suspended particles in the air perform this function.

Question 5.
List any three human activities that you think would lead to air pollution.
Answer:
Following are the three human activities which leads to air pollution.

1. Burning of fossil fuels leads to air pollution.
2. By the destruction of forests.
3. By industrialisation.

Question 6.
Why do organisms need water?
Answer:
All cellular processes take place in a water medium. All the reactions that take place in a water medium. All the reactions that take place within our body and within the cells occur between substances that are dissolved in water.

Substances are also transported from one part of the body to the other in a dissolved form. Hence organisms need to maintain the level of water within their bodies in order to stay alive. Hence organisms need water.

Question 7.
What is the major source of freshwater in the city/town/village where you live?
Answer:
In some places, the river is the major source of fresh water. In some places, freshwater is available from borewells. Tungabhadra river is the major source for us. (Gadag district)

Question 8.
Do you know of any activity which may be polluting this water source? ‘
Answer:

1. The addition of undesirable substances to water bodies. These substances could be the fertilizers and pesticides used in farming or they could be poisonous substances like mercury salts which are used by paper industries.
2. Releasing drainage water and effluents from industries to nearby water resources.
3. Releasing superheated water to rivers.
4. Change in temperature of river water.
   These are the activity from which water gets polluted.

Question 9.
How is soil formed?
Answer:
Over long periods of time thousands and millions of years, the rocks at or near the surface of the Earth are broken down by various physical, chemical, and some biological processes. The end product of this breaking down is the fine particles of soil. The Sun, water, and wind are also factors helpful for the formation of soil.

Question 10.
What is soil erosion?
Answer:
Flowing water and blowing wind carries soil particles along with them. This is called soil erosion.

Question 11.
What are’ the methods of preventing or reducing soil erosion?
Answer:

1. The roots of plants have an important role in preventing soil erosion.
2. We should give importance to reforestation. Afforestation programmes instead of deforestation.
3. Vegetative cover on the ground has a role to play in the percolation water into the deeper layers too. In the hilly or mountainous region, we should grow more grass.
   By these methods, we can prevent or reduce soil erosion.

Question 12.
What are the different states in which water is found ‘during the water cycle?
Answer:
Water is found in solid, liquid, and gaseous states during the water cycle.

Question 13.
Name two biologically important compounds that contain both oxygen and nitrogen.
Answer:

1. Oxygen:
   • Carbon dioxide
   • Phosphorus pentoxide.
2. Nitrogen:
   • Amino acid
   • Protein.

Question 14.
List any three human activities which would lead to an increase in the carbon dioxide content of air.
Answer:

1. Breathing
2. Burning of fuels
3. Overuse of vehicles.

Question 15.
What are the greenhouse effects?
Answer:
Some gases prevent the escape of heat from the earth. An increase in the percentage of such gases in the atmosphere would cause the average temperatures to increase worldwide and this is called the greenhouse effect.

Question 16.
What are the two forms of oxygen found in the atmosphere?
Answer:
Carbon dioxide and water vapour are the two forms of oxygen found in the atmosphere.

KSEEB Solutions for Class 9 Science Chapter 14 Textbook Exercises

Question 1.
Why is the atmosphere essential for life?
Answer:
(a) to establish all forms of life.
(b) necessary vital components can be – recycled.
(c) maintenance of temperature.
(d) a suitable environment for all forms of life.
(e) ozone a protective layer.

Question 2.
Why is water essential for life?
Answer:
All organisms require water for their survival. All cellular processes take place in a water medium. All the reactions that take place within our body and within the cells occur between substances that are dissolved in water. Substances are also transported from one part of the body to the other in a dissolved form. In this way water is essential to us.

Question 3.
How are living organisms dependent on the soil? Are organisms that live in water totally independent of soil as a resource?
Answer:
Soil supplies various nutrients which are essential for all organisms. Even soil is necessary for the fixing of plants firmly and increases the level of underground water. Even organisms that live in water are also totally dependent of soil as a resource. They are not surviving without soil.

Question 4.
You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather?
Answer:
Because of scientific equipment, we come to know about the weather reports. For example, if there is a depression in a barometer, we may expect heavy rainfall. By such phenomena, we are able to predict the weather.

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:

1. Farmers should use pesticides and insecticides judiciously so that no soil pollution takes place.
2. We should not allow effluents (wastes from industries) to mix with river water. Superheated water should not mix with river water.
3. We can prevent air pollution by tree plantation programmes.
   All these are the preventive measures to avoid soil, water, and air pollution.

Question 6.
Write a note on how forests influence the quality of our air, soil, and water resources.
Answer:
Forests are useful to us in many ways. They are as follows:

1. Plants take carbon dioxide which we give out during respiration and we get oxygen from plants which are very essential to us.
2. Plants increase water content in the underground. Roots of the plants minimize soil erosion and there is an increase in the water holding capacity.
3. Because of the number of trees, no soil erosion takes place. Roots bind soil particles and more water is stored.

KSEEB Solutions for Class 9 Science Chapter 14 Additional Questions and Answers

Fill in the blanks with suitable words:
Question 1.
Nearly _________ % of earth’s surface is covered with water.
Answer:
75 %.

Question 2.
Air _______ and ______ as it goes up.
Answer:
expands, cools

Question 3.
When the temperature of the air is low enough, precipitation may occur in the form of _____, ______ or _____
Answer:
snow, sleet, hail.

Question 4.
If we intake polluted air continuously, we have to suffer from diseases like ______ and ______
Answer:
Allergy, cancer and heart disease.

Question 5.
The outer covering of the earth is called ______
Answer:
crust.

Answer the following questions:

Question 1.
What is the importance of Humans?
Answer :
It is a major factor in deciding the soil structure because it causes the soil to become more porous and allows water and air to penetrate deep underground.

Question 2.
Write the importance of topsoil?
Answer :
The topmost layer of the soil that contains humus and living organisms in addition to the soil particles is called topsoil, it is important to factor that decides biodiversity in that area.

Question 3.
Mention the two important components of biogeochemical cycles.
Answer :

• Reservoir pool
• Exchange pool

KSEEB Solutions for Class 9 Science

 

Karnataka Board Class 9 Science Chapter 15 Improvement in Food Resources

KSEEB Solutions Class 9 Science Chapter 15 Intext Questions

Question 1.
What do we get from cereals, pulses, fruits, and vegetables?
Answer:
Cereals such as wheat, rice, maize, ragi and jowar provide us carbohydrates for energy requirement.
Pulses like gram, pea, black gram, green gram, pigeon pea, lentil provide us with protein, vegetables provide us a range of vitamins and minerals in addition to small amounts of proteins, carbohydrates and fats.

Question 2
How do biotic and abiotic factors affect crop production?
Answer:
Crops production can go down due to biotic (diseases, insects, and nematodes) and abiotic (drought, salinity, waterlogging, heat, cold and frost) stresses under different situations varieties resistant to these stresses can improve crop production.

Question 3.
What are the desirable agronomic characteristics for crop improvements?
Answer:
Tallness and profuse branching are desirable characters for fodder crops. Deafness is desired in cereals so that less nutrients are consumed by these crops. Thus developing varieties of desired agronomic characters help give higher productivity.

Question 4.
What are macro-nutrients and why are they called macro-nutrients?
Answer:
The nutrients which are required to plants in large quantities are called macro-nutrients. Eg. Nitrogen, phosphorus, potassium, calcium magnesium, sulphur. They are required in large quantities and are therefore called macro-nutrients.

Question 5.
How do plants get nutrients?
Answer:
Nutrients are supplied to plants
by air, water and soil. There are sixteen nutrients which are essential for plants. Air supplies carbon and oxygen, hydrogen comes from water and soil supplies the other thirteen nutrients, six are required in large quantities and are therefore called macronutrients. The other seven nutrients are used by plants in small quantities and are therefore called micro-nutrients.

Question 6.
Compare the use of manure and fertilizers in maintaining soil fertility.
Answer:

Manure	Fertilizer
1. Produced by natural phenomena.	Produced from the chemical processes.
2. Dosage is not. fixed.	The dosage is fixed.
3. Maintain the pH of soil.	the pH of soil may be altered.
4. Rich with microbes nutrients and do not pollute the soil.	4. Rich with phosphorous potassium and pollute the soil.

Question 7.
Which of the following conditions will give the most benefits? Why?
a) Farmers use high-quality seeds, do not adopt irrigation or use fertilizers.
b) Farmers use ordinary seeds, adopt irrigation and use fertilizers.
c) Farmers use quality seeds, adopt irrigation, use fertilizer and use crop protection measures.
Answer:
(c) This condition will give the most benefits. Because it includes all agricultural methods. Hence yield is more.

Question 8.
Why should preventive measures and biological control methods be preferred for protecting crops?
Answer:
Because insects, rodents, fungi, miles and bacteria, moisture and temperature cause degradation in quality, loss in weight, poor germinability, discolouration of produce, all leading to poor marketability of crops. Therefore By preventive measures and biological control methods crops should be protected.

Question 9.
What factors may be responsible for losses of grains during storage?
Answer:
Moisture and temperature are the factors which are responsible for losses of grains during storage.

Question 10.
Which method is commonly used for improving cattle breeds and why?
Answer:
Exotic or foreign breeds (eg: Jersey) are selected for long lactation periods, while local breeds (eg: Red sindhi) show excellent resistance to diseases. The two can be cross-breed to get animals with both the desired qualities.

Question 11.
Discuss the implications of the following statement:
“It is interesting to note that poultry is India’s most efficient converter of low fiber foodstuff (which is unfit for human consumption) into highly nutritious animal protein food.”
Answer:
This statement is correct. The rotation (daily food requirement) for broilers is protein-rich with adequate fat. The level of vitamins A and K is kept high in the poultry feeds.

Question 12.
What management practices are common in dairy and poultry farming?
Answer:
Broiler chickens are fed with vitamin-rich supplementary feed for good growth rate and better feed efficiency. Care is taken to avoid mortality and to maintain feathering and carcass quality. They are produced as broilers and sent to the market for meat purposes.

For the good production of poultry birds, good management practices are important. These include maintenance of temperature and hygienic conditions in housing and poultry feed, as well as prevention and control of diseases and pests. The ration (daily food requirement) for broilers is protein-rich with adequate fat. The level of vitamins A and K is kept high in the poultry feeds.

Question 13.
What are the differences between broilers and layers and in their management?
Answer:

Broilers	Layers
These provide a good quality of meat.	These are producing good quality of eggs.

The housing, nutritional and environmental requirements of broilers are somewhat different from those of egg layers. The ration (daily food requirement) for broilers is protein-rich with adequate fat. The level of vitamins A and K is kept in the poultry feeds.

Question 14.
How are fish obtained?
Answer :
There are two ways of obtaining fish, one is from natural resources, which is called capture fishing. The other way is by fish farming which is called culture fishery.

Question 15.
What are the advantages of composite fish culture?
Answer:

1. By composite fish culture, the food available in all the parts of the pond is used.
2. This increases the fish yield from the pond.
   These are the advantages of composite fish culture.

Question 16.
What are the desirable characters of bee varieties suitable for honey production?
Answer:
The value or quality of honey depends upon the pasturage, or the flowers available to the bees for nectar and pollen collection. In addition to an adequate quantity of pasturage, the kind of flowers available will determine the taste of the honey.

Question 17.
What is pasturage and how is it related to honey production?
Answer:
Varieties of plants, trees belonging to different species altogether are called pasturage. It should be rich in the environment for obtaining good quality of honey. Because the value or quality of honey depends upon the pasturage, or the flowers available to the bees for nectar and pollen collection. In addition to an adequate quantity of pasturage, the kind of flowers available will determine the taste of the honey.

KSEEB Solutions for Class 9 Science Chapter 15 Textbook Exercises

Question 1.
Explain any one method of crop production which ensures high yield.
Answer:
Inter-cropping is growing two or more crops simultaneously on the same field in a definite pattern. A few rows of one crop alternate with a few rows of second crop, for example, soyabean+maize, or finger millet (bajra) + cowpea (lobia). The crops are selected such that their nutrient requirements are different. This ensures maximum utilization of the nutrients supplied and also prevents pests and diseases from spreading to all the plants belonging to one crop in a field. This way both crops can give better returns.

Question 2.
Why are manure and fertilizers used in fields?
Answer:
Fertilizers supply nitrogen, phosphorus and potassium. They are used to ensure good vegetative growth (leaves, branches and flowers) giving rise to healthy plants. Fertilizers are a factor in the higher yields of high-cost farming. Soil fertility is maintained properly by the application of fertilizers.

Question 3.
What are the advantages of intercropping and crop rotation?
Answer:

1. Intercropping: In this method, the crops are selected such that their nutrient requirements are different. This ensures maximum utilisation of nutrients supplied and also prevent pests and diseases from spreading to all the plants belonging to one crop in a field. This way both crops can give better returns.
2. Crop rotation: The growing of different crops on a piece of land in a preplanned succession is known as crop rotation. If crop rotation is done properly then two or three crops can be grown in a year with a good harvest.

Question 4.
What is genetic manipulation? How is it useful in agricultural practices?
Answer:
The desirable characteristics of the crop can be genetically modified to establish progeny.
(a) Resources can be conserved.
(b) Period of yield can be maintained,
(c) Availability of crop of all seasons . can be produced and maintained

Question 5.
How do storage grain losses occur?
Answer:
Storage losses in agricultural produce can be very high. Factors responsible for such losses are biotic- insects, fungi, mites and bacteria, and abiotic – inappropriate, moisture, and temperatures in the place of storage. These factors cause degradation in quality, loss in weight, poor germinability, discolouration of produce.

Question 6.
How do good animal husbandry practices benefit farmers?
Answer:
Animal-based farming includes cattle, goat, sheep, poultry and fish farming. As the population increases and as living standards increase, the demand for milk, eggs and meat is also going up. Also the growing awareness of the need for the humane treatment of livestock has brought in new limitations in livestock farming.

Cattle husbandry is done for two purposes – milk and draught labour for agricultural work such as tilling, irrigation and carting. Milk-producing females are called milch animals (dairy animals) while the ones used form labour are called draught animals.

Question 7.
What are the benefits of cattle farming?
Answer:
Cattle husbandry is done for two purposes – milk and draugt labour for agricultural work such as tilling, irrigation and carting. Milk producing females are called milch animals (dairy animals), while the ones used for farm labour are called draught animals. These are the benefits of cattle farming.

Question 8.
For increasing production, what is common in poultry, fisheries and bee-keeping?
Answer:
For increasing production of all these there should be proper cleaning, sanitation and spraying of disinfectants are regular intervals. Appropriate vaccination can prevent the occurrence of infectious diseases.

Question 9.
How do you differentiate between capture fishing, mariculture and aquaculture?
Answer:

1. Capture fishing: Capture fishing is also done in such inland water bodies, the yield is not high.
2. Mari-culture fishing: Marine fish are caught using many minds of fishing nets from fishing boats. Yield are increased by locating large schools of fish in the open sea using satellites and echo-sounders.
3. Aquaculture: Aquaculture is a method of fishing in ponds, canals, dams and rivers. In this method yield is less.

KSEEB Solutions for Class 9 Science Chapter 15 Additional Questions and Answers

I. Answer the following questions:

Question 1.
Name the major groups of activities for improving crop yields.
Answer:

1. Crop variety improvement.
2. Crop production improvement.
3. Crop protection and management.

Question 2.
How is soil fertility is maintained to get higher yield?
Answer:
By using compost manure vermicompost and green manure we can maintain soil fertility. We get the maximum yield.

Question 3.
What do you mean by intercropping?
Answer:
Inter-cropping is growing two or more crops simultaneously on the same field in a definite pattern.

Question 4.
What are the weeds? Give examples.
Answer:
Weeds are unwanted plants which are grown on along with crops.
Eg: Xanthium, parthenium, etc.

Question 5.
Name two processes of obtaining fish.
Answer:

1. From natural process.
2. By fish farming which is called culture fishery.

KSEEB Solutions for Class 9 Science

Karnataka Board Class 9 Science Chapter 9 Force and Laws of Motion

KSEEB Solutions Class 9 Science Chapter 9 Intext Questions

Question 1.
Which of the following has more inertia
(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five rupees coin, and a one rupee coin?
Answer:
a) a stone
b) a train,
c) Five rupees coin.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.”
Also, identify the agent supplying the force in each case.
Answer:
Given below are the number of times at which the velocity of the ball changes. Whenever a force will be applied on the ball then the velocity of the ball will change.

1. When the first player kicks the ball towards another player of his team the velocity of the ball will change because the first player applied some force on the ball.
2. When another player kicks the ball towards the goal, then the velocity of the ball will change, here again, the force is applied on the ball.
3. When goalkeeper of the opposite team collects the ball, then the velocity of ball change. It becomes zero. Here the goalkeeper applies some force on the ball to stop.
4. When the goalkeeper kicks the ball towards his own team, the velocity of the ball changes because the goalkeeper applies some force on the ball.

Question 3.
Explain why some of the leaves may get detached from a tree we vigorously shake its branch.
Answer:
We know that the leaves are attached to the tree and are in a state of rest initially. But when the tree is vigorously shaken, the branches of the tree come in the state of motion but the leaves tend to maintain their state of rest, as a result, the leaves get detached from the tree and fall down.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall back when it accelerates from rest?
Answer:
Initially, when the bus is at rest our body also follows the same state. But all of a sudden when the bus starts to move the lower part of our body tends to move with the motion of the bus but the upper part rejects this state of motion and continues to be in a state of rest. This results in a sudden jerk backward when a bus moves. Also, a person standing in a bus will be in a state of motion, and when the brakes are applied the lower part of our body comes to the state of rest but our upper part is in state of motion. Hence, we tend to fall forward when the bus applies brakes.

Question 5.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
In order to pull the cart, the horse pushes the ground with its foot in the backward direction by pressing the ground. As a reaction of this force, the ground pushes the horse in a forward direction. As a result, the cart is pulled by the horse in the forward direction.

Question 6.
Explain Why is it difficult for a firement to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
From Newton’s third law, we know that every action has an equal and opposite reaction. So when a fireman holds a hose-pipe which is ejecting a large amount of water at high-velocity experiences a backward push due to the force of water flowing in the forward direction. Hence making it clear that due to the action of force in the forward direction the force is applied on the pipe in the backward direction, thus making the fireman difficult to hold the hose-pipe.

Question 7.
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms^-1, calculate the initial recoil velocity of the rifle.
Answer:
Mass of the gun M = 4 kg.
Mass of bullet M = 50 g = 50 × 10^-3kg.
Initial velocity of the bullet v = 35ms^-1
Let recoil velocity of the gun be vms^-1
Before firing the bullet both gun and bullet were in rest so total momentum of gun and bullet is zero, After firing
Momentum of bullet = mv,
Momentum of gun = Mv
Total momentum of bullet and gun after firing = mv + Mv
Since there is no external force applied to the system.
So total momentum after firing = total momentum before firing
mv + Mv = 0
Mv = -mv

Here negative sign shows that there recoil velocity of the gun is in the direction opposite to the velocity of the bullet.

Question 8.
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2ms^-1 and 1ms^-1 respectively. They collide and after the collision, the first object moves at a velocity of 1.67ms^-1. Determine the velocity of the second object.
Answer:
Before collision

After collision
m₁ = 100 g = 0.1 Kg
m₂ = 200 g = 0.2 Kg
V₁ = 1.67 ms^-1
Let velocity of second object after collision is v₂ms^-1 Since there is no external force on the system.

KSEEB Solutions for Class 9 Science Chapter 9 Textbook Exercises

Question 1.
An object experiences a zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no provide a reason.
Answer:
According to Newton’s 1st law of motion, no net force is required to move an object which is moving with a constant velocity. So when an object experiences a net zero external unbalanced force, then it can move with a non-zero velocity. Also if an object is initially at rest not net force acts upon it, thus the object may not move all.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When a carpet is beaten with a stick, then the fibers of the carpet attain the state of motion while the dust particles remain in rest due to the inertia of rest. Therefore the dust particles fall down.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
Because even though bus comes to rest, luggage are in motion, They fall down. So, it is tied with a rope to prevent falling.

Question 4.
A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because.
a) The batsman did not hit the ball hard enough.
b) Velocity is proportional to the force exerted on the ball.
c) There is a force on the ball opposing the motion.
d) There is no unbalanced force on the ball, so the ball would want to come to rest.
Answer:
c) because the ball slows down to rest as the force of friction acting between the ground and the ball opposes the motion of the ball.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tons
(Hint: 1 ton = 1000 kg.)
Answer:
The truck starts from rest, so initial velocity u = 0
Distance s = 400 m, t = 20s
m = 7 tons = 7 × 1000 = 7000 Kg.
From the 2nd equation of motion

From Newton’s 2nd law of motion, the force acting on the Duck
= 7000 × 2
= 14000 N
= 1.4 × 10⁴ N.

Question 6.
A stone of 1 kg is thrown with a velocity of 20ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
Mass of the stone m = 1 kg.
initial velocity u = 20 ms^-1
Final velocity v = 0 (Therefore the stone comes to rest distance travelled S = 50 m)
From third equation of motion
v² = u² + 2as
(0)² = (20)² + 2a(50)
100 a = -400
∴ a = -4ms^-2
Here negative sign shows that there is relation in the motion of stone.
Force of friction between stone and ice = Force required to stop the stone
= ma
= 1 × -4 = -4N or 4N.

Question 7.
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000N then calculate
a) the net accelerating force.
b) the acceleration of the train and
c) the force of wagon 1 on wagon 2
Answer:
a) Net accelerating force = Force exerted by the engine – Frictional force.
(Here the frictional force is substracted because it opposes the motion)
= 40000 – 5000 = 35000 N
= 3.5 × 104 N.

b) From Newton’s second law of motion
Accelerating force = mass of the train × Acceleration of train
a =\(\frac{\mathrm{F}}{\mathrm{m}}\)
Mass of train = Mass of engine + Mass of all wagons
= 8000 + 5 × 2000
= 8000 + 10000
= 18000 Kg.
∴ Acceleration = \(\frac{35000}{18000}=\frac{35}{18}\)
= 1.95 ms^-2.

c) Force of wagon 1 on wagon 2
= Mass of the engine × Acceleration
= 800 × 1.95 = 15600 N.

Question 8.
An automobile vehicle has a mass of 1500 Kg. what must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms^-2
Answer:
Mass m = 1500 Kg.
Acceleration a = -1.7 ms^-2
From Newton’s second law of motion
F = ma
= 1500 × (-1.7)
= -2550 N.

Question 9.
What is the momentum of an object of mass m, moving with a velocity V?
(a) (mv)²
(b) mv²
(c) \(\frac{1}{2}\) mv²
(d) mv.
Answer:
(d) P = mv.

Question 10.
Using a horizontal force of 200N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
The cabinet will move across the floor with constant velocity if there is no net external force of 200N that should be applied on the cabinet in opposite direction.
Thus, the frictional force = 200N
(Frictional force always acts in the direction opposite to the direction of motion)

Question 11.
Two objects, each of mass 1.5 kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5ms^-1 before the collision during which they stick together is what will be the velocity of combined object after collision?
Answer:
Mass of the first object, m₁ = 1.5 Kg.
Mass of second object m₂ = 1.5 Kg.
Velocity of first object before the collision, u₁ = 2.5 ms^-1
velocity of second object before collision, u₂ = -2.5 ms^-1
Here negative sign is taken because the second object is moving in the direction opposite to the direction of motion of first object.
Mass of combined object after collision
M = m₁ + m₂ (Since they stick together)
= 1.5 + 1.5 = 3 Kg.
Let velocity of combined object after collision be vms^-2
Here there is no external force, so from law of conservation of momentum
Momentum after collision = Momentum before collision
mv = m₁u₁ + m₂u₂
3 × v = 1.5 × 2.5 + 1.5 (-2.5) =0
∴ v = 0.

Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
The logic given by students is not correct because two equal and opposite forces cancel each other only in the case if they act on the somebody. According to Newton’s third law action and reaction force always act on two different bodies, so they cannot cancel each other. When a massive truck is pushed then the truck may not the force applied is not sufficient to move the truck.

Question 13.
A hockey ball of mass 200 g travelling at 10ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5ms^-1 Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Mass of hockey ball m = 200 g = 0.2 Kg
Initial velocity u = 10ms^-1
Final velocity v = 5 ms^-1
(Since the momentum of the ball = Final momentum – initial momentum
= mv – mu
= m (v – u)
= -0.2 (-5 – 10)
= 0.2 × -15 = -3 Kgms^-1.

Question 14.
A bullet of mass 10 g travelling horizontally with a velocity of 150ms’1 strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Mass of bullet m = \(\frac{10}{1000}\) = kg
= 0.001 Kg.
Initial velocity, u = 150ms^-1
Final velocity v = 0 (since the bullet comes to rest)
Time t = 0.03 s.
From equation of motion
v = u + at
0 = 150 + a × 0.03
∴ a = \(\frac{-150}{0.03}\)
= -5000 ms^-2
Magnitude of the force applied by the bullet on the block F = ma
= 0.01 × -5000 = -50N.

Question 15.
An object of mass 1Kg travelling in straight line with a velocity of 10ms’1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact Also calculates the velocity of the combined object.
Answer:
Mass of object m₁ = 1 Kg,
Velocity, u₁ = 10ms^-1
Mass of wooden block m₂ = 5Kg,
velocity u₂ = 0 (since wooden block is rest)
= 1 × 10 + 5 × 0 = 10 Kgms^-1
According to law of conservation of momentum
Momentum after impact = Momentum before impact
(m₁ + m₂)v = 10
Where v = velocity of combined object
(1 + 5) v = 10
v = \(\frac{10}{6}\) = 1.67ms^-1.

Question 16.
An object of mass 100 Kg is accelerated uniformly from a velocity of 5ms^-1 to 8ms^-1 in 6s. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object.
Answer:
Mass of the object, m = 100 Kg
Initial velocity, u = 5 ms^-1
Final velocity, v = 8 ms^-2
(i) Initial momentum = mv = 100 × 5 = 500 Kgms^-1
Final momentum = mv = 100 × 8 = 800 Kgms^-1

(ii) From Newton’s second law
Force exerted on the object = Rate of change of momentum

Question 17.
Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the momentum of the motor car (because the change in the velocity of the insect was much more than that of the motor car). Akthar said that since the motor car was moving with larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motor car and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
The explanation given by Rahul is correct, as there is no external force on the system. So both the insect and motorcar experienced the same force and a change in momentum. .

Question 18.
How much momentum will a dumb-bell of mass 10 Kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10ms^-2.
Answer:
m = 10 Kg, u = 0
Acceleration s = 80 cm = 0.8 m.
Velocity a = 10 ms^-2
v² = u² + 2as
v² = 0 + 2 × 10 × 0.8 =16
∴ v = \(\sqrt{16}\) = 4ms^-1
Momentum of dumb-bell just before it touches the floor
P = mv
= 10 × 4
= 40 Kgms^-1.
When the dumbbell touches the floor its velocity becomes zero. Thus the total momentum of the dumbbell is transferred to the floor. Hence the momentum transferred to floor = 40 Kgms^-2.

KSEEB Solutions for Class 9 Science Chapter 9 Additional Questions and Answer

Question 1.
The following is the distance-time table of an object in motion.

a) What conclusion can you draw about the acceleration? Is it constant increasing, decreasing or zero?
b) What do you infer about the forces acting on the object?
Answer:
a) Here, a = 0,

In this way, acceleration is increasing

b) Since acceleration is increasing. So-net unbalanced force is acting on the object.

Question 2.
Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motor car can be pushed by three persons to produce an acceleration of 0.2ms^-2 with what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)
Answer:
Mass of motor car m = 1200 Kg.
Acceleration produced a = 0.2 ms^-2
Therefore force applied to the car by three persons = 1200 × 0.2 = 240 N.
∴ Force applied on the car by one person = \(\frac{240}{3}\) = 80 N,

Question 3.
A hammer of mass 500g moving at 50ms^-1 strikes a nail. The nail stops the hammer in a very short time of 0.61s what is the force of the nail on the hammer?
Answer:
Mass of the hammer = 500g = 0.05 Kg.
Initial velocity of hammer u = 50 ms^-1
The final velocity of the hammer, v = 0 (Since the hammer stops)

Question 4.
A motor car of mass 1200 kg is moving along a straight line with a uniform velocity of 90km/h. Its velocity is slowed down to 18 km/h in 4s by an imbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Answer:
Mass m = 1200 kg.

(Here negative sign indicates decrease in velocity)

ii) Change in momentum = Final momentum – initial momentum
= mv – mu = m (v – u)
= 1200 (5 – 25)
= 1200 × (-20) = -24000Kgms^-1

iii) Magnitude of the force required = Rate of change momentum

KSEEB Solutions for Class 9 Science

Karnataka Board Class 9 Science Chapter 10 Gravitation

KSEEB Solutions Class 9 Science Chapter 10 Intext Questions

Question 1.
State the universal law of gravitation.
Answer:
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.

Question 2.
Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:

Question 3.
What do you mean by free fall?
Answer:
Whenever objects fall towards the earth under the gravitational force alone, we say that the objects are in free fall.

Question 4.
What do you mean by the acceleration due to gravity?
Answer:
Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore this acceleration is called the acceleration due to the gravitational force of the earth.

Question 5.
What are the differences between the mass of an object and its weight?
Answer:

Mass	Weight
Mass of an object is the measure of the inertia.	The weight of an object is the force with which it is attracted to the earth.

Question 6.
Why is the weight of an object on the moon l/6th its weight on the earth?
Answer:
Mass of moon is lesser than that of earth. Due to this the moon exerts lesser force of attraction on objects. Hence weight of an object on the moon l/6th its weight on the earth.

Question 7.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a starp made of a thin and strong string because the area and starp is small. Hence large pressure is exerted by the starp on the fingers.

Question 8.
What do you mean by buoyancy?
Answer:
The upward force exerted by the water on the bottle is known as upthrust or buoyant force.

Question 9.
Why does an object float or sink when placed on the surface of the water?
Answer:
Upward thrust of liquids, (different-in liquid densities) Sink of an object depends upon density. The density of an object is more than water then it sinks. Object density is less than that of water it floats.

Question 10.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
Our mass is more than 42 kg.

Question 11.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the others. Can you say which one is heavier and why?
Answer:
Iron bar is heavier than the cotton bag because for the mass the iron bar will have a lesser surface area and apply more force, so the iron bar is heavier than a cotton bag.

KSEEB Solutions for Class 9 Science Chapter 10 Textbook Exercises

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
Force of gravitation between two objects.

i.e. the force of gravitation becomes 4 times than the original value.

Question 2.
Gravitational force acts on all objects in proportion, to their masses. Why then a heavy object does not fall faster than a light object?
Answer:
Each object falls towards the earth with the acceleration equal to the acceleration due to gravity, which is constant (9.8ms^-2) and does not depend on the mass of the object. So heavy object does not fall faster than the light object.

Question 3.
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 x 10²³ kg and radius of the earth
6.4 x 10⁶m)
Answer:
Gravitational force between the earth and an F = \(\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}^{2}}\)
Where G = Gravitational constant = 6.67 × 10^-11 Nm²kg^-2.
M = Mass of earth = 6 × 10²⁴ kg.
R = Radius of earth = 6.4 × 10⁶m
m = Mass of object = 1 kg.
Hence F = \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^{6}\right)^{2}}\)
= 9.8N.

Question 4.
The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Answer:
When two objects attract each other then the gravitational force of attraction applied by first object on the second object is same as the force applied by the second object on the first object. So both earth and moon attract each other by the same gravitational force of attraction.

Question 5.
If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
The earth does not move towards the moon inspite of the attraction by moon because the force with which moon attracts the earth is same by which earth attracts the moon. So net force on the system becomes zero.

Question 6.
What happens to the force between two objects, if
i) the mass of one object is doubled?
ii) the distance between the objects is doubled and tripled?
iii) The masses of both objects are doubled?
Answer:
Force of attraction between two objects.
F = \(\frac{\mathrm{G} \mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{r}^{2}}\)
where m₁, m₂ = masses of the objects
r = Distance between the objects.
G = Gravitational constant.

(i) If mass of one object is doubled, then the new force

∴ force becomes double.

(ii) If the distance between the objects is tripled, then new force

∴ force becomes one-ninth.

iii) If masses of both objects are doubled, then new force .

i.e. force becomes four times.

Question 7.
What is the importance of universal law of gravitation?
Answer:
The importance of universal law of gravitation can be understood from the following points:

• All the planets revolve around the sun due to gravitational force between the sun and planets. Hence gravitational force is responsible for the existence of solar system.
• Tides in oceans are formed due to the gravitational force between the moon and water in oceans.
• The atmosphere of the earth is attached with it due to gravitational force of earth.
• Artificial satellites revolve around the earth due to gravitational force of earth.
• We can stand on the surface of earth due to gravitational force of earth.

Question 8.
What is the acceleration of free fall?
Answer:
The acceleration produced in the motion of an object when it falls freely towards the earth is termed acceleration of free fall. It is also called acceleration due to gravity. Its value on the earth’s surface is 9.8 ms^-2.

Question 9.
What do we call the gravitational force between the earth and an object?
Answer:
The gravitational force between the earth and an object is called the force of gravity.

Question 10.
Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator will the friend agree with the weight of Gold bought? If not, why? (Hint: The value of g is greater at the poles than at the equator)
Answer:
Weight of an object, W = mg.
where g = acceleration due to gravity.
The value of g is greater at the poles than at the equator. So the weight of the gold at the equator will be less than the weight of the gold at poles. So Amit’s friend will find the weight of the gold less than the weight told by Amit.

Question 11.
Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer:
The sheet of paper falls slower than that is compelled into a ball because in the first case the area of the sheet is more. So it experiences a large opposing force due to air. while the sheet crumpled into a ball experience less opposing force due to small area.

Question 12.
The gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?
Answer:
Mass of the object m = 10 kg.
Weight on the earth W = mg.
= 10 × 9.8 = 98 N
Weight on the moon = \(\frac{1}{6}\) of the weight on the earth.
= 1 × 98 = 16.33 N
= 16.33 N.

Question 13.
A ball is thrown vertically upwards with a velocity of 49m/s. Calculate.
i) the maximum height to which it rises.
ii) the total time it takes to return to the surface of the earth.
Answer:
Initial velocity u = 49 ms^-1
(i) At the maximum height, velocity becomes zero.
∴ Final velocity v = 0
From equation of upward motion
v² = u² – 2gh
0 = (49)² – 2 × 9.8 × h
∴ h = \(\frac{(49)^{2}}{2 \times 9.8}\) = 1225 m
Maximum height attained = 122.5 m.

(ii) Time is taken by the ball to reach the maximum height.
From equation of upward motion
v = u – gt
0 = 49 – 9.8 × t
∴ t = \(\frac{49}{9.8}\) = 5s
In the motion against gravity, the time of descent is same as the time of ascent.
So time taken by the ball to fall from maximum height = 5s.
∴ Total time taken by the ball to return to surface of the earth = 5 + 5 = 10s.

Question 14.
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
h = 19.6 m.
Initial velocity u = 0 (∵ It starts from rest)
From equation of motion.

Final velocity of stone before touching the ground = 19.6 ms^-1.

Question 15.
A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10m/s, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:
Initial velocity u = 40 ms^-1
At maximum height, final velocity becomes zero i.e., v = 0
From equation of motion
v² = u² – 2gh
(0)² = (40)² – 2 × 10 × h
0 = 1600 – 20h
Total time taken by the ball to return to surface of the earth = 5 + 5 = 10s.
∴ h = \(\frac{1600}{20}\) = 80 m
Maximum height reached by the stone = 80 m
After reaching the maximum height, the stone will fall towards the earth and will reach the earth surface covering the same distance.
So distance covered by the stone
= 80 + 80 = 160 m
Displacement of the stone = 0
(because the stone starts from the earth surface and finally reaches the earth surface again i.e the initial and final position of the stone are same

Question 16.
Calculate the force of gravitation between the earth and the sun, given that the mass of the earth = 6 × 10²⁴kg and of the sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹m)
Answer:
The force of attraction between the earth and the sun F = \(\frac{\mathrm{GM}_{1} \mathrm{M}_{2}}{\mathrm{r}^{2}}\)
where G = 6.67 × 10^-11Nm²kg^-2
Given mass of the Sun M₁= 2 × 10³⁰ kg
Mass of the earth M₁ = 6 × 10²⁴ kg.
Average distance between the earth and the sun = 1.5 × 10¹¹M

Question 17.
A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from
the ground with a velocity of 25m/s. Calculate when and where the two stones will meet.
Answer:
Let after time t both stone meet and s be the distance travelled by the stone dropped from the top of tower at which the stones will meet. Distance travelled by the stone dropped = S
∴ Distance travelled by the stone projected upwards (100 -s) m
For the stone dropped fromt the tower.

(u = 0 because stone is dropped ie it starts from rest)
s = 5t² …………. (i)
For the stone projected upward
s = ut – \(\frac{1}{2}\) gt²
(Due to upward motion negative sign is taken)

The stones will meet after; 4s
s = 5t² = 5 × (4)² = 80 m
The stones will be at a distance of 80 m from the top of tower or 20 mt (100 m – 80 m) from the base of the tower 20m (100m – 80m).

Question 18.
A ball thrown up vertically returns to the thrower after 6s. Find
a) the velocity with which it was thrown up
b) the maximum height it reaches, and
c) its position after 4s.
Answer:
Total time taken = 6 s
Time taken to reach the maximum height = \(\frac{6}{2}\) = 3s
(∵ Time of ascent = time)
a) From equation of motion v = u – gt
(Negative sign is taken due to upward motion)
0 = 4- 9.8 × 3 (∵ at maximum height v=0)
0 = u – 9.8 × 3
0 = u – 9.8 × 3
u = 29.4 ms^-1.
(∵ at maximum height v = 0)
u = 29.4

b) From equation of motion
v² = u² – 2gh
0 = (29.4)² – 2 × 9.8 × h
∴ h = \(\frac{(29.4)^{2}}{2 \times 9.8}\) = 44.1 m
Maximum height attained by the ball is 44.1 m

c) In initial 35. the ball will rise then in next 35 it falls towards the earth
∴ The position after 4s(3 + 1)
= Distance covered in Is in the download motion.
From the equation of motion

∴ ie the ball will be at 4.9m from the top of the tower.

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
The buoyant force on an object immersed in a liquid always act in upward direction.

Question 20.
Why does a block of plastic released under water come up to the surface of water.
Answer:
The buoyant force on an object immersed in a liquid always act in upward direction.

Question 21.
The volume of 50g of a substance is 20 cm³. If the density of water is 1gcm^-3 will the substance float or sink?
Answer:
Mass of a substance = 50g
Volume of substance = 20 cm³

= \(\frac{50}{20}\) = 2.5 g cm^-3
ie the density of the substance is greater than the density of water. So it will sink in water.

Question 22.
The volume of 500g sealed packet is 350 cm^-3, will the packet float or sink in water if the density of water is 1gmcm^-3? What will be the mass of the water displaced by this packet?
Answer:
Mass of packet = 500g
Volume of packet = 350 cm^-3

∴ ie the density of packet is greater than density of water, so it will sink in water.
Mass of water displaced by the packet = Volume of packet × Density of water = 350 × 1 = 350 gm.

KSEEB Solutions for Class 9 Science Chapter 10 Additional Questions and Answer

Fill in the blanks with suitable words.
Question 1.
The force that causes acceleration and keeps the body moving along the circular path acting towards the centre is called the _________
Answer:
Centripetal force.

Question 2.
Force is the product of and _________
Answer:
Mass, acceleration.

Question 3.
The ________ of an object is the force with which it is attracted towards the earth.
Answer:
weight.

Question 4.
The force acting on an object perpendicular to the surface is called the ___________
Answer:
thrust.

Question 5.
The upward force exerted by the water on objects is known as ________ force.
Answer:
buoyant.

Question 6.
The objects of density greater than that of a liquid _________ in the liquid.
Answer:
sink.

Question 7.
The value of ‘g’ in the poles is _________ than the equator.
Answer:
greater.

Answer the following questions:

Question 1.
Why is Newton’s law of Gravitation called universal law?
Answer :
It is applicable to all bodies either they ‘ are small or big. Whether they are celestial or terrestrial.

Question 2.
Mention two factors on which the buoyant force acting on an object depends?
Answer :

• The volume of the object, immersed in the liquid.
• The density of the fluid in which the object is immersed.

Question 3.
Why are big boulders can be moved easily by floods?
Answer :
Big boulders weigh less while in water and as such as easily moved by floods.

KSEEB Solutions for Class 9 Science