Class 9

Students can Download Adisthana Padavali Unit 3 Chapter 3 Kaviyum Samuhajivi Questions and Answers, Summary, Notes Pdf, Activity, Adisthana Padavali Malayalam Standard 9 Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Adisthana Padavali Malayalam Standard 9 Guide Unit 3 Chapter 3 Kaviyum Samuhajivi

Kaviyum Samuhajivi Questions and Answers, Summary, Notes

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Kerala Padavali Malayalam Standard 9 Guide Unit 5 Chapter 3 Matilerikkanni

Matilerikkanni Questions and Answers, Summary, Notes

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Kerala Padavali Malayalam Standard 9 Guide Unit 5 Chapter 1 Ambadiyilekku

Ambadiyilekku Questions and Answers, Summary, Notes

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Circle Measures Questions and Answers in Malayalam

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Kerala State Syllabus 9th Standard Maths Solutions Chapter 8 Polynomials

Polynomials Textual Questions and Answers

Textbook Page No. 123

Polynomial Root Calculator solves the roots of polynomials easily.

Polynomials Class 9 Kerala Syllabus Chapter 8 Question 1.
In rectangles with one side 1 centimetre shorter than the other, take the length of the shorter side as x centimetres.
i. Taking their perimeters as p(x) centimetres, write the relation between p(x) and x as an equation,
ii. Taking their areas as a(x) square centimetres, write the relation between a(x) and x as an equation.
iii Calculate p(l), p(2), p(3), p(4), p(5). Do you see any pattern?
iv. Calculate a(l), a(2), a(3), a(4), a(5). Do you see any pattern?
Answer:
Let x be the shorter side, then the other side will be (x+ 1).
i. Perimeter = 2[x + (x + 1)] = 2(2x + 1) = 4x + 2
That is, p(x) = 4x + 2

ii. Area = x(x + 1)
a(x) = x² + x
Area, a(x) = x² + x

iii. p(x) = 4x + 2
p(1) = 4 × 1 + 2 = 6
p(2) = 4 × 2 + 2 = 10
p(3) = 4 × 3 + 2= 14
p(4) = 4 × 4 + 2= 18
p(5) = 4 × 5 + 2 = 22
Perimeter is a sequence increasing by 4.

iv. a(x) = x² + x
a(1) = 1² + 1 = 2 = 1 × 2 = 2
a(2) = 2² + 2 = 6 = 2 × 3 = 6
a(3) = 3² + 3 = 12 = 3 × 4 = 12
a(4) = 4² + 4 = 20 = 4 × 5 = 20
a(5) = 5² + 5 = 30 = 5 × 6 = 30
Area is the product of x and the number one more than x.

Polynomial Root Calculator solves the roots of polynomials easily.

Polynomials Class 9 State Syllabus Chapter 8 Question 2.
From the four corners of a rect-angle, small squares are cut off and the sides are folded up to make a box, as shown below:

i. Taking a side of the square as x centimetres, write the dimensions of the box in terms of x.
ii. Taking the volume of the box as vfojcubic centimetres, write the relation between v(x) and x as an equation.
iii. Calculate \(\mathrm{V}\left(\frac{1}{2}\right) \quad, \mathrm{V}(1), \quad \mathrm{V}\left(1 \frac{1}{2}\right)\)
Answer:
If 1cm is the length of the small squares they are cut off, then the length of the maked box by folding it up = 7 – 1 – 1 = 5 cm Width = 5 – 1 – 1 = 3 cm, Height = 1 cm
If 2 cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – 2 – 2 = 3 cm Width = 5 – 2 – 2 = 1 cm, Height = 2cm
i. If x cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – x – x = (7 – 2x) cm
Width = 5 – x – x = (5 – 2x) cm
Height = x cm

ii. If volume of the box be v(x), then
v(x) = Length × Width × Height v(x) = (7 – 2x) (5 – 2x)x cm3

v (1) = (7 – 2 × 1) (5 – 2 × 1) 1 = (7 – 2) (5 – 2) × 1 = 5 × 3 × 1 = 15

Hss Live Guru 9th Maths Kerala Syllabus Chapter 8 Question 3.
Consider all rectangles that can be made with a 1-metre long rope. Take one of its sides as x centimetres and the area enclosed as a(x) square centimetres.
i. Write the relation between a(x) and x as an equation.
ii. Why are the numbers a(10) and a(40) equal?
iii. To get the same number as a(x), for two different numbers as x, what must be the relation between the numbers?
Answer:
i. 1 m = 100 cm
If one side is x cm, then the other side is 50 – x.
Area of rectangle = a(x) = x(50 – x)
= 50x – x² cm²
a(x) = 50x – x²

ii. a(10) = 50 × 10 – 10² = 500 – 100 = 400
a(40) = 50 × 40 – 40² = 2000 – 1600 = 400
10 and 40 are the sides of rectangle, so a(10) and a(40) are same .

iii. Numbers must be sides of rectangle. If we add the two numbers together we get the sum as half the length of the wire (50 cm) used to make it.

Textbook Page No. 126

Free Is Polynomial Calculator – Check whether a function is a polynomial step-by-step.

Hsslive Guru 9th Maths Kerala Syllabus Chapter 8 Question 1.
Write each of the relations below in algebra and see if it gives a polynomial. Also, give reasons for your conclusion.
i. A 1-metre wide path goes around a square ground. The relation between the length of a side of the ground and the area of the path.

ii A liquid contains 7 litres of water and 3 litres of acid. More acid is added to it. The relation between the amount of acid added and the change in the percentage of acid in the liquid.

iii. Two poles of heights 3 metres and 4metres are erected upright on the ground, 5 metres apart. A rope is to be stretched from the top of one pole to some point on the ground and from there to the top of the other pole:

The relation between the distance of the point on the ground from the foot of a pole and the total length of the rope.
Answer:
i. Let the side of the ground be x then the side of the square including the path is x + 2 metres.
Area of path = Area of large square – Area of small square.
a(x) = (x + 2)² – x² = x² + 4x +4 – x²
= 4x + 4
Here a(x) is a polynomial. Here x is multiplied by 4 and 4 is added to it.

ii. Ratio of acid in the first fluid = 3/10 litre that is 30%
If x litres of acid is added to it, then change in the amount of acid is
\(=\frac{3+x}{10+x}\)
Change in the percentage of acid is = \(\frac{3+x}{10+x} \times 100 \%\)
\(\mathrm{b}(\mathrm{x})=\frac{300+100 x}{10+x} \%\)

iii.

In figure, AB = x m, BD = 5 – x
Total length of the wire = BC + BE

It is not a polynomial since it involves square root of variable x.

Sum and difference of cubes calculator, multiplying and dividing rational exponents not possible.

Kerala Syllabus 9th Standard Maths Notes Chapter 8 Question 2.
Write each of the operations below as an algebraic expression, find out which are polynomials and explain why.
i. Sum of number and it’s reciprocal
ii. Sum of a number and its square root.
iii. Product of the sum and difference of a number and its square root.
Answer:
i. Let the number be x, then the reciprocal is 1/x
sum = \(x+\frac{1}{x}\)
This is not a polynomial, because here the operation of reciprocal is involved.

ii. Let the number be x, then the square root is √x
sum = x + √x
This not a polynomial because here the square root is taken.

iii. Let x be the number
(x + √x) (x – √x) = x²– x
This is a polynomial.

Std 9 Maths Notes Kerala Syllabus Chapter 8 Question 3.
Find p(1) and p(10) in the following polynomials,
i. p(x) = 2x + 5
ii. p(x) = 3x² + 6x + 1
iii. p(x) = 4x³ + 2x² + 3x + 7
Answer:
i. p(x) = 2x + 5
p(1) = 2 × 1 + 5 = 2 + 5 = 7
p(10) = 2 × 10 + 5 = 20 + 5 = 25

ii. p(x) = 3x² + 6x + 1
p(1) = 3 × 12 + 6 x 1 + 1
= 3 + 6 + 1 = 10
p(10) = 3 × 10² + 6 × 10 + 1 = 300 + 60 + 1 = 361

iii. p(x) = 4x³ + 2x² + 3x + 7
p(1) = 4 × 13 + 2 × 12 + 3 × 1 +7
=4 + 2 + 3 + 7 = 16
p(10) = 4 × 10³ + 2 × 10² + 3 × 10 + 7
= 4000 + 200 + 30 + 7 = 4237

Hss Live Maths 9th Kerala Syllabus Chapter 8 Question 4.
Find p(0), p(1) and p(-1) in the fol-lowing polynomials,
i. p(x) = 3x + 5
ii. p(x) = 3x² + 6x + 1
iii. p(x) = 2x² – 3x + 4
iv. p(x) = 4x³ + 2×2 + 3x + 7
v. p(x) = 5x³ – x2 + 2x – 3
Answer:
i. p(x) = 3x + 5
p(0) = 3 × 0 + 5 = 5
p(1) = 3 × 1 + 5 = 8
p(-1) = 3 x-1 + 5 = 2

ii. p(x) = 3x² + 6x + 1
p(0) = 3 × 0² + 6 × 0 + 1 = 1
p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(-1) = 3 × (-1)² + 6 × (-1) + 1 = -2

iii. p(x) = 2x² – 3x + 4
p(0) = 2 × 0² – 3 × 0 + 4 = 4
p(1) = 2 × 1² – 3 × 1 + 4 = 3
p(-1) =2 × (-1)² – 3 × (-1) + 4 = 9

iv. p(x) = 4x³ + 2x² + 3x + 7
p(0) = 4 × 0³ + 2 × 02 + 3 × 0 +7 = 7
p(1) = 4 × 1³ + 2 × 12 + 3 × 1 + 7 = 16
p(-1) = 4 × (-1)³ + 2 × (-1)2 + 3 × (-1) +7 =2

v. p(x) = 5x³ – x² + 2x – 3
p(0) = 5 × 0³ – 0² + 2 × 0 – 3 =-3
p(1) = 5 × 1³ – 1² + 2 × 1 – 3 = 3
p(-1) = 5 × (-1)³ – (-1)² + 2 × (-1) – 3 = -11

Scert Class 9 Maths Solutions Kerala Syllabus Chapter 8 Question 5.
Find polynomials p(x) satisfying
each set of conditions below.
i. First degree polynomials with p(1) = 1 and p(2) = 3
ii. First degree polynomials with p(1) = -1 and p(-2) = 3
iii. Second degree polynomials with p(0) = 0, p(1) = 2 and p(2) = 6.
iv. Three different second degree polynomials with p(0) = 0 and p(1) = 2.
Answer:
i. General form of a first degree poly-nomial is
p(x) = ax + b
Let p(1) = 1, then a × 1 + b = l
a + b = 1 ………. (1)
Let p(2) = 3, then a × 2 + b = 3
2a + b = 3 …….. (2)
(1) × 2, 2a + 2b = 2 …… (3)
(3) – (2), b = -1
From (1), a + -1 = 1,
a = 1 + 1 = 2 Polynomial p(x) = 2x – 1

ii. General form of a first degree poly-nomial is p(x) = ax + b
Let p(1) = -1, then a × 1 + b = -l
a + b = -1 ……….. (1)
Let p(-2) =3 , then a × (-2) + b = 3
-2a + b = 3 ………. (2)
(1) × 2, 2a + 2b = -2 ………. (3)
(2) + (3), 3b = 1, b = 1/3
From (1), a = 1/3 = -1

iii. General form of a second degree polynomial is
p(x) = ax² + bx + c
Letp(0) = 0, then a × 0² + b × 0 + c = 0
0a + 0b + c = 0.
c = 0 (1)
Letp(1) = 2 ,then a × 1² + b × 1 + c = 2
a + b + 0 = 2
a + b = 2 ………. (2)
Letp(2)= 6, then a × 2² + b × 2 + c = 6
4a + 2b = 6
2a + b = 3 (3)
(3) – (2), a = 1
From (2), 1 + b = 2, b = 2 – 1 = 1
Polynomial p(x) = x² + x

iv. General form of a second degree polynomial is p(x) = ax² + bx + c
Letp(0) = 0, then a x 0 + b x 0 + c = 0
0 + 0 + c = 0
c = 0
Letp(1) = 2, then a × 1² + b × 1 + c = 2
a + b + c = 2
a + b = 2
Selecting a and b such that a + b = 2 will give different polynomials.
a = 1, b = 1
a= 3, b =-l
a=4, b =-2
Three different second degree poly¬nomials are
p(x) = x² + x
p(x) = 3x² – x
p(x) = 4x² – 2x

Polynomials Exam Oriented Questions and Answers

Hss Live 9th Maths Kerala Syllabus Chapter 8 Question 1.
In a polynomial p(x) = 2x³ + ax² – 7x + b.
p(1) = 3, p(2) = 19. Then find the value of ‘a’ and ‘b’?
Answer:
p(x) = 2x³ + ax² – 7x + b
We have p(1) = 3.
p(1) = 2 x 1³ + a x 1² – 7 x 1 + b = 3
2 + a – 7 + b = 3
a + b – 5 = 3
a + b =8 ………. (1)
We have p(2) = 19.

p(2 ) = 2 x 2³ + a x 2²– 7 x 2 + b = 19
16 + 4a – 14 + b = 19
4a + b + 2 = 19
4a + b = 17…….. (2)
From equation (1), (2)
(2) – (1) 4a + b = 17
\(\frac{a+b=8}{3 a=9}\)
a = 9/3 = 3
From equation (1)
a + b = 8
3+b =8
b =5

Hsslive Class 9 Maths Kerala Syllabus Chapter 8 Question 2.
In a polynomial p(x) = 2x³ + 9x² + kx + 3, p(-2) = p(-3). Find the value of k.
Answer:
p(x) = 2x³ + 9x² + kx + 3
p(-2) = 2(-2)³ + 9(-2)² + k(-2) + 3
= -16 + 36 – 2k + 3 = 23 – 2k
p(-3) = 2(-3)³ + 9(-3)² + k(-3) + 3
= -54 + 81 – 3k + 3 = 30 – 3k
We have p(-2) = p(-3),
23 -2k = 30 -3k
-2k + 3k = 30 – 23
k = 7

Maths Kerala Syllabus Std 9 Notes Chapter 8 Question 3.
From the polynomial p(x) = 2x – 3x +1, find p(0), p(1) and p(-1).
Answer:
p(x) = 2x² – 3x +1
p(0) = 2(0)² – 3(0) + 1 = 1
p(1) = 2(1)² – 3(1) +1 = 2 – 3 + 1 = 0
p(-1) = 2(-1)² – 3(-1) + 1 = 2 + 3 + 1 = 6

Kerala Syllabus 9th Standard Maths Notes Pdf Chapter 8 Question 4.
In a polynomial p(x) = 2x³ – 7x² + kx + 20,
p(2) = p(3).
a. Find the value of k.
b. Using the value of k, write the polynomial.
c. Find p(1).
Answer:
a. p(2) = 2 × 2² – 7 × 2² + k × 2 + 20
= 16 – 28 + 2k + 20 = 8+ 2k
p(3) = 2 × 3³ – 7 × 3²+ k × 3 + 20
= 54 – 63 + 3k + 20 = 11 +3k
P(2) = P(3)
8 + 2k = 11 + 3k
k = -3

b. p(x) = 2x³ – 7x² -3x + 20

c. p(1) = 2 – 7 – 3 + 20 = 12

Chapter 8 Polynomials Answers Kerala Syllabus Question 5.
Simplify the followi ng
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
ii. (3x + 4)² – (2x – 1) (3x + 4)
Answer:
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
= (3x + 4) [(2x + 1 + 4x + 3)]
= (3x + 4)[6x + 4]
= (3x x (6x) + (3x) x (4)+ 4x(6x) + 4 x 4
= 18x² + 12x + 24x + 16
= 18x² + 36x + 16

ii. (3x + 4)² – (2x – 1) (3x + 4)
= (3x + 4) [3x + 4 – (2x – 1)]
= (3x + 4)[3x + 4 – 2x + 1] = (3x + 4) (x + 5)
= 3x² + 15x + 4x + 20
= 3x² + 19x + 20

Hsslive Guru Maths 9th Kerala Syllabus Chapter 8 Question 6.
7x³ – 4x² – x + 4 is a polynomial.
a. Write the terms of the polynomial.
b. Write the coefficient of x².
c. Write the constant terms of the polynomial.
d. What is the degree of the polynomial ?
Answer:
a. Terms = 7x², -4x², -x, 4
b. Coefficient of x² = -4
c. Constant term = 4
d. Degree of the polynomial = 3

Hss Live Guru Class 9 Maths Kerala Syllabus Chapter 8 Question 7.
In the polynomial p(x)=3x² – ax + 1,
Find ‘a’ satisfying p(1) = 2.
Answer:
p(x)=3x² – ax + 1
p(1) = 3(1)² – (a × 1) + 1 = 3 – a + 1 =4 – a
Given p(1) = 2
That is, 4 – a = 2 , a = 4 – 2 = 2

Hsslive Maths Class 9 Kerala Syllabus Chapter 8 Question 8.
If p(x)= x³ + 2x² – 3x + 1 and q(x) = x³ – 2x² + 3x + 5
a) Find p(x) + q(x). What is its de-gree?
b) Find p(x) – q(x). What is its de-gree?
Answer:
a) p(x) = x³ + 2x² – 3x + 1,
q(x) = x³ – 2x² + 3x + 5
p(x) + q(x)
= x³ + 2x² – 3x + 1 + x³ – 2x² + 3x + 5
= 2x³ + 6
Degree = 3

b) p(x) – q(x)
= x³ + 2x² – 3x + 1 – (x³ – 2x² + 3x + 5)
= x³ + 2x² – 3x + 1 -x³ + 2x² – 3x – 5
= 4x² – 6x – 4
Degree = 2

9th Std Maths Notes Kerala Syllabus Chapter 8 Question 9.
A right-angled triangle of perpendicular sides 3 cm and 4 cm are ex-tended equally then get another large right-angled triangle. Write the algebraic form of the hypotenuse of the large right-angled triangle.
Answer:
Hypotenuse 2 = base 2 + height 2
(According to pythagorus theorem)
Let the perpendicular side be x cm Length of other sides = 3 + x cm, 4 + x cm

Question 10.
In the polynomial p(x)= 3x² – 4x + 7, check whether p(1) + p(2) = p(3) and P(2) × p(3) = p(6).
Answer:
p(x) = 3x² – 4x + 7
p(1) = 3(1)² – 4(1) + 7 = 3 – 4 + 7 = 6
p(2) = 3(2)² -4(2) + 7 = 12 – 8 + 7 = 11
p(3) = 3(3)² – 4(3) +7 = 27 – 12 + 7 = 22
p(l) + p(2) ≠ p(3)
p(6) = 3(6)² -4(6) + 7 = 108 – 24 + 7 = 91
P(2) × p(3) ≠ p(6)

Question 11.
In the polynomial p(x)= 2x² + ax² – 7x + b,
p(1) = 3 and p(2) = 19. Find a and b.
Answer:
p(x)=2x² + ax² – 7x + b
p(l)=2 x 13+ a x 12 – 7 x 1 +b
=2 + a – 7 + b =a + b + 2 – 7 = a + b – 5
p(1) = 3, a + b – 5 = 3
a + b = 8……… (1)
p(2) = 2 x 2³ + a x 2² – 7 x 2 + b
= 16 + 4a – 14 + b = 4a + b + 2
p(x) = 19
4a + b + 2 = 19
4a + b = 17 ……….. (2)
(2) – (1)
3a = 9, a = 3 a + b – 5 = 3
From equation (1),
a + b = 8
3 + b = 8
b = 5

Question 12.
If p(x) = x² + 3x + 1 and q(x) = 2x – 4, then
i. Find the degree of the polyno-mial p(x) q(x).
ii. If the degree of p(x) × r(x) is 5, then find the degree of the polynomial r(x) .
iii. If p(x) is a third degree poly-nomial and q(x) is a fourth de-gree polynomial then find the degree of p(x) × q(x).
Answer:
i. p(x) = x² + 3x + 1 , q(x) = 2x – 4
p(x) × q(x) = (x² + 3x + 1) (2x – 4)
The degree of p(x) x q(x) is 3.

ii. p(x) = x² + 3x + 1 is a second degree polynomial.
If the degree of the polynomial p(x) × r(x) is 5, then p(x) × r(x) must have a term of x⁵.
x² × x³ = x⁵
So, r(x) is a third-degree polynomial.

iii. If p(x) is a third-degree polynomial, then p(x) must have a term of x³.
q(x) is a fourth-degree polynomial, then q(x) must have a term of x⁴
Then in p(x) × q(x), must have a term of x³ × x⁴= x⁷
So p(x) × q(x) is a seventh-degree polynomial. In general,
If p(x) is an m^th degree polynomial and q(x) is an n,h degree polynomial then p(x) × q(x) is an (m + n),h degree polynomial.

Question 13.
If p(x) = 4x² + 3x + 5 and q(x) = 3x² – x – 7, then find p(x) + q(x) and p(x) – q(x).
Answer:
p(x) + q(x) = (4x² + 3x + 5) + (3x² – x – 7)
= 4x² + 3x² + 3x – x + 5 – 7
= 7x² + 2x – 2
p(x) – q(x) = (4x² + 3x + 5) – (3x² – x – 7)
= 4x² + 3x + 5 – 3x² + x + 7
= x² + 4x + 12

Question 14.
In polynomial p(x) = ax³+ bx² + cx + d, p(l) = p(-l). Prove that a + c = 0.
Answer:
p(x) = ax³ + bx² + cx + d
p(l) = a(1)³ + b(1)² + c(1) + d
= a+ b + c + d
p(-1) = a(-1)³ + b(-1)² +c(-1) + d
= -a + b – c + d
p(1) = p(-1) is given ie, a + b + c + d = -a + b – c + d
2a + 2c = 0; 2(a + c ) = 0;
ie, a + c = 0

Question 15.
If we divide a polynomial by (x-2) we get quotient as x² + 1 and remainder as 5. Find the polynomial.
Answer:
Polynomial = (x² + 1) (x – 2) + 5
= x³ + x – 2x² – 2 + 5 ,
= x³ – 2x² + x + 3

Students can Download Maths Chapter 7 Similar Triangles Questions and Answers, Summary, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 9th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles in Malayalam Medium

Similar Triangles Questions and Answers in Malayalam

Students can Download Chemistry Part 2 Chapter 1 Acids, Bases, Salts Questions and Answers, Summary, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 9th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 5 Acids, Bases, Salts in Malayalam Medium

Acids, Bases, Salts Textual Questions and Answers in Malayalam

You can Download दौड़ Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

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दौड़ Textual Questions and Answers

दौड़ विधात्मक प्रश्न

Hsslive Guru 9th Hindi प्रश्ना 1.
बारिश का अनुभव सुहाना होता है। आपको बारिश कैसे महसूस होता है? इसपर एक छोटा-सा संस्मरण लिखें।

उत्तर:
बारिश मेरे लिए बहुत खुशी का अवसर था। बारिश देखने के लिए मैं अपने घर की खिड़की के पास खड़ा होता था। बारिश में छाता लेकर स्कूल जाना मुझे बहुत पसंद था। पीठ पर बस्ता टाँगकर, एक हाथ में छाता पकड़कर, दूसरे हाथ से छाते के कोने से आता पानी. को छुआ करता था। खेल घंटी के समय अगर बारिश होता तो कागज़ से नाव बनाकर पानी रख देते थे। बहती नाव के साथ हम भी दौडते थे। बारिश की ठंड में चादर ओढ़कर सोना भी मुझे पसंद था।

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Kerala Syllabus 9th Standard Physics Notes Chapter 2 Question 1.
Draw Position – Time graph. What is the nature of the grap?

Answer:

The graph is straight line.

Class 9 Physics Chapter 2 Kerala Syllabus Question 2.
Using the data given below, draw a position-time graph

Answer:

Physics Class 9 Chapter 2 Equations Of Motion Kerala Syllabus Question 3.
The position-time graph regarding the motion of a car is given. Find out from the graph the distance traveled by the car in 8 s.

Answer:
Draw a perpendicular to the position-time graph at the eighth second. From the point where the perpendicular meets the graph, draw another perpendicular to the Y-axis. The point at which this perpendicular meets the Y-axis is the distance traveled by the car in 8 sec. ie 40m.
The distance covered by the car is 8s = 40 m

9th Class Physics Chapter 2 Equations Of Motion Kerala Syllabus Question 4.
Complete the table using the data from the following position-time graph related to the motion of a car.

Change the scale and draw another graph.

Displacement of the body in 5s from both graphs is 25 m
We use scales while drawing a graph, to contain the given measurements on a graph paper. The size of the graph decreases as we increase the scale considered. But the value doesn’t change.
Motion of a car is shown below using a diagram.

Complete the table with the help of the figure

Position – time graph of a body moving with uniform speed will be a stright line. The body will be moving with nonuniform speed when the graph is not in a straight line.

9th Class Physics Notes Kerala Syllabus  Question 5.
Using the graph find out the displacement of the car in 3 s.
Answer:
Displacement = 11 m

Kerala Syllabus 9th Standard Physics Notes Question 6.3
From the graph find out the time taken by the car to travel a distance of 45 m.
Answer:
Time = 8.4s

Velocity – Time Graph

With the given data in the table, draw a velocity-time graph.

The displacement of the object between the 2^nd and 8^th second
= Velocity × time
= AB × AD = 10 × 6 = 60 m
= Area of ABCD
= Area of the portion under the part BC of the graph
Displacement of a body within a definite interval of time is equal to the area of the portion under velocity-time graph.

Equations Of Motion

The velocity-time graph of an object traveling with uniform acceleration (freely falling stone) is given below.

PS = AR
= u
QR = v
= t² – t¹
AQ = QR – AR
= v- u
Formatin of equation showing velocity – time relation.
Acceleration = \(\frac{\text { Change in velocity }}{\text { time }}\)

v- u = at
v = u + at.
This is the first equation of motion.
Formation of equation showing Position- Time Relation
Displacement = area of trapezium PQRS

S = ut + \(\frac { 1 }{ 2 }\) at² this is known as second equation of motion.
Formation of equation showing Position – Velocity Relation
Displacement = The area of the quadrilateral

This equation helps us to calculate the final velocity v of an object using u, a and s, even if the time taken to travel is unknown.
v² = u² + 2as
Equations of motion
v = u+at
s = ut + \(\frac { 1 }{ 2 }\) at²
v² = u² + 2as

9th Class Physics Equations Of Motion Kerala Syllabus Question 7.
The velocity of a body starting from rest is 20 m/s in the 4^th second and 40 m/s in the 8^th second. What is the distance traveled by the body between the 4^th and 8^th second?
Answer:
Velocity at the 4^th second u = 20 m/s
Velocity at the 8^th second v = 40 m/s

9th Physics Notes Kerala Syllabus Question 8.
A car came to rest when brake was applied for 4s to get a retardation of 3 m/s². Calculate how far the car would have traveled after applying the brake.
Answer:
a = -3 m/s²
t = 4s
v = 0
v = u+at
-u = -3 × 4 + 0
u =12 m/s
Displacement of the car

Kerala Syllabus 9th Standard Physics Notes Chapter 1 Question 9.
If the velocity of a car moving with uniform velocity changes from 20 m/s to 40 m/s in 5s
a) What is the acceleration of the car?
b) What is the displacement by the car during this time interval?
Answer:
a) u = 20 m/s,
v = 40 m/s,
t= 5 s

Kerala Syllabus 9th Standard Physics Notes English Medium Question 10.
If the velocity of a train starting from rest becomes 72 km/h in 10 minutes.
a) What is the acceleration?
b) Calculate the distance traveled by the train within this time interval
Answer:
a) Here , u = 0,
v= 72 km/h

Kerala Syllabus 9th Standard Physics Notes Chapter 2 Malayalam Medium Question 11.
A car starting from rest travels 100 m in 5s with uniform acceleration. Find the acceleration of the car.
Answer:
u = 0,
t= 5s,
s = 100 m,
a=?

Kerala Syllabus 9th Standard Physics Textbook Solutions Question 12.
An object starting from rest travels with an acceleration of 5 m/s². What will be its velocity after 3 s?
Answer:
u=0,
a= 5m/s²,
t= 3s,
v=?
v= u + at,
v= 0 + 5 × 3,
v= 15 m/s
Complete the table using the graph shown below. Compare.


Answer:

In a velocity-time graph when we draw perpendiculars to the graph from the time specified, we get a geometrical shape. The area of this geometrical shape will give displacement in the specified time interval.

Let Us Assess

Question 1.
Draw position-time graph

Answer:

Question 2.
Draw speed – time graph

Answer:

Question 3.
Examine the graph and answer the following questions.

a) Is the motion of the object uniform/nonuniform?
b) Say whether the acceleration from O to A is uniform? What about from A to B?
Answer:
a) Non- uniform
b) O → A. uniform acceleration (0.4 m/s²)
A → B Uniform retardation (1m/s²)

Question 4.
Complete the table by analyzing the graph


Answer:

Question 5.
If a velocity of a train which starts from rest is 72 km/h (20 m/s) after 5 minutes, find out its acceleration and the distance traveled by the train in this time.
Answer:

Question 6.
A car attains a velocity of 54 km/h (15 m/s) within 5 seconds from an initial velocity of 18 km/h (5m/s). Calculate its acceleration and displacement.
Answer:
u = 5 m/s .
t = 5s
v = 15 m/s

= 2 m/s²
s = ut + 1/2 at²
= 5 × 5 + 1/2 × 2 × 5²
= 25 + 25
= 50 m

Question 7.
Analyze the graphs given below.

a) Which graph indicates uniform velocity?
b) Which graph indicates nonuniform acceleration?
c) Which graph indicates the motion of freely falling stone?
Answer:
a) graph (2)
b) graph (4)
c) graph (3)

Equations of Motion More Questions and Answers

Question 1.
Write the equations of motion. What does each letter indicate?
Answer:
v = u + at
s = ut +1/2 at²
v² = u² + 2as
Where
u – Initial velocity
v – final velocity
a – acceleration
t – time
s – Distance (Displacement)

Question 2.
An object starting form rest travels with a uniform acceleration of 5 m/s². Calculate the velocity and distance traveled after 1 minute?
Answer:
v = u + at
u = 0
= 0+ 5m/s² x 60 s
a = 5m/s²
= 0+ 300 m/s
t = 1
mt = 60 sec
= 300 m/s
s =ut + 1/2 at²
= 0 × 60 + 1/2 × 5 × 60²
= 9000 m

Question 3.
Draw a velocity-time graph on the basis of the given table.

Answer:

X – axis – 1cm = 5s
Y – axis – 1cm = 5 m/s