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Students can Download Chemistry Chapter 3 Reactivity Series and Electrochemistry Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 3 Reactivity Series and Electrochemistry Questions and Answers Malayalam Medium

SCERT 10th Standard Chemistry Textbook Chapter 3 Solutions Malayalam Medium

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SSLC Physics Chapter 7 Energy Management Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 7 Solutions Malayalam Medium

You can Download Polynomials and Algebra Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 10 Polynomials and Algebra Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 10 Polynomials and Algebra Notes

Textbook Page No. 237

Free Degree and Leading Coefficient Calculator – Find the leading coefficient of a polynomial function step-by-step.

Polynomials Class 10 Kerala Syllabus  Questions 1.
Write the second degree polynomials. given below as the product of two first degree polynomials. Find also the solutions of the equation p(x) = 0 in each.
i. p(x) = x² – 7x+12
ii. p(x) = x² + 7x + 12
iii.p (x) = x² – 8x +12
iv. p(x) = x² + 13x +12
v. p (x) = x² + 12x – 13
vi. p (x) = x² – 12x – 13
Answer:
i. p (x) = x² – 7x + 12
a + b = –7, ab = 12
a = –3, b = –4
x² – 7x + 12 = (x – 3) (x – 4)
x² – 7x + 12 = 0
(x – 3) (x – 4) = 0
x – 3 = 0, x – 4 = 0
x = 3, x = 4
ii. p(x) = x² + 7x + 12
a + b = 7, ab= 12 a = 3, b = 4
x² + 7x + 12 = (x + 3) (x + 4)
x² + 7x + 12 = 0
(x + 3)(x + 4) = 0
x = 3, x = 4
iii. p(x) = x^{2 –} 8x + 12
a + b = 8, ab= 12
a = 6, b = –2
x² – 8x + 12 = (x – 6) (x – 2)
(x – 6) (x – 2) = 0
x = 6, x = 2
iv. p(x) =x² + 13x + 12
a+b = 13, ab = 12
a =12, b=1
(x² +13x + 12) = (x + 12) (x + 1)
x²+13x + 12 = 0
(x+ 12) (x+ 1) = 0
x+ 12 = 0, x+ 1 = 0
x = -12 ,x = –1
v. p(x) = x² + 12x – 13
a = –13, b = 1
x² + 12x – 13 = (x + 13)(x – 1)
x + 13 = 0, x – 1 =0
x = –13, x= 1 .
vi. p(x) = x² – 12x – 13
x² – 12x – 13 = (x – a) (x – b)
= x² – (a + b) x + ab
a + b= 12 ab = –13
(a – b)² =(a + b)² – 4ab
= (12)² – 4x – 13 = 196
a – b = 14
a + b= 12
a= 13; b = –1
x² – 12x – 13 = (x – 13)(x + 1)
x – 13 = 0, x + 1 =0
x= 13 ,x = –1

Textbook Page No. 240

Is this a Polynomial Calculator is an online tool that helps to calculate the result of addition, subtraction, multiplication, and division of two polynomials.

Sslc Maths Chapter 10 Kerala Syllabus Questions 1.
In each pair of polynomials given below, find the number to be subtracted from the first to get a polynomial for which the second is as factor. Find also the second factor of the polynomial got on subtracting the number.
i. x² – 3x + 5, x – 4
ii. x² – 3x + 5, x + 4
iii. x² + 5x – 7, x – 1
iv. x² – 4x – 3, x – 1
Answer:
p(x) = x² – 3x + 5
If x – 4 is a factor, p(4) = 0
p(4) = (4)² – 3 x 4 + 5 = 9
For x – 4 to become a factor of p (4) must be equal to zero.
For p (4) = 0 here we have to subtract 9 from p(x).
That is, add -9 to p(x) for (x – 4) become a factor.
∴ p(x) = x² – 3x + 5 – 9 = x² – 3x – 4
x² – 3x – 4 = (x – a) (x – b)
= x² – (a + b) x + ab
a + b = 3
ab = –4
(a – b)² = (a + b)² – 4 ab
= (3)² – 4x – 4 = 25
a – b = 5
a + b = 3
a = 4; b = –1
x² – 3x – 4 = (x – 4)(x + 1)
Second factor is (x +1)
ii. p(x) = x² – 3x + 5
If x + 4 is a factor, p(–4) = 0
p(–4) = (–4)² – 3x – 4 + 5 = 33
For x + 4 to become a factor of p (–4)
must be equal to zero.
For p (–4) = 0 here we have to subtract 33 from p(x).
That is, add –33 to p(x) for (x + 4) become a factor.
∴p(x) = x² – 3x + 5 – 33 = x² – 3x – 28
x² – 3x – 28 = (x – a) (x – b)
= x² – (a+b) x + ab
a + b = 3
ab = -28
(a – b)² = (a + b)² – 4ab
= (3)² – 4x – 28 = 121
a – b= 11
a + b = 3
a = 7;
b = -4 x² – 3x – 28 = (x – 7)(x + 4)
Second factor is (x – 7)
iii. p(x) = x² + 5x – 7
If x – 1 is a factor, p(1) = 0
p(1) = (1)² +5 x 1 – 7 = –1
For x – 1 to become a factor of p (1) must be equal to zero.
For p (1) = 0 here we have to subtract –1 from p(x).
That is, add 1 to p(x) for (x – 1) become a factor.
p(x) = x² + 5x – 7 + 1 = x² + 5x – 6
x² + 5x – 6 = (x – a) (x – b)
a + b = 5
ab = –6
a = –6; b= 1
x² + 5x – 6 = (x + 6)(x – 1)
Second factor is (x + 6)
iv. p(x) = x² – 4x – 3
If x – 1 is a factor, p(1) = 0
p(1) = (1)² – 4 x 1 – 3 = –6
For x – 1 to become a factor of p (1) must
be equal to zero.
For p (1) = 0 here we have to subtract -6 from p(x).
That is, add 6 to p(x) for (x – 1) become a factor.
∴ p(x) = x² – 4 x – 3 + 6 = x² – 4x +3
x² – 4x + 3 = (x – a) (x – b)
a + b = –4 ab = 3
a =1; b = 3
x² – 4x +3 = (x – 1)(x – 3)
Second factor is (x – 3)

Still, using a computation tool like a factor complex polynomials calculator, you can streamline things for you.

Polynomials Class 10 State Syllabus Questions 2.
In the polynomial x² + kx + 6, what number must be taken ask to get a polynomial for which x –1 is a factor? Find also the other factor of that polynomial.
Answer:
p (x) = x² + kx + 6
If (x – 1) is a factor of p(x)
then p(1) = 0
p(1) = 12 + k x 1 + 6 = 7 + k
7 + k = 0
k= –7
∴ p(x) = x² – 7x + 6
a + b = 7
ab = 6
a= 1, b = 6
factors are (x – 1 )(x – 6)
Second factor is (x – 6)

Polynomials Class 10 Kerala Syllabus Questions 3.
In the polynomial kx² + 2x – 5, what number must be taken ask to get a polynomial for which x –1 is a factor?
Answer:
p(x) = kx² + 2x – 5
If (x – 1) is a factor, then p(1) = 0
p(1) = k(l)² + 2 x 1 – 5
= k + 2 – 5 = k – 3
k – 3 = 0 k = 3

Textbook Page No. 242

Sslc Polynomials Questions And Answers Kerala Syllabus Question 1.
Write the second degree polynomials given below as die product of two first degree polynomials:
i. x² – 20x + 91
ii. x² – 20x + 51
iii. x² + 5x – 84
iv. 4x² – 16x +15
v. x² – x – 1
Answer:
p(x) = x² – 20x + 91
We must solve the equation p(x) = 0
x² – 20x + 91 = 0

Sslc Maths Polynomials Solutions Kerala Syllabus Question 2.
Prove that none of the polynomials below can be factored into a product of first degree polynomials:
i. x² + x + l
ii. x² – x + l
iii. x² + 2x + 2
iv. x² + 4x + 5
Answer:
i. x² + x+1
We must solve the equation p(x) = 0
x² + x +1 =0

no solutions
p(x) doesn’t have any first degree factors.
ii. x² – x + 1
We must solve the equation p(x) = 0
x² – x + 1 = 0

no solutions
p(x) doesn’t have any first degree factors
iii. x² + 2x + 2
We must solve the equation p(x) = 0
x² + 2x + 2 = 0

no solutions
p(x) doesn’t have any first degree factors.
iv. x² + 4x + 5
We must solve the equation p(x) = 0
x² + 4x + 5 = 0

no solutions
p(x) doesn’t have any first degree factors.

Sslc Maths Chapter 10 Solutions Kerala Syllabus Question 3.
In the polynomial p(x) = x² + 4x + k, up to what number can we take ask, so that p(x) can be factorized as a product of two first degree polynomials?
Answer:
p(x) = x² + 4x + k
p(x) = 0
x² + 4x + k = 0

p(x) can be factorized as a product of two first degree polynomials, the \(\sqrt{16-4 k}>0\) So, k can take value up to 4.

Polynomials Orukkam Questions & Answers

Worksheet 1

Polynomials Class 10 Hsslive Kerala Syllabus Question 1.
Write the product (x – 1) x (x + 1)
Find the product of (x – 1),(x + 1),(x + 2) If the poduct is p(x)find p(1), (-1), p(-2) Write the solution of the equation p(x) = 0.
Answer:
(x – 1)x (x + 1) = x² – 1
(x – 1) (x + 1) (x + 2)=(x² – 1) (x + 2)
= x³ – x + 2x² – 2 = x³ + 2x² – x – 2 = 0
p(x) = x³ + 2x² – x – 2 = 0
p(1) =1 + 2 – 1 –  2 = 0
p(–1) = – 1 + 2 + 1 – 2 = 0
p(–2) = – 8 + 8 + 2 – 2 = 0
1, – 1, – 2 are the solutions of the equation p(x) = 0.

Hss Live Guru 10th Maths Kerala Syllabus Question 2.
Expand (x – a)(x – b). If x² – 7x + 12 = (x – a)(x – b) then find a+b.
Also find ab Calculate the values of a, b. Write the factors of (x² – 7x +12 ).
Find the solutions of (x² – 7x + 12).
Answer:
(x – a) (x – b) = x² – bx – ax + ab
= x² – x (a + b) + ab
= x² – 7x + 12 = (x – a)(x – b)
a + b = 7
ab = 12
(a + b)² – 4ab = 7² – 4 x 12 = 49 – 48 = 1 = a – b
a + b = 7
a – b = 1, 2a = 8 a = \(\frac { 8 }{ 2 }\) =4
b = 7 – 4 = 3 a = 4 and b = 3.
Solution of x² – 7x + 12 is x² – 7x + 12 = (x – 4) (x – 3)
Solutions = 4, 3

Maths Questions And Answers For Class 10 Kerala Syllabus Question 3.
If p(x) = x3 – 6×2 + 11x – 1 then find p(1), p(2), p(3). Find p(x) – p(1), p(x) – p(2), p(x) – p(3), p(x) – p(1). Write the solutions of p(x) – p(1) = 0
Polynomials Class 10 Worksheet with Answer:
p(x) = x³ – 6x² + 11x – 1
p(1) = 1 – 6 + 11 – 1 = 5
p(2) = 8 – 24 + 22 – 1 = 5
p(3) = 27 – 54 + 33 – 1 = 5
p(x) – p(1) = x³ – 6x² + 11x – 6
p(x) – p(2) = x³ – 6x² + 11x – 6
p(x) – p(3) = x³ – 6x² + 11x – 6
p(x) – p(1) = x³ – 6x² + 11x – 6 = 0
If x = 1, 2, 3 then p(x) – p(1) = 0.
Factors of the equations are (x – 1), (x – 2), (x – 3).
Solutions of the equations are 1, 2, 3.

Hss Live Maths 10th Kerala Syllabus Question 4.
When p(x)is divided by (ax + b), the quotient is q(x)and the remainder is c. p(x) = (ax + b) x q(x) + c
When does the value of p(x) equal to c \(p\left(\frac{-b}{a}\right)=\left(a \times \frac{-b}{a}+b\right) \times q\left(\frac{-b}{a}\right)+c\)
What is the remainder when p(x)is divided by ax + b. When does(ax + b)becomethe factor of p(x).
Answer:
The remainder obtained when p(x) is divided by (ax + b) = \(p\left(\frac{-b}{a}\right)\)
When \(p\left(\frac{-b}{a}\right)\) = 0, then (ax + b) will be a factor of p(x).

Worksheet 2

Hsslive Maths 10th Kerala Syllabus Question 5.
Write the following as the product of first degree polynomials
1. x² + 7x + 12
2. x² + 3x + 2
3. x² – 9x – 22
4. 2x² + 5x – 3
Answer:
1. x² + 7x + 12 = (x + 4)(x + 3)
2. x² + 3x + 2 = (x + 2)(x + 1)
3. x² – 9x – 22 =(x – 11)(x + 2)
4.2x² + 5x – 3 =\(\left(x-\frac{1}{2}\right)\)(x+3)=(2x-l)(x+3)

Hss Live Maths 10 Kerala Syllabus Question 6.
Write a polynomial p(x) in which p(1) = 0, P(-2) = 0, p(2) = 0.
Answer:
If p( 1) = 0, then x – 1 is a factor.
If p(–2) = 0, then x + 2 is a factor.
If p(2) = 0, then x – 2 is a factor.
p(x) = (x – 1) (x + 2) (x – 2) = x³ – x² – 4x + 4

Class 10 Kerala Syllabus Maths Solutions Question 7.
Write a second degree polynomial p(x) in which \(p(\sqrt{2}+1)=p(\sqrt{2}-1)=0\)
Answer:

Hsslive 10th Maths Kerala Syllabus Question 8.
Prove that x² + 2x +2 cannot be written as the product of first degree polynomials
Answer:
b2 – 4ac = 22 – 4 x 1 x 2 = –4 < 0
∴ x² + 2x + 1 cannot be written as the product of first degree polynomials

Worksheet 3

Polynomials Class 10 In Malayalam Kerala Syllabus Question 9.
Find the remainder and quotient obtained by dividing x³ – 5x² + 7x + 3by (x + 2).
Answer:
Let quotient = x² + ax + b and remainder be c, then
x³ – 5x² + 7x + 3 =(x + 2)(x² + ax + b) + c +
= x³ – 5x² + 7x + 3 = x³ + ax² + bx + 2x² + 2ax + 2b + c
= x³ +( a + 2) x² + (b + 2a)x + 2b + c From this equation,
a + 2 = –5,
b + 2a = 7, 2b + c = 3
i.e., a = –7
b= 21
c = -39
quotient = x² – 7x + 21,
remainder = –39

Maths Polynomials Class 10 State Syllabus  Question 10.
Given x – 1 is a factor of x² + ax + b. Prove that (a + b = –1)
Answer:
Let (x – 1) be a factor, then p(1)=0
p(1) = 12 + a x 1 + b = 0
i.e., a + b = –1

Hss Guru Maths 10 Kerala Syllabus Question 11.
p(x) = (4x² – 1)(x + 2)Write p(x) as the product of first degree factors.Write p(x) in the form of a trird degree polynomial What is the remainder obtained by dhpding 4x³ + 6x² – x + 2by (x + 2) .What is the re¬mainder obtained by dividing 4x³ + 6x² – x + 1 by (2x – 1).
Answer:

Polynomials SCERT Questions & Answers

Question 12.
Write the second-degree polynomial p(x)= x² + x – 6 as the product of first-degree polynomials. Find also the solution of the equation p (x)=0 [Score: 4, Time: 7 minute]
Answer:

Question 13.
For what values of x, the polynomial 2x² – 7x – 15 is equal to zero? Write this polynomial as the product of two first degree polynomials. [Score: 4, Time: 7 minute]
Answer:

Question 14.
Write the polynomial p(x) = x² + 4x + 1 as the product of two first degree polynomials. Find the solution of the equation p (x) = 0. [Score: 4, Time: 5 minute]
Answer:
x² + 4x + 1 = (x – a) (x – b) = x² – (a + b) x + ab(1)
a + b = –4, ab = 1, a – b = 2√3
a = –2 + √3, b = –2 – √3 (1)
x² + 4x + 1 =(x + 2+ √3 ) (x + 2 – √3 ) (1)
x² + 4x + 1 = 0 =>(x + 2 + √3 )(x + 2 –√3 ) = 0 (1)
x = –2 – √3 ,or x = –2 + √3

Question 15.
In the polynomial p (x) = x²+ ax + b p (3 + √2 )= 0, p (3 – √2 ) = 0, write this polynomial after finding a and b. [ Score: 4, Time: 5 minute]
Answer:
p (x) = x² + ax + b
p(3 + √2) = 0, (x – 3 – √2) is a factor (1)
p(3 – √2 ) = 0, ( x – 3 + √2) is a factor (1)
p(x) = x² + ax + b = (x – 3 – √2) (x – 3 + √2)
= (x – 3)² – ( √2)² (1)
x² + ax + b = x² – 6x + 7 (1)

Question 16.
What number should be added to the polynomial p(x) = x² + x – 1, so that (x – 2)is a factor of the new polynomial. [Score: 4, Time: 6 minute]
Answer:
p (x) = x² + x – 1, remainder p(2) (1)
p (2) = (2)² + 2 – 1 = 5 (1)
For x – 2 to become a factor of p (2) must be equal to zero.
For p (2) = 0 here we have to substract 5 ffomp(x). (1)
That is, add –5 to p(x) for (x – 2) become a factor. (1)

Question 17.
What is the smallest natural number k, for which the polynomial 2x² + kx + 6 can be written as a product of two first degree polynomials? Write down the polynomial using k and express it as the product of two first degree polynomials [Score: 4, Time: 8 minutes]
Answer:

Question 18.
Method to check whether(x – a), and (x + a)are factors of a polynomial P(x).
Check whether (x + 2) and (x – 5) are factors of the polynomial p(x) = x² + 7x + 10 [Score: 4, Time: 6 minute]
Answer:
When a polynomial p(x) is divided by (x – a), if p(a) = 0 then (x – a) is a factor of p(x). When a polynomial p(x) is divided by (x + a), if p(–a) = 0 then (x + a) is a factor of p(x).
p(x) = x² + 7x + 10
p(–2) = 4 – 14 + 10 = 0 (1)
∴ x + 2 is a factor (1)
Remainder p(5) = (5)² + 7(5) +10 (1)
= 25 + 35 + 10 ≠ 0 (1)
∴ x – 5 is not a factor

Question 19.
When dividing x² + ax + b by (x – 2) and (x – 3) the remainder is zero. What are the numbers a and b. [Score: 3, Time: 5 minute]
Answer:
p(x) = x² + ax + b = (x – 3)(x – 2)
= x² – 5x + 6
a = –5, b = 6 (3)

Polynomials Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 20.
Check whether x – 1 is a factor of 3x³ – 2x² – 3x + 2.
Answer:
P(1)=3 x 13 – 2 x 12 – 3 x 1 + 2 = 0
Therefore x – 1 is a factor.

Question 21.
If (x – 1) is to be a factor of p(x) = a²x² – 4ax + 4a – 1. What should be the value of ‘a’ ?
Answer:
p(1) = 0
a2 – 4a + 4a – 1 = 0
a2 – 1 =0
a = +1 or –1

Question 22.

Answer:

Question 23.
Prove that (x – 1) is a factor of x¹³ – 1
Answer:
p(x) = x¹³ – 1
p(1)= 1¹³ – 1 = 1 – 1=0
p(1) = 0
(x – 1) is a factor of p(x)

Question 24.
Write the solution of polynomials p(x) = x² – 7x + 12.
Answer:
x² – 7x + 12 = (x – 4)(x – 3)
p(x) = 0 (x – 4) ( x – 3) = 0
x = 4, x = 3

Question 25.
The quotient is x2 – 5x + 6 when the polynomial p(x) is divided by (x – 1), and the remainder is 7. Then
a. p(x) = (………. ) ( ……… ) + 7. Complete it
b. Find P(2).
Answer:
a. p(x) = (x² – 5x + 6) (x – 1) + 7
b. p(2) = 4 – 10 + 6 + 7 = 7

Short Answer Type Questions (Score 3)

Question 26.
a. Find the remainder when
x³ – 4x³ + 12x – 45 is divided by (x – 2)?
b. Find the value of k if the remainder is zero on dividing 2x³ + 4x² – 10x + k by (x – 1)
Answer:
a. p(x) = x³ – 4x² – 12x – 45
remainder = p(2) = 23 – 4(2)² + 12 x 2 – 45 = 32 – 61 = –29
b. p(x) = 2x³ + 4x² – 10x + k; p(l) = 0
=> 2 x 13 + 4 x 12 – 10 x 1 + k = 0
2 + 4 – 10 + k = 0 – 4 + k = 0; k = 4

Question 27.
Factorise 3x² + 5x + 2 completely.
Answer:

Question 28.
Which first-degree polynomial is added to the polynomial 5x³ + 3x² to get x² – 1 as a factor.
Answer:
p(x) = 5x³ +3x² + ax + b
∴ x² – 1 isafoctorthen p(1), p(–1) will be zero.
p(1) = 5 x 1 + 3 x 1 + a x 1 + b = 0
= 5x – 1 + 3 x 1 + ax – 1 + b = 0
a + b = –8 (1)
= 5x – 1 + 3 x 1 + ax – 1 + b = 0
–a + b = 2 (2)
Find the solutions of the equation
b = –3, a = –5 added polynomial = –5x – 3

Long Answer Type Questions (Score 4)

Question 29.
a. Find the value of k if remainder when 5x³ + 4x – 11x + k is divided by (x – 1) is 0.
b. When x³ – 2x² + kx + 7 is divided by (x – 4) remainder is 11. Find k.
Answer:

Question 30.
a Show that the polynomial x + x + 1 has no first degree factors,
b. What is the remainder when the polynomial (x – 1) (x – 2) (x – 3) is divided by (x – 1)?
c. When (x – 1) (x – 2) (x – 3) + 2x + k is divided by (x – 1), the remainder is 10.
Then find out the remainder when it is divided by (x – 2).
Answer:

Long Answer Type Questions (Score 5)

Question 31.
Write 2x² + 5x + 3 as a product of two first degree polynomials.
Answer:

Question 32.
If(x + 1)and(x – 1) are factors of x3 + 2x² + px +q, find p and q.
Answer:
.

Polynomials Memory Map

Students can Download Maths Chapter 6 Coordinates Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 6 Coordinates Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 6 Solutions Malayalam Medium

Students can Download Maths Chapter 9 Geometry and Algebra Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 9 Geometry and Algebra Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 9 Solutions Malayalam Medium

You can Download Refraction of Light Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Physics Chapter 5 Refraction of Light Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 5 Refraction of Light Solutions

Textbook Page No. 103

Sslc Physics Chapter 5 Kerala Syllabus Question 1.
Do we observe the objects underwater at their original position?
Answer:
No.

Sslc Physics Chapter 5 Notes Kerala Syllabus Question 2.
Observe the path of light. Do you observe anything out of the ordinary? Can you de¬pict it in the science diary?
Answer:
The light rays below the water undergo deviation at the surface of water.

Textbook Page No. 104

Refraction Of Light Class 10 Kerala Syllabus Question 3.
Which are the media involved here?
Answer:
Air & Water.

Sslc Physics Refraction Of Light Kerala Syllabus Question 4.
What happens to the path of the light?
Answer:
Path of light undergoes deviation.

Sslc Physics Chapter 5 Refraction Of Light  Question 5.
Where does the deviation of the ray take place?
Answer:
At surface of water.

Physics Class 10 Chapter 5 Kerala Syllabus Question 6.
Why does the ray of light undergo a deviation here?
Answer:
The difference of speed of light rays in different media.

Physics Chapter 5 Class 10 Kerala Syllabus Question 7.
Does light pass through all the media at the same speed?
Answer:
No

Class 10 Physics Chapter 5 Kerala Syllabus Question 8.
As the optical density of a medium increases, the speed of light through it decreases.
What if the optical density decreases?
Answer:
Speed of light increases.

Class 10 Physics Chapter 5 Notes Kerala Syllabus Question 9.
Can the media given in the table be arranged in the increasing order of their optical densities?
Air<……….. <……… <……….
Answer:
Air < Water < Glass < Diamond

Textbook Page No. 105

10th Class Physics Chapter 5 Notes Kerala Syllabus Question 10.
Which is the incident ray on the surface of separation CD?
Answer:
QR

Refraction Of Light Class 10 State Board Kerala Syllabus Question 11.
The angle between the incident ray and the normal is called the angle of incidence. If so, can you explain what is angle of refraction??
Answer:|
∠r Between the refracted ray and the normal

Class 10 Physics Chapter 5 Kerala Syllabus Question 12.
Is the angle of refraction greater or lower than the angle of incidence when it goes from air to glass?
Answer:
Lower

Sslc Physics Chapter 5 Questions And Answers Kerala Syllabus Question 13.
What about from glass to air?
Answer:
Greater

Chapter 5 Physics Class 10 Kerala Syllabus Question 14.
Which is of greater optical density – air or glass?
Answer:
Glass

Physics Chapter 5 Class 10 Notes Kerala Syllabus Question 15.
While going from air to glass, the refracted ray deviates towards the normal/deviates away from the normal.
Answer:
Deviates towards the normal. While entering from air to glass (from a medium of lower optical density to that of a greater one) the refracted ray deviates away from the normal.

Textbook Page No. 106

Chapter 5 Physics Class 10 Notes Kerala Syllabus Question 16.
What happens while it goes from glass to air?
Answer:
Deviates away from the normal. Normal. While entering from glass to air (from a medium of greater optical density to that of a lower one) the refracted ray deviates away from the normal.

Physics Class 10 Refraction Of Light Solution Kerala Syllabus Question 17.
Are the angle of incidence, angle of refraction and the normal at the point of incidence on the same plane?
Answer:
Yes. The angle of incidence, angle of refraction and the normal at the point of incidence are in the same plane.

10th Class Physics Chapter 5 Kerala Syllabus Question 18.
Does refraction take place for a ray while entering a glass slab normal to it?
Answer:
No

Sslc Physics Chapter 5 Solutions Kerala Syllabus Question 19.
Examine using a ray of light from a laser torch. Ray diagrams of a light ray passing through different media are depicted. Find out the appropriate figures by observing these figures and also based on the concepts you have developed from the textbook.


Answer:

Textbook Page No. 107

Sslc Physics Chapter 5 Notes Pdf Kerala Syllabus Question 20.
Its an experiment, can you find out the path of light through a triangular prism using a laser torch? Record it in the science diary.

Answer:

10th Class Physics Refraction Of Light Kerala Syllabus Question 21.
What change occurs in the angle of refrac¬tion with the increase in the angle of inci¬dence when a ray enters a medium?
Answer:
Increases

Physics 10th Class Chapter 5 Kerala Syllabus Question 22.
Can you analyze the table and find out the change? What other conclusions can you arrive at from the table?
Answer:
When light passes through different pairs of media, the angle of refraction increases with the angle of incidence. The ratio of the sine of the angle of incidence to the sine of the angle of refraction \(\left(\frac{\sin i}{\sin r}\right)\) will be a constant. This constant is known as refractive index.

Textbook Page No. 108

Reflection Of Light Class 10 Kerala Syllabus  Question 23.
What specialty is observed in the value of ratio of sine of the angle of incidence to the sine of the angle of refraction, \(\left(\frac{\sin i}{\sin r}\right)\)?
Answer:
Will be a constant.

Textbook Page No. 109

Physics Class 10 Chapter 5 Notes Kerala Syllabus Question 24.
What will be refractive index n₁₂? Refractive index, n₁₂ =
Answer:

Textbook Page No. 110

Question 25.
Complete the table 5.6 (a)

Answer:

Question 26.
If the speed of light in water is 2.25 x 10⁸ m/s
a. Calculate the speed of light in vacuum
b. Calculate the speed of light in glass.
Answer:
a. Speed of light in vacuum = Speed of light in water × Refractive index

Question 27.
Place a pencil in an inclined position in a glass trough and fill three fourth of the trough with water. Depict in your science diary, the changes after adding water. What shall be the reason for the change?
Don’t you see that the position of the portion of the pencil underwater has changed? What may be the reason? Discuss.

Answer:
The light rays from pencil undergo deviation at the surface of water.

Textbook Page No. 111

Question 28.
Does the ray of light coming after reflection from the pencil undergo a deviation? What is the reason?
Answer:
Due to refraction

Question 29.
Now, will you be able to explain why a frog was obtained through the arrow was aimed at the fish?
Answer:
Light rays from fish and frog undergo deviation at surface of water. So they don’t see on exact place.

Question 30.
What is observed here?
Answer:
The coin seems to be lifted up.

Question 31.
What is the reason for this observation?
Answer:
As light from the coin comes from denser to rarer medium if bends away from normal so the position of the coin seems to be lifted up due to refraction.

Question 32.
Draw a thick line on a paper using a pen. Place a glass over it and observe as suggested below.
a Look from one side as shown in Fig. 5.8 (a)

b. Look from one side as shown in Fig. 5.8(b)

c. Look from one side as shown in Fig. 5.8 (c)

Answer:
a. The line seems to be rising on glass slab.
b. The line seems to be the surface of glass slab.
c. The line seems to be below the glass slab.

Textbook Page No. 112

Question 33.
Can you easily pick up the coin?
Answer:
We can’t easily pick up the coin.

Question 34.
What is the reason for the failure?
Answer:
Because coin’s actual place is not there. The light rays from the coin comes from denser to rarer medium it bends away from the normal, So the position of coin seems to be lifted up due to refraction.

Question 35.
Find out more examples of refraction from daily life.
Answer:
The stars in the sky appear glittering

• We can see the sun when it falls below the horizon.
• The bottom of water appears raised.
• We cannot locate the position of fish in the water

Question 36.
A person who looks at an aquarium as shown in Fig. 5.10 can see the base on the surface of water. What is the reason?
Answer:
The light rays from the bottom of the aquar¬ium reflect without undergoing refraction (Due to total internal reflection)

Textbook Page No. 113

Question 37.
What will be the angle of refraction when the refracted ray passes along the surface of water?
Answer:
Angle of refraction to becomes 90°.

Question 38.
Allow light to fall at an angle of incidence greater than the critical angle. What do you observe?
Answer:
The ray is reflected back to the same medium without undergoing refraction.

Textbook Page No. 114

Question 39.
Which are the figures that show total internal reflection?
Answer:

Question 40.
What is the critical angle of glass?
Answer:
42°

Question 41.
Will total internal reflection take place when light passing through water is inci¬dent on the surface of separation with air at an angle of incidence of 45°? Why?
Answer:
No. The critical angle of water 48.6°. You have already realized that total internal re-flection takes place only if the angle of incidence is greater than the critical angle. The total internal reflection takes place when in the incident ray is more than 48.6°

Question 42.
Find out the practical applications of total internal reflection in our day to day life.
1. Medical field – Endoscope
2. In the field of communication – Optical fiber cables.
Answer:
1. In the field of treatment- In the endoscope
2. In the field of telecommunication, optical fiber cables are used here.
3. Prism is used in the binoculars
4. Prism is used in the periscopes instead of mirrors.

Textbook Page 117

Question 43.
Complete the table 5.7

Answer:

Textbook Page 119

Question 44.
Where will the rays coining parallel to the principal axis converge?
Answer:
At principal focus

Question 45.
Where is the image formed?
Answer:
At principal focus

Question 46.
Write down the characteristics of the image.
Answer:
Position of the image: Between F and 2F
Nature of the image: Real, Inverted
Size of the image: Diminished

Textbook Page 120

Question 47.
Object at 2F

Answer:
Position of the image: At 2F on the other side Characteristic of the image: Inverted Size of the image: Same size of the object.

Question 48.
Object between F and 2F

Answer:
Position of the image: Beyond 2F
Characteristic of the image: Inverted
Size of the image: Bigger than the object

Question 49.
Object at F

Answer:
Position of the image: At the infinity
Characteristic of the image: Inverted
Size of the image: Big

Textbook Page 121

Question 50.
Object between F and Lens

Answer:
Position of the image: On the same side of the object
Characteristic of the image: direct image
Size of the image: Big in size

Question 51.
Have you observed images formed by a concave lens? What is the nature of the image?

Answer:
Images is smaller, virtual and direct

Textbook Page 122

Question 52.
Record the measurement shown in the figure as per the Cartesian System.

a. Distance of the object from the lens (u) = ………….
b. Distance of the image from the lens (v) = ………….
c. Height of object (OB) = …………..
d. Height of image (IM) = …………..
Answer:
a. u = -25cm
b. v = + 100cm
c. OB = +1cm
d. IM = -4 cm

Question 53.
Complete the Table 5.8

Answer:

Average f = 19.99 cm

Textbook Page 123

Question 54.
Is there any relation between the height of an object and the height of its image?
Can it be related to the ratio between the distance to the object and that to the image?
Answer:
The height of the object height of the image depends on the ratio of the distance to the object and distance to the image. When the object is kept at different positions, the height of the image can be varied.

Question 55.
When an object is placed at different positions in front of a lens, isn’t there a change in the height of the image?
Answer:
Yes

Question 56.
Calculate the magnification of the image formed by convex lens in Fig. 5.33
Answer:
Height of the image = (h_i) = -4 cm
Height of the object = (h₀) = +1 cm
Magnification = \(\frac{h_{1}}{h_{0}}=\frac{-4}{+1}=-4\)
or in another way
m = \(\frac { v }{ 4 }\)
v = +100, u = -25
m = \(\frac { 100 }{ -25 }\) = -4

Textbook Page 125

Question 57.
Find out the uses of lenses in our day to day life and record them in the science dairy.
Answer:

• In telescope
• In eye lenses
• In-camera
• In the microscope
• In the magnifying a lenses

Question 58.
What has the doctor indicated in the prescription?
Answer:
The doctor indicated about the type and power of lens.

Question 59.
Calculate the power of a lens of focal length + 25 cm.
Answer:

Question 60.
You can guess what the +2D in the prescription stands for. What is its focal length? What kind of lens is it?
Answer:
p = + 2D
It is a convex lense
f = \(\frac { 1 }{ p }\) \(\frac { 1 }{ 2 }\) = 50cm

Question 61.
You might have seen the twinkling of stars at night. But planets do not twinkle. Why?
Answer:
Light coming from distant stars passes through different layers of air. Each layer differs from the other in their optical densities. Hence light undergoes successive refraction. Since stars, at a greater distance, they appear like a point source. The rays of light appear to come from different points on reaching the eye after refraction. This is the reason for the twinkling of stars.

Let Us Assess

Question 1.
Refractive indices of different materials are given. Find out the medium through which light passes with maximum speed.
Answer:
Water

Question 2.
The nature of images formed by two lenses are given.
i. An erect and magnified virtual image.
ii. An erect and diminished virtual image
a. What type of lens is used in each case?
b. By using which type of lens will we get an image having the same size as the object? What is the position of the object?
Answer:
a. i. Convex lense
ii. Concave lense
b. Convex lense. Object will be at 2F.

Question 3.

a. MN represents a lens. What type of lens is this?
b. What are the characteristics of the image?
c. Copy the ray diagrams in the science diary and complete it
Answer:
a. Convex
b. Magnified, real, inverted

Question 4.
What do you mean by power of a lens? What is the SI unit of the power of a lens? Calculate the power of a concave lens of focal length 25 cm.
Answer:
Power is a term related to the focal length of a lens. Power of a lens is the reciprocal of focal length expressed in meters. Power
p = 1/f
Unit of power is dioptre. It is represented by

Question 5.
Observe the figure. Light falling on two different media are shown.

a. Which medium has greater optical density? Why?
b. Which medium has greater refractive index?
Answer:
a. Medium 1. Angle of incidence is same, but angle of refraction is different. When a ray of light passes from a medium of lower density to greater density the refracted ray deviates towards the normal. Medium 1 has lowest angle of refraction compared to medium 2.
b. Medium 1

Question 6.
An object of height 3 cm is placed in front of a convex lens of focal length 20 cm at a distance of 30 cm.
a. What is the distance to the image formed?
b. What is the nature of the image?
c. What is the height of the image?
Answer:

b. Magnified, inverted, real

Question 7.
In the table, the absolute refractive indices of certain transparent media are given.

a. Find out from the table the medium of highest and lowest optical densities,
b. If the speed of light in air is 3 x 108 m/s, what will be the speed of light through kerosene?
c. W ill a ray of light deviates towards the normal or away from the normal when it enters from air to diamond obliquely?
d. The refractive index of diamond is 2.42. What do you mean by this? Calculate the speed of light through diamond.
Answer:
a. Highest – Diamond
Lowest – air

c. Towards the normal
d. The speed of light in air (Vaccum) is 2.42 times greater than the speed of light in diamond or other hands. Speed of light in diamond is 2.42 times lesser than speed of light in air or vacuum.
Speed of light through diamond
\(=\frac{3 \times 10^{8}}{2.42}=1.25 \times 10^{8} \mathrm{m} / \mathrm{s}\)

Extended Activities

Question 1.
Half portion of a convex lens is wrapped with a black paper. Can this lens give a complete real image of an object? Explain.

Answer:
Yes. This lens gives a complete real image of an object. However, the intensity of image may be less.

Question 2.
The refractive indices of different media are given. Analyze the table and answer the following.
In which medium will light have the highest speed?
Answer:
Water

Question 3.
Glycerine, water and sunflower oil are taken in two beakers. A glass rod is dipped in one and a pyrex glass rod is dipped in the other. Do they appear in the same way? In which media are they visible? Justify.

Answer:
Objects are visible because they reflect some of light that shines on them. Refractive index of glycerine, sunflower oil, and pyrex glass are same. Glass rod clearly visible in them. But pyrex glass is not clearly visible because light ray are passed without deviation through medium having same refractive index.

Question 4.

Take a clean mineral water bottle and fill it with water. Put a hole on one side of the bottle. Allow water to flow out while passing a laser ray through it What do you observe? Why?
Answer:
As water throws out of the opening observe the case beam remains trapped in the water stream because of total inemal reflection. As light tries to pass from the more dense water to the less dense air. it bends. In narrow column of water, the light wave continues bouncing off the boundaries of the stream of water but cannot pass through into the air.

Refraction of Light Exam Oriented Questions & Answers

Very Short Answer Type Questions (Score 1)

Question 1.
The power of a convex lens of focal length 2 m is …………..
Answer:
\(P=\frac{1}{f}=\frac{1}{2}=0.5 \text { dioptre }\)

Question 2.
What change occurs for power of a lens when it’s focal length increases?
Answer:
Power decreases

Question 3.
An object is placed at 2F in the case of a convex lens what is its magnification? (greater than 1, less than 1, zero, one)
Answer:
one

Question 4.
Figure shows a ray of light obliquely travels from air to glass. Find out the correct one.

Answer:
Figure iii
It bent towards the normal

Question 5.
Find out the relationship in the first pair and fill in the second pair appropriately.
Focal length – meter
Power of the lens – …………..
Answer:
Dioptre

Very Short Answer Type Questions (Score 2)

Question 6.
The image is direct and is above principal axis an object is kept at a distance of 30 cm away from a lens. An image of same size is formed. What is the focal length of the lens?
Answer:
Image of size is formed when the object is kept at 2F
∴ 2f = 30 cm
∴ f = 15 cm
∴ \(\frac { 30 }{ 2 }\) = 15cm

Question 7.
Write two situations when angle of refraction is 90°
Answer:
i. The angle of incidence should be equal to critical angle.
ii. Rays of light should travel from the medium of greater optical density to lesser optical density

Question 8.
Represent the image formation by a concave lens. Why the focus of this lens is called virtual?

Answer:
The rays of light pass through a concave lens do not really meet. They appear to meet. So the focus is called virtual.

Question 9.
What are the situation where total internal reflection take place?
Answer:
i. Rays of light travel from medium of greater optical density to that of lesser one.
ii. Angle of incidence should be greater than critical angle.

Question 10.
An object ‘OB’ is placed in front of lens is given below.

a. Name the lens?
b. Complete the figure, then find out the position of the image.
c. Identify the two nature of the image.
Answer:
a. Convex lens

Images formed in between F and 2F.
b. Real and inverted images formed.

Question 11.
Complete the figure then find out the principal focus of the concave lens.

Answer:

Question 12.
Terms related to lens are given below? Complete the questions using these terms?
(Pole, Focal length, Centre of curvature, Principal axis)
a. The midpoint of a lens is called …………..
b. The distance between pole to focus is called ……………
c. The center of imaginary spear of two sides of the lens is called ……………
d. Imaginary line passing through the center of curvature of the lens is called …………..
Answer:
a. Pole
b. Focal length
c. Centre of curvature
d. Principal axis

Question 13.
Corrected the wrong statement from the following.
a. Optical density of two mediums are different, it is reason for refraction.
b. Velocity of light is very high in optically denser medium.
c. Optical density of glasses lower than that of water
d. Velocity of light in vacuum is 3 × 108 m/s
Answer:
Wrong statements – b, c
b. Velocity of light is comparatively low in optically denser medium.
c. Optical density of glasses higher than that of Water.

Question 14.
Calculate the power of the lens of focal length 10 cm?
Answer:
f = 10 cm = 0.1 m
\(P=\frac{1}{f}=\frac{1}{0.1}=+10 \text { dioptre }\)

Question 15.
‘IM’ is an image of the object ‘OB’. Find out the given values related to new cartesian sign conventions.

a. Distance between lens to object (u) ……………..
b. The distance between lens to image (v) …………
c. Height of the object (OB) ………….
d. Height of the image (IM) …………
Answer:
a. u = -40 cm
b. v = +24 cm
c. OB = +2 cm
d. IM = -1cm

Question 16.
The statements associated with a lens are given by the new cartesian symbol. Write the correct one?
a. All distance are measured from the focus.
b. All distances measured along the direction of incident light is positive.
c. Lightray is assumed to travel from right to left.
d. X-axis is taken as the origin.
Answer:
b. All distances measured along the direction of incident light is positive
d. X-axis is taken as the origin

Short Answer Type Questions (Score 3)

Question 17.
Observe the figure. Light rays enter from medium 1 to medium 2

a. Which medium has greater optical density?. How do you reach in such a conclusion?
b. How do optical density and velocity of light related?
c. What is the speed of light in vacuum?
Answer:
a. medium 1
When the light rays enter from medium 1 to 2, they more away from the normal. The speed of light is also greater in medium 2 then medium.
b. When optical density increases the speed of light decreases.
c. 3 × 10⁸ m/s

Question 18.
A virtual image of an object is formed at 30 cm away from a lens when an object is kept at a distance 20 cm away from it. What is the focal length of the lens? Find the magnification of the image?
Answer:
u = -20 cm, v = -30 cm
(as the image is virtual, it is formed at the same side)

Question 19.
It is written by the doctor for the eye lens as -2.50.
a. What the defect of eye of the patient?
b. What is the focal length of the lens?
c. What type of lens is it?
Answer:
a. near sightedness
b. p = -2.5 D
\(\mathrm{f}=\frac{1}{\mathrm{p}}=\frac{1}{2.5}=\frac{-10}{25}\)
= -0.4 m = -40 cm
The focal length of the lens = – 40 cm
c. Concave lens

20. Observe the figure given below.

a. Why the ray PQ reflected back as in the above figure?
b. Name this phenomenon?
c. What happens to the ray of light when angle of incident is 30°.
Answer:
a. Angle of incidence is greater than that of critical angle.
b. Total internal reflection
c. Refraction takes place when it travels from water to air.

Question 21.
Nature of some images are given below. Find out the real and virtual images and then write it in separate column.
a. Inverted image
b. Do not obtained on a screen
c. Obtained on a screen
d. If the light rays really intersect to each other.
e. Erect image
f. The image distance cannot directly
Answer:
Real images: a, c, d
Virtual images: b, e, f

Question 22.
A burning candle is placed in front of a convex lens to obtain an image on the screen.
Find out the positions of the object in each of the conditions given below?
a. We get the size of the image is equal to the size of the object.
b. The size of the image is smaller than that of the object.
c. We get large real image than the object.
Answer:
a. 2F or C
b. Beyond 2F
c. In between 2F and F.

Question 23.
A person uses a spectacle of power of lens -1.25D.
a. What type of lens is this?
b. What is meant by power of a lens?
c. Find the focal length of the lens?
Answer:
a. Power of the lens is negative. So it is a concave lens.
b. Power of a lens is the reciprocal of its focal length.

Question 24.
Observe the figure carefully OB is an object placed in front of the convex lens.

a. Complete the ray diagram then find out the position of the image.
b. An object is placed at 2F, in which position image is formed?
c. Where we have to place an object to get a virtual image from a convex lens?
Answer:

a. Beyond 2F
b. Image is formed at 2F
c. An object is placed in between optic center and focus

Short Answer Type Questions (Score 4)

Question 25.
Combine suitable

Answer:
1 – (c)
2- (a)
3 – (e)
4 – (d)

Question 26.
Light obliquely travels from glass to air, critical angle of Glass is 42°.
a. What is mean by critical Angle?
b. Angle of incidence is 42° find its refracted angle?
c. In which phenomenon we observe the incident angle is 40°? Explain the phenomenon?
d. What happens to the ray of light when the angle of incidence is 45°? Explain the phenomenon?
Answer:
a. When a Ray of light passes from a medium of greater optical density to that of lower optical density, the angle of incidence at which the angle of refraction becomes 90° is the critical angle.
b. 90°
c. Reflection takes place when a Ray of light entering obliquely from one transfer in the medium to another its path undergoes a deviation at the surface of separation. This is called refraction of light.
d. Total internal reflection. When a Ray of light passes from a medium of higher optical density to a medium of lower optical density at an angle of incidence greater than the critical angle, the ray reflected back to the same medium without undergoing refraction this phenomenon is known as total internal reflection.

Students can Download Physics Chapter 6 Vision and the World of Colours Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Physics Chapter 6 Vision and the World of Colours Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 6 Solutions Malayalam Medium

You can Download Vision and the World of Colours Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Physics Chapter 6 Vision and the World of Colours Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 6 Vision and the World of Colours Solutions

Text Book Page No. 133

Sslc Physics Chapter 6 Kerala Syllabus Question 1.
Place a lighted candle at a distance 13 cm from the lens. Do you get a clear image of the candle on the screen?
Answer:
No

Sslc Physics Chapter 6 Notes Kerala Syllabus Question 2.
Place lenses of focal length 10 cm, 15cm and 20cm on the lens holder without changing the distance between the lens and the screen. On using which lens is the image clear?
Answer:
Lens of focal length 10 cm

Text Book Page No. 134

Vision And The World Of Colours Kerala Syllabus Question 3.
Given below are the ray diagrams of image formation in the eye.
i. Where is the image formed in each case?
ii. In which case is the image formed on the retina itself ?

Answer:
(i) a) In front of retina
b) Behind the retina
c) On retina
ii.

World of Vision Question 4. Will an image be formed on the retina when the object is at the near point

Will it be possible to see a clear image?
Answer:
No. It will not be possible to see a clear image.

Textbook Page No. 135

Physics Chapter 6 Class 10 Kerala Syllabus Question 5.
Will an image be formed on the retina when the object is at a far distance

Will a clear image be formed?
Answer:
Yes. A clear image will be formed.

Sslc Physics Chapter Wise Questions And Answers Question 6.
What shall be the reasons behind this defect?
1. Can you find out a reason based on the size of the eyeball?
Size larger/ smaller?
Answer:
Size larger

2. What if it is related to the focal length (or power)?
Power is high/ low
Answer:
Power low

Physics Class 10 Chapter 6 Kerala Syllabus Question 7.
What is the remedy for long – sight-edness?
Answer:
This can be rectified by using a convex lens of suitable power.

Text Book Page No. 136

Class 10 Physics Chapter 6 Kerala Syllabus Question 8.
Where is the image formed in these cases? Write down the answer analysing.

1. When the object at O is away from the eye, where will be the image formed? Can the object be seen clearly?
Answer:
In front of retina. Objects cann’t be seen clearly.

2. Can the object be seen clearly when it is at O₁?
Answer:
Yes

3. Why is it not possible to see objects placed at long distances?
Answer:
dmages are formed in front of retina.

4. What is its remedy?
Answer:
This can be overcome by using concave lens of suitable power.

Physics Chapter 6 Question 9. When a person suffering from problem in vision met a doctor, he wrote in his prescription the following figures. +1.5 D, -2D

1. What has the doctor indicated in the prescription?
Answer:
+1.5 D, -2D indicates the power of lense they required. The power of a lens is defined as the ability of the lens to converge or diverge a beam of light falling on it.

2. Which are the types of lenses prescribed here?
Answer:
The power of a convex lens is positive while the power of a concave lens is neg-ative.
+1.5D represents convex lense.
-2D represents concave lense.

Sslc Physics Chapter 6 Notes Pdf Kerala Syllabus Question 10.
For a healthy vision, what is the dis-tance to the near point?
Answer:
25cm

Text Book Page No. 137

Kerala Syllabus 10 Physics Chapter 6 Question 11.
Pass sunlight through a prism and allow it to fall on a screen. What are the colours seen on the screen?

Answer:
Violet, Indigo, Blue, Green, Yellow, Orange, Red.

Text Book Page No. 138

Hss Live Guru 10th Physics Kerala Syllabus Question 12.
The beam of white light from a torch is allowed to fall obliquely on the prism.

1. Which are the colours formed on the screen?
Answer:
Violet, Indigo, Blue, Green, Yellow, Orange, Red.

2. Aren’t these colours the same as the component colours obtained from the sunlight?
Answer:
Yes

Hsslive Guru 10th Physics Kerala Syllabus Question 13.

1. Which colour deviates the most due to dispersion?
Answer:
Violet

2. Which colour deviates at least?
Answer:
Red.

3. What may be the reason behind this difference in deviation?
Answer:
Differences in wavelength

Physics 10th Class Notes Kerala Syllabus Question 14.

1. Which colour has the shortest wavelength?
Answer:
Violet.

2. Which one has the longest?
Answer:
Red

Question 3.
When light passes through the prism, as the wavelength increases, how does the deviation change? Will it increase or decrease?
Answer:
The deviation of colours is minimum for the colours with more wavelength when the composite light passes through the prism. When the wavelength of the colour decreases, the deviation increases.

Text Book Page No. 139

Question 15.
When is the rainbow formed?
Answer:
In the morning and in the evening.

Question 16.
Where will be the sun when the rainbow is seen in the east?
Answer:
In the opposite direction (west)

Question 17.
Where will be the sun when the rainbow is seen in the west?
Answer:
East

Question 18.
How many times does a ray of light undergo refraction when it passes through a water droplet?
Answer:
The sunlight undergoes two times refraction in the water drop.

Question 19.
What about internal reflection?
Answer:
One time

Question 20.
What is the colour seen at the upper edge of the rainbow?
Answer:
Red

Question 21.
What is the colour seen at the lower edge?
Answer:
Violet

Text Book Page No. 140

Question 22.
What happened to the light when it passed through the first prism?
Answer:
The white light separates into its component colours when it passes through the first prism

Question 23.
What happened when it passed through the second one?
Answer:
The colours formed by the first prism recombine to form white light when it passes through the second prism.

Text Book Page No. 141

Make a Newton’s disc by painting the constituent colours of white col-our of white light in the same order and proportion.

Question 24.
In which colour does the disc appear when rotated fast?
Answer:
Very near to the white colour

Question 25.
Give Reason.
Answer:
Light from the seven colours of the colour disc falls continuously on the retina and due to the persistence of vision, disc appears as white.

Question 26.
The disc appeared white due to persistence of vision. Find out more examples of persistence of vision and write them down.
Answer:

• A torch rotated rapidly appears as an illuminated circle.
• Sparkler’s trail effect.
• Colour – top
• Thaumatrope.
• Kaleidoscopic colour-top.
• Rubber pencil trick.
• LED POV displays
• Revolving wheels,
• Rotation of fan leafs

Question 27.
During sunset, you might have no- I ticed that the western horizon becomes reddish. Why is it so?
Answer:
During sunset, the sunlight travels I maximum distance through the atmosphere to reach in our eyes. So red also I undergo scattering. So the horizon appears red.

Text Book Page No. 142

Question 28.
Is the reflection of light here regular or irregular?
Answer:
Irregular

Question 29.
Is the distribution of sunlight to all regions made possible by this type of scattering. Discuss.
Answer:
Yes, scattering is the partial and irregular reflection of light. The colours in sunlight undergo scattering when they fall on the particles in the atmosphere. Sunlight reaches in the rooms and under the trees by this method.

Text Book Page No. 143

Question 30.
When hydrochloric acid was added to the solution, which colour did spread first in the solution?
Answer:
Blue colour spreads first in the solution when acid was added.

Question 31.
Write down the colour changes observed on the screen in the order it occurred.
Answer:
Violet → indigo
blue → green → yellow
orange → red

Question 32.
What was the final colour observed on the screen?
Answer:
Red colour is seen at the end.

Question 33.
Which component colour in white light does undergo maximum scattering?
Answer:
Blue has the maximum scattering

Question 34.
During sunset, the horizon appears to have red colour. What may be the reason?
Answer:
During sunrise and sunset, light reaching us from the horizon has to travel long distances through the atmosphere. During this long journey, colours of shorter wavelength would be almost fully lost due to scattering. Then, the red light which undergoes only less amount of scattering decides the colour of the horizon. That is why the sun appears red during sunset and sunrise.

Text Book Page No. 144

Question 35.
Which are the occasions when sun-light has to travel greater distance through the atmosphere before reaching the eyes of ait observer on the earth?
Answer:
Morning and evening.

Question 36.
As sunlight passes through the atmosphere, which colour in it undergoes maximum scattering? Which colour undergoes minimum scattering?
Answer:
Colour in it undergoes maximum scattering: Violet
Colour in it undergoes minimum scattering: Red

Question 37.
When light reaches the observer after travelling long distances through the atmosphere, which colour reaches the eye? What is the reason?
Answer:
Red, it has highest wavelength and least scattering.

Question 38.
The western horizon remains reddish for some more time even after sunset. Why?
Answer:
During sunset, the sunlight travels maximum distance through the atmosphere to reach in our eyes. So red also undergoes scattering. So the horizon appears red.

Question 39.
Can you now guess why red colour has been given to the tail lamps of vehicles and signal lights? Discuss and note it down in your science diary.
Answer:
The primary reason why the colour red is used for danger signals is that red light is scattered the least by air molecules. So red light is able to travel the longest distance through fog, rain, and the like.

Text Book Pace No. 145

Question 40.
Write down what else can be ‘done to minimize the light pollution.
Answer:

• Start with the light switch
• Check with your power company to see if you’re paying for outdoor lighting.
• Consider replacing outdoor lights with intelligently designed, low- glare fixtures.
• Place motion sensors on essential outdoor lamps.
• Replace conventional high-energy bulbs with efficient outdoor CFLs and LED floodlights.
• Reduce the use decorative lighting.
• Use of covered bulbs that light facing downwards.
• The use of automatic systems to turn off street light at certain times.
• Have all information arid facts about light pollution.
• Use of glare-free bulbs, installing low hanging bulbs, having the lights facing downwards, and covering the bulbs to reduce bright skies at night.
• Share with family and friends.

Let Us Assess

Question 1.
How is the condition of the ciliary muscles while watching a distant object? How does this influence the focal length of the eye lens?
Answer:
While looking at far objects the ciliary muscles are relaxed and the curvature of the lens decreases. The focal length of the lens increases.

Question 2.
A child sitting at the backbench of a classroom is unable to see the letters on the board clearly. What is the defect of the eye of the child? How can it be remedied? Draw its ray diagram.
Answer:
Myopia or Nearsightedness. This can be rectified by using a concave lens of suitable power.

Question 3.
A person is not able to see objects beyond 1.3 m. What remedy can you suggest for this defect?
Answer:
Myopia or Nearsightedness. This can be rectified by using a concave lens of suitable power.

Question 4.
In what colour does the sky appear for an astronaut?
Answer:
Black

Question 5.
Red light is used as signal lamps to indicate danger. Explain.
Answer:
The primary reason why the colour red is used for danger signals is that red light is scattered the least by air molecules. So red light is able to travel the longest distance through fog, rain, and the a like.

Question 6.
What is the reason for using yellow light as fog lamps?
Answer:
Yellow lights are strictly used for fog situations, and for construction, so you can tell what is what on the road. Yellow has high wavelength, so yellow light is scattered the least by air molecules.

Question 7.
Which is the phenomenon behind dispersion of light?
a. Reflection
b. Refraction
c. Tyndal Effect
d. Scattering
Answer:
b. Refraction

Question 8.
During dispersion, different colours deviate differently. Explain why.
Answer:
The rate of refraction for different col-ours during dispersion is different due to the difference in wavelength of the colours. The rate of refraction decreases with increase in wavelength. Dispersion. is the phenomenon of splitting of light into its component colours.

Question 9.
The telescope called ‘Chandra X-ray Observatory’ is placed in the outer space. What is the advantage of placing it there? Explain with reference to the scattering of light in the atmosphere.
Answer:
Light can reach the telescope which is kept in space without scattering as there are no gases and tiny particles in space. So we can observe the heavenly bodies clearly there. The pictures may not be clear on earth due to the scattering of light.

Extended Activities

Question 1.
White light is allowed to fall on the bright side of a compact disc (CD). The reflected light is allowed to fall on a white wall. Observe the colours available in the spectrum and write them down in your science diary.
Answer:
Violet, indigo, blue, green, yellow, Orange, Red.

Vision and the World of Colours Orukkam Questions and Answers

Question 1.
Observe the figure

a. Which are the colours formed on the sc-reen?
b. Explain the phenomenon that causes the formation of array of colours
c. Which are the colours denoted by ‘A’ and ‘B’?
d. Which colour deviates the most?
e. Which colour deviates at least?
f. Different colours undergo different deviation. Why?
Answer:
a. Violet, Indigo, Blue, Green, Yellow, Orange, Red
b. Dispersion. The phenomenon of splitting, of a composite light into its constituent colours is known as dispersion of light.
c. A – Violet B – Red
d. Violet
e. Red
f. Different colours undergo different deviation due to difference in wavelength

Question 2.
Newton’s colour disc is made by painting the constituent colours of white light in the same order and proportion.
a. In which colour does the disc appear when rotated fast?
b. The reason behind such an appearance is a phenomenon related to our eyes. What is it explain?
c. If the disc is rotated slowly, what will be the observation?
d. Why does a torch rotated appear as an illuminated circle?
Answer:
a. White colour
b. Persistence of vision. When a person views an object, its image remains in the retina of the eye for a time interval of 0.0625 s
c. If the disc is rotated slowly it is easy to identify the different colours.
d. Due to the Persistence of vision.

Evaluation Activities

Question 3.
The figure shows a ray of sunlight falls obliquely on a water drop.

a. Complete the figure
b. How many times does the ray of light undergo refraction?
c. A natural phenomenon is there that connected with this figure. Name it?
Answer:

b. Twice
c. Rainbow

Question 4.
Match columns. A, B and C suitably.

Answer:
a.Magenta
b. White light
c. Cyan

Question 5.
The telescope ‘Chandra’ is placed in the space. What is the advantage of placing it there?
Answer:
Light can reach the telescope which is kept in space without scattering as there are no gases and tiny particles in space. So we can observe the heavenly bodies clearly there. The pictures may not be clear on. earth due to the scattering of light.

Vision and the World of Colours SCERT Questions and Answers

Question 6.
The figure shows a ray of light falling obliquely on a droplet of water for the formation of rainbow.

a. Copy the figure and complete the unfinished part.
b. What is the phenomenon that took place inside the droplet?
c. What is the colour of ‘X’?
d. What is the reason behind the red app-earing at the outer edge of a rainbow?
Answer:

b. Internal reflection
c. Violet
d. Sunlight undergoes refraction and internal reflection while passing through drops of water. All the drops that appear in the same colour are seen in the same arc. Thus red having highter wavelength is seen in the outer edge at higher angle.

Question 7.
Analyse the following statements and find out the reason behind them.
a. Stars can be seen even in day time while viewed from the moon
b. Raindrops falling down during rain appear like a glass rod.
c. A rainbow is seen in the shape of a circle when viewed from a height.
d. The sky of cities mostly appears in grey colour.
Answer:
a. There is no scattering for the light around the moon since there is no atmosphere around it. Hence the sky of moon appears dark.
b. Raindrops come down faster during rain. The distance travelled by a drop in 1/16 of a second appears like a glass rod due to persistence of vision.
c. When viewed from a height the observer can see points at 42.70 upwards, downwards as well as sideways. Hence rainbow is seen as a circle.
d. In cities there will be large particles in the atmosphere. Hence all colours of light scatter equally.

Question 8.
Two teams A and Bare taking part in a volley ball competition held on a ground illuminated by a sodium vapour lamp (yellow light). Team A is wearing white jersey and blue shorts while Team B is wearing yellow jersey and black shorts.
a. Can you distinguish between the teams based on the colour of their dress?
b. Justify your answer.
Answer:
a. Cannot be distinguished
b. In yellow light the white jersey and yell-ow jersey appear in yellow colour. This is because both white and yellow surfaces can reflect yellow light. The black shorts appears black itself and blue shorts appears dark since both of them absorb yellow light.

Question 9.
A disc painted with different colours is shown

a. Which colour is X if the disc appears white on rotating fast?
b. Why did the disc appear white on being rotated fast?
c. In what colour will the disc appear if green light alone is made to fall on the disc when it is being rotated fast?
Answer:
a. Cyan
b. Due to persistence of vision
c. In green colour
On rotating the disc faster, the disc app-ears white due to persistence of vision. The cyan appears as green since it reflects green light.

Question 10.
Colour filters are materials that will allow only certain colours of light to pass.
a. What is the colour of the filter that can transmit blue and red colours of light?
b. Which are the radiations that will be absorbed completely by an infrared filter?
c. Write down:
(i) Two specialities of infrared rays
(ii) Two uses of infrared rays
Answer:
a. Magenta
b. Visible light, ultra violet radiations
c. (i) Higher wavelength than that of visible light; becomes the reason for the heat . of the sunlight.
(ii) Used to take photograph of distant objects, remote sensing, secret signalling, for controlling robots.

Question 11.
The figure shows sunlight falling on a white screen after being passed through yellow and cyan filters.

a. Supply the omitted portion and complete the diagram
b. Which colours is‘X’?
c. (i) What is the complementary colour of this colour?
(ii) What do you mean by complementary colours?
Answer:
a.

b. Green
c. (i) Magenta
(ii) If a primary colour and a secondary colour can given white light on mixi-ng, then they are complementary colours.

Question 12.
Match the columns A, B and C suitably

Answer:

Question 13.
A spectrum is obtained by passing white light from a torch through a glass prism.
a. Write down the steps of this experiment.
b. Draw a figure that represents the experiment.
Answer:
a. Fix a black paper in from of a torch. Put a small hole in the middle of the paper. Arrange a screen on the other side. Make the beam of light fall obliquely on the prism.

Question 14.
The telescope ‘Chandra’ is installed on the outer space. Identify the correct statements related to it.
a. There is no scattering of light in the outer space
b. There is a greater scattering in the outer space
c. The vision is more accurate and clear
d. Presence of dust in the outer space helps in better vision and clarity
Answer:
Correct statements: a and c

Question 15.
Which of the following does not belong to the group?
a. red, yellow, blue, green
b. visible light, sound, X-rays, radio waves
Answer:
a. Yellow – it is a secondary colour
The rest are primary colours
b. Sound- it is a mechanical wave The rest are electromagnetic waves

Question 16.
Find out the relation between the first pair and complete the other accordingly:
a. Black: absorbs all colours of light white:
Answer:
a. Reflects all colours of light.

Vision and the World of Colours Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 17.
Find the odd one in the group and write the reason. [Red, Green, Blue, Yellow]
Answer:
Yellow. Others are primary colours

Question 18.
Using the relation from the first pair, complete the other.
Tyndal effect – scattering
Dispersion – ………………
Answer:
Refraction

Question 19.
Which of the following causes skin cancer,
a. Ultraviolet
b. Infrared
c. Radio
d. Microwave
Answer:
Ultraviolet

Question 20.
What is the colour seen at the inner edge of a rainbow?
Answer:
Violet

Question 21.
Fill in the blanks.
Green + ………. = Yellow
Answer:
Red –

Question 22.
What is the change in the deviation of those rays having lesser wavelength compared to those rays having higher wavelength? (more, less, no change)
Answer:
less

Question 23.
Name the phenomenon responsible for dispersion of light.
Answer:
Refraction

Question 24.
In the given figure which is correct.

Answer:

Question 25.
Distance to the near point of a healthy person is? (10 cm, 50 cm, 100 cm, 25 cm)
Answer:
25cm

Short Answer Type Questions (Score 2)

Question 26.
Arrange the following colours in the in-creasing “order of their wavelengths.
Violet, Red, Green, Blue
Answer:
Violet, Blue, Green, Red

Question 27.

Yellow light from a torch falls on a green paper and then falls on a white screen.
a. Which colour is seen on the screen?
b. Write the reason in two or three sentences.
Answer:
a. Green colour
b. When yellow light falls on a green paper it absorbs yellow and reflects green. So . the green colour is seen on the screen.

Question 28.
Myopia and Hypermetropia are the eye defect of human beings, identify the given statement then separate the reason for Myopia and Hypermetropia.
a. Image is formed behind the retina
b. Images formed in front of the retina
c. Power of the eye lens decreases
d. Power of the eye lens increases
e. Suitable power of concave lens is used to solve this problem
f. Suitable power of convex lens is used to solve this problem
Answer:
Myopia -b, d, e
Hypermetropia -a, c, f

Question 29.
Why concave lens always create virtual and erect image of the object?
Answer:
In this case, refracted Ray do not actually intersect to each other. It appears to intersect the images formed the same side of the lens.

Short Answer Type Questions (Score 3)

Question 30.

a. Complete the ray diagrams in the figures A and B when the rays of light passes through the prisms and reach out of them with the changes occurred to them.
b. If there are differences between the figures explain the reason
Answer:

b. There is only one colour in laser. Sunlight is the composite light of seven colours and so undergoes refraction

Question 31.
Explain the reason for the following.
a. At the time of sunset, western horizon is seen red.
b. Sky in the moon ha» a dark colour
Answer:
a. At the time of sunset, Light rays have to travel very long distances through air. Hence Light rays are scattered. Light rays like blue with shorter wavelength get lost due to scattering but rays with longer wavelength remain because it gets less scattering.
b. In moon there is no atmosphere and therefore scattering of light does not take place.

Question 32.
The figure shows a ray of light falling obliquely on a drop of water in atmosphere

a. Copy the diagram and complete it show-ing the internal reflection and refractions.
b. How does the sunlight appear as rainbow in water droplets?
Answer:

b. All the water droplets of the same colour appear to be in a same arc of a circle. The rays of light incident on the water droplets must be parallel to the line of vision. Each colour ray emerging from the water drop makes a definite angle from 40.8° to 42.7°. Red makes the higher angle of 42.7° and violet ma-kes a lower angle of 40.8°. Hence red colour is seen at the outer edge and violet colour at the inner edge. The other colours are seen in between depending on their wavelengths.

Question 33.
The image formation of a defected eye is given below

a. In which position images formed on a normal eye?
b. What is this eye defect?
c. How to solve this defect? Draw the diagram.
Answer:
a. On the retina.
b. Near-sightedness(Myopia).
c. Suitable power of concave lenses is used to solve this problem.

Short Answer Type Questions (Score 4)

Question 34.
Find the appropriate terms from the box for the statements given below.
Red, Scattering, Refraction, Dispersion
a. Phenomenon responsible for Rainbow.
b. Phenomenon responsible for Tyndal effect.
c. Colour seen at the outer edge of the rainbow.
d. Phenomenon responsible for Dispersion of Light.
Answer:
a. Dispersion
b. Scattering
c. Red
d. Refraction

Question 35.
Rearrange the table given below correctly.

Answer:

Question 36.
Analyse the picture and answer the questions.

a. What is the colour of the light falling on the white screen?
b. What colour is obtained on the screen if blue filter is used instead of green filter?
Answer:
a. Green
b. When blue filter paper is used, no light will fall on the screen. The yellow, green and red colours coming out of the yellow filter will not pass through blue filter paper.

Question 37.
When Newton’s colour disc rotates fast it appears white.
a. Which phenomenon is responsible for this?
b. Define this phenomenon.
c. Write another example related to this phenomenon.
Answer:
a. Persistence of Vision.
b. When an object is viewed by a person, the image remains in the retina for a time interval of 1/16 second. This phenomenon is called persistence of vision.
c. Raindrops look like glass rods during rain.

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SCERT 10th Standard Physics Textbook Chapter 4 Solutions Malayalam Medium

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SSLC Chemistry Chapter 7 Chemical Reactions of Organic Compounds Questions and Answers Malayalam Medium

SCERT 10th Standard Chemistry Textbook Chapter 7 Solutions Malayalam Medium

You can Download Statistics and Algebra Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 11 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 11 Statistics and Algebra Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 11 Statistics and Algebra Notes

Textbook Page No. 245

Statistics Class 10 Kerala Syllabus Questions 1.
The distance covered by an athlete in long jump practice are
6.10, 6.20, 6.18, 6.20, 6.25, 6.21, 6.15, 6.10
in meters. Find the mean and median. Why is it that there is not much difference between these?
Answer:
Mean = \(\begin{array}{l}{=\frac{6.10+6.20+6.18+6.20+6.25+6.21+6.15+6.10}{8}} \\ {=\frac{49.39}{8}=6.17}\end{array}\)
If distances are written in ascending order
6.10, 6.15, 6.18, 6.20, 6.21, 6.25
Median = \(\frac{6.18+6.20}{2}=6.19 \mathrm{m}\)
Mean and median are gives the average dis¬tance covered by a person. Hence they will not have much difference..

Polynomial in Descending Order Calculator is a free online tool that determines the descending order of a polynomial in just a few taps.

Sslc Maths Statistics Kerala Syllabus Questions 2.
The table below gives the rainfall during one week of September 2015 in various districts of Kerala.

Calculate the mean and median rainfall in Kerala during this week. Why is the mean less than median?
Answer:
Mean = Total amount of rain / No. of districts
= \(\frac { 700 }{ 14 }\) = 50
In ascending order:
13.5, 20.5, 30.5, 33.5, 45.5, 50.3, 53.4, 56.3, 56.4, 56.9, 56.9, 66.7, 70.6, 89
Median \(=\frac{53.4+56.3}{2}=54.85\)
The mean is less than median because the number contains are far small and large numbers than mean.

Use this number sequence calculator to easily calculate the n-th term of an arithmetic, geometric or fibonacci sequence.

Sslc Statistics Solutions Kerala Syllabus Questions 3.
Prove that for a set of numbers arithmetic sequence, the mean and median are equal.
Answer:
Let a, a+d, a+3d, a+4d are the numbers of an arithmetic sequence, then median = \(\frac{a+a+4 d}{2}=\frac{2 a+4 d}{2}=a+2 d\)
Median will be the term which is at center = a + 2d
∴ A set of numbers in arithmetic sequence, the mean and median are equal.

Textbook Page No. 248

Sslc Maths Statistics Notes Kerala Syllabus  Questions 1.
35 households in a neighborhood are sorted according to their monthly income in the table below.

Calculate the median income.
Answer:

In the table monthly income up to 18th place be 6000 rupees.
That is 18^th place family also includes in the middle of total number of families.
∴ Median of the income = Rs. 6000

Sslc Maths Chapter 11 Statistics Kerala Syllabus Questions 2.
The table below shows the workers in a factory sorted according to their daily wages.

Calculate the median daily wage.
Answer:

Total number of workers = 30
Half= 15
So median is the wage of 15th worker.
The daily wages between the place 11 and 18 = 700 rupees
Median of daily wages = 700 rupees

Sslc Maths Chapter 11 Kerala Syllabus Questions 3.
The table below gives the number of babies born in a hospital during a week, sorted according to their birth weight.

Calculate the median birth-weight
Answer:

Total number of babies = 70
Half =35
So median is the weight of 35^th baby.
The weight of 35 111 child be in between 29 and 40 placed child, its weight will be 3 kg.
∴ Median of the weight = 3 kg.

Textbook Page No. 254

Sslc Statistics Notes Kerala Syllabus Questions 1.
The table shows some households sorted according to their usage of electricity:

Calculate the median usage of electricity.
Answer:

Half of the number of houses = 20
We have to find the electricity usage of the 20th house. According to this, we can divide

Statistics Class 10 State Syllabus Question 2.
Answer:

Total number of children = 34
It is an even number, so we will take the half of sum of weight of the children, those are in the 17 and 18 positions. According to this child between 13 and 22 have weight between 50.5 and 55.5. Our required children (between and 17 and 18) are in these positions. Divide 5 years from 50.5 to 55.5 into 10 equal, parts. Let consider each part have one child.

Weight of each part = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\).
Hence weight of a child in 13th place is in middle of 50.5 and 51. That is 50.75. Further, each student’s weight is increased by 5.
Hence weight of the man in the 17th position

Median \(=\frac{52.75+53.25}{2}=53\)

Statistics 10th Standard State Syllabus Question 3.
The workers of a company are arranged as given below. Calculate median

Answer:

Total number of workers = 30
Half = 15
So median is the wage of 15th worker.
The daily wage between the place 10 and 18 = 600 rupees
Median of daily wage = 600 rupees

Statistics SCERT Questions & Answers

Scert Class 10 Maths Statistics Kerala Syllabus Question 5.
10 households in a neighborhood are sorted according to their monthly income are given below
16500, 21700, 18600, 21050, 19500, 17000, 21000, 18000, 22000, 75000
a. What is the mean income of these 10 families?
b. How many families have monthly incomes less than the mean income? Prove that in such situation this average is suitable or not?
Answer:
a. Mean = \(\frac { sum }{ number }\) = \(\frac { 248000 }{ 10 }\) = 24800
b. 9 families have monthly income less than the mean income. So in this situation this is not a suitable average.

Sslc Maths Statistics Questions And Answers Kerala Syllabus Question 6.
Number of members in 10 families, collected by mathematics club survey are given. Calculate mean; median and explain. Which is the suitable average?.
4, 2, 3, 5, 4, 3, 2, 20, 4, 3
Answer:
a. Mean = 5
b. Median = 3.5
Suitable average median = 3.5

Sslc Statistics Questions Kerala Syllabus Question 7.
Weekly Wages of 9 persons Working in a factory are given. Find the median 2100; 3500, 2100, 2500, 2800, 4900, 2300, 2200, 3300
Answer:
Write the number in order.
2100, 2100, 2200, 2300, 2500, 2800, 3300, 3300, 3500 (1)

Median = 2500

Statistics Class 10 State Board Kerala Syllabus Question 8.
The table shows the workers doing different jobs in a factory according to their daily wages.

Calculate median of daily wages.

Answer:

The worker in the 12th position to 20th position has daily wage 270 ie, Median

Sslc Maths Chapter 11 Solutions Kerala Syllabus Question 9.
The table below shows the 60 children in a class sorted according to their heights

Find the median height?
Answer:

Sslc Maths Chapter Statistics Kerala Syllabus  Question 10.
Answer:
a. 100
b. 25
c.

Mean = \(\frac { 17840 }{ 100 }\) = 178.4

Hss Live Guru 10th Maths Kerala Syllabus Question 11.
The mean of the frequency table given below is 50. Then find out the values of a and b.

Answer:

mean = 50
∴ \(\frac{3480+30 a+70 b}{120}=50\)
30a + 70b = 6000 – 3480
30a + 70b = 2520 (1)
17 + 32 + 19 + a + b = 120
68 + a + b = 120,
a + b = 52 (2)
(2) x 30 => 30a + 30b = 1560
30a + 70b = 2520,
30a + 30b = 1560,
40b = 960,
b = \(\frac { 960 }{ 40 }\) = 24,
a + 24 = 52
a = 52 – 24 = 28,
∴ a = 28, b = 24

Long Answer Type Questions (Score 5)

10th Standard Maths Statistics Kerala Syllabus Question 21.
The table below shows groups of children in a class according to their heights:

a. If the children are lined up according to their heights, the median is the height of the child in which position?
b. According to the table, the height of this child is between what limits?
c. What are the assumptions used to compute the median?
d. What is the median height according to these assumptions?
Answer:

a. Height of the 21st child is the median height.
b. Height of the 21st child is between 145 cm and 150 cm.
c. Methods to find the median are.
1. Divide 5cm in between 145 cm and 150 cm into 10 equal sections.
2. Consider that the height of each sub-group is exactly on the midpoint of the subgroup.
Height of the 14th child is in between 145 cm and 145 \(\frac { 5 }{ 10 }\) cm.
i.e., 145 \(\frac { 5 }{ 20 }\) cm.
Similarly, the height of the 15th student is in between 145 \(\frac { 5 }{ 10 }\) cm and

Hence height of each child can be increased by \(\frac { 5 }{ 10 }\) cm.
There are 7 children to reach the 21st child from 14th child.
i.e., 14 th term is 145 \(\frac { 5 }{ 20 }\) and common difference is \(\frac { 5 }{ 10 }\).
Mean is the 21 st term of the arithmetic sequence.

Use our simple online Limit Of Sequence Calculator to find the Limit with step-by-step explanation.

Statistics Menton Map

Arithmetic mean is the sum divided by the number of terms.

Mean = \(\frac { sum of terms }{ Number of terms }\)

When the numbers are arranged in a ascending order, then the middle term is the median.
i.e., half of the total frequency will give the median.

You can Download Circles Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 2 Circles Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 2 Circles Notes

Textbook Page No, 42

Circles Class 10 Kerala Syllabus Question 1.
Suppose we draw a circle with the bottom side of the triangles in the picture as diameter. Find out whether the top corner of each triangle is inside the circle, on the circle or outside the circle.

Answer:
Angle of the first triangle =110°
As 110° > 90° the top comer will be inside the circle
Angle of second triangle = 90°
∴ The top comer will be on the circle.
Angle of the third triangle = 70°
70° > 90°
∴ The top comer will be outside the circle

Sslc Maths Circles Questions And Answers Question 2.
For each diagonal of the quadrilateral shown, check whether the other two corners are inside, on or outside the circle with that diagonal as diameter

Answer:
The fourth angle in ABCD
= 360 – (110 + 105 + 55) = 360

Drawing diagonal AC and taking it as diameter of a circle As, ∠D = 90°
D will be on the circle. ∠B = 55° (90 > 55°)
∴ B will be outside the circle
Drawing diagonal BD and taking it as diameter of a circle.
∠A = 105°, ∠C = 110°
As both angles are greater than 90, they lie inside the circle.

Sslc Maths Chapter 2 Kerala Syllabus Question 3.
If circles are drawn with each side of a triangle of sides 5 centimetres, 12 centimetres and 13 centimetres, as diametres, then with respect to each circle, where would be the third vertex?
Answer:
As the sides are 5, 12, 13 cm and also
52 + 122 = 25 + 144 = 169 = 132
∴ ABC is a right triangle

Taking BC as diameter and drawing a circle, ∠A (<90°), A will be outside the circle.
Taking AB as diameter and drawing a circle ∠C (<90°) C will be outside the circle.
Taking AC as diameter and drawing a circle, ∠B = 90°, B will be on the circle.

Circles Class 10 Scert Solutions Kerala Syllabus Question 4.
In the picture, a circle is drawn with a line as diameter and a smaller circle with half the line as diameter. Prove that any chord of the larger circle through the point where the circles meet is bisected by the small circle.

Answer:
∠ADO = ∠APB = 90°
(angle subtended by diameter is always 90°)
⇒ OD\\PB

AO = OB (Radius of bigger circle)
(If in a triangle, the line drawn from midpoint of one side, is parallel to another side, then the line will bisect the third side)
Therefore AD = DP
(AB’s midpoint is ‘O’ and OD\\PB)

Sslc Maths Circles Questions And Answers Pdf Question 5.

Use a calculator to determine up to two decimal places, the perimeter and the area of the circle in the picture.
Answer:

Maths Chapter 2 Class 10 Kerala Syllabus Question 6.
The two circles in the picture cross each other at A and B. The points P and Q are the other ends of the diameters through A.

i. Prove that P, B, Q lie on a line.
ii. Prove that PQ is parallel to the line joining the centres of the circles and is twice as long as this line.
Answer:

i. ∆ PAB (angle subtended by semicircle)
∠PBA = 90°
∆ ABQ be a triangle on the semi¬circle of centre D.
∴ ∠ABQ = 90°
∠PBA + ∠ABQ = 180 (Linear pair)
As AP, AQ are diameters of the circle. PQ be the line drawn through B per¬pendicularly to AB. Therefore P, B, Q lies on the same line.

ii.

Sslc Maths Circles Notes Kerala Syllabus Question 7.
Prove that the two circles drawn on the two equal sides of an isosceles triangle as diameters pass through the midpoint of the third side.
Answer:

∠ADB= 90° (AABD angle subtended by semicircle)
∠CDA = 90°
∴ ∠ADB +∠CDA = 180° (linear pair)
∆ ABD ∆ ADC are right angled triangles.
In ∆ ABD
BD² = AB² – AD² ( AB = AC )
= AC² – AD² = DC²
BD = CD

Sslc Maths Chapter 2 Circles Questions And Answers Question 8.
Prove that all four circles drawn with the sides of a rhombus as diameters pass through a common point.

Prove that this is true for any quadrilat¬eral with adjacent sides equal, as in the picture.

Answer:

ABCD is a rhombus so diameter are perpendicular bisectors.
∠AOD = 90°
O be on the circle having diameter AD.
∠AOB = 90°, therefore
O be on the circle having diameter AB.
∠BOC= 90°, therefore
O be on the circle having diameter BC
∠DOC= 90°, therefore
O be on the circle having diameter DC
O be the common point on the circle.
∠A0D = ∠AOB and
∠COD = ∠BOC and,
AD = AB,
AO be the common side.

Δ AOD, Δ AOB are equal triangles.
OD = OB
Both the circles can passed through O.
A BCD is an isosceles triangle.
Those circles having diameters CD and BC are passing through midpoint of BD.
∴ O be common for the four circles. (Diameter)

Sslc Maths Chapter 2 Circles Notes Kerala Syllabus Question 9.
A triangle is drawn by joining a point on a semicircle to the ends of the diameter. Then semicircles are drawn with the other two sides as diameter.

Prove that the sum of the areas of the blue and red crescents in the second picture is equal to the area of the triangle.
Answer:

Area of triangle = \(\frac { 1 }{ 2 }\) × 2r × h = rh
Area of the semicircle = \(\frac{\pi r^{2}}{2}\)
Area of the rest of the figure

The diameters of the red and blue semi-circles are the sides of the two triangles.
∴ Their areas

Area of red and blue crescents

= area of the triangle

Textbook Page No. 53

Circles Class 10 Notes Pdf Kerala Syllabus  Question 1.
In all the pictures given below, O is the centre of the circle and A, B, C are points on it. Calculate all angles of Δ ABC and Δ OBC in each.

Answer:
a.

OA = OB (radii)
∴ ∠OAB = 20° ; (∠OAB = ∠OBA)
OC = OA (radii) ;
∠OAC = 30°
∠BAC = ∠OAB + ∠OAC
= 20 + 30 = 50°
∠BOC = 2x ∠BAC = 100°
OB = OC (radii)
∴ ∠OBC = ∠OCB = 40°

Angles of triangle ABC are ∠BAC = 50° ∠ABC = 60°, ∠ACB = 70°
Angles of ∆ BOC are
∠OBC = 40°, ∠OCB = 40°, ∠BOC = 100°

b.

Angles of ∆ ABC are ∠ABC = 50°, ∠BAC = 60°, ∠BCA = 70°
Angles of ∆ AOC are
∠OAC = 40°, ∠AOC = 100°, ∠OCA = 40°

c.

∠ACB = 180 – 55 = 125° ; OA = OC
∠CAO = ∠ACO = 70° ; ∠OBC = ∠BCO = 55° Angles of A OBC are
∠OBC = 55° ∠COB = 70° ∠BCO = 55°
Angles of ∆ ABC

Class 10 Maths Chapter 2 Circles Kerala Syllabus Question 2.
The numbers 1,4,8 on a clock’s face are joined to make a triangle.

Calculate the angles of this triangle.
How many equilateral triangles can we make by joining numbers on the clock’s face?
Answer:
The angle subtended by two adjacent numbers at the centre of the clock is 30°

We can make 4 equilateral triangles by joining the numbers on the clock (1. 5, 9), (2, 6, 10), (3. 7, 11), (4, 8, 12)

10th Class Maths Chapter 2 Circles Kerala Syllabus Question 3.
In each problem below, draw a circle and a chord to divide it into two parts such that the parts are as specified.
i. All angles on one part 80°.
ii. All angles on one part 110°.
iii. All angles on one part half of all angles on the other.
iv. All angles on one part, one and a half times the angles on the other.
Answer:
i. ∠AOB = 160°
Therefore all angles in the are ACB arc 80°.

ii. Draw angle as central angle 220° so angle on the small arc AB will be 110°.

iii. Draw angle as central angle 120° or 240° All angles on one part will be 120°, and other part be 60°.

iv. Draw a circle and draw central angle 144° All angles on the part APB will be 120° and All angles on the part AQB will be 108°.

Sslc Maths Chapter 2 Circles Kerala Syllabus Question 4.
A rod bent into an angle is placed with its corner at the centre of a circle and it is found that \(\frac { 1 }{ 10 }\) of the circle lies within it. If
it is placed with its corner on another circle, what part of the circle would be within it?

Answer:

Circles Class 10 Notes Kerala Syllabus Question 5.
In the picture, O is the centre of the circle and A, B, C are points on it. Prove that
∠OAC + ∠ABC = 90°

Answer:

Class 10 Maths Chapter 2 Kerala Syllabus Question 6.
Draw a triangle of circumradius 3 centimetres and two of the angles 32\(\frac { 1° }{ 2 }\) and 37\(\frac { 1° }{ 2 }\)
Answer:
Draw a circle with radius 3 cm and central angle 65°.
Half of 65° is 32\(\frac { 1° }{ 2 }\) Similarly we can draw 75°
Join the points A, B and C Halfof75°is 373\(\frac { 1° }{ 2 }\) Complete the triangle.

Circles Class 10 Notes State Syllabus Question 7.
In the picture, AB and CD are mutually perpendicular chords of the circle. Prove that the arcs APC and BQD joined together would make half the circle.

Answer:
If ∠ADC = x
∠AOC =2x
(Angle subtended on the alternate arc is half of the central angle of arc )
If ∠BAD = y
∠BOD= 2y
∠AOC + ∠BOD = 2x + 2y
= 2 (x + y) = 2 × 90 = 180°
∴ The arcs APC and BQD joined together will make half the circle.

Kerala Syllabus 10th Standard Maths Chapter 2  Question 8.
In the picture, A, B, C, D are points on a circle centred at O. The lines AC and BD are extended to meet at P. The line and BC intersect at Q. Prove that the angle which the small are AB makes at O is the sum of the angles it makes at P and Q.

Answer:

Textbook Page No. 59

Kerala Syllabus 10th Standard Maths Circles Question 1.
Calcula te the angles of the quadrilateral in the picture and also the angles between their diagonals:

Answer:
Since
∠ACD = 30°
∠ABD = 30°
(Angle in the same segment of a circle)
Since ZCBD = 45°
∠CAD = 45°
Since ZBDC = 50°
∠BAC = 50°
∠ABC + ∠ADC = 180 (cyclic quadrilateral)
∠ABC = 75°
∴ ∠ADC = 180 – 75 = 105°

∠ADB = 105 – 50 = 55°
As, ∠BAD = 95°
∠DCB = 180 – 95 = 85°
∴ ∠ACB = 85 – 30 = 55°

Circles Class 10 State Syllabus Kerala Syllabus Question 2.
Prove that any outer angle of a cyclic quadrilateral is equal to the inner angle at the opposite vertex.
Answer:

PQRS is cyclic Quadrilateral
∠PSR + ∠PQR = 180°
(sum of opposite angles)
∠PQR + ∠RQT = 180° (linear pair)
From this we get ∠PSR = ∠RQT

Maths Circles Class 10 State Syllabus Question 3.
Prove that a parallelogram which is not a rectangle is not cyclic.
Answer:
PQRS is cyclic quadrilateral.
∠P + ∠R = 180
Also in a parallelogram opposite angles will be equal.
∠P + ∠R = 180°
∠P = ∠R = 90°
This means that PQRS must be a rectangle, otherwise, it is not cyclic.

Sslc Maths Chapter 2 Notes Kerala Syllabus Question 4.
Prove that any non-isosceles trapezium is not cyclic.
Answer:

opposite angles are not supplementary.
∴ ABCD is not cyclic. Non-isosceles trapezium is not cyclic.

10th Class Maths Notes Kerala Syllabus Question 5.
In the picture, bisectors of adjacent angles of the quadrilateral ABCD intersect at P, Q, R, S.

Prove that PQRS is a cyclic quadrilateral.
Answer:

2x + 2y + 2z + 2w = 360°;
x + y + z + w = 180°
Δ DPC; ∠DPC = 180 – (w + z)
Δ ARB ; ∠ARB =180 – (x + y)
∠R + ∠P = 360 – (x + y + z + w) = 360 – 180 = 180°
In ΔBQC ZQ = m – (w + x)
In Δ ASD ∠S = 180 – (r + y)
Similarily ∠S +∠Q = 180.
PQRS is a cyclic quadrilateral.

Question 6.
i) The two circles below intersect at P, Q and lines through these points meet the circles at A, B, C, D. The lines AC and BD are not parallel. Prove that if these lines are of equal length, then ABDC is a cyclic quadrilateral.

ii) In the picture, the circle on the left and right intersect the middle circle at P, Q, R, S; the lines joining them meet the left and right circles at A, B, C, D. Prove that ABDC is a cyclic quadrilateral.

Answer:
i.

∴ ABCD is a cyclic quadrilateral.

ABCD is a cyclic quadrilateral.

Question 7.

In the picture, points P, Q, R are marked on the sides BC, CA, AB of AABC and the circumcircles of ΔAQR and ΔBRP are drawn. M is a point where these circles intersect.

Prove that the circumcircle of ΔCPQ also passes through M.
Answer:

Let M be a common point which three circles can passed.

Therefore the circumcircle of ACPQ also passes through M

Textbook Page No. 67

Question 1.
In the picture, chords AB and CD of the circle are extended to meet at P.

i. Prove that the angles of Δ APC and Δ PBD, formed by joining AC and BD, are the same.
ii. Prove that PA × PB = PC × PD.
iii. Prove that if PB = PD, then ABDC is an isosceles trapezium.
Answer:

i. As ABCD is a cyclic quadrilateral.
If ∠BAC = x° then
∠BDC = 180 – x ∠BDP = x°
If ∠ACD = y° then ∠PBD = y°
As ∠APC is common angle.
Angles of Δ APC and Δ PBD are same

iii. If PB = PD ; AP= PC
In ABCD
ABDC is a cyclic quadrilateral, so their opposite angles are supplementary.
If AP = PC, in ∆ APC
∠A =∠C

As AB = CD
AC || BD
Adjacent angles are supplementary
∴ ABCD will be an isosceles trapezium

Question 2.
Draw a rectangle of width 5 centimetres and height 3 centimetres.
i. Draw a rectangle of the same area with width 6 centimetres.
ii. Draw a square of the same area.
Answer:
i. Draw a rectangle of length 5 cm and width 3cm.

Extend AB up to 6cm.
(AE = 6cm) Draw an arc having radius as AE and A as centre. Extend DA and mark the point F.

Extend BA towards left. Mark G as AD = AG
Draw Δ GFB.

Circum circle of Δ GFB meets AD at D.
∴ AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.

ii. Draw a rectangle of length 5cm and width 3cm. Area = 5 x 3 = 15 cm2.
Therefore side of a square will be √15.

Draw a semicircle of diameter AH.
Extend BC, and mark the point F.
AB × BH = 5 × 3 = 15
AB × BH = BF2 ; BF = √5 cm
Area of BEGF = √15 × √15 = 15 cm²

Question 3.
Draw a square of area 15 square centimetres.
Answer:
Draw a rectangle of length 5cm and width 3cm. Area = 3 × 5 cm² = 15 cm². Side of the square is √15.
Draw a semicircle of diameter AH.
BC can touch the point F.
AB × BH = BF2 = √152 = 15 cm².
Area of BEGF = 15 cm².

Question 4.
Draw a square of area 5 square centimetres in three different ways. (Recall Pythagoras theorem)
Answer:
i. Draw a rectangle of length 5cm and width 1 cm.

Draw a semicircle of diameter AE. Extend BC up to F. √5 is the side of the square BGHF.

ii. Draw a right-angled triangle of perpendicular sides 2 cm and 1cm.

Hypotenuse will be y is cm. The area of the square ACDE is 5 cm², because here we take the hypotenuse as sides of the square.

iii. Draw a right-angled triangle of hypotenuse 3 cm and one side 2cm.
Third side = \(\sqrt{3^{2}-2^{2}}=\sqrt{5} \mathrm{cm}\)
Draw a square having side BC, then area of the square BEDC will be 5cm²

Question 5.
In the picture, a line through the centre of a circle cuts a chord into two parts:

What is the radius of the circle?
Answer:

The intersection may be with in the circle.
Chords AB & CD intersect at P i. e.,
AP × PB = CP × PD
(The intersection will be inside the circle)
AP × PB = CP × PD
4 × 6 = CP × (OP + OD)
24 = CP × (OP + OC)
24 = CP × (OP + OP + CP)
24 = CP × (5 + 5 + CP)
24 = CP × (10 + CP)
CP =2
Radius = CP + OP = 2 + 5 = 7cm

Question 6.
In the picture, a line through the centre of a circle meets a chord of the circle:

What are the lengths of the two pieces of the chord?
Answer:


CP= 13 – PB
Therefore equation (1)
(13 – PB)PB = 40
PB = 5, 8
If PB = 5cm then PC = 8cm
If PB = 8cm then PC = 5cm
In figure PB > PC
PB = 8 cm, PC = 5 cm

Circles Orukkam Questions & Answers

Worksheet 1

Question 1.
In triangle ABC, AB = 8cm, BC = 6cm , AC = 10cm.
1. What kind of triangle is this?
2. What is the position of B based on the circle with AC as the diameter? Why?
3. What is the position of A based on the circle with BC as the diameter? Why?
4. What is the position of the point C based on the circle with diameter AB?
Answer:
In Δ ABC
AB2 + BC2 = 82 + 62 = 64 + 36 = 100 = AC2
Δ ABC is a right angled triangle.
If we draw a circle taken in a AC as diameter, ∠B = 90°, Therefore the point B on the circle.
Δ ABC is a right-angled triangle.
If we draw a circle taking BC as diameter, ∠A < 90°, Therefore the position of point A will be outside the circle.
If we draw a circle taking AB as diameter, ∠C < 90°, Therefore the position of point C will be outside the circle.

Question 2.
Three vertices of a parallelogram are on a circle and the fourth vertex is at the center. Find the angles of the parallelogram.

Mark a point P on the top of the figure on the circle, join AP and CP. If angle AP C = x then write? AOC
Write ABC?
Write ∠ABC + ∠APC ?
What is APC?
Find the angles
Answer:

The angles of a parallelogram are 60, 120, 60 and 120.

Question 3.
In triangle ABC ,AB = AC, angle BAC = 30, BC = 5cm Find the radius of ABC
Draw the figure
Mark the center, BO and CO
Find the measure of angle BOC
Write the angles of triangle OBC
What kind of angle is triangle OBC
Write the radius of the circumcircle
Answer:

∠OBC= 75 – 15 = 60 = OCB (∵OB = OC)
Angle of Δ OBC are: 60, 60, 60
ΔOBC is an equilateral triangle.
Radius of circum circle of Δ OBC = OB = OC = BC = 5 cm.

Question 4.
P QRS is cyclic.
∠P = 3x, ∠Q = y, ∠R = x, ∠ = 5 v
Find the angles
Draw circle , mark P, Q, R, S on it, complete PQRS Enter the given angles.
What is 3x + x? Find x
What is y + 5y? Find y
Find the angles
Answer:

3x + x = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180″)
=> 4x = 180° => 4x = 45°
y + 5y = 180 (Sum of the opposite angles of a cyclic quadrilateral is 180°)

Question 5.
In the figure ABC D is a trapezium. If the vertices are on a circle, prove that it is an isosceles trapezium draw figure
What is ∠A + ∠C?

What is ∠B + ∠C?
Write the relation between A and B. Write the conclusion.
Answer:

∠A + ∠C = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180°)
∠B + ∠C = 180° (AB || CD)
(Sum of the alternate angles of a cyclic quadrilateral is 180″)
∠A + ∠C = ∠B + ∠C
∴ ∠A = ∠B
∴ ABC’D is an isoceles trapezium

Question 6.
In ABC AB = AC . P is the midpoint of AB and Q is the midpoint of AC. Prove that BPQC is cyclic?
Draw figure. Mark PQ and complete BP QC?
Is PQ parallel to BC?
Note that ∠B = ∠C?
What is ∠C + ∠Q?
What is ∠B + ∠Q?
Write conclusion.
Answer:

PQ || BC
(The line joining midpoints of two sides of a circle will be parallel to the third side).
∠B = ∠C (1)
(∵ AB = AC are given )
∠C + ∠Q = 180° (2)
(∵ PQ || BC, QC is the bisector, so the sum of alternate angles are 180°)
∠B + ∠Q = 180° .
From (1) and (2) we conclude that BPQC is an cyclic trapezium

Worksheet 2

Question 7.
Prove that ABCD given in the figure is cyclic

Draw figure and mark PQ
If ∠BAP =xthen
what is ∠BQP?
Find ∠PQD
Find ∠PDC? Why?
What is ∠A + ∠C?
Answer:
i. Quadrilateral ABPQ is cyclic
If ∠A = x, then ∠BQP = 1 80 – x
If ∠B = y, then ∠APQ = 180 – y
Quadrilateral PQCD is cyclic, SO
∠QCD = 180 – x (∠DPQ = x )
∠PDC = 180 – x (∠PQC = y)
∴ ABCD is a cyclic quadrilateral.

Worksheet 3

Question 8.
In the figure AB, C Dare extended and intersect at P. If AB = 5, BP = 3, P D = 2 then find CD? Draw the figure.

Write the relation between PA, PB, PC, PD
Find C D
Answer:
PA x PB = PCxPD
If CD = x, then
⇒ 8 × 3 = (2 + x)2 ⇒ 24 = 4 + 2x
⇒ 2x = 20 ⇒ x = \(\frac { 20 }{ 2 }\) = 10
∴ CD= 10

Question 9.
In the figure AB is the diameter and CD is parallel to the diameter. AB = 8cm,BD = 2cm, find CD

Answer:
If we draw a perpendicular DP to AB from D. Then PAxPB = PD2.
Here PB = x, then PA = 8 – x.

x(8 – x) = PD², 2² = x² + PD²
x(8 – x) = 4 – x², 8x – x² = 4 – x²
8x = 4, x = \(\frac { 1 }{ 2 }\)
Similarly, let us draw a perpendicular CQ to AB from C
AQ = ,PQ = 8– \(\left(\frac{1}{2}+\frac{1}{2}\right)\) = 7
CD = 7cm

Question 10.
Draw a rectangle of length 6cm and width 4cm. Draw another rectangle whose area equal to area of the first rectangle and one of the sides is 8cm.
Ans: Draw ABCD as in the given measurement. Mark E by extending AB 2cm more. AE = 8 will be 8cm. WithAas centre and AE radius draw an arc. This arc cut DA produced at F. Extend BA such that AD = AG and mark G. Draw triangle GF B and construct a circumcircle. The circle meets AD at H . Complete the rectangle AHIE

Worksheet 4

Question 11.
Draw an equilateral triangle of height 3cm. What is the length of a side?

Write the principle of construction. Students are advised to construct as in the steps given below.
Answer:
Draw a circle of radius 2cm and mark a diameter AB which is 4cm. Mark a point P from one end A is 3 cm apart on the diameter.
Draw a chord C D perpendicular to AB. Complete triangle C AD.
Using PA x PB = PD², PD= √3.
Now we get AD = AC = C D = 2√3
Height AP = 3cm.

Worksheet 5

Question 12.
In the figure, PA is a tangent and O is the centre of the circle. P A = 17, ∠OPA = 30° then calculate the radius of the circle and distance from centre to the point P Triangle OAP with 30°, 60°, 90° is right triangle. Using the property of this special right triangle find the radius and the distance.

Answer:
Δ OAP is a right-angled triangle having sides 30°, 60°and 90°.
Length of side which is opposite to the angle 90°, is twice the side which is opposite to the angle 30°.
Length of side which is opposite to the
angle 60°, is √3 times of the side which is opposite to the angle 30°
That is radius of the circle , OA = \(\frac { 17 }{ √3 }\)
Distance from centre to the point P

Worksheet 7

Question 13.
Draw a circle and construct 30°, 150° angles on it.
Answer:

Question 14.
Draw a circle and construct \(22 \frac{1}{2}^{0}\) on it.
Answer:

Question 15.
In triangle ABC the radius of the circumcircle is 6 cm, ∠A = 70°, ∠B = 80°. Construct the triangle
Answer:

Question 16.
Draw a rectangle of length 7cm, and width 5cm and construct a square whose area is same as the area of this rectangle.
Answer:
Draw a rectangle of length 7cm, and width 5cm. Area = 7 × 5 = 35 cm².
Therefore length of one side of the square is √35.

Draw a semicircle taking AH as diameter Extend BC, and mark a point F.
AB × BH = 7× 5 = 35
AB × BH = BF2 ;
BF = √35cm
Area of BEGF= √35 × √35 = 35 cm²

Question 17.
Draw a rectangle of one side 5cm, width 7cm. Construct another rectangle whose one side is 8cm and area equal to the area of the first rectangle.
Answer:
Draw a rectangle of length 7cm and width 5cm.

Extend AB up to 8 cm (AE = 8cm) Draw an arc taking A as centre and AE as radius. Extend DA and mark the point F.

Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB

Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.

Question 18.
Draw a square of side 5cm and construct a rectangle having one side 7cm and area equal to area of the square.
Answer:
Draw a square of side 5cm.

Extend AB to 7 cm
(AE = 7cm) Draw an arc taking A as centre and AE a radius. Extend DA and mark the point F.

Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB.

Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE a width AH.

Question 19.
What is the position of the vertex of an equilateral triangle with opposite side as the diameter?
Answer:

Angles of an equilateral triangle is 60° each. Position of the vertex of triangle with opposite side as the diameter is outside of the circle, because the angle is less than 90°.

Circles SCERT Questions & Answers

Question 20.
In the figure “ ABC is a right triangle
a. If a circle is drawn with AC as diameter find the position of B.
b. If a circle is drawn with BC as diameter, find the position of A. [ Score: 3 Time: 5 minutes]

Answer:
a. On the circle (1)
∠B = 90°

b. Outside the circle (1)
Position of the vertex of an triangle with opposite side as the diameter is outside the circle, because. ∠A <9o°.

Question 21.
A circle is drawn with AB as diameter. Find the positions of the points C, D, E related to the circle.

[ Score: 3 Time: 5 minute]
Answer:
C inside the circle (1)
∠C > 90°.
D on the circle (1)
∠D = 90°.
E outside the circle (i)
∠E <90°.

Question 22.
In Δ ABC and Δ PQR, BC = QR, ∠ A = ∠P, ∠Q = 90°, QR = 5 cm, PQ = 12 cm.

Find the diameter of the circumcircle of Δ ABC. [ Score: 4 Time: 6 minute]
Answer:
QR = BC (1)
∠A = ∠P (1)
PR Diameter of the circumcircle of Δ ABC (1)
Diameter = PR= \(\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{cm}\) .(1)

Question 23.
PQ and RS are two mutually prependicular chords of a circle. < QPR=50° find< PQS. [ Score: 3, Time: 6 minute]
Answer:
∠PRS = 90 – 50 = 40° (1)
∠PQS = 40° (1)

Question 24.
O is the centre of the circle. If ∠BOC = 130° and ∠AOB = 110°. What is ∠AOC?
Find all angles of Δ ABC
[ Score: 3, Time: 3 minute]
Answer:

Question 25.
Find all angles of the hexagon ABCDEF
[ Score: 4 Time:5 minute]
Answer:
∠ EFD = ∠EAD = 30°
∠ FE A = ∠FDA = 40°
∠FDE = ∠FAE = 35° (1)
∠BAC= ∠BDC = 45°
∠ABD= ∠ACD = 62°
∠ACB = ∠ADB= 35° (1)
∠A = 1480, ∠B = 100°
∠C = 97° ∠D= 155° (1)
∠E= 115° ∠F= 105° (1)

Question 26.
O is the centre of the circle and AB is a chord. AC is the bisector of ∠OAB. ∠OAB = 56°.
a. Prove that OC and AB are parallel,
b. Find ∠ABC and ∠OBE.

Answer:

Question 27.
O is the centre of the circle. AD and BC are perpendicular to XY. CB cuts the circle at E. Prove that –CE = AD.

Answer:
∠AEB = 90° (Angle in a semi circle) ( 1)
∠AEC=90°
∴ AECD is a rectangle
∴ AD = CE (2)

Question 28.
ABCD is a parallelogram. A, B, E, F are the points on a circle. ∠DEF = 80° Find out the angles of the quadrilateral AEFB.
[ Score: 4, Time: 4 minute]
Answer:
∠AEF = 180 – 80 = 100° (1)
∠ABF = 180 – 100 = 80° (1)
∠A = 180 – 80 = 100° (1)
∠EFB = 180 – 100 = 80° (1)

Question 29.

O is the centre of the circle. ? ∠OCA = x °.
a. Find ∠OAC
b. Prove that ∠OCA + ∠ABC = 90°
c. Prove that ∠ADC – ∠OCA = 90° [Score: 4, Time: 4 minute]
Answer:

Circles Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 30.
In the figure AB is the diameter. PC is perpendicular to AB. PC = 6cm, PB = 3cm. Find the radius of the semi-circle.

Answer:
AP × PB = PC²
AP × 3 = 62
AP = 36/3 = 12 .
AB = 12 + 3 = 15
ie Radius = 15/2 = 7.5 cm

Question 31.
In Δ PQR, ZP = 60°, ∠R = 30° find whether the vertex Q on the circle with PR as diameter.
Answer:
∠P + ∠R = 60 + 30 = 90°.
∴ ∠Q = 180 – 90 = 90°.
So point Q on the circle.

Question 32.
In Δ ABC, ∠A = 60°, ∠B = 70°. Find whether the vertex C is inside or outside the circle with AB as diameter.
Answer:
The vertex C is outside the circle Since ∠C = 180 – (60 + 70) = 50° ∠90°

Question 33.
In the diagram, the central angle of arc ABC is 100° and ∠OAD is 30°. Find ∠OCD.
Answer:

∠D = 50°
As ∠OAD
is an isosceles triangle.
∠ODA = 30°
∠ODC = 20°
∴∠OCD = 20°

Question 34.
In the figure, find ∠PQB, O is the centre.

Answer:

Short Answer Type Questions (Score 3)

Question 35.
The central angle of arc ABC is 60°, then find out the following,
i) ∠D
ii) Central angle of arc AEC,
iii) ∠B

Answer:
i. ∠D = 30°
ii. Then central angle of arc AEC = 300°.
iii. ∠B = 150°

Question 36.
Show that the arcs APC, BQD when joined make a semicircle?

Answer:
∠ADC is the half of the central angle of arc APC
The central angle of are APC = 2 ∠ADC
The central angle of arc BQD = 2 ∠DAB
In triangle AOD, CD ⊥ AB, ∠AOD = 90°,
∠DAO + ∠ADO = 180 – 90° =90°,
ie, ∠DAB + ∠ADC = 90°
2( ∠DAB + ∠ADC ) = 90° × 2 = 180°

Question 37.
The central angle of the complementary arc of a circle is 40° more than 3 times the central angle of the arc. Find out the central angles of each arc?

Answer:
x + 3x + 40 = 360
4x + 40 = 360
4x = 360 – 40 = 320
x = \(\frac { 320 }{ 4 }\) = 80
∴ central angle of arc
ABC = 80
central angle of arc ADC = 360 – 80 = 280°

Question 38.
ABCD in the diagram is a rectangle. Then find out the area of the circle.

Answer:
ABCD is a rectangle ∠B = ∠D = 90°
AC in the diameter of the circle
AC = \(\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10\)
∴ radius = 5cm
∴ Area of the circle = πr² = π × 5² = 25πcm²

Question 39.
In the figure, AD = 16cm, BD = 6cm, CD = 2cm. Find the length EF.

Answer:

Question 40.
In the given figure, O is the centre of the circle. If ∠OAP = 35° and ∠OBP = 40°, find the value of ∠x.

Answer:
Join OP
Since OA = OP and
∠APO = ∠OAP=35°
Similarly, OB = OP and ∠OPB = ∠OBP = 40°
∠APB = 35°+ 40° = 75°
∠AOB = 2 × 750 = 150°

Long Answer Type Questions (Score 4)

Question 41.
In the figure find ∠APB,∠ABQ ; where O is the centre of the circle ∠OAP = 32° and  ∠OBP = 47° .

Answer:
JoinOP.
In OAP, OA = OP = Radius
∠OAP = ∠OPA = 32°
In OPR, OB = OP = radius
∠OBP = ∠OPB =47°
∠APB = 32°+ 47°= 79°
∠AQB = 180° – 79°= 10°

Question 42.
Draw a line of √7 cm.
Answer:

Question 43.
Ois the centre of the circle as shown in the figure.
Find ∠CBD.

Answer:

Takeapoint E on the circle, join AE and CE.
∠AEC= \(\frac { 100 }{ 2 }\) = 50°
∠AEC + ∠ABC = 180° (Opposite angles of a cyclic quadrilaterals)
∠ABC = 130°
∠ABC + ∠CBD = 180° (linearpair)
130°+ ∠CBD = 180°
∠CBD = 50°

Long Answer Type Questions (Score 5)

Question 44.
In the figure O is the centre of the circle. Central angle of arc AXB is 60°, arc CYD is 80°. Then find all the angles of ΔAPD.

Answer:
Central angle of arc AXB = 60°
i.e., ∠AOB = 60°
i.e., ∠ADP = 30°
Central angle of arc CYD = 80°
i.e., ∠COD = 80°
i e., ∠DPA = 40°
i e., ∠APD = 180° – (30 + 40) = 110°
Angles of = 30°, 40°, 110°

Question 45.
‘O’ is the centre of the circle ∠D = 80°, find the following measurements.

a. ∠C
b. ∠ABC
c. ∠BAC
d. ∠F
Answer:
∠D = 80°
a. ∠C = 80° (∠D and ∠C are angled on a same arc So, both are equal)

b. ∠ABC = 90°
(AC is diameter, Angle of a hemisphere is right)

c. ∠BAC = 180 – (80 + 90)
= 180 – 170 = 100

d. ∠F= 180 – 80 = 100° (Opposite angles of a cyclic quadrilateral are equal)

Circles Memory Map

Angle in a semicircle is right:
The angle made by any arc of a circle on the alternate arc is half the angle made at the centre.

All angles in an arc is equal:

If AB, CD are two chords, then
PA × PB = PC × PD

The area of the rectangle formed of parts into which a diameter of a circle is cut by a perpendicular chord is equal to the area of the square formed by half the chord.
PA × PB = PC²

Students can Download Chemistry Chapter 2 Gas Laws Mole Concept Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 2 Gas Laws Mole Concept Questions and Answers Malayalam Medium

SCERT 10th Standard Chemistry Textbook Chapter 2 Solutions Malayalam Medium

You can Download Reactivity Series and Electrochemistry Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 3 Reactivity Series and Electrochemistry Textbook Questions and Answers

SCERT Class 10th Standard Chemistry Chapter 3 Reactivity Series and Electrochemistry Solutions

Reactivity Series And Electrochemistry Kerala Syllabus Text Book Page No: 48

→ Which metal reacts vigorously?
Answer:
Sodium.

→ Which gas is formed as a result of this reaction?
Answer:
Hydrogen.

→ Write down its chemical equation.
Answer:
2Na + 2H₂O → 2NaOH + H₂

→ Complete the table (3.1) given below.

Answer:

→ Based on your observation, arrange these metals in the decreasing order of their reactivity.
Answer:
Sodium > Magnesium > Copper
→ 2Mg + O₂ →
Answer:
2Mg + O₂ → 2MgO

Sslc Chemistry Chapter 3 Kerala Syllabus Text Book Page No: 49

→ Which metal among magnesium, copper, gold, sodium and aluminium, loses its lustre at a faster rate?
Answer:
Sodium

→ List the above metals in the decreasing order of their reactivity with air and thereby losing lustre
Answer:
Sodium > Magnesium > Aluminium > Copper > Gold.

Text Book Page No: 50

→ What happened to the Zn rod?
Answer:
Before the experiment the Zn rod was colourless. After the experiment Zn rod became blue due to the deposition of copper.

→ What is the reason for this?
Answer:
When the Zn rod is dipped in CuSO₄ solution, the Cu²⁺ ions in the solution get deposited at the Zn rod as Cu atoms.

→ What is the reason for the change in intensity of the colour of CuSO₄ solution?
Answer:
The blue colour of CuSO₄ solution is due to the presence of Cu²⁺ ions. The change in intensity of the colour of CuSO₄ solution because when the Zn rod is dipped in CuSO₄ solution, the Cu²⁺ ions in the solution get, deposited at the Zn rod as Cu atoms.

→ Which is the metal that gets displaced here?
Answer:
Copper

→ Which is more reactive Zn or Cu?
Answer:
Zn

→ On the basis of the position of Zn and Cu in the reactivity series, can you explain why Cu had been displaced?
Answer:
Zn is placed above Cu in the reactivity series because Zn has a higher reactivity than Cu.

→ Isn’t it due to the higher reactivity of zinc (Zn) when compared to copper (Cu)?
Answer:
Yes.

Chemistry Class 10 Chapter 3 Kerala Syllabus Text Book Page No: 51

→ Is this reaction oxidation or reduction? Why?
Answer:
Oxidation. Because the losing of electrons is called oxidation.

→ The change that happened to Cu²⁺
Answer:
Cu^–2+ + 2e^– → Cu

→ What is the name of this reaction? Why?
Answer:
Reduction. The gaining of electrons is called reduction.
→

Complete this chemical equation by assigning oxidation numbers.
Answer:
2 Ag⁺¹ NO₃^1–+ Cu°→ Cu²⁺ (NO₃)^–1₂ + 2Ag⁰

→ Which metal was oxidised in this case? Which metal was reduced?
Answer:
Metal which was oxidised: Cu
Metal ion which was reduced: Ag⁺

→ Write equations showing oxidation and reduction.
Answer:
Oxidation : Cu⁰ → Cu²⁺ + 2e^–
Reduction : Ag⁺+ le^– → Ag⁰

Reactivity Series And Electrochemistry Sslc NotesText Book Page No: 52

→ Complete the table 3.3.

Answer:

Sslc Chemistry Chapter 3 Notes Pdf Kerala Syllabus Text Book Page No: 53

→ Which electrode has the ability to donate electrons in a cell constructed using these metals?
Answer:
Zn

→ Which one can gain electrons?
Answer:
Cu

→ Identify the chemical reaction that takes place at the Zn electrode. Tick ✓ the right one.
Answer:
Zn(s) → Zn²⁺ (aq) + 2e^–  (✓)
Zn²⁺(aq) + 2e^– → Zn(s)   (✘)

→ What is the reaction taking place here?
Answer:
Oxidation.

→ Write the chemical equation for the reaction taking place at the Cu electrode.
Answer:
Cu²⁺ (aq) + 2e^– → Cu(s)

→ Sketch the cell constructed.
Answer:

→ Note down the reaction of the Galvanic cell.
Answer:
Cu(s) + 2Ag⁺(aq) → Cu²⁺ (aq) + 2Ag(s)
Cu(s) → Cu²⁺(aq) + 2e^– (Anode)
Ag⁺(aq) + le^– → Ag(s) (Cathode)

→ Direction of flow of electrons From Cu to Ag
Answer:
Mark the direction of electron how in the cell illustrated.

→ write the reactions taking place at cathode and anode.
Answer:
At cathode : Ag⁺ + le^– → Ag
At anode : Cu → Cu²⁺ + 2e^–

Hss Live Guru 10th Chemistry Kerala Syllabus Text Book Page No: 55

→ You have used three metals Zn, Cu and Ag. How many cells can be produced using these?
Answer:
Three.

→ Complete the Table 3.4 by writing anode and cathode in each.

Answer:

→ What are the substances obtained when electricity is passed through acidified water?
Answer:
Hydrogen, Oxygen.

→ Do such type of chemical changes happen when electricity is passed through metals?
Answer:
Yes.

Sslc Chemistry Chapter 3 Questions And Answers Kerala Syllabus Text Book Page No: 56

→ To which electrodes are the positive ions attracted during electrolysis?
Answer:
Towards negative electrodes(Cathode)

→ To which electrodes are the negative ions attracted?
Answer:
Towards positive electrodes(Anode),

→ What changes happen to the ions which N are attracted to cathode?
Answer:
Reduction

→ What about the changes happening to the ions attracted to anode?
Answer:
Oxidation

→ Which ion is attracted to the positive electrode (anode)?
Answer:
Chloride ion (Cl^–)

→ What is the chemical reaction taking place there?
Answer:
2Cl^– (aq) → Cl₂(g) + 2e^–

Text Book Page No: 57

→ Which is the gas liberated at the anode?
Answer:
Chlorine(Cl₂)

→ Which is the ion attracted to the negative electrode (cathode)? Write the change happening to it?
Answer:
Na⁺ ions. These ions accept one electron and changes to sodium atom. That is sodium ions are reduced.

→ Which is the metal deposited at the cathode?
Answer:
Sodium (Na)

→ Which are the ions attracted to the positive electrode?
Ans.
Cl^–,OH^–

→ Which are the ions attracted to the negative electrode?
Answer:
Na⁺,
H₃O⁺,
H₂O.

Text Book Page No: 59

→ Which metal is connected to the negative terminal of the battery?
Answer:
Iron.

→ Which metal is connected to the positive terminal of the battery?
Answer:
Copper.

→ Which solution is used as the electrolyte?
Answer:
Copper sulphate solution.

→ What happens to Cu²⁺ ions at the cathode? Complete the equation.
Answer:
Cu²⁺ + 2e → Cu

→ What happened to the copper ions? Oxidation/Reduction?
Answer:
Reduction.

→ Complete the equation given below.
Answer:
Cu → Cu²⁺ + 2e^–

Text Book Page No: 60

→ Find out more examples and extend the list.
Answer:

• Chromium plating is used in motor car etc.
• To make metal coating easily corroding metals to prevent corrosion.
• In ICs (Integrated Circuits) coating of gold /silver is made by electroplating.

Reactivity Series and Electrochemistry Let Us Assess

Kerala Genetics Question 1.
The solutions of ZnSO₄, FeSO₄, CuSO₄ and AgNO₃ are taken in four different test tubes. Suppose, an iron nail is kept immersed in each one
In which test tube the iron nail undergoes a colour change?
What is the reaction taking place here?
Justify your answer. (Refer reactivity series of metals).
Answer:
Iron nail immersed in solution of CuSO₄ and AgNO₃ undergoes a colour change.
i. Fe(s) + CuSO₄(aq) →
FeSO₄ (aq) + Cu (s).
ii. Fe(s) + 2AgNO₃ (aq) →
Fe(NO₃)₂(aq) + 2Ag(s).
Iron displaces Cu from CuSO₄ and Ag from AgNO₃ because Fe has higher reactivity than Cu and Ag.

Genetic Engineering Syllabus Question 2.
Compare the electrolysis of molten potassium chloride and solution of potassium chloride. What are the processes taking place at the cathode and the anode?
Answer:
Molten KCl
KCl (s) → K⁺ + Cl^–
At the negative electrode:
K⁺ + le^– → K (reduction – cathode)
At the positive electrode:
2Cl^– → Cl₂ + 2e^– (oxidation- anode)
Solution of potasium chloride.
At the negative electrode:
2H₂O + 2e^– → H₂ + 2OH^– (cathode).
At the positive electrode:
2Cl^– → Cl₂ + 2e^– (anode).

Future Diary 10th Question 3.
You are given a solution of AgNO₃, a solution of MgSO₄, a Ag rod and a Mg ribbon. How can you arrange a Galvanic cell using these? Write down the reactions taking place at the cathode and the anode.
Answer:
At anode,
Mg (s) → Mg²⁺ (aq) + 2e^–
At cathode,
Ag (aq) + le^– → Ag (s)

Reactivity Series and Electrochemistry Extended Activities

The Reactivity Question 1.
1. Keep two carbon rods immersed in copper sulphate solution. Then pass electricity through the solution.
i. At which electrode does colour change occur anode or cathode?
ii. Is there any change in the blue colour of the copper sulphate solution?
iii. Write down chemical equations for the changes occurring here.
Answer:
i. At cathode
ii. Colour fades
iii.At cathode: Cu²⁺ (aq) + 2e^– → Cu (s)
At anode: 2H₂O → O₂(g) + 4H⁺ (aq) + 2e^–

SSLC Chemistry Chapter 4 Question 2.
When acidified copper sulphate solution is electrolysed oxygen is obtained at the anode. What arrangements are to be made for this? Find the element deposited at the cathode.
Answer:

Element deposited at the cathode: Copper.

Question 3.
a. When Galvanic cells are made using the metals like Mg, Cu, Zn and Ag, what will be the nature of reactions in each cell?
(Reactivity: Mg > Zn > Cu >Ag)
b. How many Galvanic cells can be made by using the metals Ag, Cu, Zn and Mg?
Chapter 7 Biology test Answers:
a. i. Cu-Ag cell
Anode: Cu (s) → Cu²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag(s)
ii.Zn-Ag cell
Anode: Zn (s) → Zn²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag (s)
(iii) Mg-Ag cell
Anode: Mg (S) → Mg²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag (s)
(iv) Zn – Cu cell
Anode: Zn (s) → Zn²⁺ (aq) + 2e^–
Cathode: Cu²⁺ (aq)+ 2e^– → Cu(s)
(v) Mg-Cu cell
Anode: Mg (s) → Mg²⁺ (aq) + 2e^–
Cathode: Cu²⁺ (aq) + 2e^– → Cu (s)
(vi) Mg-Zn cell
Anode: Mg (s) → Mg²⁺ (aq) + 2e^–
Cathode: Zn²⁺(aq) + 2e^– → Zn(s)
b. 6 cells

Reactivity Series and Electrochemistry Orukkam Questions and Answers

Scope of Genetic Engineering Question 1.
1. Take cold water and Hot water in two test tubes, Add one or two drops of phenolphtha lein in it. Drop equally sized Mg ribbon in it.
a. In which test tube pink colour occured sharply?
b.Why did pink colour appear in that test tube so early?
c. Which gas evolved out from both test tubes?
d. Write balanced equation for the above mentioned reaction.
Answer:
a. Pink colour occurred sharply in the test tube with hot water.
b. Temperature is a factor that affects the rate of a reaction. When heated the kinetic energy of molecules increases and hence the rate of chemical reaction also increases which causes the pink colour to appear early.
c. Hydrogen
d. Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g)

SCERT Question Pool 2017 Question 2.
Cut a small sodium metal piece into two, watch it.
a. What change occurred on the surface of sodium metal?
b.Write one word for the process of this type of decomposition.
c. Write down the equations for this.
Answer:
a. After some time the cut piece of sodium will turn dull.
b.Corrosion.
c. 4Na(s) + O₂(g) → 2Na₂O
2Na(s) + 2H₂O(l) → 2NaOH(s) + H₂(g)
2NaOH (s) + CO₂(g) → Na₂CO₃(s) + H₂O(l)

Question 3.
Take equal quantities of dil HCl in five test tubes. Drop Mg, Zn, Fe, Cu in each test tube. Watch carefully.
a. Arrange metals in decreasing order of reactivity.
b.Write balanced equations for each reaction.
SCERT Question Pool Answer:
a. Mg > Zn > Fe > Cu
b. Zn + 2HCl → ZnCl₂ + H₂
Fe + 2HCl → FeCl₂ + H₂
Cu + HCl → No reaction
Mg + 2HCl → Mg Cl₂ + H₂

SCERT Question Pool Question 4.
Some metals and metallic compounds are given in the table. If the metal substitute the metal in the compound put a tick mark in the corresponding column and otherwise a cross mark in the column. Write down correct answer based on the table given below.

a. Correct the table if necessary.
b. Is it possible to substitute lower positioned metals by top positioned metals in the reactivity series?
c. What type of reaction is this?
d. Write down balanced equations for all the true sign given in the table.
Answer:
a.

b.Yes. It is possible.
c. Substitution reactions.
d.CuSO₄ + Mg → Cu + MgSO₄
CuSO₄ + Zn → ZnSO₄+ Cu
CuSO₄ + Fe → FeSO₄ + Cu
ZnSO₄ + Mg → MgSO₄ + Zn
AgNO₃ + Mg → Ag + MgNO₃
AgNO₃ + Cu → CuNO₃ + Ag
AgNO₃ + Zn → ZnNO₃ + Ag
AgNO₃ + Fe → FeNO₃ + Ag

Human Insulin Gene Question 5.
Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
a. Complete the table based on the figures drawn.

B. Write down the general names used for an electrode which gives electrons,
c. Metals in that electrode in the reactivity series is (in the Top, Bottom)
d. General name of the Electrode which accepts electron.
e. Process of giving electron is
f. Process of Accepting electron is
g.Direction of the flow of Electron
h.Write down the balanced equation taking place in both electrodes.

Answer:

Zn-Ag cell is also possible.
a.

b. Anode.
c. In the top.
d. Cathode.
e. Oxidation.
f. Reduction.
g. From Anode to Cathode.

Chemistry Reactivity Series Question 6.
Take Cupric chloride (CuCl₂) solution in a beaker. Dip two graphic rod in it. Pass 5V electricity through it.
a. Why electricity passes through cupric chloride solution?
b.Which gas evolved out through positive electrode? How did you identify that gas?
c. Which product formed in negative electrode?
d. In which electrode oxidation and reduction take place?
e. Write one word for the process of chemical change happening in a Electrolyte while passing Electricity?
Answer:
a. Electrolytes are substances which conduct electricity in molten states or in aqueous solutions. In molten state ions of CuCl₂ can more freely. These ions are responsible for the Conduction of electricity by electrolytes.
b.Chlorine.
c. Copper.
d.Oxidation takes place at the positive electrode and Reduction takes place at the negative electrode.
e. Electrolysis.

Reactivity Series Class 10 Question 7.
Take 25 ml water in a beaker and the pass electricity through it. Then add little sulphuric acid in it.
a. Why electricity didn’t pass through pure water?
b. What happens when dil H₂SO₄ is added?
c. Which type of ion formed more when sulphuric acid is added in water?
d. Complete the equation of the Ionization 0f H₂SO₄
H₂SO₄ → 2H⁺ +
Based on the equation given below write down the correct answers.
2H⁺ +  + 2H₂O → 2H₃O +  + SO₄^2–
e. Complete the equation.
f. Write down the name of H3O⁺ ion?
g.Which ion is moving towards negative ion?
h.Complete the reaction taking place in the negative electrode.
2H₃O⁺(aq) + 2e^– →  +
i. Whicfrion is having highest oxidation potential?
j. Complete the reaction taking place in positive electrode.
2H₂O →  + 4H⁺
k. Ions remain in the beaker after the electrolysis are
I. What product form when these two combined together
Answer:
a. Since the number of ions is less, pure water does not allow the passage of electricity.
b. When dil H₂SO₄ is added the water becomes a good electrolyte. Hence electricity passes through it. When a little dilute sulphuric acid is mixed with water large quantity of hydronium ions are formed.
c. Hydronium ion (H₃O⁺)
d. H₂SO₄ → 2H⁺ + SO₄^2–
e. 2H⁺ + S0₄^2– + 2H₂O → 2H₃O⁺ + SO₄^2–
f. Hydronium ion
g. H₃O⁺
h. 2H₃O⁺(aq) + 2e^– → H_2(g) + 2H₂O
i. H₂O has the highest oxidation potential when compared to SO₄^2–
j. 2H₂O → O₂(g) + 4H⁺ + 4e^–
k. 4H⁺, SO₄^2–ions.
l. H₂SO₄ is formed when these 2 combine together.

Reactivity Series Chemistry Question 8.
a. Complete the table based on the Electrolysis of molten sodium chloride.

b. Write down the reaction taking place in each electrodes and products formed in the electrolysis of sodium chloride solution.

c. Why is hydrogen formed in the cathode instead of sodium?
d. Write one word for a solution undergoes chemical change when electricity passes through it.
e. Write the name of the above process.
f. Write down the uses of above type of reaction.
Answer:

c. When Na+ ion and water are compared reduction occurs to water. Hence H₂ is liberated at cathode.
H⁺+ e → H,
H + H → H2.
d. Electrolyte.
e. Electrolysis.
f. Electroplating, Production of chemicals, Purification of metals.

Reactivity Series and Electrochemistry Evaluation Questions

Take little water in a test tube add two drops of phenolphthalein in it Same quantity of Kerosene is added to the mixture and a small piece of sodium is dipped in it.

Insulin Gene Question 1.
What kind of colour formed in the test tube? Why?
Answer:
Water become pink in colour. Because when phenolphthalein is added to water it becomes alkaline.

Question 2.
Which gas is bubbled on the surface of sodium metal?
Answer:
Hydrogen

Question 3.
Write balanced equation of the reaction between sodium and water.
Answer:
2Na(s) + 2H₂O(l) → 2NaOH (aq) + H₂(g).

Question 4.
What products occurs when Iron reacts with water vapour?
Answer:
Fe₃O₄ (Iron Oxide) and Hydrogen gas.

Question 5.
Lustre of magnesium disappeared fast when it placed in open space why?
Answer:
When Magnesium is kept in open space it reacts with atmosphere air and a light coat of magnesium oxide is formed. This is the reason for Magnesium to lose its lustre.

Question 6.
Verdigris formed on copper utensils disappeared after some days why?
Answer:
The copper in copper utensils react with atmospheric air and forms copper oxide. Due to this verdigris are formed on Copper utensils.

Question 7.
Lustre of aluminium utensils disappear after some days. Why?
Answer:
Aluminium reacts with atmospheric air and Aluminium oxide is formed. This process takes place slowly. Hence Aluminium utensils loose their lustre.

Question 8.
Write down the equation for the reaction between CuSO₄ and iron nail? What type of reaction is this?
Answer:
CuSO₄ + Fe → FeSO₄ + Cu.
Redox reaction.

Reactivity Series and Electrochemistry SCERT Questions and Answers

Question 1.
5ml water is taken in 3 test tubes. Copper, sodium and magnesium of equal mass are dropped in different test tubes. Test tubes having copper and magnesium are heated.
a. Write the observations in the heated test tubes.
b. Write the equation for the reaction in the test tube in which sodium is dropped,
c. Arrange these metals in the decreasing order of their reactivity.
Answer:
a. Mg reacts with hot water liberating hydrogen, Copper does not react with hot water.
b. 2Na + 2H₂O → 2NaOH + H₂.
c. Na > Mg > Cu.

Question 2.
a. Which metal among copper, aluminium and gold loose its metallic lustre at a faster rate? Write the equation of the reaction.
b. Sodium is kept in Kerosene. Why?
Answer:
a. Aluminium, Al + 3O₂ → 2Al₂O₃
b. Na reacts with air (oxygen) and water.

Question 3.
An experimental setup is made to compare the reactions of Mg, Zn and Cu with dilute hydrochloric acid.
a. Write the procedure and observation of the reaction.
b. Which is the gas evolved when zinc reacts with dilute hydrochloric acid?
Answer:
a. Take Mg, Zn and Cu in different test tubes and add dilute HCl to each.
Observation: Magnesium and Zinc reacts with dilute hydrochloric acid copper does not react with the acid.
b.Hydrogen.

Question 4.

a. What are the changes that can be observed with the iron rod and the colour of copper sulphate solution?
b. Write the equations of the oxidation and reduction reactions.
c. What will be the change if silver rod is used instead of iron rod? What is the reason?
Answer:
a. Copper is deposited on iron and the blue.colour of copper sulphate solution decreases.
b. Cathode – Cu²⁺(aq) + 2e^– → Cu (s) (Reduction)
Anode – Fe (s) → Fe²⁺(aq) + 2e^– (Oxidation)
c. No change occurs, Reactivity of silver is less than that of copper. In the reactivity series, the position of silver is below copper.

Question 5.
Sodium reacts with water.
a. Identify the gas evolved in the reaction h If two drops of phenol[ihthaloin is added to the water, what will be colour change of the resultant solution? Explain the reason?
Answer:
a. Hydrogen.
b. Colour changes to pink. Due to the presence of sodium hydroxide (alkaline nature).

Question 6.
Three Galvanic cells are given.

a. Find out the most reactive metal and least reactive metal among them.
b. In cell, which electrode undergoes oxidation why?
c. Write the equation of the redox reaction occurring in cell 3 (Valency of A, B are 2.)
Answer:
a. Most reactive metal A, Least reactive metal B.
b. A Reactivity of A is higher than B.
c. A+ 2e^– → A²⁺,
C²⁺ + 2e^– → C ,
A + C²⁺ → A²⁺ C .

Question 7.
Some metals and salt solutions are given (Cu, Zn, Ag, ZnSO₄, AgNO₃, MgCl₂)
a. Draw the diagram of a Galvanic cell that can be made using these substances.
b. Find out the anode and cathode of this cell and write the chemical equation for the reaction at cathode.
Answer:

b. Anode – Zn,
Cathode – Ag
2Ag⁺+ 2e^–→ 2Ag

Question 8.
Give reasons for the following.
a. Iron vessels are not used a boilers that are used to boil water.
b. Blue vitriol solution is not kept in iron vessels.
Answer:
a. Iron reacted with steam-heated to high temperature

Question 9.
Examine the given electrolytic cell.

a. Which gas is evolved at the positive electrode?
b. Write the oxidation and reduction reactions of this cell.
c. What is the difference in the energy transformation of a Galvanic cell and an electrolytic cell?
Answer:
a. Chlorine / Cl₂
b. 2Cl^– → Cl₂ + 2e^–
Cu²⁺ + 2e^– → Cu
c. Galvanic cell – Chemical energy is converted to electrical energy.
Electrical Cell – Electrical energy is changed to chemical energy.

Question 10.
The solutions in the given table electrolyzed.

List any two areas in which electrolysis is made use of?
Answer:
a. i. Hydrogen.
ii. Chlorine.
iii.Chlorine.
iv. Hydrogen.
b.

• Purification of metals.
• Electroplating.
• Production of chemicals.

Question 11.
The position of iron is below that of zinc in reactivity series. The cell formed by them is given. Correct the mistakes and redraw.

Answer:

Question 12.
Sodium chloride solution is electrolysed using platinum electrodes.
a. Write the chemical equation of the reaction at cathode.
b. What happens when phenolphthalein is added to the solution? State the reason?
Answer:
a. 2H₂O + 2e^– → H₂ + 2OH
b. Colour turns pink. Presence of sodium hydroxide in the solution.

Question 13.
The anode and cathode of two Galvanic cells are given.

A. Mg → Mg²⁺ + 2e^–
B. Zn²⁺+2e^– → Zn
C. Ag⁺ +le^– → Ag
D. Zn → Zn²⁺ + 2e^–
E. Ag → Ag⁺ + le^–
F. Mg²⁺ +2e^– → Mg
a. Find out the reactions at the anode and cathode for each cell from the above.
b. Which metal can act only as cathode? Why?
Answer:
a. Cell 1: Anode Mg → Mg²⁺ + 2e^–
Cathode Zn²⁺ + 2e^– → Zn
Cell 2: Anode Zn → Zn²⁺ + 2e^–
Cathode Ag⁺ + le^– → Ag
b. Ag. Lesser tendency to give up electron that is it is a less reactive metal.

Question 14.
The chemical reactions of various Galvanic cells are given as incomplete in the table. Complete them.

Answer:
a. Zn → Zn²⁺+ 2e^–
b. Zn+Cu²⁺ → Zn²⁺ + Cu
c. Fe → Fe²⁺ + 2e^–
d. 2Ag⁺ + 2e- → 2Ag
e. Pb²⁺+ 2e^– → Pb
f. Mg + Pb²⁺ → Mg²⁺ + Pb

Question 15.
The picture of a Galvanic Cell is given below.

a. Identify A and B.
b. Give the direction of electron flow?
c. Write the chemical equation at the anode and cathode.
Answer:
a. A – Zn,
B – FeSO₄Solution
b. From Zn to Fe
c. Anode Zn → Zn²⁺ + 2e^–
Cathode Fe²⁺ + 2e^– → Fe

Question 16.
An incomplete table about the electrolysis of different electrolytes are given below. Complete it.

Answer:
a. Cu²⁺,
Cl^–,
H₂O.
b. 2H₂O → O₂ + 4H⁺ + 4e^–
c. Na⁺,
Cl^–
d. Na⁺ + le^– → Na
e. 2Cl → Cl₂ + 2e^–
f. 2H₂O + 2e^– → H₂ + 2OH^–

Question 17.
5mI AgNO₃ is taken in a test tube and a copper rod is dipped into
a. Identify the changes occurring with the copper rod and the solution?
b. Complete the equation of the reaction.
Cu + 2AgNO₃ →  +
c. Write the equations of the oxidation and reduction reactions.
Answer:
a. Silver is deposited in copper rod. Colour of solution changes to blue.
b. Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2Ag
c. 2Ag⁺ + 2e^– → 2Ag (Reduction)
Cu → Cu²⁺ + 2e^– (Oxidation)

Reactivity Series and Electrochemistry Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
The symbols of certain metals are given below. Arrange them as they are given in the reactivity series.
Mg, Pb, Ag, Cu, Zn, Fe, Au, Sn.
Answer:
Mg, Zn, Fe, Sn, Pb, Cu, Ag, Au.

Question 2.
Analyse the table given below and answer the questions.

a. Find out the metals which are likely to be A, B and C from the box given below.

b. Write down the chemical equation between metal ‘B’ and water.
Answer:
a. A – Mg,
B – Cu,
C – Ca
b. Cu (s) + 2H₂O (l) → Cu OH₂ (aq) + H₂ (g)

Very Short Answer Type Questions (Score 2)

Question 3.
Certain metals are given below:
Ag, Zn, Pb, Sn, Fe
a. When a galvanic cell is constructed using these metals, which one acts only as anode? Give the reason.
b. Draw Zn-Fe cell. Mark the direction of electron flow and write the chemical equation anode.
Answer:
a. Zn. Because Zn has a higher reactivity than the other 4 metals.

Direction of electron flow from Zn to Fe Chemical reaction at anode:
Zn(s) → Zn²⁺(aq) + 2e^–

Question 4.
Zn (s) + 2AgNO₃ (aq) →
Zn(NO₃)₂ (aq) + 2Ag
a. Write the oxidation state of each element in this displacement reaction.
b. Write the chemical equation for oxidation and reduction.
Answer:
a. Zn° (s) + 2Ag¹⁺N⁵⁺ O₃^2– (aq)
→ Zn²⁺(N⁵⁺O^2–₃)₂ (aq) + 2Ag°

b.Oxidation:
Zn° (s) → Zn²⁺ (aq) + 2e^–
(Oxidation means losing of electrons)

Reduction:
Ag¹⁺ (aq) + le^– → Ag (s)
(Reduction means gaining of electrons)

Question 5.
Based on the reactions given below, answer the following questions.
i. Aqueous solution of CuCl₂ undergoes electrolysis using graphite rods.
ii. Molten KCl undergoes electrolysis.
iii. Aqueous solution of NaCI undergoes electrolysis.
a. In which all reactions Cl₂ gas is formed? At which electrode is Cl₂ gas formed?
b. In which reaction H₂ gas is formed? Write the chemical equation of this reaction.
Answer:
a. Cl₂ gas is formed in all the three reactions. Chlorine gas is formed at anode.

b. When aqueous solution of NaCI undergoes electrolysis, hydrogen gas is formed at cathode.
Equation: 2H₂O + 2e^– → H₂ + 2OH^–

Very Short Answer Type Questions (Score 3)

Question 6.
Take water in 4 different beakers and add a small piece of sodium, lead, iron and copper in each.
a. In which all solutions gas bubbles will be formed? Which gas is formed?
b. Which solution will turn pink on adding phenolphthalein? Why?
c. Write the chemical equation between this metal and water.
Answer:
a. Gas bubbles are formed in the beaker containing sodium. Hydrogen is tlie gas formed.

b. Beaker containing sodium turns pink because sodium reacts with water and forms an alkali, sodium hydroxide. Alkalies turn phenolphthalein pink.

c. 2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂ (g)

Question 7.
Analyse the picture given below.

a. Identify ‘A’.
b. Write the chemical equation at ‘B’.
c. Add few drops of phenolphthalein to the remaining solution after electrolysis. What change will take place? Why?
Answer:
a. A – cathode.

b. Chemical equation at ‘B’:
2Cl^– → Cl₂ + 2e^–

c. Solution turns pink. Because, after electrolysis K⁺ and OH^– ions are present in the remaining solution. That means KOH, an alkali is formed.

Question 8.
The flow of electron in certain galvanic cells are given below:
i. Cu → Ag
ii. Ag → Zn
iii.Na → Mg
iv.Fe → K
a. Choose the incorrect ones.
b. Explain your answer.
Answer:
a. ii. Ag → Zn,
iv. Fe → K are the incorrect ones.

b. Zn has a higher reactivity than Ag. Hence Zn undergoes oxidation (loses electrons). So the direction of electron flow is from Zn to Ag. Similarly, K has higher reactivity than Fe. Hence direction of electron flow is from K → Fe.

Question 9.
Which of the chemical reactions given below are wrong? Explain the reason.
a. Cu (s) + 2HCl (aq) →
CuCl₂(aq) + H₂ (g)
b. Mg(s) + Pb(NO₃)₂ (aq) →
Mg(N0₃)₂ (aq) + Mg (s)
c. 3Fe (s) + 4H₂O (l) →
Fe₃O₄(s) + 4H₂(g)
Answer:
Equations (a) and (c) are wrong.
a. Copper cannot displace hydrogen from acids because it is placed below hydrogen in the reactivity series.

c. Fe reacts only with superheated steam. Fe does not react with water in liquid state.

Question 10.
Write the chemical equation of the electrolysis of water to which little sulphuric acid (H₂SO₄) is added.
Answer:
H₂SO₄ (l) + 2H₂O (l) → 2H₃O (aq) + SO₂^2–(aq)
at cathode:
2H₃O+ (aq) + 2e^– → H₂(g) + 2H₂O (l) (reduction)
at anode:
2H₂O (l) → O₂(g) + 4H^– (aq) + 4e^– (oxidation)

Long Answer Type Questions (Score 4)

Question 11.
Analyse the reactions and answer the following questions:

a. Which among the following test tubes will undergo a chemical reaction?
b. What are these chemical reactions called?
c. Explain the oxidation and reduction reactions taking place here including the chemical equation?
Answer:
a. Chemical reaction takes place in test tube (ii) only.
b. These chemical reactions are called displacement reactions. .
c. Chemical equation: (including oxidation states)
Mg° (s) + Fe²⁺S⁶⁺O4^2– (aq) →
Mg²⁺S⁶⁺O^2–₄ (aq) + Fe° (s)
At Mg : Mg°(s) → Mg²⁺(aq) + 2e^– (oxidation)
At Fe²⁺, Fe²⁺ (aq) + 2e^– →  Fe°(s) (reductipn)

Question 12.
A students observation is given below:
i. When Zn is put in salt solution, Na gets deposited over Zn.
ii. Au reacts with water vapour and hydrogen gas is formed.
iii. Al reacts with acid and forms hydrogen gas.
iv. Mg reacts with hot water and forms hydrogen gas.
a. Which statements are incorrect?
b. Give reason for your answer.
Answer:
a. Statements (i) and (ii) are wrong.

b. i. Zn cannot displace sodium. Because sodium has a higher reactivity than Zn.
ii. Au does not react with water vapour.

Question 13.
“Sodium cannot be kept open in atmospheric air, and cannot be stored in water. So it is stored in kerosene.” Give explanation for the above statement with its chemical equation.
Answer:
Sodium which is placed above in the reactivity series reacts vigorously with water and oxygen.
4Na (s) + O₂ (g) → 2Na₂O (s)
2Na (s) + 2H₂O (l) →
2NaOH (aq) + H₂(g)
This NaOH reacts with CO₂ in the atmospheric air and forms sodium carbonate.
2NaOH (s) + CO₂ (g) → Na₂CO₃ + H₂O (l)
To avoid the reaction with O₂, H₂O and CO₂, sodium is stored in kerosene.

Question 14.
The direction of electron flow in certain galvanic cells are given below. (Symbols are not real)
i. B → A,
ii. E → C,
iii.D → E,
iv.A → D.
a. Arrange the metals A, B, C, D and E in the decreasing order of their reactivity.
b. Choose the reaction taking place at ‘C’ in cell (ii) E → C. Give the reason.
i. C⁺ (aq) + le^– → C(s)
ii. C (s) → C⁺ (aq) + le^–
iii.C (s) → C⁺ (aq) + le^–
Answer:
a. B,
A,
D,
E,
C.
b.C⁺ (aq) + le^– → C (s) is the correct equation. Because ‘E’ has higher reactivity than. So ‘C’undergoes reduction.

Question 15.
Give reason for the following.
a. CuSO₄ solution is not stored in iron vessels.
b. Buttermilk is not stored in aluminium vessels.
Ans.
a. Fe displaces copper from copper sulphate solution and forms FeSO₄.
b. Aluminium reacts with acid in the buttermilk.

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